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PUBLISHED BY THE PRESS SYNDICATE OF THE UNIVERSITY OF CAMBRIDGEThe Pitt Building, Trumpington Street, Cambridge, United Kingdom

CAMBRIDGE UNIVERSITY PRESSThe Edinburgh Building, Cambridge CB2 2RU, UK http://www.cup.cam.ac.uk

40 West 20th Street, New York, NY 10011-4211, USA http://www.cup.org10 Stamford Road, Oakleigh, Melbourne 3166, Australia

Ruiz de Alarc6n 13, 28014 Madrid, Spain

© Cambridge University Press 2000

This book is in copyright. Subject to statutory exceptionand to the provisions of relevant collective licensing agreements,

no reproduction of any part may take place withoutthe written permission of Cambridge University Press.

First Edition Published 1986Second Edition Published 2000

Printed in the United States of America

Typeface Times Roman 10/13 pt. System I-6TEX2E [TB]

A catalog record for this book is available from the British Library.

Library of Congress Cataloging in Publication Data

Aschbacher, Michael, 1944-

Finite group theory / M. Aschbacher. - 2nd ed.

p. cm. - (Cambridge studies in advanced mathematics ; 10)

Includes bibliographical references and index.

ISBN 0-521-78145-0 (hb) - ISBN 0-521-78675-4 (pbk.)

1. Finite groups. I. Title. II, Series.

QA177.A82 2000

512'.2 - dc2l 99-055693

ISBN 0 521 78145 0 hardbackISBN 0 521 78675 4 paperback

PUBLISHED BY THE PRESS SYNDICATE OF THE UNIVERSITY OF CAMBRIDGE

The Pitt Building, Trumpington Street, Cambridge, United Kingdom

CAMBRIDGE UNIVERSITY PRESS

The Edinburgh Building, Cambridge CB2 2RU, UK http://www.cup.cam.ac.uk 40 West 20th Street, New York, NY 1001 1-421 1, USA http://www.cup.org

10 Stamford Road, Oakleigh, Melbourne 3166, Australia Ruiz de Alarcdn 13,28014 Madrid, Spain

@ Cambridge University Press 2000

This book is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements,

no reproduction of any part may take place without the written permission of Cambridge University Press.

First Edition Published 1986 Second Edition Published 2000

Printed in the United States of America

Typeface Times Roman 1011 3 pt. System I 4 T s 2~ [TB]

A catalog record for this book is available from the British Library.

Library of Congress Cataloging in Publication Data

Aschbacher, Michael, 1944-

Finite group theory / M. Aschbacher. - 2nd ed.

p. cm. - (Cambridge studies in advanced mathematics ; 10)

Includes bibliographical references and index.

ISBN 0-521-78145-0 (hb) - ISBN 0-521-78675-4 (pbk.)

1. Finite groups. 1. Title. 11. Series.

QA177 .A82 2000

512'.2 - dc21 99-055693

ISBN 0 521 78145 0 hardback ISBN 0 521 78675 4 paperback

To PamTo Pam

Contents

Preface

1 Preliminary results1 Elementary group theory2 Categories3 Graphs and geometries4 Abstract representations

2 Permutation representations5 Permutation representations6 Sylow's Theorem

3 Representations of groups on groups7 Normal series8 Characteristic subgroups and commutators

9 Solvable and nilpotent groups

10 Semidirect products11 Central products and wreath products

4 Linear representations12 Modules over the group ring

13 The general linear group and special linear group

14 The dual representation

5 Permutation groups15 The symmetric and alternating groups16 Rank 3 permutation groups

6 Extensions of groups and modules17 1-cohomology18 Coprime action

7 Spaces with forms19 Bilinear, sesquilinear, and quadratic forms

20 Witt's Lemma21 Spaces over finite fields

22 The classical groups

8 p-groups

23 Extremal p-groups24 Coprime action on p-groups

Contents

Preface

1 Preliminary results 1 Elementary group theory 2 Categories 3 Graphs and geometries 4 Abstract representations

2 Permutation representations 5 Permutation representations 6 Sylow's Theorem

3 Representations of groups on groups 7 Normal series 8 Characteristic subgroups and commutators 9 Solvable and nilpotent groups

10 Semidirect products 11 Central products and wreath products

4 Linear representations 12 Modules over the group ring 13 The general linear group and special linear group 14 The dual representation

5 Permutation groups 15 The symmetric and alternating groups 16 Rank 3 permutation groups

6 Extensions of groups and modules 17 1-cohomology 18 Coprime action

7 Spaces with forms 19 Bilinear, sesquilinear, and quadratic forms 20 Witt's Lemma 21 Spaces over finite fields 22 The classical groups

8 p-groups 23 Extremal p-groups 24 Coprime action on p-groups

viii Contents

9 Change of field of a linear representation 117

25 Tensor products 117

26 Representations over finite fields 123

27 Minimal polynomials 127

10 Presentations of groups 138

28 Free groups 138

29 Coxeter groups 141

30 Root systems 148

11 The generalized Fitting subgroup 156

31 The generalized Fitting subgroup 157

32 Thompson factorization 162

33 Central extensions 166

12 Linear representations of finite groups 177

34 Characters in coprime characteristic 178

35 Characters in characteristic 0 181

36 Some special actions 192

13 Transfer and fusion 197

37 Transfer 197

38 Alperin's Fusion Theorem 200

39 Normal p-complements 202

40 Semiregular action 205

14 The geometry of groups of Lie type 209

41 Complexes 209

42 Buildings 215

43 BN-pairs and Tits systems 218

15 Signalizer functors 229

44 Solvable signalizer functors 229

16 Finite simple groups 242

45 Involutions in finite groups 243

46 Connected groups 245

47 The finite simple groups 249

48 An outline of the Classification Theorem 260

Appendix 269

References 297

List of Symbols 299

Index 301

Contents

9 Change of field of a linear representation 25 Tensor products 26 Representations over finite fields 27 Minimal polynomials

10 Presentations of groups 28 Free groups 29 Coxeter groups 30 Root systems

11 The generalized Fitting subgroup 3 1 The generalized Fitting subgroup 32 Thompson factorization 33 Central extensions

Linear representations of finite groups Characters in coprime characteristic Characters in characteristic 0 Some special actions

Transfer and fusion Transfer Alperin's Fusion Theorem Normal p-complements Semiregular action

The geometry of groups of Lie type Complexes Buildings BN-pairs and Tits systems

Signalizer functors Solvable signalizer functors

Finite simple groups Involutions in finite groups Connected groups The finite simple groups An outline of the Classification Theorem

Appendix

References

List of Symbols

Index

Preface

Finite Group Theory is intended to serve both as a text and as a basic referenceon finite groups. In neither role do I wish the book to be encyclopedic, soI've included only the material I regard as most fundamental. While suchjudgments are subjective, I've been guided by a few basic principles which Ifeel are important and should be made explicit.

One unifying notion is that of a group representation. The term representa-tion is used here in a much broader sense than usual. Namely in this book arepresentation of a group G in a category 6 is a homomorphism of G into theautomorphism group of some object of -C Among these representations, thepermutation representations, the linear representations, and the representationsof groups on groups seem to be the most fundamental. As a result much of thebook is devoted to these three classes of representations.

The first step in investigating representations of finite groups or finite di-mensional groups is to break up the representation into indecomposable orirreducible representations. This process focuses attention on two areas ofstudy: first on the irreducible and indecomposable representations themselves,and second on the recovery of the general representation from its irreducibleconstituents. Both areas receive attention here.

The irreducible objects in the category of groups are the simple groups. Iregard the finite simple groups and their irreducible linear and permutationrepresentations as the center of interest in finite group theory. This point ofview above all others has dictated the choice of material. In particular I feelmany of the deeper questions about finite groups are best answered through thefollowing process. First reduce the question to a question about some class ofirreducible representations of simple groups or almost simple groups. Secondappeal to the classification of the finite simple groups to conclude the groupis an alternating group, a group of Lie type, or one of the 26 sporadic simplegroups. Finally invoke the irreducible representation theory of these groups.

The book serves as a foundation for the proof of the Classification Theorem.Almost all material covered plays a role in the classification, but as it turns outalmost all is of interest outside that framework too. The only major result treatedhere which has not found application outside of simple group theory is the Sig-nalizer Functor Theorem. Signalizer functors are discussed near the end of thebook. The last section of the book discusses the classification in general terms.

Preface

Finite Group Theory is intended to serve both as a text and as a basic reference on finite groups. In neither role do I wish the book to be encyclopedic, so I've included only the material I regard as most fundamental. While such judgments are subjective, I've been guided by a few basic principles which I feel are important and should be made explicit.

One unifying notion is that of a group representation. The term representation is used here in a much broader sense than usual. Namely in this book a representation of a group G in a category +?is a homomorphism of G into the automorphism group of some object of 6 Among these representations, the permutation representations, the linear representations, and the representations of groups on groups seem to be the most fundamental. As a result much of the book is devoted to these three classes of representations.

The first step in investigating representations of finite groups or finite dimensional groups is to break up the representation into indecomposable or irreducible representations. This process focuses attention on two areas of study: first on the irreducible and indecomposable representations themselves, and second on the recovery of the general representation from its irreducible constituents. Both areas receive attention here.

The irreducible objects in the category of groups are the simple groups. I regard the finite simple groups and their irreducible linear and permutation representations as the center of interest in finite group theory. This point of view above all others has dictated the choice of material. In particular I feel many of the deeper questions about finite groups are best answered through the following process. First reduce the question to a question about some class of irreducible representations of simple groups or almost simple groups. Second appeal to the classification of the finite simple groups to conclude the group is an alternating group, a group of Lie type, or one of the 26 sporadic simple groups. Finally invoke the irreducible representation theory of these groups.

The book serves as a foundation for the proof of the Classification Theorem. Almost all material covered plays a role in the classification, but as it turns out almost all is of interest outside that framework too. The only major result treated here which has not found application outside of simple group theory is the Sig- nalizer Functor Theorem. Signalizer functors are discussed near the end of the book. The last section of the book discusses the classification in general terms.

x Preface

The first edition of the book included a new proof of the Solvable SignalizerFunctor Theorem, based on earlier work of Helmut Bender. Bender's proofwas valid only for the prime 2, but it is very short and elegant. I've come tobelieve that my extension to arbitrary primes in the first edition is so compli-cated that it obscures the proof, so this edition includes only a proof of theSolvable 2-Signalizer Functor Theorem, which is closer to Bender's originalproof. Because of this change, section 36 has also been truncated.

In some sense most of the finite simple groups are classical linear groups.Thus the classical groups serve as the best example of finite simple groups.They are also representative of the groups of Lie type, both classical and ex-ceptional, finite or infinite. A significant fraction of the book is devoted to theclassical groups. The discussion is not restricted to groups over finite fields.The classical groups are examined via their representation as the automorphismgroups of spaces of forms and their representation as the automorphism groupsof buildings. The Lie theoretic point of view enters into the latter representationand into a discussion of Coxeter groups and root systems.

I assume the reader has been exposed to a first course in algebra or itsequivalent; Herstein's Topics in Algebra would be a representative text forsuch a course. Occasionally some deeper algebraic results are also needed;in such instances the result is quoted and a reference is given for its proof.Lang's Algebra is one reference for such results. The group theory I assume islisted explicitly in section 1. There isn't much; for example Sylow's Theoremis proved in chapter 2.

As indicated earlier, the book is intended to serve both as a text and as abasic reference. Often these objectives are compatible, but when compromiseis necessary it is usually in favor of the role as a reference. Proofs are moreterse than in most texts. Theorems are usually not motivated or illustratedwith examples, but there are exercises. Many of the results in the exercises areinteresting in their own right; often there is an appeal to the exercises in thebook proper. In this second edition I've added an appendix containing solutionsto some of the most difficult and/or important exercises.

If the book is used as a text the instructor will probably wish to expand manyproofs in lecture and omit some of the more difficult sections. Here are somesuggestions about which sections to skip or postpone.

A good basic course in finite group theory would consist of the first eightchapters, omitting sections 14, 16, and 17 and chapter 7, and adding sections28, 31, 34, 35, and 37. Time permitting, sections 32, 33, 38, and 39 could beadded.

The classical groups and some associated Lie theory are treated in chapter7, sections 29 and 30, chapter 14, and the latter part of section 47. A differentsort of course could be built around this material.

x Preface

The first edition of the book included a new proof of the Solvable Signalizer Functor Theorem, based on earlier work of Helmut Bender. Bender's proof was valid only for the prime 2, but it is very short and elegant. I've come to believe that my extension to arbitrary primes in the first edition is so complicated that it obscures the proof, so this edition includes only a proof of the Solvable 2-Signalizer Functor Theorem, which is closer to Bender's original proof. Because of this change, section 36 has also been truncated.

In some sense most of the finite simple groups are classical linear groups. Thus the classical groups serve as the best example of finite simple groups. They are also representative of the groups of Lie type, both classical and ex- ceptional, finite or infinite. A significant fraction of the book is devoted to the classical groups. The discussion is not restricted to groups over finite fields. The classical groups are examined via their representation as the automorphism groups of spaces of forms and their representation as the automorphism groups of buildings. The Lie theoretic point of view enters into the latter representation and into a discussion of Coxeter groups and root systems.

I assume the reader has been exposed to a first course in algebra or its equivalent; Herstein's Topics in Algebra would be a representative text for such a course. Occasionally some deeper algebraic results are also needed; in such instances the result is quoted and a reference is given for its proof. Lang's Algebra is one reference for such results. The group theory I assume is listed explicitly in section 1. There isn't much; for example Sylow's Theorem is proved in chapter 2.

As indicated earlier, the book is intended to serve both as a text and as a basic reference. Often these objectives are compatible, but when compromise is necessary it is usually in favor of the role as a reference. Proofs are more terse than in most texts. Theorems are usually not motivated or illustrated with examples, but there are exercises. Many of the results in the exercises are interesting in their own right; often there is an appeal to the exercises in the book proper. In this second edition I've added an appendix containing solutions to some of the most difficult andlor important exercises.

If the bookis used as a text the instructor will probably wish to expand many proofs in lecture and omit some of the more difficult sections. Here are some suggestions about which sections to skip or postpone.

A good basic course in finite group theory would consist of the first eight chapters, omitting sections 14, 16, and 17 and chapter 7, and adding sections 28, 31, 34, 35, and 37. Time permitting, sections 32, 33,38, and 39 could be added.

The classical groups and some associated Lie theory are treated in chapter 7, sections 29 and 30, chapter 14, and the latter part of section 47. A different sort of course could be built around this material.

Preface xi

Chapter 9 deals with various concepts in the theory of linear representationswhich are somewhat less basic than most of those in chapters 4 and 12. Much ofthe material in chapter 9 is of principal interest for representations over fieldsof prime characteristic. A course emphasizing representation theory wouldprobably include chapter 9.

Chapter 15 is the most technical and specialized. It is probably only ofinterest to potential simple groups theorists.

Chapter 16 discusses the finite simple groups and the classification. Thelatter part of section 47 builds on chapter 14, but the rest of chapter 16 is prettyeasy reading. Section 48 consists of a very brief outline of the proof of the finitesimple groups makes use of results from earlier in the book and thus motivatesthose results by exhibiting applications of the results.

Each chapter begins with a short introduction describing the major resultsin the chapter. Most chapters close with a few remarks. Some remarks ac-knowledge sources for material covered in the chapter or suggest referencesfor further reading. Similarly, some of the remarks place certain results in con-text and hence motivate those results. Still others warn that some section inthe chapter is technical or specialized and suggests the casual reader skip orpostpone the section.

In addition to the introduction and the remarks, there is another good wayto decide which results in a chapter are of most interest: those results whichbear some sort of descriptive label (e.g. Modular Property of Groups, FrattiniArgument) are often of most importance.

Preface xi

Chapter 9 deals with various concepts in the theory of linear representations which are somewhat less basic than most of those in chapters 4 and 12. Much of the material in chapter 9 is of principal interest for representations over fields of prime characteristic. A course emphasizing representation theory would probably include chapter 9.

Chapter 15 is the most technical and specialized. It is probably only of interest to potential simple groups theorists.

Chapter 16 discusses the finite simple groups and the classification. The latter part of section 47 builds on chapter 14, but the rest of chapter 16 is pretty easy reading. Section 48 consists of a very brief outline of the proof of the finite simple groups makes use of results from earlier in the book and thus motivates those results by exhibiting applications of the results.

Each chapter begins with a short introduction describing the major results in the chapter. Most chapters close with a few remarks. Some remarks ac- knowledge sources for material covered in the chapter or suggest references for further reading. Similarly, some of the remarks place certain results in con- text and hence motivate those results. Still others warn that some section in the chapter is technical or specialized and suggests the casual reader skip or postpone the section.

In addition to the introduction and the remarks, there is another good way to decide which results in a chapter are of most interest: those results which bear some sort of descriptive label (e.g. Modular Property of Groups, Frattini Argument) are often of most importance.

Preliminary results

I assume familiarity with material from a standard course on elementary alge-bra. A typical text for such a course is Herstein [He]. A few deeper algebraicresults are also needed; they can be found for example in Lang [La]. Section 1lists the elementary group theoretic results assumed and also contains a list ofbasic notation. Later sections in chapter 1 introduce some terminology and no-tation from a few other areas of algebra. Deeper algebraic results are introducedwhen they are needed.

The last section of chapter 1 contains a brief discussion of group representa-tions. The term representation is used here in a more general sense than usual.Namely a representation of a group G will be understood to be a group homo-morphism of G into the group of automorphisms of an object X. Standard useof the term representation requires X to be a vector space.

1 Elementary group theoryRecall that a binary operation on a set G is a function from. the set productG x G into G. Multiplicative notation will usually be used. Thus the image ofa pair (x, y) under the binary operation will be written xy. The operation isassociative if (xy)z = x(yz) for all x, y, z in G. The operation is commutativeif xy = yx for all x, y in G. An identity for the operation is an element 1 inG such that x 1 = lx = x for all x in G. An operation possesses at most oneidentity. Given an operation on G possessing an identity 1, an inverse for anelement x of G is an element y in G such that x y = yx = 1. If our operation isassociative and x possesses an inverse then that inverse is unique and is denoted

by x-1 in multiplicative notation.A group is a set G together with an associative binary operation which

possesses an identity and such that each element of G possesses an inverse.The group is abelian if its operation is commutative. In the remainder of thissection G is a group written multiplicatively.

Let X E G and n a positive integer. x" denotes the product of x with itselfn times. Associativity insures x" is a well-defined element of G. Define x_n

to be (x-1)" and x° to be 1. The usual rules of exponents can be derived from

Preliminary results

I assume familiarity with material from a standard course on elementary algebra. A typical text for such a course is Herstein [He]. A few deeper algebraic results are also needed; they can be found for example in Lang [La]. Section 1 lists the elementary group theoretic results assumed and also contains a list of basic notation. Later sections in chapter 1 introduce some terminology and notation from a few other areas of algebra. Deeper algebraic results are introduced when they are needed.

The last section of chapter 1 contains a brief discussion of group representations. The term representation is used here in a more general sense than usual. Namely a representation of a group G will be understood to be a group homomorphism of G into the group of automorphisms of an object X. Standard use of the term representation requires X to be a vector space.

1 Elementary group theory Recall that a binary operation on a set G is a function from the set product GxG into G. Multiplicative notation will usually be used. Thus the image of a pair (x, y) under the binary operation will be written xy. The operation is associative if (xy)z = x(yz) for all x, y, z in G. The operation is commutative if xy = yx for all x, y in G. An identity for the operation is an element 1 in G such that xl = lx = x for all x in G. An operation possesses at most one identity. Given an operation on G possessing an identity 1, an inverse for an element x of G is an element y in G such that xy = yx = 1. If our operation is associative and x possesses an inverse then that inverse is unique and is denoted by x-' in multiplicative notation.

A group is a set G together with an associative binary operation which possesses an identity and such that each element of G possesses an inverse. The group is abelian if its operation is commutative. In the remainder of this section G is a group written multiplicatively.

Let x E G and n a positive integer. xn denotes the product of x with itself n times. Associativity insures xn is a well-defined element of G. Define x-" to be (x-')" and x0 to be 1. The usual rules of exponents can be derived from

2 Preliminary results

this definition:

(1.1) Let G be a group, x E G, and n and m integers. Then

(1) (xn)(xm) = xn+m = (xm)(xn).

(2) (xn)m = xnm

A subgroup of G is a nonempty subset H of G such that for each x, y E H, xyand x-1 are in H. This insures that the binary operation on G restricts to abinary operation on H which makes H into a group with the same identity asG and the same inverses. I write H < G to indicate that H is a subgroup of G.

(1.2) The intersection of any set of subgroups of G is also a subgroup of G.

Let S C G and define

(S) = n HscH<G

By 1.2, (S) is a subgroup of G and by construction it is the smallest subgroup ofG containing S. The subgroup (S) is called the subgroup of G generated by S.

(1.3) Let S C G. Then

(S) = {(sl)E1 ... (S,)"': Si E S, e = +1 or -1}.

(1.4) Let X E G. Then (x) = {xn: n c 1].

Of course 1.4 is a special case of 1.3. A group G is cyclic if it is generated bysome element x. In that case x is said to be a generator of G and by 1.4, Gconsists of the powers of x.

The order of a group G is the cardinality of the associated set G. Write IGIfor the order of a set G or a group G. For X E G, Ix I denotes I (x) I and is calledthe order of x.

A group homomorphism from a group G into a group H is a function a : G -*H of the set G into the set H which preserves the group operations: thatis for all x, y in G, (xy)a = xaya. Notice that I usually write my maps onthe right, particularly those that are homomorphisms. The homomorphisma is an isomorphism if a is a bijection. In that case a possesses an inversefunction a`1: H -* G and it turns out a-1 is also a group homomorphism. Gis isomorphic to H if there exists an isomorphism of G and H. Write G = H toindicate that G is isomorphic to H. Isomorphism is an equivalence relation. His said to be a homomorphic image of G if there is a surjective homomorphismof G onto H.


this definition:

(1.1) Let G be a group, x E G, and n and m integers. Then (1) (xn)(xm) = xn+m = (xrn)(x"). (2) (xn)m = xnm.

A subgroup of G is a nonempty subset H of G such that for each x, y E H, xy and x-' are in H. This insures that the binary operation on G restricts to a binary operation on H which makes H into a group with the same identity as G and the same inverses. I write H ( G to indicate that H is a subgroup of G.

(1.2) The intersection of any set of subgroups of G is also a subgroup of G.

Let S E G and define

By 1.2, ( S ) is a subgroup of G and by construction it is the smallest subgroup of G containing S. The subgroup ( S ) is called the subgroup of G generated by S.

(1.3) Let S g G. Then

( S ) = { ( s ~ ) ~ ' . . . ( s ~ ) ~ " : s ~ E S, & = + I or-1).

(1.4) Let x E G. Then ( x ) = {xn: n E Z].

Of course 1.4 is a special case of 1.3. A group G is cyclic if it is generated by some element x. In that case x is said to be a generator of G and by 1.4, G consists of the powers of x.

The order of a group G is the cardinality of the associated set G. Write IGl for the order of a set G or a group G. For x E G, Ix I denotes I (x) 1 and is called the order of x.

A group homomorphism from a group G into a group H is a function a : G 4

H of the set G into the set H which preserves the group operations: that is for all x, y in G, (xy)cr = xaya. Notice that I usually write my maps on the right, particularly those that are homomorphisms. The homomorphism a is an isomorphism if cr is a bijection. In that case a possesses an inverse function a-': H + G and it turns out a-' is also a group homomorphism. G is isomorphic to H if there exists an isomorphism of G and H. Write G Z H to indicate that G is isomorphic to H. Isomorphism is an equivalence relation. H is said to be a homomorphic image of G if there is a surjective homomorphism of G onto H.

Elementary group theory 3

A subgroup H of G is normal if g-lhg c- H for each g E G and h c- H.Write H a G to indicate H is a normal subgroup of G. If a: G X is agroup homomorphism then the kernel of a is ker(a) _ {g E G: ga = 1} and itturns out that ker(a) is a normal subgroup of G. Also write Ga for the image{ga: g E G} of Gin X. Ga is a subgroup of X.

Let H < G. For X E G write Hx = {hx: h E H} and xH = {xh: h E H}. Hxand xH are cosets of H in G. Hx is a right coset and xH a left coset.

To be consistent I'll work with right cosets Hx in this section. G/H denotesthe set of all (right) cosets of H in G. G/H is the coset space of H in G. Denoteby I G : H I the order of the coset space G/H. As the map h H hx is a bijectionof H with Hx, all cosets have the same order, so

(1.5) (Lagrange's Theorem) Let G be a group and H < G. Then G I _ H II G: H 1. In particular if G is finite then I H I divides I G I.

If H a G the coset space G/H is made into a group by defining multiplicationvia

(Hx)(Hy) = Hxy x, y E G

Moreover there is a natural surjective homomorphism rr: G -* G/H definedby rr: x H Hx. Notice ker(rr) = H. Conversely if a: G -* L is a surjectivehomomorphism with ker(a) = H then the map ,B: Hx i-+ xa is an isomorphismof G/H with L such that ir,B = a. The group G/H is called the factor group ofG by H. Therefore the factor groups of G over its various normal subgroupsare, up to isomorphism, precisely the homomorphic images of G.

(1.6) Let H a G. Then the map L H L/H is a bijection between the set of allsubgroups of G containing H and the set of all subgroups of G/H. Normalsubgroups correspond to normal subgroups under this bijection.

For x, y E G, set xy = y-lxy. For X CG set Xy _ {xY: x E X). XY is theconjugate of X under y. Write XG for the set {X8: g c G} of conjugates of Xunder G. Define

NG(X)={gEG:XB=X).

NG(X) is the normalizer in G of X and is a subgroup of G. Indeed if X < Gthen NG(X) is the largest subgroup of G in which X is normal. Define

CG(X) _ {g E G: xg = gx for all x E X}.

CG(X) is the centralizer in G of X. CG(X) is also a subgroup of G.


A subgroup H of G is normal if g-'hg E H for each g E G and h E H. Write HI! G to indicate H is a normal subgroup of G. If a: G + X is a group homomorphism then the kernel of a is ker(a) = {g E G: ga = 11 and it turns out that ker(a) is a normal subgroup of G. Also write G a for the image {ga: g E GI of G in X. Ga is a subgroup of X.

Let H 5 G. Forx E G write Hx = {hx: h E HI andxH = {xh: h E HI. Hx and xH are cosets of H in G. Hx is a right coset and xH a left coset.

To be consistent I'll work with right cosets Hx in this section. GIH denotes the set of all (right) cosets of H in G. GIH is the coset space of H in G. Denote by IG : HI the order of the coset space G/H. As the map h t+ hx is a bijection of H with Hx, all cosets have the same order, so

(1.5) (Lagrange's Theorem) Let G be a group and H i G. Then IGI = IHJ (G : H 1. In particular if G is finite then IH ( divides \GI.

If H 5 G the coset space GIH is made into a group by defining multiplication via

Moreover there is a natural surjective homomorphism n : G + GIH defined by n : x t+ Hx. Notice ker(n) = H. Conversely if a : G + L is a surjective homomorphism with ker(a) = H then the map /3: Hx t+ x a is an isomorphism of GIH with L such that np = a. The group GIH is called the factor group of G by H. Therefore the factor groups of G over its various normal subgroups are, up to isomorphism, precisely the homomorphic images of G.

(1.6) Let H I! G. Then the map L t+ LIH is a bijection between the set of all subgroups of G containing H and the set of all subgroups of GIH. Normal subgroups correspond to normal subgroups under this bijection.

For x, y E G, set X Y = y-lxy. For X g G set XY = {xY:x E X). XY is the conjugate of X under y. Write xG for the set {Xg: g E G) of conjugates of X under G. Define

NG(X) = {g E G:Xg = X).

NG(X) is the normalizer in G of X and is a subgroup of G. Indeed if X 5 G then NG(X) is the largest subgroup of G in which X is normal. Define

CG(X) = {g E G: xg = gx for all x E X).

CG(X) is the centralizer in G of X. CG(X) is also a subgroup of G.

For X, Y C G define XY = {xy: x E X, y E Y). The set XY is the productof X with Y.

(1.7) Let X, Y < G. Then(1) XYis a subgroup of G if and only if XY = YX.(2) If Y < NG(X) then XY is a subgroup of G and XY/X = Y/(Y n X).(3) IXYI = )XHHYl/IX n Y.

(1.8) Let H and K be normal subgroups of G with K < H. Then G/K/H/KG/H.

Let G1, ... , G. be a finite set of groups. The directproduct G1 x ... x Gn =

l 1n=1 Gi of the groups G1, ..., Gn is the group defined on the set productG1 x . . . x Gn by the operation

(x1,...,Xn)(y1,...,yn)=(xiyi,...,xnyn) xi,yi c Gi

(1.9) Let G be a group and (Gi: 1 < i < n) a family of subgroups of G. Thenthe following are equivalent:

(1) The map (xi, ..., xn) H x1 ... x, is an isomorphism of G with G1 x . xG.

(2) G = (Gi: 1 < i <n) and for each i, 1 < i < n, Gi 4 G and Gin (GG: j

(3) Gi a G for each i, 1 < i < n, and each g E G can be written uniquely asg = x1 ... xn with xi E G,.

If any of the equivalent conditions of 1.9 hold, G will be said to be.the directproduct of the subgroups (Gi: 1 < i < n).

(1.10) Let G = (g) be a cyclic group and Z the group of integers under addi-tion. Then

(1) If H is a nontrivial subgroup of Z then H = (n), where n is the leastpositive integer in H.

(2) The map a: 71 -± G defined by ma = g' is a surjective homomorphismwith kernel (n), where n = 0 if g is of infinite order and n = min {m > 0:gm = 11 if g has finite order.

(3) If g has finite order n then G = {gi:0 < i < n} and n is the leastpositive integer m with g' = 1.

(4) Up to isomorphism 71 is the unique infinite cyclic group and for eachpositive integer n, the group 71n of integers modulo n is the unique cyclic groupof order n.

Preliminary results

For X, Y G define XY = {xy: x E X, y E Y } . The set XY is the product of X with Y .

(1.7) Let X , Y 5 G. Then (1) XY is a subgroup of G if and only if XY = YX. (2) If Y 5 NG(X) then XY is a subgroup of G and XY/X 2 Y/(Y n X).

(3) IXYt = tXIIYIIIX n YI.

(1.8) Let H and K be normal subgroups of G with K 5 H. Then G/K/H/K G

G/H.

Let G 1 , . . . , G, be a finite set of groups. The directproduct G I x - . . x G, = ny=, Gi of the groups G I , . . . , G, is the group defined on the set product G1 x . . . x G, by the operation

(1.9) Let G be a group and (Gi: 1 5 i 5 n ) a family of subgroups of G. Then the following are equivalent:

(1) The map ( x l , . . . , x,) I+ xl . . . x, is an isomorphism of G with GI x . . x

G, . (2) G = (Gi: 1 5 i 5 n ) and for each i , 1 5 i 5 n , Gi 9 G and Gi n (Gj : j

# i ) = 1. (3 ) Gi 9 G for each i, 1 5 i 5 n , and each g E G can be written uniquely as

g = x l ... x, withxi E Gi.

If any of the equivalent conditions of 1.9 hold, G will be said to be the direct product of the subgroups (Gi: 1 5 i 5 n).

(1.10) Let G = (g) be a cyclic group and Z the group of integers under addition. Then

(I) If H is a nontrivial subgroup of Z then H = (n ) , where n is the least positive integer in H.

(2) The map a: Z -+ G defined by ma = g" is a surjective homomorphism with kernel (n ) , where n = 0 if g is of infinite order and n = min {rn > 0: gm = 1 ) if g has finite order.

(3) lf g has finite order n then G = (g" 0 5 i < n) and n is the least positive integer rn with gm = 1. (4) Up to isomorphism Z is the unique infinite cyclic group and for each

positive integer n, the group Z, of integers modulo n is the unique cyclic group of order n.

(5) Let IgI = n. Then for each divisor m of n, (g"/') is the unique subgroupof G of order m. In particular subgroups of cyclic groups are cyclic.

(1.11) Each finitely generated abelian group is the direct product of cyclicgroups.

Let p be a prime. A p-group is a group whose order is a power of p. Moregenerally if Tr is a set of primes then a rr-group is a group G of finite ordersuch that 7r (G) C jr, where 7r (G) denotes the set of prime divisors of I GI. p'denotes the set of all primes distinct from p. An element x in a group G is a7r -element if (x) is a 7r -group. An involution is an element of order 2.

(1.12) Let 1 : G be an abelian p-group. Then G is the direct product ofcyclic subgroups G, = Z pe; ,1 e2 > ... > e > 1. Moreover theintegers n and (e1: 1 < i < n) are uniquely determined by G.

The exponent of a finite group G is the least common multiple of the ordersof the elements of G. An elementary abelian p-group is an abelian p-groupof exponent p. Notice that by 1.12, G is an elementary abelian p-group oforder p" if and only if G is the direct product of n copies of TLp. In particularup to isomorphism there is a unique elementary abelian p-group of order p",which will be denoted by Ep,. The integer n is the p-rank of Ep». The p-rankof a general finite group G is the maximum p-rank of an elementary abelianp-subgroup of G, and is denoted by mp(G).

(1.13) Each group of exponent 2 is abelian.

If Tr is a set of primes and G a finite group, write 0, (G) for the largest normal7r-subgroup of G, and O" (G) for the smallest normal subgroup H of G suchthat G/H is a 7r-group. 0, (G) and 0' (G) are well defined by Exercise 1.1.

Define Z(G) = CG (G) and call Z(G) the center of G. If G is a p-group thendefine

Q,(G)=(xEG:x" _1)

Y(G)=(x'':xEG).For X < G define AUtG(X) = NI(X)/CG(X) to be the automizer in G of

X. Notice that by Exercise 1.3, AutG(X) < Aut(X) and indeed Autc(X) is thegroup of automorphisms induced on X in G.

A maximal subgroup of a group G is a proper subgroup of G which isproperly contained in no proper subgroup of G. That is a maximal subgroup is


(5) Let Ig ( = n. Then for each divisor m of n, (gn/m) is the unique subgroup of G of order m. In particular subgroups of cyclic groups are cyclic.

(1.11) Each finitely generated abelian group is the direct product of cyclic groups.

Let p be a prime. A p-group is a group whose order is a power of p. More generally if n is a set of primes then a n-group is a group G of finite order such that n(G) g n , where n(G) denotes the set of prime divisors of /GI. p' denotes the set of all primes distinct from p. An element x in a group G is a n-element if (x) is a n-group. An involution is an element of order 2.

(1.12) Let 1 # G be an abelian p-group. Then G is the direct product of cyclic subgroups Gi 2 Zpe,, 1 _( i _( n, el 2 ez 2 . . . L. en > 1. Moreover the integers n and (ei: 1 5 i 5 n) are uniquely determined by G.

The exponent of a finite group G is the least common multiple of the orders of the elements of G. An elementary abelian p-group is an abelian p-group of exponent p. Notice that by 1.12, G is an elementary abelian p-group of order pn if and only if G is the direct product of n copies of Z,. In particular up to isomorphism there is a unique elementary abelian p-group of order pn, which will be denoted by Ep. . The integer n is the p-rank of Epa . The p-rank of a general finite group G is the maximum p-rank of an elementary abelian p-subgroup of G, and is denoted by mp(G).

(1.13) Each group of exponent 2 is abelian.

If n is a set of primes and G a finite group, write 0, (G) for the largest normal n-subgroup of G, and OK(G) for the smallest normal subgroup H of G such that G/H is a n-group. O,(G) and OK(G) are well defined by Exercise 1.1.

Define Z(G) = CG(G) and call Z(G) the center of G. If G is a p-group then define

Q,,(G) = (x E G: xp" = 1)

Un(G) = (xp":x E G).

For X ( G define AutG(X) = NG(X)/CG(X) to be the automizer in G of X. Notice that by Exercise 1.3, AutG(X) _( Aut(X) and indeed Autc(X) is the group of automorphisms induced on X in G.

A maximal subgroup of a group G is a proper subgroup of G which is properly contained in no proper subgroup of G. That is a maximal subgroup is

a maximal member of the set of proper subgroups of G, partially ordered byinclusion.

If a: S -+ T is a function and R C S then a)R denotes the restriction of ato R. That is a 4R: R - * T is the function from R into T agreeing with a.

Here's a little result that's easy to prove but useful.

(1.14) (Modular Property of Groups) Let A, B, and C be subgroups of a groupG with A < C. Then AB F) C = A(B F) C).

If G is a group write G# for the set G - (1) of nonidentity elements of G. Onthe other hand if R is a ring define R# = R - (O).

Denote by C, R, and 0 the complex numbers, the reals, and the rationals,respectively. Often 1 will denote the integers.

Given a group G, a subgroup H of G, and a collection C of subgroups of G,I'll often write c n H for the set of members of C which are subgroups of H.

I'll use the bar convention. That is I'll often denote a homomorphic imageGa of a group G by G (or G* or G) and write g (or g* or g) for ga. This willbe done without comment.

Other notation and terminology are introduced in later chapters. The List ofSymbols gives the page number where a notation is first introduced and defined.

2 CategoriesIt will be convenient to have available some of the elementary concepts andlanguage of categories. For a somewhat more detailed discussion, see chapter 1of Lang [La].

A category i' consists of

(1) A collection Ob(i) of objects.(2) For each pair A,B of objects, a set Mor(A,B) of morphisms from A to B.(3) For each triple A, B, C of objects a map

Mor(A, B) x Mor(B, C) Mor(A, C)

called composition. Write f g for the image of the pair (f, g) under thecomposition map.

Moreover the following three axioms are required to hold:

Cat (1) For each quadruple A, B, C, D of objects, Mor(A, B) f1 Mor(C, D) isempty unless A = C and B = D.

Cat (2) Composition is associative.Cat (3) For each object A, Mor(A, A) possesses an identity morphism 1A such

that for all objects B and all f in Mor(A, B) and g in Mor(B, A),lAf=fandglA=g.


a maximal member of the set of proper subgroups of G, partially ordered by inclusion.

If a: S + T is a function and R G S then ~ I R denotes the restriction of a to R. That is alR: R + T is the function from R into T agreeing with a.

Here's a little result that's easy to prove but useful.

(1.14) (Modular Property of Groups) Let A, B, and C be subgroups of a group GwithA 5 C.ThenABnC = A ( B n C ) .

If G is a group write G' for the set G - (1) of nonidentity elements of G. On the other hand if R is a ring define R' = R - (0).

Denote by C, R, and Q the complex numbers, the reals, and the rationals, respectively. Often Z will denote the integers.

Given a group G, a subgroup H of G, and a collection C of subgroups of G, I'll often write C n H for the set of members of C which are subgroups of H.

I'll use the bar convention. That is I'll often denote a homomorphic image Ga of a group G by G (or G* or G ) and write g (or g* or 8 ) for ga . This will be done without comment.

Other notation and terminology are introduced in later chapters. The List of Symbols gives the page number where a notation is first introduced and defined.

2 Categories It will be convenient to have available some of the elementary concepts and language of categories. For a somewhat more detailed discussion, see chapter 1 of Lang [La].

A category fi? consists of

(1) A collection Ob(&) of objects. (2) For each pair A,B of objects, a set Mor(A,B) of morphisms from A to B. (3) For each triple A, B, C of objects a map

called composition. Write f g for the image of the pair (f, g) under the composition map.

Moreover the following three axioms are required to hold:

Cat (1) For each quadruple A, B, C, D of objects, Mor(A, B) n Mor(C, D) is empty unless A = C and B = D.

Cat (2) Composition is associative. Cat (3) For each object A, Mor(A, A) possesses an identity morphism 1.4 such

that for all objects B and all f in Mor(A, B) and g in Mor(B, A), 1.4 f = f and g1.4 = g.

Graphs and geometries 7

Almost all categories considered here will be categories of sets with structure.That is the objects of the category are sets together with some extra structure,Mor(A, B) consists of all functions from the set associated to A to the setassociated to B which preserve the extra structure, and composition is ordinarycomposition of functions. The identity morphism IA is forced to be the identitymap on A. Thus we need to know the identity map preserves structure. Wealso need to know the composition of maps which preserve structure alsopreserves structure. These facts will usually be obvious in the examples weconsider.

We'll be most interested in the following three categories, which are allcategories of sets with structure.

(1) The category of sets and functions: Here the objects are the sets andMor(A, B) is the set of all functions from the set A into the set B.

(2) The category of groups and group homomorphisms: The objects are thegroups and morphisms are the group homomorphisms.

(3) The category of vector spaces and linear transformations: Fix a field F. Theobjects are the vector spaces over F and the morphisms are the F-lineartransformations.

Let f be a morphism from an object A to an object B. An inverse for f in -'is a morphism g e Mor(B, A) such that 1A = f g and 1 B = gf. The morphismf is an isomorphism if it possesses an inverse in w. An automorphism of A isan isomorphism from A to A. Denote by Aut(A) the set of all automorphismsof A and observe Aut(A) forms a group under the composition in i.

If a: A Bisanisomorphismdefinea*:Mor(A, A) Mor(B, B) bya`1 fla and observe a* restricts to a group isomorphism of Aut(A) with Aut(B).

Let (Ai: i e I) be a family of objects in a category -i'. A coproduct of thefamily is an object C together with morphisms c, : A; -- C, i E I, satisfying theuniversal property: whenever X is an object and at: Al --+ X are morphisms,there exists a unique morphism a: C -+ X with cia = ai for each i e I. As aconsequence of the universal property, the coproduct of a family is determinedup to isomorphism, if it exists.

The product of the family is defined dually. That is to obtain the definitionof the product, take the definition of the coproduct and reverse the direction ofall arrows.

Exercise 1.2 gives a description of coproducts and products in the threecategories listed above.

3 Graphs and geometriesThis section contains a brief discussion of two more categories which willmake occasional appearances in these notes.

Graphs and geometries 7

Almost all categories considered here will be categories of sets with structure. That is the objects of the category are sets together with some extra structure, Mor(A, B) consists of all functions from the set associated to A to the set associated to B which preserve the extra structure, and composition is ordinary composition of functions. The identity morphism lA is forced to be the identity map on A. Thus we need to know the identity map preserves structure. We also need to know the composition of maps which preserve structure also preserves structure. These facts will usually be obvious in the examples we consider.

We'll be most interested in the following three categories, which are all categories of sets with structure.

(1) The category of sets and functions: Here the objects are the sets and Mor(A, B) is the set of all functions from the set A into the set B.

(2) The category of groups and group homomorphisms: The objects are the groups and morphisms are the group homomorphisms.

(3) The category of vector spaces and linear transformations: Fix a field F. The objects are the vector spaces over F and the morphisms are the F-linear transformations.

Let f be a morphism from an object A to an object B. An inverse for f in ff? is a morphism g E Mor(B, A) such that lA = f g and lB = g f . The morphism f is an isomorphism if it possesses an inverse in ff?. An automorphism of A is an isomorphism from A to A. Denote by Aut(A) the set of all automorphisms of A and observe Aut(A) forms a group under the composition in 6'.

If a : A + B is an isomorphism define a*: Mor(A , A) + Mor(B, B) by B + a - l ~ a and observe a* restricts to a group isomorphism of Aut(A) with Aut(B).

Let (Ai: i E I ) be a family of objects in a category ff?. A coproduct of the family is an object C together with morphisms ci: Ai + C , i E I, satisfying the universal property: whenever X is an object and ai: Ai -+ X are morphisms, there exists a unique morphism a : C + X with cia = ai for each i E I. As a consequence of the universal property, the coproduct of a family is determined up to isomorphism, if it exists.

The product of the family is defined dually. That is to obtain the definition of the product, take the definition of the coproduct and reverse the direction of all arrows.

Exercise 1.2 gives a description of coproducts and products in the three categories listed above.

3 Graphs and geometries This section contains a brief discussion of two more categories which will make occasional appearances in these notes.

A graph fi= (V, *) consists of a set V of vertices (or objects or points) to-gether with a symmetric relation * called adjacency (or incidence or somethingelse). The ordered pairs in the relation are called the edges of the graph. I writeu * v to indicate two vertices are related via * and say u is adjacent to v. A pathof length n from u to v is a sequence of vertices u = uo, u 1, ... , u, = v suchthat ui.ui+l for each i. Denote by d(u, v) the minimal length of a path from uto v. If no such path exists set d(u, v) = no. d(u, v) is the distance from u to v.

The relation - on V defined by u - v if and only if d(u, v) < no is anequivalence relation on V. The equivalence classes of this relation are calledthe connected components of the graph. The graph is connected if it has justone connected component. Equivalently there is a path between any pair ofvertices.

A morphism a: 9 i' of graphs is a function a: V -+ V' from the vertexset V of 9 to the vertex set V' of " which preserves adjacency; that is if u andv are vertices adjacent in 07then ua is adjacent to va in 9'.

So much for graphs; on to geometries. In this book I adopt a notion of ge-ometry due to Tits. Let I be a finite set. A geometry over I is a triple (F, r, *)where F is a set of objects, r: r -+ I is a type function, and * is a symmetricincidence relation on r such that objects u and v of the same type are incidentif and only if u = v. r(u) is the type of the object u. Notice (F, *) is a graph.I'll usually write r for the geometry (F, r, *).

A morphism a: F -+ F' of geometries is a function a: F -+ F' of the asso-ciated object sets which preserves type and incidence; that is if u, v c F withu * v then r(u) = r'(ua) and ua *' va.

A flag of the geometry r is a set T of objects such that each pair of objectsin T is incident. Notice our one (weak) axiom insures that a flag T possessesat most one object of each type, so that the type function r induces an injectionof T into I. The image r (T) is called the type of T. The rank and corank of Tare the order of r (T) and I - r (T ), respectively. The residue rT of the flag Tis (v c F - T: v * t for all t c T J regarded as a geometry over I - r(T).

The geometry r is connected if its graph (I', *) is connected. F is residuallyconnected if the residue of every flag of corank at least 2 is connected and theresidue of every flag of corank 1 is nonempty.

Here's a way to associate geometries to groups. Let G be a group and= (Gi: i E I) a family of subgroups of G. Define F(G, ) to be the geo-

metry whose set of objects of type i is the coset space G/Gi and with objectsGix and Gay incident if Gix fl Gay is nonempty. For J c_ I write J' forthe complement I - J of J in I and define G j = fl G3. Observe that forxEG,Sj,X=(Gjx:jEJ}isaflag

of automorphisms of a geometry r is said to be flag transitive ifH is transitive on flags of type J for each subset J of I.


A graph B = (V, *) consists of a set V of vertices (or objects or points) together with a symmetric relation * called adjacency (or incidence or something else). The ordered pairs in the relation are called the edges of the graph. I write u * v to indicate two vertices are related via * and say u is adjacent to v. Apath of length n from u to v is a sequence of vertices u = ug, u 1, . . . , U, = v such that u,* u,+ for each i. Denote by d(u, v) the minimal length of a path from u to v . If no such path exists set d (u , v) = oo. d (u , v) is the distance from u to v.

The relation - on V defined by u - v if and only if d(u, v) < oo is an equivalence relation on V. The equivalence classes of this relation are called the connected components of the graph. The graph is connected if it has just one connected component. Equivalently there is a path between any pair of vertices.

A morphism a : B -, g' of graphs is a function a: V + V' from the vertex set V of $?to the vertex set V' of @'' which preserves adjacency; that is if u and v are vertices adjacent in @'then ua is adjacent to va in i?'.

So much for graphs; on to geometries. In this book I adopt a notion of geometry due to Tits. Let I be a finite set. A geometry over I is a triple ( r , t., *) where is a set of objects, t.: r + I is a type function, and * is a symmetric incidence relation on l? such that objects u and v of the same type are incident if and only if u = v. t.(u) is the type of the object u. Notice ( r , *) is a graph. I'll usually write r for the geometry ( r , t., *).

A morphism a: r + r' of geometries is a function a: r + r' of the associated object sets which preserves type and incidence; that is if u, v E r with u * v then t.(u) = ~ ' ( u a ) and ua *' va.

A$ag of the geometry r is a set T of objects such that each pair of objects in T is incident. Notice our one (weak) axiom insures that a flag T possesses at most one object of each type, so that the type function t. induces an injection of T into I. The image t.(T) is called the type of T. The rank and corank of T are the order of t.(T) and I - t.(T), respectively. The residue rT of the flag T is (v E r - T: v * t for all t E T] regarded as a geometry over I - t.(T).

The geometry r is connected if its graph ( r , *) is connected. r is residually connected if the residue of every flag of corank at least 2 is connected and the residue of every flag of corank 1 is nonempty.

Here's a way to associate geometries to groups. Let G be a group and 9 = (G,: i E I ) a family of subgroups of G. Define r (G , F) to be the geometry whose set of objects of type i is the coset space GIG, and with objects G,x and G,y incident if G,x n G, y is nonempty. For J G I write J' for the complement I - J of J in I and define G J = nJEJ G,. Observe that for x E G, Sj,, = (GJx: j E J ) is a flag of r (G , g) of type J .

A group H of automorphisms of a geometry r is said to beflag transitive if H is transitive on flags of type J for each subset J of I.

Abstract representations 9

4 Abstract representationsLet 6' be a category. A -2-representation of a group G is a group homomor-phism n: G -f Aut(X) of G into the group Aut(X) of automorphisms of someobject X in 6. (Recall the definition of Aut(X) in section 2.) We will be mostconcerned with the following three classes of representations.

A permutation representation is a representation in the category of sets andfunctions. The group Aut(X) of automorphisms of set X is the symmetricgroup Sym(X) of X. That is Sym(X) is the group of all permutations of Xunder composition.

A linear representation is a representation in the category of vector spacesand linear transformations. Aut(X) is the general linear group GL(X) of thevector space X. That is GL(X) is the group of all invertible linear transforma-tions of X.

Finally we will of course be interested in the category of groups and grouphomomorphisms. Of particular interest is the representation of G via conjuga-tion on itself (cf. Exercise 1.3).

Two '-representations nl : G -+ Aut(X1 ), i = 1, 2, are said to be equiva-lent if there exists an isomorphism a: X1 -+X2 such that n2 = Trla*, wherea*: Aut(X1) -+ Aut(X2) is the isomorphism described in section 2. The map ais said to be an equivalence of the representations. 6-representations n, : Gi -+Aut(Xl) i = 1, 2, are said to be quasiequivalent if there exists a group isomor-phismQQ,B: G2-+ G1 of groups and a 6-isomorphism a: X1 -+ X2 such that

n2 = N7rla*Equivalent representations of a group G are the same for our purposes.

Quasiequivalent representations are almost the same, differing only by an au-tomorphism of G.

A representation 7r of G is faithful if 7r is an injection. In that event n inducesan isomorphism of G with the subgroup G7r of Aut(X), so G may be regardedas a group of automorphisms of X via 7T .

Let ni: G -+ Aut(XZ ), i = 1, 2, be 6-representations. Define a G-morphisma: X1 -+ X2 to be a morphism a Of X1 to X2 which commutes with the actionof G in the sense that (gal)a = a(g7r2) for each g E G. Write MorG(X1, X2)for the set of G-morphisms of X1 to X2. Notice that the composition of G-morphisms is a G-morphism and the identity morphism is a G-morphism.Similarly define a G-isomorphism to be a G-morphism which is also an iso-morphism. Notice the G-isomorphisms are the equivalences of representationsof G.

One focus of this book is the decomposition of a representation 7r into smallerrepresentations. Under suitable finiteness conditions (which are always presentin the representations considered here) this process of decomposition mustterminate, at which point we have associated to Tr certain indecomposable


4 Abstract representations Let B be a category. A &-representation of a group G is a group homomorphism n : G + Aut(X) of G into the group Aut(X) of automorphisms of some object X in 6'. (Recall the definition of Aut(X) in section 2.) We will be most concerned with the following three classes of representations.

A permutation representation is a representation in the category of sets and functions. The group Aut(X) of automorphisms of set X is the symmetric group Sym(X) of X. That is Sym(X) is the group of all permutations of X under composition.

A linear representation is a representation in the category of vector spaces and linear transformations. Aut(X) is the general linear group GL(X) of the vector space X. That is GL(X) is the group of all invertible linear transformations of X.

Finally we will of course be interested in the category of groups and group homomorphisms. Of particular interest is the representation of G via conjugation on itself (cf. Exercise 1.3).

Two B-representations ni: G + Aut(Xi), i = 1,2, are said to be equivalent if there exists an isomorphism a: X1 + X2 such that n2 = nla*, where a*: Aut(X1) + Aut(X2) is the isomorphism described in section 2. The map a is said to be an equivalence of the representations. &-representations ni: Gi + Aut(Xi) i = 1, 2, are said to be quasiequivalent if there exists a group isomorphism /3: G2 + G1 of groups and a &-isomorphism a : X1 + X2 such that n2 = /3n1a*.

Equivalent representations of a group G are the same for our purposes. Quasiequivalent representations are almost the same, differing only by an automorphism of G.

A representation n of G is faithful if n is an injection. In that event n induces an isomorphism of G with the subgroup Gn of Aut(X), so G may be regarded as a group of automorphisms of X via n .

Let ni: G + Aut(Xi), i = 1,2, be &-representations. Define a G-morphism a : X1 + X2 to be a morphism a of X1 to X2 which commutes with the action of G in the sense that (gnl)a = a(gnz) for each g E G. Write Morc(X1, X2) for the set of G-morphisms of X1 to X2. Notice that the composition of G- morphisms is a G-morphism and the identity morphism is a G-morphism. Similarly define a G-isomorphism to be a G-morphism which is also an isomorphism. Notice the G-isomorphisms are the equivalences of representations of G.

One focus of this book is the decomposition of a representation n into smaller representations. Under suitable finiteness conditions (which are always present in the representations considered here) this process of decomposition must terminate, at which point we have associated to n certain indecomposable

or irreducible representations which cannot be broken down further. It willdevelop that the indecomposables associated to rr are determined up to equiv-alence. Thus we are reduced to a consideration of indecomposable representa-tions.

In general indecomposables are not irreducible, so an indecomposable rep-resentation it can be broken down further, and we can associate to rr a set ofirreducible constituents. Sometimes these irreducible constituents are deter-mined up to equivalence, and sometimes not. Even when the irreducible con-stituents are determined, they usually do not determine jr. Thus we will also beconcerned with the extension problem: Given a set S of irreducible represen-tations, which representations have S as their set of irreducible constituents?There is also the problem of determining the irreducible and indecomposablerepresentations of the group.

It is possible to give a categorical definition of indecomposability (cf. Ex-ercise 1.5). There is also a uniform definition of irreducibility for the classesof representations considered most frequently (cf. Exercise 1.6). I have cho-sen however to relegate these definitions to the exercises and to make theappropriate definitions of indecomposability and irreducibility for each cat-egory in the chapter discussing the elementary representation theory of thecategory. This process begins in the next chapter, which discusses permutationrepresentations.

However one case is of particular interest. A representation of a group G onitself via conjugation (in the category of groups and group homomorphisms)is irreducible if G possesses no nonidentity proper normal subgroups. In thiscase G is said to be simple. To my mind the simple groups and their irreduciblelinear and permutation representations are the center of interest in finite grouptheory.

Exercises for chapter 11. Let G be a finite group, jr a set of primes, 0 the set of normal 7r -subgroups

of G, and r the set of normal subgroups X of G with G/X a rr-group. Prove(1) If H, K E 0 then HK E 0. Hence (S2) is the unique maximal member

of0.(2) If H, K E P then H fl K E P. Hence

I IHEP H is the unique minimalmember of P.

2. Let e, be the category of sets and functions, '2 the category of vec-tor spaces and linear transformations, and 03 the category of groups andhomomorphisms. Let F = (A,: 1 < i < n) be a family of objects in 1k.Prove

(1) Let k = 1. Then the coproduct C of F is the disjoint union of thesets A, with c1: A, -). C the inclusion map. The product P of F is


or irreducible representations which cannot be broken down further. It will develop that the indecomposables associated to n are determined up to equivalence. Thus we are reduced to a consideration of indecomposable representations.

In general indecomposables are not irreducible, so an indecomposable representation n can be broken down further, and we can associate to n a set of irreducible constituents. Sometimes these irreducible constituents are determined up to equivalence, and sometimes not. Even when the irreducible constituents are determined, they usually do not determine n . Thus we will also be concerned with the extension problem: Given a set S of irreducible representations, which representations have S as their set of irreducible constituents? .

There is also the problem of determining the irreducible and indecomposable representations of the group.

It is possible to give a categorical definition of indecomposability (cf. Ex- ercise 1.5). There is also a uniform definition of irreducibility for the classes of representations considered most frequently (cf. Exercise 1.6). I have chosen however to relegate these definitions to the exercises and to make the appropriate definitions of indecomposability and irreducibility for each category in the chapter discussing the elementary representation theory of the category. This process begins in the next chapter, which discusses permutation representations.

However one case is of particular interest. A representation of a group G on itself via conjugation (in the category of groups and group homomorphisms) is irreducible if G possesses no nonidentity proper normal subgroups. In this case G is said to be simple. To my mind the simple groups and their irreducible linear and permutation representations are the center of interest in finite group theory.

Exercises for chapter 1 1. Let G be a finite group, n a set of primes, 52 the set of normal n-subgroups

of G, and r the set of normal subgroups X of G with G/X a n-group. Prove (1) If H, K E then H K E a. Hence (a) is the unique maximal member

of a. (2) If H, K E r then H n K E r. Hence nH,, H is the unique minimal

member of r. 2. Let 8, be the category of sets and functions, 82 the category of vec-

tor spaces and linear transformations, and ti?, the category of groups and homomorphisms. Let F = (Ai: 1 _( i _( n) be a family of objects in gk. Prove (1) Let k = 1. Then the coproduct C of F is the disjoint union of the

sets Ai with ci: Ai + C the inclusion map. The product P of F is

the set product Al x ... x An with pi: P -+ Ai the projection mapPi:(a1,...,an)Hai.

(2) Let k = 2. Then C = P = i 1 Ai is the direct sum of the subspacesAi, with ci : Ai -). C defined by ai ci = (0, . . . , ai, . . . , 0) and pi: PAi the projection map.

(3) Let k = 3. Then the product P of F is the direct product A 1 x . . . x An.with pi: P Ai the projection map. (The coproduct turns out to bethe so-called free product of the family.)

3. Let G be a group, H < G, and for g E G define g7r: H -). H byx(gir) = xg, x E H. Let -' be the category of groups and homomorphisms.Prove it is a -'-representation of G with kernel CG(H). it is the repre-sentation by conjugation of G on H. If H = G, the image of G under itis the inner automorphism group of G and is denoted by Inn(G). ProveInn(G) <Aut(G). Define Out(G) = Aut(G)/Inn(G) to be the outer auto-morphism group of G.

4. Let -' be a category, F = (A1: 1 < i < n) a family of objects in -t', C andP the coproduct and product of the family with canonical maps ci : Ai C

and pi: P Ai, respectively. For ai E Aut(Ai) define ai to be the uniquemember of Mor(C, C) with ai ci = ci ai and ci = ci ai for all i j. Defineai E Mor(P, P) dually. Prove the map

n

0: fAut(Ai) Aut(X)i=1

(a1,...,CIO H a1 ... an

is an injective group homomorphism for X = C and P.5. Assume the hypothesis and notation of Exercise 1.2, and let 7r: G -> Aut(X)

be a -'k-representation, where X = C if k = 1 or 2, and X = P if k = 3.Prove the following are equivalent:(a) There exist -1'k-representations r1: G Aut(Ai ), 1 < i < n, such that

7r where 0 is the injection of Exercise 1.4 andn

G F1 Aut(Ai) is defined by g* = (girl , ... , gJrn).i-1

(b) jr is decomposable.(If k = 1, transitivity is the same as indecomposability. See chapter 2 forthe definition of transitivity. See chapters 5 and 4 for the definitions ofdecomposability when k = 2 and 3.)

6. Assume the hypothesis and notation of Exercise 1.2, let X be an object in-'k, and p: G Aut(X) a -'k-representation. A -ik-equivalence relationon X is an equivalence relation on X such that - is preserved by theoperations on X if k = 2 or 3 (i.e. if y is an n-ary operation on X and

Abstract representations 1 1

the set product Al x . . - x A, with pi: P + Ai the projection map pi :(al, .. . ,a,) H ai.

(2) Let k = 2. Then C = P = @:='=,Ai is the direct sum of the subspaces Ai, with ci: Ai + C defined by aici = (0,. . . , ai, . . . ,0 ) and pi: P + Ai the projection map.

(3) Let k = 3. Then the product P of F is the direct product Al x . . . x A,. with pi: P + Ai the projection map. (The coproduct turns out to be the so-called free product of the family.)

3. Let G be a group, H II G, and for g E G define gn: H + H by x(gn) = xg, x E H. Let & be the category of groups and homomorphisms. Prove n is a &-representation of G with kernel CG(H). n is the representation by conjugation of G on H. If H = G, the image of G under n is the inner automorphism group of G and is denoted by Inn(G). Prove Inn(G) <I Aut(G). Define Out(G) = Aut(G)/Inn(G) to be the outer automorphism group of G.

4. Let 6' be a category, F = (Ai: 1 5 i 5 n) a family of objects in 8, C and P the coproduct and product of the family with canonical maps ci: Ai + C and pi: P + Ai, respectively. For ai E Aut(Ai) define Ei to be the unique member of Mor(C, C) with ai ci = ciEi and c j = c jEi for all i # j. Define Ei E Mor(P, P ) dually. Prove the map

(a1, . . . , a,) H El . . . En

is an injective group homomorphism for X = C and P. 5. Assume the hypothesis and notation of Exercise 1.2, and let n : G + Aut(X)

be a gk-representation, where X = C if k = 1 or 2, and X = P if k = 3. Prove the following are equivalent: (a) There exist &k-representations ni: G + Aut(Ai), 1 5 i 5 n, such that

n = $4, where 4 is the injection of Exercise 1.4 and n

$ : G + n Aut(Ai) is defined by g$ = (gnl, . . . , gn,).

(b) n is decomposable. (If k = 1, transitivity is the same as indecomposability. See chapter 2 for the definition of transitivity. See chapters 5 and 4 for the definitions of decomposability when k = 2 and 3.)

6. Assume the hypothesis and notation of Exercise 1.2, let X be an object in -ek, and p: G + Aut(X) a &k-representation. A &k-equivalence relation on X is an equivalence relation - on X such that - is preserved by the operations on X if k = 2 or 3 (i.e. if y is an n-ary operation on X and


xi ^- yi then y(xl, ... , xn) - y(yl,... , y,,)). Define Grr to preserve ^-if x ^- y implies xgTr yglr for each g c G. Prove that (a) and (b) areequivalent:

(a) Gzr preserves no nontrivial k-equivalence relation on X.(b) is an irreducible -'k-representation.

(See chapters 5 and 4 for the definition of an irreducible -'k-repre-sentation when k = 2 and 3. A wl 1-representation is irreducible if it isprimitive, and primitivity is defined in chapter 2.)

7. Let n, or: G -* Aut(X) be faithful -'-representations. Prove rr is quasi-equivalent to or if and only if G7r is conjugate to Ga in Aut(X).

8. Let G be a group and 9 = (Gi : i E I) a family of subgroups of G. Prove(1) The geometry F = 1'(G, O is connected if and only if G = (J9,_).(2) For g c G, define gir: F - F by (Gix)g'r = Gixg. Prove it is a re-

presentation of G as a group of automorphisms of F.


xi -- yi then y(x1,. . . , x,) -- y(y1,. . . , y,)). Define G n to preserve - if x - y implies xgn -- ygn for each g E G. Prove that (a) and (b) are equivalent: (a) G n preserves no nontrivial gk-equivalence relation on X. (b) is an irreducible gk-representation.

(See chapters 5 and 4 for the definition of an irreducible gk-representation when k = 2 and 3. A &l-representation is irreducible if it is primitive, and primitivity is defined in chapter 2.)

7. Let n, 0 : G + Aut(X) be faithful &-representations. Prove n is quasiequivalent to o if and only if G n is conjugate to G o in Aut(X).

8. Let G be a group and 9 = (Gi: i E I) a family of subgroups of G. Prove (1) The geometry r = r(G, 5F) is connected if and only if G = (F). (2) For g E G, define gn : r + r by (Gix)gn = Gixg. Prove n is a re-

presentation of G as a group of automorphisms of r.

Permutation representations

Section 5 develops the elementary theory of permutation representations. Thefoundation for this theory is the notion of the transitive permutation represen-tation. The transitive representations play the role of the indecomposables inthe theory. It will develop that every transitive permutation representation of agroup G is equivalent to a representation by right multiplication on the set ofcosets of some subgroup of G. Hence the study of permutation representationsof G is equivalent to the study of the subgroup structure of G.

Section 6 is devoted to a proof of Sylow's Theorem. The proof supplies anice application of the techniques developed in section 5. Sylow's Theorem isone of the most important results in finite group theory. It is the first theoremin the local theory of finite groups. The local theory studies a finite group fromthe point of view of its p-subgroups and the normalizers of these p-subgroups.

5 Permutation representationsIn this section X is a set, G a group, and 7r: G -* Sym(X) is a permutation rep-resentation of G. Recall Sym(X) is the symmetric group on X; that is Sym(X)is the group of all permutations of X. Thus Sym(X) is the automorphism groupof X in the category of sets and functions, and 7r is a representation in thatcategory.

For x c X and a E Sym(X) write xa for the image of x under a. Noticethat, by definition of multiplication in Sym(X):

x(ap) = (xa)p x E X, a, P E Sym(X).

I'll often suppress the representation 7r and write xg for x(g7r), x E X, g E G.One feature of this notation is that:

x(gh) = (xg)h x E X, g, h E G.

The relation - on X defined by x - y if and only if there exists g e G withxg = y is an equivalence relation on X. The equivalence class of x under thisrelation is

xG=fxg:geG}

and is called the orbit of x under G. As the equivalence classes of an equivalencerelation partition a set, X is partitioned by the orbits of G on X.


Section 5 develops the elementary theory of permutation representations. The foundation for this theory is the notion of the transitive permutation representation. The transitive representations play the role of the indecomposables in the theory. It will develop that every transitive permutation representation of a group G is equivalent to a representation by right multiplication on the set of cosets of some subgroup of G. Hence the study of permutation representations of G is equivalent to the study of the subgroup structure of G.

Section 6 is devoted to a proof of Sylow's Theorem. The proof supplies a nice application of the techniques developed in section 5. Sylow's Theorem is one of the most important results in finite group theory. It is the first theorem in the local theory of finite groups. The local theory studies a finite group from the point of view of its p-subgroups and the normalizers of these p-subgroups.

5 Permutation representations In this section X is a set, G a group, and n: G + Sym(X) is a permutation representation of G. Recall Sym(X) is the symmetric group on X; that is Sym(X) is the group of all permutations of X. Thus Sym(X) is the automorphism group of X in the category of sets and functions, and n is a representation in that category.

For x E X and a E Sym(X) write x a for the image of x under a . Notice that, by definition of multiplication in Sym(X):

x(ap) = (xa)p x E X, a , p E Sym(X).

I'll often suppress the representation n and write xg for x(gn), x E X, g E G. One feature of this notation is that:

The relation -- on X defined by x -- y if and only if there exists g E G with xg = y is an equivalence relation on X. The equivalence class of x under this relation is

and is called the orbit of x under G. As the equivalence classes of an equivalence relation partition a set, X is partitioned by the orbits of G on X.

14 Permutation representations

Let Y be a subset of X. G is said to act on Y if Y is a union of orbits of G.Notice G acts on Y precisely when yg E Y for each y E Y, and each g E G.Further if G acts on Y then g IY is a permutation of Y for each g E G, and therestriction map

G -f Sym(Y)

gHglyis a permutation representation with kernel

Gy={g(=-G:yg=y forall yEY}.

Hence Gy < G when G acts on Y. Even when G does not act on Y, we canconsider

G(Y) = {g E G: Yg = Y},

where Yg = { yg: y E Y}. GY and G(Y) are subgroups of G called the pointwisestabilizer of Y in G and the global stabilizer of Y in G, respectively. G(Y) isthe largest subgroup of G acting on Y. Write GY for the image of G(Y) underthe restriction map on Y. We have seen that:

(5.1) The restriction map of G(Y) on Y is a permutation representation of G(Y)with kernel Gy and image GY - G(Y)/Gy, for each subset Y of X.

For X E X write Gx for G{x}. Next for S c G define

Fix(S) = {x E X: xs = x f o r all s E S}.

Fix(S) is the set of fixed points of S. Notice Fix(S) = Fix((S)). Also

(5.2) If H a G then G acts on Fix(H). More generally G permutes the orbitsof H of cardinality c, for each c.

For Y C X, I'll sometimes write CG(Y) and NG(Y) for Gy and G(Y), respec-tively, and I'll sometimes write Cx(G) for Fix(G). Usually this notation willbe used only when X possesses a group structure preserved by G.

(5.3) Let P be the set of all subsets of X. Then a: G - Sym(P) is a per-mutation representation of G on P where ga: Y Yg for each g E G andYcX.

7r is a transitive permutation representation if G has just one orbit on X;equivalently for each x, y E X there exists g E G with xg = y. G will also besaid to be transitive on X.

Let Y be a subset of X. G is said to act on Y if Y is a union of orbits of G. Notice G acts on Y precisely when yg E Y for each y E Y, and each g E G. Further if G acts on Y then gly is a permutation of Y for each g E G, and the restriction map

G + Sym(Y)

g I+ glr

is a permutation representation with kernel

G y = { g € G : y g = y forall ~ E Y ) .

Hence Gy L] G when G acts on Y. Even when G does not act on Y, we can consider

G(Y) = (g E G: Yg = Y),

where Yg = (yg: y E Y). Gy and G(Y) are subgroups of G called thepointwise stabilizer of Y in G and the global stabilizer of Y in G, respectively. G(Y) is the largest subgroup of G acting on Y. Write G~ for the image of G(Y) under the restriction map on Y. We have seen that:

(5.1) The restriction map of G(Y) on Y is apermutationrepresentation of G(Y) with kernel Gy and image GY 2 G(Y)/Gy, for each subset Y of X.

For x E X write G, for G{,). Next for S G G define

Fix(S) = {x E X: xs = x for all s E S).

Fix(S) is the set offuedpoints of S. Notice Fix(S) = Fix((S)). Also

(5.2) If H 9 G then G acts on Fix(H). More generally G permutes the orbits of H of cardinality c, for each c.

For Y X, I'll sometimes write CG(Y) and NG(Y) for Gy and G(Y), respectively, and I'll sometimes write Cx(G) for Fix(G). Usually this notation will be used only when X possesses a group structure preserved by G.

(5.3) Let P be the set of all subsets of X. Then a : G + Sym(P) is a permutation representation of G on P where ga: Y + Yg for each g E G and Y E X.

IT is a transitive permutation representation if G has just one orbit on X; equivalently for each x, y E X there exists g E G with xg = y. G will also be said to be transitive on X.

Permutation representations 15

Here's one way to generate transitive representations of G:

(5.4) Let H < G. Then a: G - Sym(G/H) is a transitive permutation repre-sentation of G on the coset space G/H, where

ga: Hx i-+ Hxg.

a is the representation of G on the cosets of H by right multiplication. H is thestabilizer of the coset H in this representation.

We'll soon see that every transitive representation of G is equivalent to arepresentation of G by right multiplication on the cosets of some subgroup.But first here is another way to generate permutation representations of G.

(5.5) The map a: G - Sym(G) is a permutation representation of G on itself,where

ga:xHxg x,gEG.a is the representation of G on itself by conjugation. For S c G, the globalstabilizer of S in G is NG(S), while CG(S) is the pointwise stabilizer of S.

Notice that 5.5 is essentially a consequence of Exercise 1.3. Recall NG(S) =(g E G: Sg = S}, Sg = (sg: s E S}, and CG(S) = (g E G: sg = s for all s ES}. By 5.3, G is also represented on the power set of G, and evidently, forS C G, the set SG = (S9: g E G} of conjugates of S under G is the orbit of Sunder G with respect to this representation.

(5.6) Let Y C X and g E G. Then G(Yg) = G(Y)9 and Gyg = (GY)g.

(5.7) Assume 7r is a transitive representation and let x E X and H = Gx . Then

ker(7r) = n Hg = kerH(G)gEG

is the largest normal subgroup of G contained in H.

(5.8) Assume it is a transitive permutation representation, let x E X, H = G,and let a be the representation of G on the cosets of H by right multiplication.Define

8: G/H - XHg i-+ xg.

Then P is an equivalence of the permutation representations a and 7r.


Here's one way to generate transitive representations of G:

(5.4) Let H 5 G. Then a: G + Sym(G/H) is a transitive permutation representation of G on the coset space G/H, where

ga: Hx I+ Hxg.

a is the representation of G on the cosets of H by right multiplication. H is the stabilizer of the coset H in this representation.

We'll soon see that every transitive representation of G is equivalent to a representation of G by right multiplication on the cosets of some subgroup. But first here is another way to generate permutation representations of G.

(5.5) The map a: G + Sym(G) is a permutation representation of G on itself, where

ga: x I+ xg x , g E G.

a is the representation of G on itself by conjugation. For S G, the global stabilizer of S in G is NG(S), while CG(S) is the pointwise stabilizer of S.

Notice that 5.5 is essentially a consequence of Exercise 1.3. Recall NG(S) = {g E G: Sg = S) , Sg = {sg: s E S), and CG(S) = {g E G: sg = s for all s E

S]. By 5.3, G is also represented on the power set of G, and evidently, for S G, the set sG = {Sg: g E G ) of conjugates of S under G is the orbit of S under G with respect to this representation.

(5.6) Let Y X and g E G. Then G(Yg) = G(Y)g and Gya = (Gy)g.

(5.7) Assume n is a transitive representation and let x E X and H = G,. Then

is the largest normal subgroup of G contained in H.

(5.8) Assume n is a transitive permutation representation, let x E X, H = G,, and let a be the representation of G on the cosets of H by right multiplication. Define

B: G/H + X

Hg I-+ xg.

Then B is an equivalence of the permutation representations a and n.


Proof. We must show ,B is a well-defined bijection of G/H with X and that,for each a, g E G, (Ha)O(ga) = (Ha)gnf. Both computations are straight-forward.

(5.9) (1) Every transitive permutation representation of G is equivalent to arepresentation of G by right multiplication on the cosets of some subgroup.

(2) If jr': G -> Sym(X') and rr are transitive representations, x E X, andx' E X', then n is equivalent to jr' if and only if Gx is conjugate to G,,, in G.

Proof. Part (1) follows from 5.8. Assume the hypothesis of (2). If, 8: X -> X'is an equivalence of jr and n' then Gx = GxA and, by 5.6, Gxp is conjugate toGx, in G. Conversely if Gx is conjugate to Gx in G, then by 5.6 there is y E Xwith Gy = Gx' and by 5.8 both jr and jr' are equivalent to the representationof G on the cosets of Gx' and hence equivalent to each other.

Let (Xi: i E I) be the orbits of G on X and ni the restriction of n to Xi. By5.1, ni is a permutation representation of G on Xi and, as Xi is an orbit ofG, ni is even a transitive representation. (ni: i c= I) is the family of transitiveconstituents of n and we write n = Eic] ni. Evidently:

(5.10) The transitive constituents (ni : i E I) of n are uniquely determined byn, and if n' is a permutation representation of G with transitive constituents(jr: j E J), then n is equivalent to n' if and only if there is a bijection a of Iwith J such that nta is equivalent to ni for each i E I.

So the study of permutation representations is effectively reduced to the studyof transitive representations, and 5.9 says in turn that the transitive permutationrepresentations of a group are determined by its subgroup structure.

The transitive representations play the role of the indecomposable permuta-tion representations. For example see Exercise 1.5.

(5.11) If G is transitive on X then X has cardinality IG : Gx I for each x E X.

Proof. This is a consequence of 5.8.

(5.12) Let S C G. Then S has exactly IG NG(S)I conjugates in G.

Proof. We observed earlier that G is transitively represented on the set SG ofconjugates of S via conjugation, while, by 5.5, NG(S) is the stabilizer of S withrespect to this representation, so the lemma follows from 5.11.


Proof. We must show /3 is a well-defined bijection of G/H with X and that, for each a , g E G, (Ha)B(ga) = (Ha)gnp. Both computations are straightforward.

(5.9) (1) Every transitive permutation representation of G is equivalent to a representation of G by right multiplication on the cosets of some subgroup.

(2) If n': G += Sym(X1) and n are transitive representations, x E X, and x' E X', then 7t is equivalent to n' if and only if G, is conjugate to G,t in G.

Proof. Part (1) follows from 5.8. Assume the hypothesis of (2). If B: X += X' is an equivalence of n and n' then G, = GXB and, by 5.6, GXB is conjugate to G,? in G. Conversely if G,I is conjugate to G, in G, then by 5.6 there is y E X with G, = G,I and by 5.8 both n and n' are equivalent to the representation of G on the cosets of G,. and hence equivalent to each other.

Let (Xi: i E I ) be the orbits of G on X and ni the restriction of n to Xi. By 5.1, q is a permutation representation of G on Xi and, as Xi is an orbit of G, ni is even a transitive representation. (ni: i E I ) is the family of transitive constituents of n and we write n = xi,, q. Evidently:

(5.10) The transitive constituents (xi: i E I) of n are uniquely determined by n , and if n' is a permutation representation of G with transitive constituents (nj: j E J), then n is equivalent to n' if and only if there is a bijection a! of I with J such that ni6( is equivalent to ni for each i E I.

So the study of permutation representations is effectively reduced to the study of transitive representations, and 5.9 says in turn that the transitive permutation representations of a group are determined by its subgroup structure.

The transitive representations play the role of the indecomposable permutation representations. For example see Exercise 1.5.

(5.11) If G is transitive on X then X has cardinality IG : G, J for each x E X.


(5.12) Let S G G. Then S has exactly J G : NG(S)I conjugates in G.

Proof. We observed earlier that G is transitively represented on the set SG of conjugates of S via conjugation, while, by 5.5, NG(S) is the stabilizer of S with respect to this representation, so the lemma follows from 5.11.

Let p be a prime and recall that a p-group is a group whose order is a powerof p.

(5.13) If G is a p-group then all orbits of G on X have order a power of p.

Proof. This follows from 5.11 and the fact that the index of any subgroup ofG divides the order of G.

(5.14) Let G be a p-group and assume X is finite. Then IXI - (Fix(G)I mod p.

Proof. As the fixed points of G are its orbits of length 1, 5.14 follows from 5.13.

Here are a couple applications of 5.14:

(5.15) Let G and H be p-groups with H 1 and let a: G Aut(H) be agroup homomorphism. Then CH(G) ; 1.

Proof. a is also a permutation representation of G on H. By 5.14, CHIJFix(G)I mod p. But Fix(G) = CH(G) in this representation, while IHI - 0mod p as H is a p-group with H 1. So CH(G) 1, as claimed.

(5.16) If G is a p-group with G 1, then Z(G) 1.

Proof. Apply 5.15 to the representation of G on itself by conjugation and recallZ(G) = CG(G).

The following technical lemma will be used in the next section to prove Sylow'sTheorem:

(5.17) Let X be finite and assume for each x E X that there exists a p-subgroupP(x) of G such that {x} = Fix(P(x)). Then

(1) G is transitive on X, and(2) 1 XI - 1 mod p.

Proof. Let X = Y + Z be a partition of X with G acting on Y and Z. LetY 0 and pick y E Y. For V = Y or Z and H < G denote by Fixv(H) thefixed points of H on V. By hypothesis (y} = Fix(P(y)), so 1 = (Fixy(P(y))Iand0= IFix,(P(y))I.Hence, by5.14,IY) 1modpandIZI 0modp.Butif Z is nonempty, then, by symmetry, I Y I 1 mod p, a contradiction. Thus


Let p be a prime and recall that a p-group is a group whose order is a power of p.

(5.13) If G is a p-group then all orbits of G on X have order a power of p.

Proof. This follows from 5.1 1 and the fact that the index of any subgroup of G divides the order of G.

(5.14) Let G be a p-group and assume X is finite. Then 1x1 - IFix(G)I mod p.

Proof. As the fixed points of G are its orbits of length 1,5.14 follows from 5.13.

Here are a couple applications of 5.14:

(5.15) Let G and H be p-groups with H # 1 and let a: G -+ Aut(H) be a group homomorphism. Then CH (G) # 1.

Proof. a is also a permutation representation of G on H . By 5.14, (HI -. IFix(G)I mod p. But Fix(G) = CH(G) in this representation, while IHI =- 0 mod p as H is a p-group with H # 1. So Cn(G) # 1, as claimed.

(5.16) If G is a p-group with G # 1, then Z(G) # 1.

Proof. Apply 5.15 to the representation of G on itself by conjugation and recall Z(G) = CG(G).

The following technical lemma will be used in the next section to prove Sylow's Theorem:

(5.17) Let X be finite and assume for each x E X that there exists a p-subgroup P(x) of G such that {x) = Fix(P(x)). Then

(1) G is transitive on X, and (2) /XI zz 1 modp.

Proof. Let X = Y + Z be a partition of X with G acting on Y and Z. Let Y # 0 and pick y E Y. For V = Y or Z and H _( G denote by Fixv(H) the fixed points of H on V. By hypothesis (y) = Fix(P(y)), so 1 = IFixy (P(y))l and 0 = IFix,(P(y))l. Hence, by 5.14, IYI r 1 mod p and I ZI -. 0 mod p. But if Z is nonempty, then, by symmetry, [YI r 1 mod p, a contradiction. Thus

Y = X and, as IYI =- 1 mod p, (2) holds. Since we could have chosen Y to bean orbit of G on X, (1) holds.

Let Q be a partition of X. Q is G-invariant if G permutes the members ofQ. Equivalently regard Q as a subset of the power set P of X and representG on P as in 5.3; then Q is G-invariant if G acts on the subset Q of P withrespect to this representation. In particular notice that if Q is G-invariant thenthere is a natural permutation representation of G on Q. Q is nontrivial ifQ0{{x}:xEX)andQ0{X}.

Let G be transitive on X. G is imprimitive on X if there exists a nontrivial G-invariant partition Q of X. In this event Q is said to be a system of imprimitivityfor G on X. G is primitive on X if it is transitive and not imprimitive.

(5.18) Let G be transitive on X and y E X.(1) If Q is a system of imprimitivity for G on X and y E Y E Q, then G is

transitive on Q, the stabilizer H of Y in G is a proper subgroup of G properlycontaining Gy, Y is an orbit of H on X, IXI = IYIIQI, IQI = IG : HI, andIYI = IH:GyI.

(2) If Gy < H < G then Q = {Yg: g E G) is a system of imprimitivity forG on X, where Y = yH and H is the stabilizer of Y in G.

The proof is left as an exercise. As a direct consequence of 5.18 we have:

(5.19) Let G be transitive on X and X E X. Then G is primitive on X if andonly if Gx is a maximal subgroup of G.

Let G be finite and transitive on X, let I X I > 1, and let x E X. Then there isa sequence Gx = Ho < Hl < ... < H = G with H, maximal in Hi+i. Thisgives rise to a family of primitive permutation representations: the representa-tions of Hi+I on the cosets of Hi. This family of primitive representations canbe used to investigate the representation it of G on X.

From this point of view the primitive representations play the role of irre-ducible permutation representations. See also Exercise 1.6.

I close this section with two useful lemmas. The proofs are left as exercises.

(5.20) Let G be transitive on X, X E X, and H < G. Then H is transitive onX if and only if G = GxH.

(5.21) Let G be transitive on X, X E X, H = Gx and K < H. Then NG(K) istransitive on Fix(K) if and only if KG fl H = K".


Y = X and, as IY 1 E 1 mod p, (2) holds. Since we could have chosen Y to be an orbit of G on X , (1) holds.

Let Q be a partition of X . Q is G-invariant if G permutes the members of Q . Equivalently regard Q as a subset of the power set P of X and represent G on P as in 5.3; then Q is G-invariant if G acts on the subset Q of P with respect to this representation. In particular notice that if Q is G-invariant then there is a natural permutation representation of G on Q . Q is nontrivial if

Q # { I x ) : x E XI and Q # 1x1. Let G be transitive on X . G is imprimitive on X if there exists a nontrivial G-

invariant partition Q of X . In this event Q is said to be a system of imprimitivity .

for G on X . G is primitive on X if it is transitive and not imprimitive.

(5.18) Let G be transitive on X and y E X . (1) If Q is a system of imprimitivity for G on X and y E Y E Q , then G is

transitive on Q , the stabilizer H of Y in G is a proper subgroup of G properly containing G,, Y is an orbit of H on X , 1x1 = IYllQl, lQl = I G : HI, and IYl = I H : Gyl.

(2) If Gy < H < G then Q = {Yg: g E G) is a system of imprimitivity for G on X , where Y = yH and H is the stabilizer of Y in G.

The proof is left as an exercise. As a direct consequence of 5.18 we have:

(5.19) Let G be transitive on X and x E X . Then G is primitive on X if and only if G, is a maximal subgroup of G.

Let G be finite and transitive on X , let 1x1 > 1, and let x E X . Then there is a sequence G, = Ho ( HI 5 . . ( H,, = G with Hi maximal in Hi+l . This gives rise to a family of primitive permutation representations: the representations of Hi+l on the cosets of Hi. This family of primitive representations can be used to investigate the representation n of G on X .

From this point of view the primitive representations play the role of irreducible permutation representations. See also Exercise 1.6.

I close this section with two useful lemmas. The proofs are left as exercises.

(5.20) Let G be transitive on X , x E X , and H 5 G. Then H is transitive on X if and only if G = G, H .

(5.21) Let G be transitive on X , x E X , H = G , and K 5 H. Then NG(K) is transitive on Fix(K) if and only if lCG n H = K H .

Sylow's Theorem 19

6 Sylow's TheoremIn this section G is a finite group. If n is a positive integer and p a prime, writenp for the highest power of p dividing n. np is the p-part of n.

A Sylow p-subgroup of G is a subgroup of G of order I G I p. Write Sy1p(G)for the set of Sylow p-subgroups of G.

In this section we prove:

Sylow's Theorem. Let G be a finite group and p a prime. Then(1) Sylp(G) is nonempty.(2) G acts transitively on Sylp(G) via conjugation.(3) ISylp(G)I = I G : NG(P)I - 1 mod p for P E Sylp(G).(4) Every p-subgroup of G is contained in some Sylow p-subgroup of G.

Let P be the set of all p-subgroups of G and 0 the set of all maximal p-subgroups of G; that is partially order r by inclusion and let 0 be the maximalmembers of this partially ordered set. It follows from 5.3 and 5.5 that G isrepresented as a permutation group via conjugation on the power set of G, andit is evident that G acts on F, 2, and Sy1p(G) with respect to this representation.

Let R E 0. Claim R is the unique point of 0 fixed by the subgroup R of G.For if R fixes Q E 0 then, by 5.5, R < NG (Q), so, by 1.7.2, RQ < G and,by 1.7.3, IRQI = IRIIQI/IR n Q. Thus RQ E F, so, as R < RQ ? Q andR, Q E S2, we conclude R = R Q = Q. So the claim is established.

I've shown that for each R E 0 there is a p-subgroup P(R) of G such thatR is the unique point of 0 fixed by P(R); namely R = P(R). So it followsfrom 5.17 that:

(i) G is transitive on 0, and(ii) IQ I-1mod p.

Let P E 0 and suppose I P I= I G I p. Then P E Sylp (G), so, as G is transitiveon 0 and G acts on Sylp (G), we have S2 C Sylp (G). On the other hand as I R Idivides I G I for each R E 0 it is clear Sylp(G) c 0, so SZ = Sylp(G). Thus(i) implies parts (1) and (2) of Sylow's Theorem, while (ii) and 5.12 implypart (3). Evidently each member of r is contained in a member of 0, so, asSZ = Sylp(G), the fourth part of Sylow's Theorem holds.

So to complete the proof of Sylow's Theorem it remains to show I PI = IGIp

for P E 0. Assume otherwise and let M = NA(P). By (i), (ii), and 5.12,I G : M I- 1 mod p, so I M I p= I G I p, and hence p divides I M/P I. Therefore,by Cauchy's Theorem (Exercise 2.3), there exists a subgroup R/P of M/P of

order p. But now I R I = I R/P I I P I is a power of p, so P < R E I', contradictingPES2.

This completes the proof of Sylow's Theorem.

6 Sylow's Theorem In this section G is a finite group. If n is a positive integer and p a prime, write n, for the highest power of p dividing n. n, is the p-part of n.

A Sylow p-subgroup of G is a subgroup of G of order I G I,. Write Syl, (G) for the set of Sylow p-subgroups of G.

In this section we prove:

Sylow's Theorem. Let G be a finite group and p a prime. Then (1) Syl,(G) is nonempty. (2) G acts transitively on Syl,(G) via conjugation. (3) ISyl,(G)I = IG : NG(P)I r 1 mod p for P E Syl,(G). (4) Every p-subgroup of G is contained in some Sylow p-subgroup of G.

Let r be the set of all p-subgroups of G and Q the set of all maximal p- subgroups of G; that is partially order r by inclusion and let Q be the maximal members of this partially ordered set. It follows from 5.3 and 5.5 that G is represented as a permutation group via conjugation on the power set of G, and it is evident that G acts on r, Q, and Sylp(G) with respect to this representation.

Let R E Q. Claim R is the unique point of Q fixed by the subgroup R of G. For if R fixes Q E Q then, by 5.5, R 5 NG(Q), so, by 1.7.2, RQ 5 G and, by 1.7.3, [RQl = IRIIQIIIR n Ql. Thus RQ E r, so, as R 5 RQ 2 Q and R, Q E C2, we conclude R = RQ = Q. So the claim is established.

I've shown that for each R E Q there is a p-subgroup P(R) of G such that R is the unique point of Q fixed by P(R); namely R = P(R). So it follows from 5.17 that:

(i) G is transitive on Q, and (ii) 1Q1 E 1 modp.

Let P E Q and suppose I P I = I G I,. Then P E Syl,(G), so, as G is transitive on Q and G acts on Syl,(G), we have C2 E Syl,(G). On the other hand as I RI divides JGJ for each R E Q it is clear Syl,(G) G Q, so Q = Syl,(G). Thus (i) implies parts (1) and (2) of Sylow's Theorem, while (ii) and 5.12 imply part (3). Evidently each member of r is contained in a member of Q, so, as Q = Syl,(G), the fourth part of Sylow's Theorem holds.

So to complete the proof of Sylow's Theorem it remains to show I PI = I G 1, for P E Q. Assume otherwise and let M = NG(P). By (i), (ii), and 5.12, IG :MI G 1 mod p, so /MIp = IGl,, and hence p divides [MIPI. Therefore, by Cauchy's Theorem (Exercise 2.3), there exists a subgroup RIP of MIP of order p. But now I R I = I RIP I I P I is a power of p, so P < R E r, contradicting P E Q.

This completes the proof of Sylow's Theorem.


Next a few consequences of Sylow's Theorem.

(6.1) Let P E Sylp(G). Then P a G if and only if P is the unique Sylowp-subgroup of G.

Proof. This is because G acts transitively on Sylp(G) via conjugation withNG(P) the stabilizer of P in this representation.

Lemma 6.1 and the numerical restrictions in part (3) of Sylow's Theorem canbe used to show groups of certain orders have normal Sylow groups. See forexample Exercises 2.5 and 2.6.

(6.2) (Frattini Argument) Let H a G and P E Sylp(H). Then G = HNG(P).

Proof. Apply 5.20 to the representation of G on Sylp(H), using Sylow's The-orem to get H transitive on Sylp(H).

Actually Lemma 6.2 is a special case of the following lemma, which has asimilar proof, and which I also refer to as a Frattini Argument:

(6.3) (Frattini Argument) Let K be a group, H a K, and X C H. Then K =HNK (X) if and only if XK = X H. Indeed H has I K : HNK (X) I orbits of equal

length on XK, with representatives (X-Y: y c Y), where Y is a set of cosetrepresentatives for HNK(X) in K.

(6.4) Let H a G and P E Sylp(G). Then P n H E Sylp(H).

Exercises for chapter 21. Prove Lemma 5.18.2. Prove Lemmas 5.20 and 5.21.3. Prove Cauchy's Theorem: Let G be a finite group and p a prime divisor of

G 1. Then G contains an element of order p. (Hint: Prove p divides I CG (x)for some x E G#. Then proceed by induction on

4. Let G be a finite group and p a prime. Prove:(1) If G/Z(G) is cyclic, then G is abelian.(2) If G A = p2 then G = Zp2 or Ep2.

5. (1) Let IGI = pem, p > m, p prime, (p, m) = 1. Prove G has a normalSylow p-subgroup.

(2) Let IGI = pq, p and q prime. Prove G has a normal Sylow p-subgroupor a normal Sylow q-subgroup.

Next a few consequences of Sylow's Theorem.

(6.1) Let P E SylP(G). Then P 5 G if and only if P is the unique Sylow p-subgroup of G.

Proof. This is because G acts transitively on Sylp(G) via conjugation with NG(P) the stabilizer of P in this representation.

Lemma 6.1 and the numerical restrictions in part (3) of Sylow's Theorem can be used to show groups of certain orders have normal Sylow groups. See for example Exercises 2.5 and 2.6.

(6.2) (Frattini Argument) Let H 5 G and P E Syl,(H). Then G = HNG(P).

Proof. Apply 5.20 to the representation of G on SyI,(H), using Sylow's The- orem to get H transitive on Syl,(H).

Actually Lemma 6.2 is a special case of the following lemma, which has a similar proof, and which I also refer to as a Frattini Argument:

(6.3) (Frattini Argument) Let K be a group, H 5 K, and X E H. Then K = HNK (X) if and only if x = X H . Indeed H has I K : HNK (X) 1 orbits of equal length on x K , with representatives (XY: y E Y), where Y is a set of coset representatives for HNK(X) in K.

(6.4) Let H 5 G and P E Syl,(G). Then P n H E Syl,(H).

Exercises for chapter 2 1. Prove Lemma 5.18. 2. Prove Lemmas 5.20 and 5.21. 3. Prove Cauchy's Theorem: Let G be a finite group and p a prime divisor of

(GI. Then G contains an element of order p. (Hint: Prove p divides 1 CG(x)l for some x E G'. Then proceed by induction on !GI.)

4. Let G be a finite group and p a prime. Prove: (1) If G/Z(G) is cyclic, then G is abelian. (2) If JGI = P2 then G 2 Zpz or Ep2.

5. (1) Let /GI = pem, p > m, p prime, (p, m) = 1. Prove G has a normal Sylow p-subgroup.

(2) Let I G I = pq, p and q prime. Prove G has a normal Sylow p-subgroup or a normal Sylow q-subgroup.

Sylow's Theorem 21

6. Let I G I = pq2, where p and q are distinct primes. Prove one of the followingholds:(1) q > p and G has a normal Sylow q-group.(2) p > q and G has a normal Sylow p-group.(3) I G I = 12 and G has a normal Sylow 2-group.

7. Let G act transitively on a set X, x E X, and P E Sylp(G,,). Prove NG(P)is transitive on Fix(P).

8. Prove Lemmas 6.3 and 6.4.9. Prove that if G has just one Sylow p-subgroup for each p E 7r(G), then G

is the direct product of its Sylow p-subgroups.

6. Let I G I = pq2, where p and q are distinct primes. Prove one of the following holds: (1) q > p and G has a normal Sylow q-group. (2) p > q and G has a normal Sylow p-group. (3) IG I = 12 and G has a normal Sylow 2-group.

7. Let G act transitively on a set X, x E X, and P E Syl,(G,). Prove Nc(P) is transitive on Fix(P).

8. Prove Lemmas 6.3 and 6.4. 9. Prove that if G has just one Sylow p-subgroup for each p E n(G), then G

is the direct product of its Sylow p-subgroups.

3

Representations of groups on groups

Chapter 3 investigates representations in the category of groups and homomor-phisms, with emphasis on the normal and subnormal subgroups of groups.

In section 7 the concept of an irreducible representation is defined, and theJordan-Holder Theorem is established. As a consequence, the compositionfactors of a finite group are seen to be an invariant of the group, and thesecomposition factors are simple.

The question arises as to how much the structure of a group is controlledby its composition factors. Certainly many nonisomorphic groups can have thesame set of composition factors, so control is far from complete. To investigatethis question further we must consider extensions of a group G by a group A.Section 10 studies split extensions and introduces semidirect products.

Section 9 investigates solvable and nilpotent groups. For finite groups thisamounts to the study of groups all of whose composition factors are, in the firstcase, of prime order and, in the second, of order p for some fixed prime p.

Commutators, characteristic subgroups, minimal normal subgroups, centralproducts, and wreath products are also studied.

7 Normal seriesIn this section G and A are groups, and n: A -p Aut(G) is a representationof A in the category of groups and homomorphisms. I'll also say that Aacts as a group of automorphisms on G. Observe that 7r is also a permuta-tion representation, so we can use the terminology, notation, and results fromchapter 2.

A normal series of length n for G is a series

1=Go<G1a...4G.=G.The series is A-invariant if A acts on each Gi.

(7.1) Let (Gi : 0 < i < n) be an A-invariant normal series and H an A-invariantsubgroup of G. Then

(1) The restriction 7r I H: A -a Aut(H) is a representation of A on H.(2) (Gi fl H: 0 < i < n) is an A-invariant normal series for H.(3) If H < G thenTrG/H: A-4 Aut(G/H) is a representation, where a7rG/H:

Hg HH(ga) for a E A, g E G.

Representations of groups on groups

Chapter 3 investigates representations in the category of groups and homomorphisms, with emphasis on the normal and subnormal subgroups of groups.

In section 7 the concept of an irreducible representation is defined, and the Jordan-Holder Theorem is established. As a consequence, the composition factors of a finite group are seen to be an invariant of the group, and these composition factors are simple.

The question arises as to how much the structure of a group is controlled by its composition factors. Certainly many nonisomorphic groups can have the same set of composition factors, so control is far from complete. To investigate this question further we must consider extensions of a group G by a group A. Section 10 studies split extensions and introduces semidirect products.

Section 9 investigates solvable and nilpotent groups. For finite groups this amounts to the study of groups all of whose composition factors are, in the first case, of prime order and, in the second, of order p for some fixed prime p.

Commutators, characteristic subgroups, minimal normal subgroups, central products, and wreath products are also studied.

7 Normal series In this section G and A are groups, and n : A + Aut(G) is a representation of A in the category of groups and homomorphisms. I'll also say that A acts as a group of automorphisms on G. Observe that n is also a permutation representation, so we can use the terminology, notation, and results from chapter 2.

A normal series of length n for G is a series

l = G o i G 1 g . - . a G , = G .

The series is A-invariant if A acts on each Gi .

(7.1) Let (Gi: 0 j i j n) be an A-invariant normal series and H an A-invariant subgroup of G. Then

(1) The restriction n l H : A + Aut(H) is a representation of A on H. (2) (Gi f l H: 0 j i j n) is an A-invariant normal series for H. (3) If H G then TGIH: A + Aut(G/H) is a representation, where anc/ff:

Hg t+ H(ga) for a E A, g E G.

Normal series 23

(4) If H < G then (GiH/H: 0 < i < n) is an A-invariant normal series forG/H.

(5) If X is an A-invariant subgroup of G and H < G, then x fl H is anA-invariant normal subgroup of X and XH/H is an A-invariant subgroup ofG/H which is A-isomorphic to X/(X fl H).

A subgroup H of G is subnormal in G if there exists a series H = Go < G 1< G = G. Write H < < G to indicate H is subnormal in G.

(7.2) If X =Go < G1 4 ... < G. = G, then eitherX = G or (XG) < G,_1 G,

so (Xo) G.

(7.3) Let X and H be A-invariant subgroups of G with X < < G. Then(1) There exists an A-invariant series

X=Go<G1<...

(2) X fl H < < H.(3) If H < G then XH/H < < G/H.

Proof. Part (1) follows from 7.2 and induction on the length n of a subnormalseries X = Go < G1 < < G = G, since (XG) is A-invariant. Parts (2) and(3) are straightforward.

The family of factors of a normal series (Gi: 0 < i < n) is the family of factorgroups (Gi+1/Gi: 0 < i <n). Partially order the set of normal series for G via

(Hi:0<i <m)<(Gi:O<i <n) if Hi=GI(i)

for each 0 < i < m, and some j (i ).The representation it is said to be irreducible if G and 1 are the only A-

invariant normal subgroups of G. We also say that G is A-simple. An A-composition series for G is a normal series (Gi: 0 < i < n) maximal subject tobeing A-invariant and to Gi # Gi+1 for 0 < i < n. Of particular importance isthe case A = 1. G is said to be simple if it is 1-simple. Similarly the compositionseries for G are its 1-composition series.

(7.4) If G is finite then G possesses an A-composition series.

(7.5) An A-invariant normal series is an A-composition series if and only ifeach of its factors is a nontrivial A-simple group.

(4) If H 9 G then (GiH/H: 0 5 i _< n) is an A-invariant normal series for GIH.

(5) If X is an A-invariant subgroup of G and H 9 G, then X f l H is an A-invariant normal subgroup of X and XHIH is an A-invariant subgroup of GIH which is A-isomorphic to X / ( X n H).

A subgroup H of G is subnormal in G if there exists a series H = Go 9 G1 5 . . . a G, = G. Write H 9 9 G to indicate H is subnormal in G. -

(7.2) IfX = Go 9 G1 9 . . 9 G, = G, theneitherx = G or (XG) _< Gn-1 # G, so (xG) # G.

(7.3) Let X and H be A-invariant subgroups of G with X 9 <I G. Then (1) There exists an A-invariant series

(2) X n H S H . (3) If H 9 G then XHIH 9 9 GIH.

Proof. Part (1) follows from 7.2 and induction on the length n of a subnormal series X = GO a G1 9 - . - 9 G, = G, since (XG) is A-invariant. Parts (2) and (3) are straightforward.

The family of factors of a normal series (Gi: 0 ( i ( n) is the family of factor groups (Gi+1/Gi: 0 5 i 5 n). Partially order the set of normal series for G via

for each 0 5 i _< m, and some j(i). The representation n is said to be irreducible if G and 1 are the only A-

invariant normal subgroups of G. We also say that G is A-simple. An A- composition series for G is a normal series (Gi: 0 5 i ( n) maximal subject to being A-invariant and to Gi # Gi+I for 0 ( i ( n. Of particular importance is the case A = 1. G is said to be simple if it is 1-simple. Similarly the composition series for G are its 1-composition series.

(7.4) If G is finite then G possesses an A-composition series.

(7.5) An A-invariant normal series is an A-composition series if and only if each of its factors is a nontrivial A-simple group.

24 Representations of groups on groups

Jordan-Holder Theorem. Let (Gi : 0 < i < n) and (Hi: 0 < i < m) be A-composition series for G. Then n = m and there exists a permutation a of{i:0 < i <n} such that the representations of A on GQ+1/GQ and Hi+1/Hiare equivalent for 0 < i < n.

Proof. Let H = and k = min{i : Gi H). I'll show:

(a) G/H is A-isomorphic to Gk/Gk_1, and(b) (Xi = Gia n H: 0 k-1.

Suppose (a) and (b) hold. By induction on n, n - 1 = m - 1, and there is apermutation ,8 of {i: 0 < i < n - 11 with Xip+1/Xifl A-isomorphic to Hi+1/Hi.Hence n = m and the permutation a of ti: 0 < i < n) defined below does thetrick in the Jordan-Holder Theorem:

is=i,Ba ifi<n-1,(n - 1)a = k - 1,

So it remains to establish (a) and (b). First, as Gi < H for i < k, Gi = GiaXi, so certainly Xi/X1_1, is A-isomorphic to Gi,, /Gia_1.

If Gk n H Gk_1, then, as (Gk n H)Gk_1/Gk_1 is an A-invariant normalsubgroup of the A-simple group Gk/Gk-1, we have Gk = (Gk n H)Gk_1 < H,contrary to the definition of k. So Gk n H = Gk_1. On the other hand, if j > k,then Gj H, so a similar argument using 7.3 shows G =HG1, and henceG/H =HGj/H is A-isomorphic to Gj /(G1 n H). In particular Gj /(Gj n H)is A-simple, so, if Gj n H < Gj_1, then Gj_i l(Gj n H) is a proper A-invariantnormal subgroup of the A-simple group Gj /(Gj n H), and hence G j_1 = Gj nH < H. But then j = k by definition of k. Moreover G/H is A-isomorphic toGk/(Gk n H) = Gk/Gk-1, so (a) holds.

By the last paragraph, Gj nH Gj_1 for j > k. So, as above, G3 _ (Gj n H)Gj_1, and hence Gj/G1_1 is A-isomorphic to (Gj n H)/(G1_1 n H), com-pleting the proof of (b).

The Jordan-Holder Theorem says that the factors of an A-composition seriesof G are (up to equivalence and order) independent of the series, and hence arean invariant of the representation n. These factors are the composition factorsof the representation n. If A = 1, these factors are the composition factorsof G.


Jordan-Holder Theorem. Let (Gi: 0 5 i 5 n) and (Hi: 0 5 i 5 m) be A- composition series for G. Then n = m and there exists a permutation a of {i: 0 i i c n) such that the representations of A on GiU+l/ Giu and Hi+I/Hi are equivalent for 0 5 i c n.

Proof. Let H = HmW1 and k = min{i: Gi 5 H). I'll show:

(a) G I H is A-isomorphic to Gk/Gk-l, and (b) (Xi = Gi, f l H: 0 5 i c n) is an A-composition series for H with Xi+1 /Xi

A-isomorphic to Giatl/Gia for 0 5 i ( n - 1, where i a = i if i c k - 1 a n d i a = i + l f o r i z k - 1.

Suppose (a) and (b) hold. By induction on n, n - 1 = m - 1, and there is a permutation j3 of {i: 0 5 i c n - 1) with XiS+l/Xia A-isomorphic to Hi+l/Hi. Hence n = m and the permutation a of {i: 0 5 i < n} defined below does the trick in the Jordan-Holder Theorem:

(n- l ) a = k - 1,

So it remains to establish (a) and (b). First, as Gi 5 H for i c k, Gi = Gia = Xi, so certainly Xi/XiPl, is A-isomorphic to Gia/ Gia-1.

If Gk f l H -$ Gk-1, then, as (Gk n H)Gk-l/Gk-l is an A-invariant normal subgroup of the A-simple group Gk/Gk-1, we have Gk = (Gk fl H)Gk-1 i H, contrary to the definition of k. So Gk fl H = Gk-1. On the other hand, if j 2 k, then Gj -$ H, so a similar argument using 7.3 shows G = HGj, and hence G I H = HGj/H is A-isomorphic to Gj/(Gj n H). In particular Gj/(Gj n H) is A-simple, so, if G n H 5 G j-1, then G j-1 /(G n H) is a proper A-invariant normal subgroup of the A-simple group G j/(G fl H), and hence G j-1 = G j fl

H 5 H. But then j = k by definition of k. Moreover G /H is A-isomorphic to Gk/(Gk n H) = Gk/Gkel, SO (a) holds.

By thelastparagraph, G j n H Gjel for j > k. So, as above, G j = (Gjn H) Gj-1, and hence Gj/GjP1 is A-isomorphic to (Gj n H)/(Gj-1 n H), completing the proof of (b).

The Jordan-Holder Theorem says that the factors of an A-composition series of G are (up to equivalence and order) independent of the series, and hence are an invariant of the representation n. These factors are the composition factors of the representation n. If A = 1, these factors are the composition factors of G.

Characteristic subgroups and commutators 25

(7.6) Let X be an A-invariant subnormal subgroup of the finite group G.Then

(1) The A-composition factors of X are a subfamily of the A-compositionfactors of G.

(2) If X < G then the A-composition factors of G are the union of theA-composition factors of X and G/X.

Proof. There is an A-invariant normal series containing X by 7.3, and as G isfinite this series is contained in a maximal A-invariant series. Thus there is anA-composition series through X, so that the result is clear.

8 Characteristic subgroups and commutatorsA subgroup H of a group G is characteristic in G if H is Aut(G)-invariant.Write H char G to indicate that H is a characteristic subgroup of G.

(8.1) (1) If H char K and K char G, then H char G.(2) If H char K and K < G, then H < G.(3) If H char G and K char G, then HK char G and (H fl K) char G.

A group G is characteristically simple if G and 1 are the only characteristicsubgroups of G. A minimal normal subgroup of G is a minimal member of theset of nonidentity normal subgroups of G, partially ordered by inclusion.

(8.2) If 1 G is a characteristically simple finite group, then G is the directproduct of isomorphic simple subgroups.

Proof. Let H be a minimal normal subgroup of G and M maximal subjectto M < G and M the direct product of images of H under Aut(G). NowX = (Ha: a E Aut(G)) is characteristic in G, so by hypothesis X = G. Hence,if M 0 G, there is a c Aut(G) with Ha M. As Ha is a minimal normal sub-group of G and Ha fl M < G, we conclude Ha fl M = 1. But then M < M xHa < G, contradicting the maximality of M.

So G = M. Thus G = H x K for some K < G, so every normal subgroup ofH is also normal in G. Thus H is simple by minimality of H, and the lemmais established.

(8.3) Minimal normal subgroups are characteristically simple.

Proof. This follows from 8.1.2.

Characteristic subgroups and commutators

(7.6) Let X be an A-invariant subnormal subgroup of the finite group G. Then

(1) The A-composition factors of X are a subfamily of the A-composition factors of G.

(2) If X 9 G then the A-composition factors of G are the union of the A-composition factors of X and G/X.

Proof. There is an A-invariant normal series containing X by 7.3, and as G is finite this series is contained in a maximal A-invariant series. Thus there is an A-composition series through X, so that the result is clear.

8 Characteristic subgroups and commutators A subgroup H of a group G is characteristic in G if H is Aut(G)-invariant. Write H char G to indicate that H is a characteristic subgroup of G.

(8.1) (1) If H char K and K char G, then H char G. (2) If H char K and K a G, then H 9 G. (3) If H char G and K char G, then HK char G and ( H f l K) char G.

A group G is characteristically simple if G and 1 are the only characteristic subgroups of G. A minimal normal subgroup of G is a minimal member of the set of nonidentity normal subgroups of G, partially ordered by inclusion.

(8.2) If 1 # G is a characteristically simple finite group, then G is the direct product of isomorphic simple subgroups.

Proof. Let H be a minimal normal subgroup of G and M maximal subject to M 9 G and M the direct product of images of H under Aut(G). Now X = ( H a : a E Aut(G)) is characteristic in G, so by hypothesis X = G. Hence, if M # G, there is a E Aut(G) with H a M. As H a is a minimal normal subgroup of G and H a n M 9 G, we conclude H a n M = 1. But then M < M x H a 9 G, contradicting the maximality of M.

So G = M. Thus G = H x K for some K 5 G, so every normal subgroup of H is also normal in G. Thus H is simple by minimality of H , and the lemma is established.

(8.3) Minimal normal subgroups are characteristically simple.

(8.4) (1) The simple abelian groups are the groups of prime order.(2) If G is characteristically simple, finite, and abelian, then G = EP" for

some prime p and some integer n.

For x, y c G, write [x, y] for the group element x-1 y-1xy. [x, y] is the com-mutator of x and y. For X, Y < G, define

[X, Y] = ([x, y]: x E X, y E Y).

For z c Z < G write [x, y, z] for [[x, y], z] and [X, Y, Z] for [[X, Y], Z].

(8.5) Let G be a group, a, b, c E G, and X, Y < G. Then(1) [a, b] =1 if and only if ab = ba.(2) [X, Y] = 1 if and only if xy = yx for all x E X and y c Y.(3) If a: G - H is a group homomorphism then [a, b]a = [act, bet] and

[X, Y]a = [Xa, Yet].

(4) [ab, c] = [a, c]b[b, c] and [a, be] = [a, c][a, b]'.(5) X < NG(Y) if and only if [X, Y] < Y.

(6) [X, Y] = [Y, X] :a (X, Y).

Proof. I prove (6) and leave the other parts as exercises. Notice [a, b]-1 =[b, a], so [X, Y] = [Y, X]. Further, to prove [X, Y] :a (X, Y), it will suffice toshow [x, y]Z E [X, Y] for each x E X, Y E Y, and Z E X U Y. As [x, y]-1 =[y, x], we may assume z E Y. But, by (4), [x, y]Z = [x, z]-1[x, yz] E [X, Y],so the proof is complete.

(8.6) Let G be a group, x, y E G, and assume z = [x, y] centralizes x and y.Then

(1) [xn ym] =Znm for all n, m E7L.(2) (yx)' = Zn(n-1)12ynxn for all 0 < n E 71.

Proof. Without loss G = (x, y), so z E Z(G). z = [x, y] so xy = xz. Then, forn E 1, (xn)y = (xy)n = (xZ)n = xnZn as z E Z(G). Thus [x", y] = Zn. Similarly

[x, ym] = Zm, So [xn, ym] _ [x, ym]n = Zmn, and (1) holds.Part (2) is established by induction on n. Namely (yx)n+1 = (yx)nyx =

Zn(n-1)12ynxnyx, while by (1) xny = yxfZn, so that the result holds.

(8.7) (Three-Subgroup Lemma) Let X, Y, Z be subgroups of a group G with[X,Y,Z]=[Y,Z,X]=1.Then [Z,X,Y]=1.

Proof. Let x E X, y c Y, and z c Z. A straightforward calculation shows:

(*) [x, y-1, zly =x-1y-1xz-1x-1yxy-1zy =a(x, y,z)-1a(y,

z, x),

ups on groups

(8.4) (1) The simple abelian groups are the groups of prime order. (2) If G is characteristically simple, finite, and abelian, then G 2 Epll for

some prime p and some integer n.

For x , y E G , write [x , y] for the group element X - ' ~ - ' X ~ . [ x , y] is the commutator of x and y. For X, Y 5 G , define

[ X , Y ] = ( [ x , y]: x E X, y E Y ) .

For z E Z 5 G write [ x , y, z ] for [ [x , y] , z ] and [ X , Y , Z ] for [ [ X , Y ] , Z ] .

(8.5) Let G be a group, a , b, c E G , and X, Y 5 G. Then (1) [a, b]= 1 if andonly if ab=ba. (2) [X , Y ] = 1 if and only if xy = yx for all x E X and y E Y . (3) If a: G + H is a group homomorphism then [a, b]a = [aa, ba] and

[ X , Y ] a = [Xa , Y a ] . (4) [ab, c] = [a, clb[b, C ] and [a, bc] = [a, c][a, bIC. (5 ) X 5 N c ( Y ) if and only if [ X , Y ] 5 Y . (6) [ X , Y l = [Y, XI 9 ( X , Y ) .

Proof. I prove (6) and leave the other parts as exercises. Notice [a, b]-' = [b, a ] , so [ X , Y ] = [Y, XI. Further, to prove [ X , Y ] 9 ( X , Y ) , it will suffice to show [x , ylZ E [ X , Y ] foreachx E X, y E Y , andz E XU Y . As [ x , y]-' = [y , x ] , we may assume z E Y . But, by (4), [ x , ylZ = [ x , z]- '[x, yz] E [ X , Y ] , so the proof is complete.

(8.6) Let G be a group, x , y E G , and assume z = [x , y] centralizes x and y. Then

(1) [xn , ym] = znm for all n, m E Z. (2) (yx)" = ~ " ( ~ - ' ) / ~ y " x " for all 0 5 n E Z.

Proof. Without loss G = ( x , y ) , so z E Z(G). z = [x , y] so xY = xz. Then, for n E Z, (xn)Y = (xY)" = (xz)" = xnzn as z E Z(G). Thus [xn , y] = zn. Similarly [x , ym] =zm, so [xn, ym]= [x , ymIn =zmn, and (1) holds.

Part (2) is established by induction on n. Namely (YX)"+' = (yx)"yx = Zn(n-1)/2 y n x n yx, while by (1) xny = yxnzn, so that the result holds.

(8.7) (Three-Subgroup Lemma) Let X, Y , Z be subgroups of a group G with [X, Y, Z ] = [Y, Z , XI = 1. Then [ Z , X , Y ] = 1.

Proof. Let x E X, y E Y , and z E Z. A straightforward calculation shows:

(*) [x , y-', zlY = x - ' ~ - ~ x z - ~ x - ' ~ x ~ - ' z ~ =a(x , y, z ) - ' ~ ( ~ , z , x ) ,

Solvable and nilpotent groups 27

where a (u, v, w) = uw u-1 vu. Applying the permutations (x, y, z) and (x, z, y)to (*) and taking the product of (*) with these two images, we conclude:

[x, Y-'z]y[Y, z-1, x]Z[z, x-1, Y]X =1.

As [X, Y, Z] _ [Y, Z, X] =1, also [x, y-1, z] _ [y, z-1, x] = 1, so by (**) weget [z, x-1, y] = 1. Finally as [Z, X] is generated by the commutators [z, x-1],z E Z, X E X, it follows from 8.5.1 that y centralizes [Z, X]. But then, by 8.5.2,[Z, X, Y]=1.

The commutator group or derived group of a group G is the subgroup GM =[G, G]. Extend the notation recursively and define G(") = [G(n-1), G(n-1)] forn > 1. Define G(0) = G and G-= f°_ 1 G(O

(8.8) Let G be a group and H < G. Then(1) H(n) <G(n).

(2) If a: G X is a surjective group homomorphism then G(n)a = X (n).(3) G(n) char G.(4) GM < H if and only if H a G and G/H is abelian.

A group G is perfect if G = G(1).

(8.9) Let X and L be subgroups of a group G with L perfect and [X, L, L] = 1.Then [X, L] = 1.

Proof. [L, X, L] = [X, L, L] and by hypothesis both are 1. So by the Three-Subgroup Lemma, [L, L, X] =1. Butby hypothesis L = [L, L], so [L, X] = 1.

9 Solvable and nilpotent groupsA group G is solvable if it possesses a normal series whose factors are abelian.

(9.1) A group G is solvable if and only if Gin) = 1 for some positive integer n.

Proof. If G(n) = 1, then (G(n-`): 0 < i < n) is a normal series with abelian fac-tors by 8.8.4. Conversely if (G1: 0 < i < n) is such a series then, by 8.8.4 andinduction on i, G(`) < Gn_i, so G(n) = 1.

(9.2) A finite group is solvable if and only if all its composition factors are ofprime order.

Solvable and nilpotent groups

wherea(u, v , w)=uwu-'vu.~pplyingthepermutations(x, y, z)and(x, z, y) to (*) and taking the product of (*) with these two images, we conclude:

As [X, Y, Z] = [Y, Z, XI = 1, also [x, y-', z] = [y, z-', x] = 1, so by (**) we get [z, x-', y] = 1. Finally as [Z, XI is generated by the commutators [z, x-'1, z E Z, x E X, it follows from 8.5.1 that y centralizes [Z, XI. But then, by 8.5.2, [Z, X, Y] = 1.

The commutator group or derived group of a group G is the subgroup G(') = [G, GI. Extend the notation recursively and define G(") = [G("-'), G("-')I for n > 1. Define G(O) = G and Gm = nzl G(').

(8.8) Let G be a group and H 5 G. Then (1) H(") 5 G("). (2) If a: G + X is a surjective group homomorphism then ~ ( " ) a = x("). (3) G(") char G. (4) G(') 5 H if and only if H 9 G and G/H is abelian.

A group G is pe$ect if G = G(').

(8.9) Let X and L be subgroups of a group G with L perfect and [X, L, L] = 1. Then [X, L] = 1.

Proof. [L, X, L] = [X, L , L] and by hypothesis both are 1. So by the Three- Subgroup Lemma, [L, L, XI = 1. But by hypothesis L = [L , L], so [L , XI = 1.

9 Solvable and nilpotent groups A group G is solvable if it possesses a normal series whose factors are abelian.

(9.1) A group G is solvable if and only if G(") = 1 for some positive integer n.

Proof. If G(") = 1, then (G("-'): 0 5 i 5 n) is a normal series with abelian factors by 8.8.4. Conversely if (Gi: 0 5 i 5 n) is such a series then, by 8.8.4 and induction on i, G(') 5 GnWi, so G(") = 1.

(9.2) A finite group is solvable if and only if all its composition factors are of prime order.

Proof. If all composition factors are of prime order then a composition se-ries for G is a normal series all of whose factors are abelian. Conversely if(Gi : 0 < i < n) is such a series then the composition factors of the abeliangroup Gi+i / Gi are also abelian, and hence, by 8.4.1, of prime order. Then, by7.6, the composition factors of G are of prime order.

(9.3) (1) Subgroups and homomorphic images of solvable groups aresolvable.

(2) If H < G with H and G/H solvable, then G is solvable.

(9.4) Solvable minimal normal subgroups of finite groups are elementaryabelian p-groups.

Proof. Let G be finite and M a solvable minimal normal subgroup of G. By 9.1and solvability of M, M(1) M. Next, by 8.3, M is characteristically simple.So, as MM char M, we conclude M(l) = 1. Thus M is abelian by 8.8. Then8.4.2 completes the proof.

Define L 1(G) = G, and, proceeding recursively, define Ln(G) = [L,-1(G), G]for 1 < n c Z. G is said to be nilpotent if Ln (G) = 1 for some 1 < n E Z. Theclass of a nilpotent group is m - 1, where m =min{i: Li(G) = 11.

(9.5) (1) Ln(G) char G for each 1 <n (=- Z.

(2) Ln+1(G) Ln(G)(3) Ln(G)/Ln+1(G) < Z(G/Ln+1(G))

Proof. Part (1) follows from 8.5.3 by induction on n. Then (1) and 8.5.5 imply(2) while 8.5.1 and 8.5.3 imply (3).

Define Z0(G) =1 and proceeding recursively define Zn (G) to be the preimagein G of Z(G/Zn-1(G)) for 1 <, n E Z. Evidently Zn(G) char G.

(9.6) G is nilpotent if and only if G = Zn(G) for some 0 <n E Z. If G isnilpotent then the class of G is k = min{n: G = Zn(G)}.

Proof. I first show that if G is nilpotent of class m then L,n+i-i (G) < Zi (G)for 0 0and L.+2-i(G) < Zi-i(G) then [L,n+i_j(G), G] <Zi_i(G), so,by 8.5, Lm+i-i(G)Zj-i(G)/Zi-i(G) < Z(G/Zi-1(G)) = Z1(G)/Zi-1(G), andhence L,n+i_i (G) < Zi (G). So the claim follows by induction on i, and we seeZ,n(G)=L1(G)=G, sok<m.


Proof. If all composition factors are of prime order then a composition series for G is a normal series all of whose factors are abelian. Conversely if (Gi: 0 i 5 n) is such a series then the composition factors of the abelian group Gi+1/Gi are also abelian, and hence, by 8.4.1, of prime order. Then, by 7.6, the composition factors of G are of prime order.

(9.3) (1) Subgroups and homomorphic images of solvable groups are solvable.

(2) If H 9 G with H and G/H solvable, then G is solvable.

(9.4) Solvable minimal normal subgroups of finite groups are elementary abelian p-groups.

Proof. Let G be finite and M a solvable minimal normal subgroup of G. By 9.1 and solvability of M, M(') # M. Next, by 8.3, M is characteristically simple. So, as M(') char M, we conclude M(') = 1. Thus M is abelian by 8.8. Then 8.4.2 completes the proof.

Define L1(G) = G, and, proceeding recursively, define L,(G) = [L,-l (G), GI for 1 < n E Z. G is said to be nilpotent if L,(G) = 1 for some 1 5 n E Z. The class of a nilpotent group is m - 1, where m = min{i : Li (G) = 1).

(9.5) (1) L,(G) char G for each 1 5 n E Z. (2) Ln+l(G) i Ln(G). (3) Ln(G)/Ln+l (GI i Z(G/Ln+l (GI).

Proof. Part (1) follows from 8.5.3 by induction on n. Then (1) and 8.5.5 imply (2) while 8.5.1 and 8.5.3 imply (3).

Define Zo(G) = 1 and proceeding recursively define Z, (G) to be the preimage in G of Z(G/Z,-l (G)) for 1 < n E Z. Evidently Z,(G) char G.

(9.6) G is nilpotent if and only if G = Z,(G) for some 0 n E Z. If G is nilpotent then the class of G is k = min(n: G = Z,(G)}.

Proof. I first show that if G is nilpotent of class m then Lm+l-,(G) 5 Zi(G) for 0 i 5 m. For i = 0 this follows directly from the definitions, while if i > 0 and Lm+2-i(G) i Zc-l(G) then [Lm+l-i(G), GI =Lm+2-i(G) 5 Zi-l(G), so, by 8.5, Lm+i-,(G)Zc-i(G)/Zl-~(G) i Z(G/ZL-i(G)) = Zz(G)/Zi-l(G), and hence Lm+l-, (G) 5 Z, (G). So the claim follows by induction on i, and we see Zm(G)= L1(G)=G, so k s m .

Semidirect products 29

Next let's see that if Zn (G) = G for some 0 <n c Z, then L;+1(G) < Zn-i (G)for each 0 0 and Li (G) <Zn_i+l (G) then Li+1(G) = {Li (G), G] < {Zn-i+l (G), G] < Zn_i (G) by 8.5.5,establishing the claim. In particular Ln+1(G) < Zo(G) = 1, so G is nilpotent ofclass m < n, so that m < k.

(9.7) 1 G is nilpotent of class m if and only if G/Z(G) is nilpotent of classm-1.

Proof This is a direct consequence of 9.6.

(9.8) p-groups are nilpotent.

Proof. Let G be a minimal counter example. Then certainly G 1, so, by 5.16,

Z(G) 1. Hence, by minimality of G, G/Z(G) is nilpotent, so, by 9.7, G isnilpotent, contrary to the choice of G.

(9.9) Let G be nilpotent of class m. Then subgroups and homomorphic imagesof G are nilpotent of class at most m.

(9.10) If G is nilpotent and H is a proper subgroup of G, then H is proper inNG (H).

Proof. Assume NC(H) = H < G. Then Z(G) < NC(H) = H, so H* < G* =G/Z(G). By 9.7 and induction on the nilpotence class of G, H*But, as Z(G) < H, NG.(H*) = NG(H)*, so H < NC(H), a contradiction.

(9.11) A finite group is nilpotent if and only if it is the direct product of itsSylow groups.

Proof. The direct product of nilpotent groups is nilpotent, so by 9.8 the directproduct of p-groups is nilpotent. Conversely let G be nilpotent; we wish to showG is the direct product of its Sylow groups. Let P E Syl,(G). By Exercise 2.9it suffices to show P < G. If not, M = NG (P) < G, so, by 9.10, M < NG(M).But, as P < M, {P) = SylP(M), so P char M. Hence NG(M) < NG(P) = M,a contradiction.

10 Semidirect productsIn this section A and G are groups and rr: A Aut(G) is a representation ofA as a group of group automorphisms of G.


Next let's see that if Z,(G) = G for some 0 5 n E Z, then Li+l (G) 5 Z,+(G) for each 0 5 i 5 n. Again the case i = 0 is trivial, while if i > 0 and Li(G) 5 zn-i+,(G) then Li+l(G) = [Li(G), GI 5 [zn-i+l(G), GI I Zn-i(G) by 8.5.5, establishing the claim. In particular Ln+~(G) 5 Zo(G) = 1, so G is nilpotent of class m 5 n, so that m _( k.

(9.7) 1 # G is nilpotent of class m if and only if G/Z(G) is nilpotent of class m - 1.

Proof. This is a direct consequence of 9.6.

(9.8) p-groups are nilpotent.

Proof. Let G be a minimal counter example. Then certainly G # 1, so, by 5.16, Z(G) # 1. Hence, by minimality of G, G/Z(G) is nilpotent, so, by 9.7, G is nilpotent, contrary to the choice of G.

(9.9) Let G be nilpotent of class m. Then subgroups and homomorphic images of G are nilpotent of class at most m.

(9.10) If G is nilpotent and H is a proper subgroup of G, then H is proper in

NG(H>.

Proof. Assume NG(H) = H < G. Then Z(G) _( NG(H) = H , so H* < G* = G/Z(G). By 9.7 and induction on the nilpotence class of G, H* < NG*(H*). But, as Z(G) 5 H, NG*(H*) = NG(H)*, so H < NG(H), a contradiction.

(9.11) A finite group is nilpotent if and only if it is the direct product of its Sylow groups.

Proof. The direct product of nilpotent groups is nilpotent, so by 9.8 the direct product of p-groups is nilpotent. Conversely let G be nilpotent; we wish to show G is the direct product of its Sylow groups. Let P E Syl,(G). By Exercise 2.9 it suffices to show P a G. If not, M = NG(P) < G, so, by 9.10, M < Nc(M). But, as P M, {P} = Syl,(M), so P char M. Hence NG(M) 5 NG(P) = M, a contradiction.

10 Semidirect products In this section A and G are groups and n: A + Aut(G) is a representation of A as a group of group automorphisms of G.

Let H a G. A complement to H in G is a subgroup K of G with G = HKand H f1 K =1. G is said to be an extension of a group X by a group Y if thereexists H a G with H - X and G/H - Y. The extension is said to split if Hhas a complement in G. The following construction can be used to describesplit extensions.

Let S be the set product A x G and define a binary operation on S by

(a,g)(b,h)=(ab,gb"h) a,bEA,g,hEG

where gb" denotes the image of G under the automorphism b7r of G. We callS the semidirect product of G by A with respect to 7r. Denote S by S(A, G, 7r).

(10.1) (1) S = S(A, G, rr) is a group.(2) The maps A -+ S and o-G: G -+ S are injective group homomor-

phisms, where orA: a i-+ (a, 1) and aG: g i-+ (1, g).(3) GaG a S and AorA is a complement to Goo in S.(4) (1,g)fa'1?=(1,ga")forgEG,aEA.

Observe that if it is the trivial homomorphism then the semidirect product isjust the direct product of A and G.

(10.2) Let H be a group, G < H, and B a complement to Gin H. Let a: B -+Aut(G) be the conjugation map (i.e., ba: g -+ gb for b E B, g E G; see Exercise1.3). Define ,B: S(B, G, a) -+ H by (b, g),B = bg. Then ,B is an isomorphism.

We see from 10.1 and 10.2 that the semidirect products of G by A are pre-cisely the split extensions of G by A. Moreover the representation defining thesemidirect product is a conjugation map.

Under the hypotheses of 10.2, I'll say that H is a semidirect product of Gby B. Formally this means the map ,B of 10.2 is an isomorphism.

(10.3) Let Si = S(A1, GI , ,ri ), i =1, 2, be semidirect products. Then there ex-ists an isomorphism 0: Sl -+ S2 with A1orA,0=A2orA2 and G1orG,0=G2o'o2if and only if 7r1 is quasiequivalent to n2 in the category of groups and homo-morphisms.

It is not difficult to see that semidirect products Si = S(A, G, 7r1) and S2 =S(G, A, ir2) can be isomorphic without Jr1 being quasiequivalent to 72. Toinvestigate just when Sl and S2 are isomorphic we need to know more about howAut(SS) acts on its normal subgroups isomorphic to G, and how the stabilizerin Aut(Si) of such a subgroup acts on its complements. This latter question isconsidered in chapter 6.

30 Representations ofgroups on groups

Let H 9 G. A complement to H in G is a subgroup K of G with G = H K and H n K = 1. G is said to be an extension of a group X by a group Y if there exists H 9 G with H Z X and G/H Z Y. The extension is said to split if H has a complement in G. The following construction can be used to describe split extensions.

Let S be the set product A x G and define a binary operation on S by

(a, g)(b, h) = (ab, gbrh) a , b E A , g, h E G

where gbk denotes the image of G under the automorphism bn of G. We call S the semidirectproduct of G by A with respect to n . Denote S by S(A, G, n).

(10.1) (1) S = S(A, G, n ) is a group. (2) The maps u ~ : A + S and u ~ : G + S are injective group homomor-

phisms, where o;l: a H (a, 1) and u ~ : g H (1, g). (3) GuG 9 S and A- is a complement to GuG in S. (4) (1, g)(atl)= (1, gar) for g E G, a E A.

Observe that if n is the trivial homomorphism then the semidirect product is just the direct product of A and G.

(10.2) Let H be a group, G 9 H , and B a complement to G in H. Let a : B + Aut(G) be the conjugation map (i.e., ba: g + gb forb E B, g E G; see Exercise 1.3). Define B: S(B, G, a ) + H by (b, g)B = bg. Then B is an isomorphism.

We see from 10.1 and 10.2 that the semidirect products of G by A are precisely the split extensions of G by A. Moreover the representation defining the semidirect product is a conjugation map.

Under the hypotheses of 10.2, I'll say that H is a semidirect product of G by B. Formally this means the map B of 10.2 is an isomorphism.

(10.3) Let Si = S(Ai, Gi, xi), i = 1,2, be semidirect products. Then there exists an isomorphism @: S1+ S2 with A i u ~ , @ = A z ~ A , and Gluc1@ = G z ~ G , if and only if nl is quasiequivalent to n2 in the category of groups and homomorphisms.

It is not difficult to see that semidirect products S1= S(A, G, nl) and S2 = S(G, A, n2) can be isomorphic without nl being quasiequivalent to n2. To investigate just when S1 and S2 are isomorphic we need to know more about how Aut(Si) acts on its normal subgroups isomorphic to G, and how the stabilizer in Aut(Si) of such a subgroup acts on its complements. This latter question is considered in chapter 6.

It's also easy to cook up nonsplit extensions, and it is of interest to generateconditions which insure that an extension splits. The following is perhaps themost important such condition:

(10.4) (Gaschutz' Theorem) Let p be a prime, V an abelian normal p-subgroupof a finite group G, and P E Sylp(G). Then G splits over V if and only if Psplits over V.

Proof. Notice V < P. Hence if H is a complement to V in G then by theModular Property of Groups, 1.14, P = P fl G = P fl VH= V(P fl H) andP fl H is a complement to V in P.

Conversely suppose Q is a complement to V in P. Let G = G/ V and observeP Q = Q. Let X be a set of coset representatives for V in G. Then the mapx H x is a bijection of X with G and I denote the inverse of this map by a H xa.Then

(i) xaxb =xaby(a, b) for a, b E G, and for some y(a, b) E V.

Next using associativity in G and G we have xabcy(a, bc)y(b, c) = xaxbCY

(b, c) = xa (xbxc) = (xaxb)xc = xab y (a, b)xc = xabxcY (a, b)x` = xabc y (ab, c)y(a, b)x from which we conclude:

(ii) y(ab, c)y(a, b)x1= y(a, bc)y(b,,c) for all a, b, c E G.

Now choose X = Q Y, where Y is a set of coset representatives for P in G.Then, for g E Q and a E G, xga = gxa, so:

(iii) xg = g and y(g, a) = 1 for all g E Q and a E G.

Now (ii) and (iii) imply:

(iv) y(gb, c) = y(b, c) for all b, c E G and g E Q.

Next for C E G define ,B(c) = I1yEpy(y, c). By (iv), ,B(c) is independent of thechoice of the set Y of coset representatives. But if b E G then Yb is another setof coset representatives for Q in G, so:

(v) NO= fl y(yb, c) for all b, c E G.yFk

As V is abelian we conclude from (ii) that

(flv(bc)) (flv(b)) _ fj Y(y, bc) (flYbc))yEY yEY yEY yEY

It's also easy to cook up nonsplit extensions, and it is of interest to generate conditions which insure that an extension splits. The following is perhaps the most important such condition:

(10.4) (Gaschiitz' Theorem) Let p be a prime, V an abelian normal p-subgroup of a finite group G, and P E Syl,(G). Then G splits over V if and only if P splits over V.

Proof. Notice V 5 P. Hence if H is a complement to V in G then by the Modular Property of Groups, 1.14, P = P n G = P fl VH = V(P fl H) and P n H is a complement to V in P.

Conversely suppose Q is a complement to V in P. Let C? = G/ V and observe P Z Q r Q. Let X be a set of coset representatives for V in G. Then the map x H f is a bijection of X with C? and I denote the inverse of this map by a H xu. Then

(i) xaxb = xaby (a, b) for a , b E G, and for some y (a, b) E V.

Next using associativity in G and G we have xabc y (a, bc)y (b, c) = xaxb, y

(b, C) =XU(X~XC) = ( ~ a ~ b ) ~ c = x a b ~ ( a , ~ ) X C = x ~ ~ x ~ Y ( ~ , b)XC =xabcy(ab, c)y (a, b)q, from which we conclude:

(ii) y(ab, c)y(a, b)XC F y(a, bc)y(b, c) for all a , b, c E G.

Now choose X = QY, where Y is a set of coset representatives for P in G. Then, for g E Q and a E G, xsa = gx,, so:

(iii) x s=g and y ( g , a ) = l f o r a l l g ~ Q and U E G .

Now (ii) and (iii) imply:

(iv) y(gb,c)=y(b,c) f o r a l l b , c ~ G and ~ E Q .

Next for c E G define p(c) = n,,y y(7, c) . By (iv), p(c) is independent of the choice of the set ? of coset representatives. But if b E G then ?b is another set of coset representatives for Q in G, so:

(v) p(c) = fl y(gb, c) for all b, c E G. ~ € 9

As V is abelian we conclude from (ii) that

and then appealing to (v) we obtain

(vi) j(c)j(b) ', =,(bc)y(b, c)' for all b, c E G, where m = jG : P1.

As P E Sylp(G), (m, p) = 1. Thus m is invertible mod I V 1. Hence we candefine a(c) = ,6(c)-'"-' , for c E G. Then taking the -m-1 power of (vi) weobtain:

(vii) a(c)a(b) , =a(bc)y(b, c)-1 for all b, c E G.

Finally define y,, = xaa(a) for a E G and set H = {y,,: a E G}. H will beshown to be a complement to V in G. This will complete the proof.

It suffices to show YbYc = ybc for all b, c E G. But ybyC =xba(b)x0a(c) _

xbxà(b)Xà(c)=xbcY(b,

c)a(b)xù(c)=ybca(bc)-ly(b, c)a(b)X°a(c). Then,

as V is abelian, (vii) implies ybYC = Ybc, as desired.

The Schur-Zassenhaus Theorem in section 18 is another useful result onsplitting.

11 Central products and wreath products(11.1) Let {G1: 1 < i < n} be a set of subgroups of G. Then the following areequivalent:

(1) G = (Gi: 1 <i <n) and [Gi, G] =1 for i 0 j.(2) Themap7r: (x1, ... , x") H x1 ... x" is asurjectivehomomorphismofG1 x

x G" = D onto G with Diir = G and Di fl ker(7r) =1, where Diconsists of those elements of D with 1 in all but the ith component.

If either of the equivalent conditions of 11.1 holds, then G is said to be acentral product of the subgroups Gi, 1 < i < n. Notice that the kernel of thehomomorphism it of 11.1 is contained in the centre of G1 x x G.

(11.2) Let (Gi : 1 < i < n) be a family of groups such that Z(G 1) - Z(G1) andAut(Z(Gi))=AutAut(G,)(Z(Gi)) for 1 <i <n. Then, up to an isomorphismmapping Gi to Gi for each i, there exists a unique central product of thegroups Gi in which Z(G1) = Z(G1) for each i.

Proof. Adopt the notation of 11.1 and identify Gi with Di. By hypothesisthere are isomorphisms ai: Z(D1) Z(Di), 1 <i <n. Let E be the subgroupof D generated by z(z-lai), z E Z(D1)1 < i <n. Observe E is a complementto Z(Di) in Z = (Z(Di ): 1 < i < n) = Z(D) for each i. Thus DIE is a centralproduct of the groups Gi with Z(G1) = Z(G1) for each i, by 11.1.


and then appealing to (v) we obtain

(vi) j ? (~) j? (b)~~ = j?(bc)y(b, c ) ~ for all b, c E G, where m = IG : PI.

As P E Syl,(G), (m, p) = 1. Thus m is invertible mod I V I. Hence we can define a(c) = j ? ( ~ ) - ~ - ' , for c E G. Then taking the -m-' power of (vi) we obtain:

(vii) a ( ~ ) a ( b ) ~ ~ =a(bc)y(b, c)-' for all b, c E G.

Finally define y, = x,a(a) for ii E G and set H = {y,: a E G). H will be shown to be a complement to V in G. This will complete the proof.

It suffices to show yby, = yb, for all b, c E G. But yby, =xba(b)xca(c) = xbx,a(b)XCa(c) = xbcy(b, c)a(b)xca(c) = Ybca(b~)-l y (b, c)a(b)xca(c). Then, as V is abelian, (vii) implies yby, = yb,, as desired.

The Schur-Zassenhaus Theorem in section 18 is another useful result on splitting.

11 Central products and wreath products (11.1) Let {Gi: 1 _( i _( n) be a set of subgroups of G. Then the following are equivalent:

(1) G = ( G i : l ( i ( n ) a n d [ G i , G j ] = l f o r i # j . (2) Themapn: (XI, . . . , x,) H xl . . . x, isasurjectivehomomorphismof G1 x

. . x G, = D onto G with Din = Gi and Di fl ker(n) = 1, where Di consists of those elements of D with 1 in all but the ith component.

If either of the equivalent conditions of 11.1 holds, then G is said to be a central product of the subgroups Gi, 1 ( i n. Notice that the kernel of the homomorphism n of 1 1.1 is contained in the centre of G1 x . . . x G,.

(11.2) Let (Gi: 1 _( i _( n) be a family of groups such that Z(G1) S Z(Gi) and Aut(Z(Gi)) = A U ~ ~ " ~ ( ~ , ) ( Z ( G ~ ) ) for 1 ( i ( n. Then, up to an isomorphism mapping Gi to Gi for each i , there exists a unique central product of the groups Gi in which Z(G1) = Z(Gi) for each i.

Proof. Adopt the notation of 11.1 and identify Gi with Di. By hypothesis there are isomorphisms ai: Z(D1) + Z(Di), 1 5 i 5 n. Let E be the subgroup of D generated by z(z-'ai), E Z(D1)l < i 5 n. Observe E is a complement to Z(Di) in Z = (Z(Di): 1 5 i 5 n) = Z(D) for each i. Thus D I E is a central product of the groups Gi with Z(Gi) = Z(G1) for each i, by 11.1.

Central products and wreath products 33

Next assume G is a central product of the Gi with Z(G 1) = Z(G,) for eachi, and let 7r: D G be the surjective homomorphism supplied by 11.1. LetPi: Z(D1) Z(Di) be the isomorphism which is the composition of irIz(o,):Z(D1) - Z(G1) and (7r(z(o;))-1: Z(G,) Z(Di). Observe

ker(7r) = (z(z-1pi): z E Z(D1), 1 < i < n) = A

is a complement to Z(Di) in Z for each i, and of course G = D/A. To com-plete the proof I exhibit y E Aut(D) with Di y = Di and Ey = A. Notice yinduces an isomorphism of DIE with D/A mapping Gi to Gi, demonstratinguniqueness.

Let 8i =(a1)`1,Bi: Z(D,) Z(D,), so that 8i E Aut(Z(Di)). By hypothesisthere is yi E Aut(Di) with yi I z(oi) = 8i. Define y: D D by

(x1,...,xn)H(x1,x2Y2,...,xnYn)

and observe y E Aut(D) with (z(z-1ai))y = z(z-1,Bi), so that Ey = A. Thusthe proof is complete.

Under the hypotheses of 11.2, we say G is the central product of the groupsG, with identified centers, and write G = G 1 * G2 * . . . * Gn.

Let L be a group and 7r: G Sym(X) a permutation representation of G on

X = {1, . . . , n}. Form the direct product D of n copies of L. G acts as a groupof automorphisms of D via the representation a defined by

ga: (xi, ... , xn) H (xig-'n, ... , xng-in).

The wreath product of L by G (with respect to 7r) is defined to be the semidirectproduct S(G, D, a). The wreath product is denoted by Lwr G or Lwr,G orLwrr G.

(11.3) Let W = Lwr, G be the wreath product of L by G with respect to Jr.Then

(1) W is a semidirect product of D by G where D = L1 x x Ln is adirect product of n copies of L.

(2) G permutes A = {Li : 1 < i < n } via conjugation and the permutationrepresentation of G on A is equivalent to r. That is (L1)g = Lign for eachgEGandI<i<n.

(3) The stabilizer Gi of i in G centralizes Li.

Exercises for chapter 31. Let G and A be finite groups with (I G 1, ( A 1) = 1, assume A is represented

on G as a group of automorphisms, and (Gi : 0 < i < n) is an A-invariantnormal series for G such that A centralizes Gi+1 /Gi for 0 < i < n. Prove

Central products and wreath products 3 3

Next assume G is a central product of the G, with Z(G1) = Z(Gi) for each i , and let n : D + G be the surjective homomorphism supplied by 11.1. Let PI: Z(Dl) + Z(Dl) be the isomorphism which is the composition of nlziD,): Z(D1) + Z(G1) and (n[Z(D,))-l: Z(G,) + Z(D,). Observe

ker(n) = (z(z-'~i): z E Z(D1), 1 5 i 5 n) = A

is a complement to Z(Di) in Z for each i, and of course G 2 DIA. To complete the proof I exhibit y E Aut(D) with Diy = Di and Ey =A. Notice y induces an isomorphism of DIE with DIA mapping Gi to Gi, demonstrating uniqueness.

Let = (ai)-lBi: Z(Di) + Z(Di), SO that ai E Aut(Z(Di)). By hypothesis there is yi E Aut(Di) with yi I z ( D , ) = ai. Define y: D + D by

and observe y E Aut(D) with ( ~ ( z - l a ~ ) ) ~ = z(z-'pi), so that E y =A. Thus the proof is complete.

Under the hypotheses of 11.2, we say G is the central product of the groups Gi with identijied centers, and write G = G1 * G2 * . . . * G,.

Let L be a group and n: G + Sym(X) a permutation representation of G on X = (1, . . . , n ) . Form the direct product D of n copies of L. G acts as a group of automorphisms of D via the representation a defined by

ga: (XI, . . . , x,) - (xlg-I,, . . . , xng-I,).

The wreath product of L by G (with respect to n ) is defined to be the semidirect product S(G, D, a). The wreath product is denoted by Lwr G or Lwr,G or Lwr,G.

(11.3) Let W = Lwr,G be the wreath product of L by G with respect to n. Then

(1) W is a semidirect product of D by G where D = L1 x . . . x L, is a direct product of n copies of L.

(2) G permutes A = {Li: 1 i 5 n ) via conjugation and the permutation representation of G on A is equivalent to n. That is (Li)g = Lign for each g E G and 1 ( i c n .

(3) The stabilizer Gi of i in G centralizes Li.

Exercises for chapter 3 1. Let G and A be finite groups with (I G I , (A 1 ) = 1, assume A is represented

on G as a group of automorphisms, and (Gi: 0 ( i I n) is an A-invariant normal series for G such that A centralizes Gi+1/Gi for 0 ( i < n. Prove

A centralizes G. Produce a counter example when (I A G j) 01. (Hint:Reduce to the case where A is a p-group and use 5.14.)

2. Let G be a finite group, p a prime, and X a p-subgroup of G. Prove(1) Either X E Sylp(G) or X is properly contained in a Sylow p-subgroupof NG(X).(2) If G is a p-group and X is a maximal subgroup of G, then X 4 G andJG:X1 =p.

3. Prove lemmas 10.1 and 10.3. Exhibit a nonsplit extension.4. Let p and q be primes with p > q. Prove every group of order pq is a split

extension of 7LP by 7Lq. Up to isomorphism, how many groups of order pqexist? (Hint: Use 10.3 and Exercise 1.7. Prove Aut(Zp) is cyclic of orderp - 1. You may use the fact that the multiplicative group of a finite field iscyclic.)

5. Let G be a central product of n copies G1 , 1 < i < n, of a perfect group L andlet a be an automorphism of G of order n permuting {G1 : 1 < i < n} transi-tively. Prove CG (a) = KZ where K = CG (aP) - L/ U for some U < Z(L)and Z=CZ(G)(a). Further NA,t(G)(GI) n C(K) <C(G1).

6. If A acts on a group G and centralizes a normal subgroup H of G then[G, A] <, CG(H).


A centralizes G. Produce a counter example when (1 A 1, (GI) # 1. (Hint: Reduce to the case where A is a p-group and use 5.14.)

2. Let G be a finite group, p a prime, and X a p-subgroup of G. Prove (1) Either X E Sylp(G) or X is properly contained in a Sylow p-subgroup of NG (X). (2) If G is a p-group and X is a maximal subgroup of G, then X a G and (G:XI=p .

3. Prove lemmas 10.1 and 10.3. Exhibit a nonsplit extension. 4. Let p and q be primes with p > q. Prove every group of order pq is a split

extension of Zp by Z,. Up to isomorphism, how many groups of order pq exist? (Hint: Use 10.3 and Exercise 1.7. Prove Aut(Z,) is cyclic of order p - 1. You may use the fact that the multiplicative group of a finite field is cyclic.)

5. Let G be a central product of n copies Gi, 1 5 i 5 n, of a perfect group L and let a! be an automorphism of G of order n permuting {Gi : 1 5 i 5 n ) transitively. Prove CG(a!) = KZ where K = ~ ~ ( a ! ) ( ' ) Z L/ u for some u 5 Z(L) and Z = CZ(G)(~). Further NAut(~)(G1) n C(K) i C(G1).

6. If A acts on a group G and centralizes a normal subgroup H of G then

[G, A1 6 CG(H).

4

Linear representations

Chapter 4 develops the elementary theory of linear representations. Linearrepresentations are discussed from the point of view of modules over the groupring. Irreducibility and indecomposability are defined, and we find that theJordan-Holder Theorem holds for finite dimensional linear representations.Maschke's Theorem is established in section 12. Maschke's Theorem saysthat, if G is a finite group and F a field whose characteristic does not dividethe order of G, then the indecomposable representations of G over F areirreducible.

Section 13 explores the connection between finite dimensional linear repre-sentations and matrices. There is also a discussion of the special linear group,the general linear group, and the corresponding projective groups. In particularwe find that the special linear group is generated by its transvections and isalmost always perfect.

Section 14 contains a discussion of the dual representation which will beneeded in section 17.

12 Modules over the group ringSection 12 studies linear representations over a field F using the group ring ofG over F. This requires an elementary knowledge of modules over rings. Onereference for this material is chapter 3 of Lang [La].

Throughout section 12, V will be a vector space over F. The group of auto-morphisms of V in the category of vector spaces and F-linear transformationsis the general linear group GL(V). Assume n: G -* GL(V) is a representationof G in this category. Such representations will be called FG-representationsand V will be called the representation module for n. Representation modulesfor FG-representations will be termed FG-modules.

Let R = F[G] be the vector space over F with basis G and define multi-plication on R to be the linear extension of the multiplication of G. Hence atypical member of R is of the form agg, where ag c- F and at most a finite

number of the coefficients ag are nonzero. Multiplication becomes

(ag)(bhh)=a h a

agbhgh.,hEG

Linear representations

Chapter 4 develops the elementary theory of linear representations. Linear representations are discussed from the point of view of modules over the group ring. Irreducibility and indecomposability are defined, and we find that the Jordan-Holder Theorem holds for finite dimensional linear representations. Maschke's Theorem is established in section 12. Maschke's Theorem says that, if G is a finite group and F a field whose characteristic does not divide the order of G, then the indecomposable representations of G over F are

Section 13 explores the connection between finite dimensional linear representations and matrices. There is also a discussion of the special linear group, the general linear group, and the corresponding projective groups. In particular we find that the special linear group is generated by its transvections and is almost always perfect.

Section 14 contains a discussion of the dual representation which will be needed in section 17.

12 Modules over the group ring Section 12 studies linear representations over a field F using the group ring of G over F. This requires an elementary knowledge of modules over rings. One reference for this material is chapter 3 of Lang [La].

Throughout section 12, V will be a vector space over F. The group of automorphisms of V in the category of vector spaces and F-linear transformations is the general linear group GL(V). Assume n: G + GL(V) is a representation of G in this category. Such representations will be called FG-representations and V will be called the representation module for n. Representation modules for FG-representations will be termed FG-modules.

Let R = F [GI be the vector space over F with basis G and define multiplication on R to be the linear extension of the multiplication of G. Hence a typical member of R is of the form xgEG agg, where a, E F and at most a finite number of the coefficients a, are nonzero. Multiplication becomes

36 Linear representations

As is well known (and easy to check), this multiplication makes R into a ringwith identity and, as the multiplication on. R commutes with scalar multipli-cation from F, R is even an F-algebra. R = F[G] is the group ring or groupalgebra of G over F.

Observe that V becomes a (right) R-module under the scalar multiplication:

v (Eagg) = Eag(v(gn)) vEV, Eagg c R

Conversely if U is anR-module

then we have a representation a: G GL(U)defined by u(ga) = ug, where ug is the module product of u c U by g c R.

Further if 7r,: G GL(V1), i = 1, 2, are FG-representations then ,B: V1 - V2is an equivalence of the representations precisely when ,B is an isomorphism ofthe corresponding R-modules. Indeed y: V1 V2 is an FG-homomorphism ifand only if y is an R-module homomorphism of the corresponding R-modules.Here y: V1 V2 is defined to bean FG-homomorphism if y is an F-linear mapcommuting with the actions of G in the sense that v(gn1)y = v y (gn2) for eachv E V1, and g c G. In the terminology of section 4, the FG-homomorphismsare the G-morphisms.

The upshot of these observations is that the study of FG-representations isequivalent to the study of modules for the group ring F[G] = R. I will takeboth points of view and appeal to various standard theorems on modules overrings. Lang [La] is a reference for such results.

Observe also that V is an abelian group under addition and n is a repre-sentation of G on V in the category of groups and homomorphisms. Indeed ninduces a representation

n': F# x G Aut(V)

in that category defined by (a, g)n': v i--> av(g7r), for a E F#, g c G.. Here F# isthe multiplicative group of F. Two FG-representations n and or are equivalentif and only if jr' and a' are equivalent, so we can use the results of chapter 3to study FG-representations. In the case where F is a field of prime order wecan say even more.

(12.1) Let F be the field of integers modulo p for some prime p and assumeV is of finite dimension. Then

(1) V is an elementary abelian p-group and n is a representation of G inthe category of groups and homomorphisms.

(2) If U is an elementary abelian p-group written additively, then U is a vec-tor space F U over F, where scalar multiplication is defined by ((p) + n)u = nu,nEl,uEU.

(3) GL(F U) is equal to the group Aut(U) of group automorphisms of U.Indeed if W is an elementary abelian p-group then the group homomorphismsfrom U into W are precisely the F-linear transformations from F U into F W W.

3 6 Linear representations

As is well known (and easy to check), this multiplication makes R into a ring with identity and, as the multiplication on R commutes with scalar multiplication from F , R is even an F-algebra. R = F[G] is the group ring or group algebra of G over F .

Observe that V becomes a (right) R-module under the scalar multiplication:

Conversely if U is an R-module then we have a representation a: G -+ GL(U) defined by u(ga) = ug, where ug is the module product of u E U by g E R.

Further if ni: G -+ GL(Vi), i = 1,2, are FG-representations then B: V1 -+ V2 is an equivalence of the representations precisely when B is an isomorphism of the corresponding R-modules. Indeed y: Vl -+ Vz is an FG-homomorphism if and only if y is an R-module homomorphism of the corresponding R-modules. Here y: Vl -+ V2 is defined to be an FG-homomorphism if y is an F-linear map commuting with the actions of G in the sense that v(gnl)y = v y (gn2) for each v E Vl, and g E G. In the terminology of section 4, the FG-homomorphisms are the G-morphisms.

The upshot of these observations is that the study of FG-representations is equivalent to the study of modules for the group ring FIG] = R. I will take both points of view and appeal to various standard theorems on modules over rings. Lang [La] is a reference for such results.

Observe also that V is an abelian group under addition and n is a representation of G on V in the category of groups and homomorphisms. Indeed n induces a representation

n': F' x G -+ Aut(V)

in that category defined by (a, g)n': v H av(gn), for a E F', g E G. Here F' is the multiplicative group of F . Two FG-representations n and a are equivalent if and only if n' and a' are equivalent, so we can use the results of chapter 3 to study FG-representations. In the case where F is a field of prime order we can say even more.

(12.1) Let F be the field of integers modulo p for some prime p and assume V is of finite dimension. Then

(1) V is an elementary abelian p-group and n is a representation of G in the category of groups and homomorphisms.

(2) If U is an elementary abelian p-group written additively, then U is a vector space F U over F , where scalar multiplication is defined by ((p) + n)u = nu, ~ E ~ , u E U .

(3) GL(FU) is equal to the group Aut(U) of group automorphisms of U. Indeed if W is an elementary abelian p-group then the group homomorphisms from U into W are precisely the F-linear transformations from F U into F W.

Modules over the group ring 37

(4) The vector space F V defined using the construction in part (2) is pre-cisely the vector space V.

As a consequence of 12.1, if F is a field of prime order, the FG-representationsare the same as the representations of G on elementary abelian p-groups.

A vector subspace U of V is an FG-submodule of V if U is G-invariant.U is an FG-submodule if and only if U is an R-submodule of the R-moduleV. From 7.1 there are group representations of F# x G on U and V/U. Theserepresentations are also FG-representations and they correspond to the R-modules U and V/ U.

V is irreducible or simple if 0 and V are the only FG-submodules. A com-position series for V is a series

O=Va<Vi<...<V,,=V

of FG-submodules such that each factor module V,+1 / Vi is a simple FG-module. This corresponds to the notion of composition series in section 7. Thefamily (Vi+1/ Vi: 0 < i < n) is the family of composition factors of the series.If V is of finite dimension then it is easy to see that V possesses a compositionseries. Appealing to the Jordan-Holder Theorem, established in section 7, andto remarks above, we get:

(12.2) (Jordan-Holder Theorem for FG-modules) Let V be a finite dimen-sional FG-module. Then V possesses a composition series and the composi-tion factors are independent (up to order and equivalence) of the choice ofcomposition series.

The restrictions ni = 7r I v,/v,-,, 0 < i < n, of 7r to the composition series (Vi: 0 <i < n) of a finite dimensional FG-representation 7r are called the irreducibleconstituents of jr. They are defined only up to order and equivalence but, sub-ject to this constraint, they are well defined and unique by the Jordan-HolderTheorem.

V is decomposable if there exist proper FG-submodules U and W of Vwith V = U ® W. Otherwise V is indecomposable. I'll write 7r = 7r1 + n2 ifV = Vl ® V2 with Vl and V2 FG-submodules of V and 7r I v; is equivalent toni. Observe that if a = al + a2 is an FG-representation with ai equivalent toni for i =1 and 2, then 7r is equivalent to a.

As in section 4, an FG-module V is said to be the extension of a module X bya module Y if there exists a submodule U of V with U = X and V/ U = Y. Acomplement to U in V is an FG-submodule W with V = U ® W. The extensionis said to split if U possesses a complement in V. As in chapter 3, we wish toinvestigate when extensions split.


(4) The vector space FV defined using the construction in part (2) is precisely the vector space V .

As a consequence of 12.1, if F is a field of prime order, the FG-representations are the same as the representations of G on elementary abelian p-groups.

A vector subspace U of V is an FG-submodule of V if U is G-invariant. U is an FG-submodule if and only if U is an R-submodule of the R-module V . From 7.1 there are group representations of F# x G on U and V/U. These representations are also FG-representations and they correspond to the R- modules U and V / U.

V is irreducible or simple if 0 and V are the only FG-submodules. A composition series for V is a series

of FG-submodules such that each factor module Vi+I/V; is a simple FG- module. This corresponds to the notion of composition series in section 7. The family (Vi+l/ Vi: 0 5 i < n) is the family of composition factors of the series. If V is of finite dimension then it is easy to see that V possesses a composition series. Appealing to the Jordan-Holder Theorem, established in section 7, and to remarks above, we get:

(12.2) (Jordan-Holder Theorem for FG-modules) Let V be a finite dimensional FG-module. Then V possesses a composition series and the composition factors are independent (up to order and equivalence) of the choice of composition series.

Therestrictions ni = n I v , / v , _ , , 0 < i 5 n, of n to thecomposition series (Vi: 0 5 i 5 n) of a finite dimensional FG-representation n are called the irreducible constituents of n. They are defined only up to order and equivalence but, subject to this constraint, they are well defined and unique by the Jordan-Holder Theorem.

V is decomposable if there exist proper FG-submodules U and W of V with V = U @ W . Otherwise V is indecomposable. I'll write n = nl + n2 if V = Vl @ V2 with Vl and V2 FG-submodules of V and nlv, is equivalent to ni. Observe that if a = a1 + a2 is an FG-representation with ai equivalent to ni for i = 1 and 2, then n is equivalent to a .

As in section 4, an FG-module V is said to be the extension of a module X by a module Y if there exists a submodule U of V with U Z X and V / U Z Y. A complement to U in V is an FG-submodule W with V = U @ W. The extension is said to split if U possesses a complement in V . As in chapter 3, we wish to investigate when extensions split.


An R-module V is cyclic if V = x R = f xr: r E R } for some x E V. Equiva-lently V = (xG) is generated as a vector space by the images of x under G.The element x is said to be a generator for the cyclic module V. Notice thatirreducible modules are cyclic.

(12.3) (1) If V =xR is cyclic then the map r i-+ xr is surjective R-modulehomomorphism from R onto V with kernel A(x) = f r E R: xr = 01.

(2) Homomorphic images of cyclic modules are cyclic, so the cyclic R-modules are precisely the homomorphic images of R.

(3) V is irreducible if and only if A(x) is a maximal right ideal of R.

Given R-modules U and V, HomR(U, V) denotes the set of all R-modulehomomorphisms of U into V. HomR(U, V) is an abelian group under thefollowing definition of addition:

u(a +,8) = ua + u,8 u E U, a B E HomR(U, V).

If R is commutative, HomR(U, V) is even an R-module when scalar multipli-cation is defined by

u(ar) = (ur)a u E U, r E R, a E HomR(U, V).

Finally HomR (V, V) = EndR (V) is a ring, where multiplication is defined tobe composition. That is

u (a ,8) _ (ua),8 u E V, a, ,8 E EndR (V ).

In the language of section 4, HomR (U, V) = MorG (U, V)-

(12.4) (Schur's Lemma) Let U and V be R-modules and a E HomR(U, V).Then

(1) If U is simple either a = 0 or a is an injection.(2) If V is simple either a = 0 or a is a surjection.(3) If U and V are simple then either a = 0 or a is an isomorphism.(4) If V is simple then EndR(V) is a division ring.

The module V is a semisimple R-module if V is the direct sum of simplesubmodules. The socle of V is the submodule Soc(V) generated by all thesimple submodules of V.

(12.5) Assume Q is a set of simple submodules of V and A C_ Q such thatV = (0) and (A) = ®AEA A. Then there exists r c 0 with A C F such that

V = ®bEr B.


An R-module V is cyclic if V = x R = {xr: r e R] for some x E V. Equiva- lently V = (xG) is generated as a vector space by the images of x under G. The element x is said to be a generator for the cyclic module V. Notice that irreducible modules are cyclic.

(12.3) (1) If V = x R is cyclic then the map r H xr is surjective R-module homomorphism from R onto V with kernel A(x) = (r E R: xr = 0).

(2) Homomorphic images of cyclic modules are cyclic, so the cyclic R- modules are precisely the homomorphic images of R.

(3) V is irreducible if and only if A(x) is a maximal right ideal of R.

Given R-modules U and V, HomR(U, V) denotes the set of all R-module homomorphisms of U into V. HomR(U, V) is an abelian group under the following definition of addition:

u(a + /?) = ua + u/? u E U, a, /? E HomR(U, V).

If R is commutative, HomR(U, V) is even an R-module when scalar multiplication is defined by

Finally HomR(V, V) = EndR(V) is a ring, where multiplication is defined to be composition. That is

In the language of section 4, H o m ~ (U, V) = Morc (U, V).

(12.4) (Schur's Lemma) Let U and V be R-modules and a E HomR(U, V). Then

(1) If U is simple either a= 0 or a is an injection. (2) If V is simple either a = 0 or a is a surjection. (3) If U and V are simple then either a = 0 or a is an isomorphism. (4) If V is simple then EndR(V) is a division ring.

The module V is a semisimple R-module if V is the direct sum of simple submodules. The socle of V is the submodule Soc(V) generated by all the simple submodules of V.

(12.5) Assume Q is a set of simple submodules of V and A Q such that V = (Q) and (A) = eAEA A. Then there exists r G SZ with A E r such that

V = a,, B .

Proof. Let S be the set of r C Q with A C_ F and (F) = ®BErB. Partiallyorder S by inclusion. Check that if C is a chain in S then UrEC F is an upperbound for C in S. Hence by Zorn's Lemma there is a maximal member r ofS. Finally prove V = (F).

(12.6) The following are equivalent:(1) V is semisimple.(2) V =Soc(V).(3) V splits over every submodule of V.

Proof. The equivalence of (1) and (2) follows from 12.5.Assume (3) holds but V Soc(V). By (3) there is a complement U to Soc(V)

in V. Let X E U#, I a maximal right ideal of R containing A (x) _ (r E R: x r = 01,and W the image of I in x R under the homomorphism of 12.3.1. By (3) thereis a complement Z to W in V. By the Modular Property of Groups, 1.14,Z fl x R = M is a complement to W in x R. Then M x R/ W, so M is simpleby 12.3.3. Hence M < Soc(V), so 0 $ M < Soc(V) fl U = 0, a contradiction.Thus (3) implies (2).

Finally assume V is semisimple and U is a submodule of V with no comple-ment in V. Now Soc(U) = ®AEoA for some set of simple submodules, so by12.4 there is a set r of simple submodules of V with A C F and V = ®BEr B.Then W = (F - A) is a complement to Soc(U) in V. Hence U # Soc(U). Bythe Modular Property of Groups, 1.14, U = Soc(U) ® (U fl W). Thus U fl Whas no simple submodules.

Choose X E (U fl W )O so that x has nonzero projection on a minimal numbern of members of r, and let A E F such that xa # 0, where u: xR -+ A is theprojection of x R onto A. For 0 0 y E x R, the set supp(y) of members of rupon which y projects nontrivially is a subset of supp(x). Thus by minimalityof n, supp(x) = supp(y). Therefore a: x R -+ A is an injection, and hence, by12.4.2, a is an isomorphism. But then x R A is simple, whereas it has already

been observed that U fl W has no simple submodules.

(12.7) Submodules and homomorphic images of semisimple modules aresemisimple.

(12.8) Assume G is finite, let U be an FG-submodule of V, and if char(F) =p > 0 assume there is an FP-complement W to U in V for some P E Sylp(G).Then V splits over U.

Proof. Let W be a vector subspace of V with V = U ® W, and if char(F) =p > 0 choose W to be P-invariant for some P E Sylp(G). If char (F) = 0 let


Proof. Let S be the set of r E C2 with A E r and (r) = @,,,B. Partially order S by inclusion. Check that if C is a chain in S then U,,, r is an upper bound for C in S. Hence by Zorn's Lemma there is a maximal member r of S. Finally prove V = (r).

(12.6) The following are equivalent: (1) V is semisimple. (2) v = Soc(V). (3) V splits over every submodule of V.

Proof. The equivalence of (1) and (2) follows from 12.5. Assume (3) holds but V # Soc(V). By (3) there is a complement U to Soc(V)

in V. Let x E u#, I a maximal right ideal of R containing A(x) = (r E R: xr = 01, and W the image of I in x R under the homomorphism of 12.3.1. By (3) there is a complement Z to W in V. By the Modular Property of Groups, 1.14, Z n x R = M isacomplementto WinxR.ThenM ZxR/W,soMissimple by 12.3.3. Hence M 5 Soc(V), so 0 + M 5 Soc(V) n U = 0, a contradiction. Thus (3) implies (2).

Finally assume V is sernisimple and U is a submodule of V with no complement in V. Now Soc(U) = @,,,A for some set of simple submodules, so by 12.4 there is a set r of simple submodules of V with A E r and V = B . Then W = (r - A) is a complement to Soc(U) in V. Hence U + Soc(U). By the Modular Property of Groups, 1.14, U ='SOC(U) CB (U fl W). Thus U fl W has no simple submodules.

Choose x E (U n w)# so that x has nonzero projection on a minimal number n of members of r , and let A E r such that x a + 0, where a: x R + A is the projection of x R onto A. For 0 + y E X R, the set supp(y) of members of r upon which y projects nontrivially is a subset of supp(x). Thus by minimality of n, supp(x) = supp(y). Therefore a: x R + A is an injection, and hence, by 12.4.2, a is an isomorphism. But thenx R Z A is simple, whereas it has already been observed that U n W has no simple submodules.

(12.7) Submodules and homomorphic images of semisimple modules are semisimple.

(12.8) Assume G is finite, let U be an FG-submodule of V, and if char(F) = p > 0 assume there is an FP-complement W to U in V for some P E Syl,(G). Then V splits over U.

Proof. Let W be a vector subspace of V with V = U @ W, and if char(F) = p > 0 choose W to be P-invariant for some P E Syl,(G). If char (F) = 0 let

P = 1. Let X be a set of coset representatives for P in G and let 7r: V -+ Ube the projection of V on U with respect to the decomposition V = U ® W.Let n = jG : P) and define 9: V -+ V by 9 = (r_xex x-17rx)/n, where the sumtakes place in EndF(V). As (p, n)=1, 1ln exists in F. Also x, x-1, and nare in EndF(V), so 0 is a well-defined member of EndF(V). As W is P-invariant, h7r = 7rh for all h E P, so if x H hx is a map from X into P then0 = (EXEX x-1(hx)-17rhxx)/n. That is 9 is independent of the choice of cosetrepresentatives X of P in G.

Claim 0 E EndR (V). As the multiplication in R is a linear extension of that inG and 0 E EndF(V), it suffices to show g0 = Og for all g E G. But gO = (rxex(xg-1)-17rxg-1)g/n = Og as Xg-1 is a set of coset representatives for P in Gand 0 is independent of the choice of X.

It remains to observe that, as 7r is the identity on U, as W = ker(7r), and as U is

G-invariant, we also have 0 the identity on U and U = V O. Hence 02 = 0. There-fore V = VO ® ker(0). As 0 E EndR(G), ker(0) is an FG-submodule. Hence,as U = V0, ker(0) is a complement to U in V. That is V splits over U.

(12.9) (Maschke's Theorem) Assume G is a finite group and char(F) doesnot divide the order of G. Then every FG-module is semisimple and everyFG-extension splits.

Proof. This is a direct consequence of 12.6 and 12.8.

Using 12.9 and notation and terminology introduced earlier in this section wehave:

(12.10) Assume G is a finite group and char(F) does not divide the order ofG. Let 7r: G --> GL(V) be a finite dimensional FG-representation. Then

(1) 7r = yi_1 7ri is the sum of its irreducible constituents (7ri: 1 < i < r).(2) If a = Ei_1 ai is another finite dimensional FG-representation with

irreducible constituents (a1: 1 < i < s) then 7r is equivalent to a if and onlyif r = s and there is a permutation a of {1, 2, ... , r} with it equivalent to aivfor each i.

So, in this special case, the study of FG-representations is essentially reducedto the study of irreducible FG-representations.

Let V be a semisimple R-module and S a simple R-module. The homoge-neous component of V determined by S is (U: U < V, U = S). V is homoge-neous if it is generated by isomorphic simple submodules.

(12.11) Let V be a semisimple R-module. Then


P = 1. Let X be a set of coset representatives for P in G and let n: V -+ U be the projection of V on U with respect to the decomposition V = U @ W. Let n = I G : PI and define 8: V -+ V by 8 = (Ex,, x-'nx)/n, where the sum takes place in EndF(V). As (p, n) = 1, l / n exists in F . Also x, x-', and n are in EndF(V), so 8 is a well-defined member of E n d ~ ( v ) . As W is P- invariant, hn = n h for all h E P, so if x H h, is a map from X into P then 6' = (Ex,, x-'(h,)-'nh,x)/n. That is 6' is independent of the choice of coset representatives X of P in G.

Claim 6 E EndR(V). As the multiplication in R is a linear extension of that in G and 8 E EndF(V), it suffices to show g6 = Bg for all g E G. But go = (Ex,, ( ~ ~ - ' ) - ' n x ~ - ' ) ~ / n = Bg as xg-' is a set of coset representatives for P in G and 8 is independent of the choice of X.

It remains to observe that, as n is the identity on U, as W = ker(n), and as U is G-invariant, we also have 8 the identity on U and U = VB. Hence B2 = 6. There- fore V = VB @ ker(6). As 8 E EndR(G), ker(6) is an FG-submodule. Hence, as U = V8, ker(8) is a complement to U in V. That is V splits over U.

(12.9) (Maschke's Theorem) Assume G is a finite group and char(F) does not divide the order of G. Then every FG-module is semisimple and every FG-extension splits.


Using 12.9 and notation and terminology introduced earlier in this section we have:

(12.10) Assume G is a finite group and char(F) does not divide the order of G. Let n : G -+ GL(V) be a finite dimensional FG-representation. Then

(1) n = E;=, ni is the sum of its irreducible constituents (xi: 1 ( i 5 r). (2) If a = xi=, ai is another finite dimensional FG-representation with

irreducible constituents (ai: 1 ( i ( s) then n is equivalent to a if and only if r = s and there is a permutation a of {1,2, . . . , r) with ni equivalent to ai,

for each i .

So, in this special case, the study of FG-representations is essentially reduced to the study of irreducible FG-representations.

Let V be a semisimple R-module and S a simple R-module. The homogeneous component of V determined by S is (U: U ( V, U Z S). V is homogeneous if it is generated by isomorphic simple submodules.

(12.11) Let V be a semisimple R-module. Then

(1) If V is homogeneous then every pair of simple submodules of V isisomorphic.

(2) V is the direct sum of its homogeneous components.

Proof. As V is semisimple, V = ®A,12 A for some set S2 of simple submodulesof V. Let T be a simple submodule of V and supp(T) the set of submodules inSZ upon which T projects nontrivially. If A E supp(T) then the projection mapa: T --> A is an isomorphism by Schur's Lemma. But if V is homogeneousthen, by 12.5, we may choose A = S for some simple R-module S and allA E 0. Hence (1) holds.

Similarly if S is a simple R-module, H the homogeneous component ofV determined by S, and K the submodule of V generated by the remaininghomogeneous components, then, as V is semisimple, V = H + K. Further ifH f1 K 0, we may choose T < H f1 K by 12.7. But now, by (1), S = T = Qfor some simple R-module Q determining a homogeneous component distinctfrom that of S, a contradiction.

(12.12) Let H < G and U a simple FH-submodule of V. Then(1) Ug is a simple FH-submodule of V for each g E G.(2) If g E CG(H) then U is FH-isomorphic to Ug.(3) If X and Y are isomorphic FH-submodules of V then Xg and Yg are

FH-isomorphic submodules for each g c G

(12.13) (Clifford's Theorem) Let V be a finite dimensional irreducible FG-module and H < G. Then

(1) V is a semisimple FH-module.(2) G acts transitively on the FH-homogeneous components of V.(3) Let U be an FH-homogeneous component of V. Then NG(U) is irre-

ducible on U and HCG(H) < NG(U).

Proof. By 12.12.1, G acts on the socle of V, regarded as an FH-module. Thus(1) holds by 12.6 and the irreducible action of G. By 12.12.3, G permutesthe homogeneous FH-components of V. Then, by 12.11.2 and the irreducibleaction of G, G is transitive on those homogeneous components. By 12.12.2,HCG(H) acts on each homogeneous component. Then 12.11.2 and the irre-ducible action of G completes the proof of (3).

Observe that 7r can be extended to a representation of R on V (that is to anF-algebra homomorphism of R into EndF(V)) via 7r: > agg H > ag(g7r).Indeed for r E R and v E V, v(r7r) = yr is just the image of v under the module

Modules over the group ring 4 1

(1) If V is homogeneous then every pair of simple submodules of V is isomorphic.

(2) V is the direct sum of its homogeneous components.

Proof. As V is semisimple, V = eAGQ A for some set S2 of simple submodules of V. Let T be a simple submodule of V and supp(T) the set of submodules in S2 upon which T projects nontrivially. If A E supp(T) then the projection map a: T -+ A is an isomorphism by Schur's Lemma. But if V is homogeneous then, by 12.5, we may choose A Z S for some simple R-module S and all A E S2. Hence (1) holds.

Similarly if S is a simple R-module, H the homogeneous component of V determined by S, and K the submodule of V generated by the remaining homogeneous components, then, as V is semisimple, V = H + K. Further if HnK#O,wemaychooseT 5 H n K b y 12.7.Butnow,by(l),SZ T Z Q for some simple R-module Q determining a homogeneous component distinct from that of S, a contradiction.

(12.12) Let H 9 G and U a simple FH-submodule of V. Then (1) Ug is a simple FH-submodule of V for each g E G. (2) If g E CG(H) then U is FH-isomorphic to Ug. (3) If X and Y are isomorphic FH-submodules of V then Xg and Yg are

FH-isomorphic submodules for each g E G'!

(12.13) (Clifford's Theorem) Let V be a finite dimensional irreducible FG- module and H 9 G. Then

(1) V is a semisimple FH-module. (2) G acts transitively on the FH-homogeneous components of V. (3) Let U be an FH-homogeneous component of V. Then NG(U) is irre-

ducible on U and HCG(H) 5 NG(U).

Proof. By 12.12.1, G acts on the socle of V, regarded as an FH-module. Thus (1) holds by 12.6 and the irreducible action of G. By 12.12.3, G permutes the homogeneous FH-components of V. Then, by 12.11.2 and the irreducible action of G, G is transitive on those homogeneous components. By 12.12.2, HCG(H) acts on each homogeneous component. Then 12.11.2 and the irreducible action of G completes the proof of (3).

Observe that n can be extended to a representation of R on V (that is to an F-algebra homomorphism of R into EndF(V)) via n : C agg H C ag(gn). Indeed for r E R and v E V, v(rn) = vr is just the image of v under the module

product of v by r in the R-module V. Further

ker(ir) = (r E R: yr = 0 for all v E V1.

V is said to be a faithful R-module if it is an injection on R. As G generates Ras an F-algebra, R,r is the subalgebra of EndF(V) generated by G7r. We callRzr the enveloping algebra of the representation n.

(12.14) EndF(V) = CEfldF(v)(R7r) = CEndF(v)(GJr).

(12.15) If G is finite and it is irreducible then Z(Gir) is cyclic.

Proof. Let E =EndF(V). As 7r is irreducible, D =EndF(V) is a divisionring by Schur's Lemma. Z = Z(G7r) < CE(G7r) = D by 12.14. Also D <CE(GTr) < CE(Z), so Z < Z(D). Thus the sub-division-ring K of D genera-ted by Z is a field. Now Z is a finite subgroup of the multiplicative group ofthe field K, and hence K is cyclic.

I conclude this section by recording two results whose proofs can be found insection 3 of chapter 17 of Lang [La].

(12.16) Let 7r: G -* GL(V) be an irreducible finite dimensional FG-representa-tion. Then Rir is isomorphic as an F-algebra to the ring of all m by m matricesover the division ring EndFG (V) = D, where m = dimD (V). Further F is in thecentre of D.

(12.17) (Burnside) Assume F is algebraically closed and 7r: G -* GL(V) isan irreducible finite dimensional FG-representation. Then EndF(V) = Rir andF=EndFG(V).

13 The general linear group and special linear groupIn this section F is a field, n is a positive integer, and V is an n-dimensionalvector space over F. Recall the group of vector space automorphisms of V isthe general linear group GL(V). As the isomorphism type of V depends only onn and F, the same is true for GL(V), so we can also write GL"(F) for GL(V).

(13.1) Let F""" denote the F-algebra of all n by n matrices over F, letX = (xi .... x") be an ordered basis for V, and for g E EndF(V) let Mx(g) _(gig) be the matrix defined by xi g = >j gig xj, gij E F. Then

(1) The map Mx: g H Mx(g) is an F-algebra isomorphism of EndF(V)with F"'Hence the map restricts to a group isomorphism of GL(V) with thegroup of all nonsingular n by n matrices over F.


product of v by r in the R-module V. Further

ker(n) = ( r E R: vr = O for all v E V).

V is said to be a faithful R-module if n is an injection on R. As G generates R as an F-algebra, Rn is the subalgebra of EndF(V) generated by Gn. We call Rn the enveloping algebra of the representation n .

(12.15) If G is finite and n is irreducible then Z(Gn) is cyclic.

Proof. Let E = EndF (V). As n is irreducible, D = EndR(V) is a division ring by Schur's Lemma. Z = Z(Gn) 5 CE(Gn) = D by 12.14. Also D 5 CE(Gn) ( CE(Z), so Z 5 Z(D). Thus the sub-division-ring K of D generated by Z is a field. Now Z is a finite subgroup of the multiplicative group of the field K, and hence K is cyclic.

I conclude this section by recording two results whose proofs can be found in section 3 of chapter 17 of Lang [La].

(12.16) Let n : G + GL(V) be an irreducible finite dimensional FG-representation. Then Rn is isomorphic as an F-algebra to the ring of all m by m matrices over the division ring EndFG(V) = D, where m = dimD(V). Further F is in the centre of D.

(12.17) (Burnside) Assume F is algebraically closed and n : G + GL(V) is an irreducible finite dimensional FG-representation. Then EndF(V) = Rn and F = EndFG(V).

13 The general linear group and special linear group In this section F is a field, n is a positive integer, and V is an n-dimensional vector space over F. Recall the group of vector space automorphisms of V is the general linear group GL(V). As the isomorphism type of V depends only on n and F , the same is true for GL(V), so we can also write GLn(F) for GL(V).

(13.1) Let Fnxn denote the F-algebra of all n by n matrices over F , let X = (xl . . . , xn) be an ordered basis for V, and for g E EndF(V) let Mx(g) = (gij) be the matrix defined by xig = Cj gijxj, gij E F. Then

(1) The map Mx: g H Mx(g) is an F-algebra isomorphism of EndF(V) with FnXn. Hence the map restricts to a group isomorphism of GL(V) with the group of all nonsingular n by n matrices over F.

The general linear group and special linear group 43

(2) Let Y = (yl, ... , be a second ordered basis of V, let h be the uniqueelement of GL(V) with xih = yi, 1 < i < n, and B = My(h). Then Mx =h*My = MyB* is the composition of h* with My and of My with B*, whereh* and B* are the conjugation automorphisms induced by h and B on EndF(V)and F'1 , respectively.

Because of 13.1, we can think of subgroups of GL(V) as groups of matricesif we choose. I take this point of view when it is profitable. Similarly an FG-representation it on V can be thought of as a homomorphism from G into thegroup of all n by n nonsingular matrices over G, by composing it with the iso-morphism Mx. it is equivalent to ir': G -* GL(V) if and only if 7r'= irh* forsome h E GL(V), and by 13.1.2 this happens precisely when Jr'Mx = Tr Mx B *for some nonsingular matrix B. This gives a notion of equivalence for `matrixrepresentations'. Namely two homomorphisms a and a' of G into the groupof all n by n nonsingular matrices over F are equivalent if there exists a non-singular matrix B with a' = a B*.

Let's see next what the notions of reducibility and decomposability corre-spond to from the point of view of matrices.

(13.2) Let 7r: G GL(V) be an FG-representation, U an FG-submodule ofV,V = V/ U, and X = (xi: 1 < i < n) a basis for V with Y = (xi: 1 < i < m)

a basis for U. Then, for g E G,

Mx(gn) =[My(g7r l u) 0

L A(g) Mx(g7r(g7r v)

for some n - m by m matrix A(g).

Of course there is a suitable converse to 13.2.

(13.3) Let ir: G -* GL(V) be an FG-representation, U and W FG-submodulesof V with V=U ® W, and X = (xi : 1 < i < n) a basis for V such that Y=(xi: 1 < i < m) and Z = (xi: m < i < n) are basis for U and W, respectively.Then for g E G,

Mx(gn)My(gnly) 0

[ fi Mz(gn)w)

Again there is a suitable converse to 13.3.Recall the notion of geometry defined in section 3. We associate a geometry

PG(V) to V, called the projective geometry of V. The objects of PG(V) arethe proper nonzero subspaces of V, with incidence defined by inclusion. If U


(2) Let Y = (yl, . . . , yn) be a second ordered basis of V, let h be the unique element of GL(V) with xih = yi , 1 i i 5 n, and B = My (h). Then Mx = h*My =My B* is the composition of h* with My and of My with B*, where h* and B* are the conjugation automorphisms induced by h and B on EndF(V) and Fnxn, respectively.

Because of 13.1, we can think of subgroups of GL(V) as groups of matrices if we choose. I take this point of view when it is profitable. Similarly an FG- representation n on V can be thought of as a homomorphism from G into the group of all n by n nonsingular matrices over G, by composing n with the isomorphism Mx. n is equivalent to n': G + GL(V) if and only if n' = nh* for some h E GL(V), and by 13.1.2 this happens precisely whennfMx = nMxB* for some nonsingular matrix B. This gives a notion of equivalence for 'matrix representations'. Namely two homomorphisms a and a' of G into the group of all n by n nonsingular matrices over F are equivalent if there exists a nonsingular matrix B with a' = a B*.

Let's see next what the notions of reducibility and decomposability correspond to from the point of view of matrices.

(13.2) Let n : G + GL(V) be an FG-representation, U an FG-submodule of v , ~ = V / U , andX=(xi: 1 5 i 5 n)a basis for V with Y =(xi: 1 5 i 5 m) a basis for U. Then, for g E G,

for some n - m by m matrix A(g).

Of course there is a suitable converse to 13.2.

(13.3) Let n : G + GL(V) be an FG-representation, U and W FG-submodules of V with V = U @I W, and X = (xi: 1 5 i n) a basis for V such that Y = (xi: 1 5 i 5 m) and Z = (xi: m < i 5 n) are basis for U and W, respectively. Then for g E G,

Again there is a suitable converse to 13.3. Recall the notion of geometry defined in section 3. We associate a geometry

PG(V) to V, called the projective geometry of' V. The objects of PG(V) are the proper nonzero subspaces of V, with incidence defined by inclusion. If U


is a subspace of V, the projective dimension of U is Pdim(U) = dimF(U) - 1.The type function for PG(V) is the projective dimension function

Pdim: PG(V) I= {0, 1, ..., n - 1).

PG(V) is said to be of dimension n - 1. The subspaces of projective dimension0, 1, and n - 2 are referred to as points, lines, and hyperplanes, respectively.

Forg EGL(V)define gP:PG(V)- PG(V)bygP: U H Ug, for U EPG(V).Evidently P: GL(V) Aut(PG(V)) is a representation of GL(V) in the cat-egory of geometries. (See the discussion in section 3.) Denote the image ofGL(V) under P by PGL(V). PGL(V) is the projective general linear group.The notation PGLn(F) is also used for PGL(V).

A scalar transformation of Visa member g of EndF(V) such that vg = avfor all v in V and some a in F independent of v. A scalar matrix is a matrixof the form aI, a E F, where I is the identity matrix.

(13.4) (1) Z(EndF(V)) is the set of scalar transformations. The image ofZ(EndF(V)) under Mx is the set of scalar matrices.

(2) Z(GL(V)) is the set of nonzero scalar transformations.(3) Z(GL(V))=ker(P).

By 13.4, the projective general linear group PGL(V) is isomorphic to the groupof all n by n nonsingular matrices modulo the subgroup of scalar matrices. Oftenit will be convenient to regard these groups as the same.

Given any FG-representation n: G GL(V), n can be composed with Pto obtain a homomorphism nP: G PGL(V). Observe that nP is a represen-tation of G on the projective geometry PG(V).

For y E EndF (V) define the determinant of y to be det(x) = det(Mx (y)). Thatis the determinant of y is the determinant of its associated matrix. Similarlydefine the trace of y to be Tr(y) = Tr(Mx (y)). So the trace of y is the traceof its associated matrix. If A is a matrix and B is a nonsingular matrix thendet(AB) = det(A) and Tr(AB) = Tr(A), so det(y) and Tr(y) are independent ofthe choice of basis X by 13.1.2.

Define the special linear group SL(V) to be the set of elements of GL(V) ofdeterminant 1. The determinant map is a homomorphism of GL(V) onto themultiplicative group of F with SL(V) the kernel of this homomorphism, soSL(V) is a normal subgroup of GL(V) and GL(V)/SL(V) = F#. Also writeSLn(F) for SL(V). The image of SL(V) under P is denoted by PSL(V) orPSLn (F). The group PSL(V) is the projective special linear group. SometimesPSLn(F) is denoted by Ln(F).


is a subspace of V, the projective dimension of U is Pdim(U) = dimF(U) - 1. The type function for PG(V) is the projective dimension function

Pdim: PG(V) + I= {O,1, . . . , n - 1).

PG(V) is said to be of dimension n - 1. The subspaces of projective dimension 0, 1, and n - 2 are referred to as points, lines, and hyperplanes, respectively.

Forg E GL(V) definegP: PG(V) + PG(V) by gP: U t-, Ug, for U E PG(V). Evidently P: GL(V) + Aut(PG(V)) is a representation of GL(V) in the category of geometries. (See the discussion in section 3.) Denote the image of GL(V) under P by PGL(V). PGL(V) is the projective general linear group. ,

The notation PGL,(F) is also used for PGL(V). A scalar transformation of V is a member g of EndF(V) such that vg = av

for all v in V and some a in F independent of v. A scalar matrix is a matrix of the form a l , a E F, where I is the identity matrix.

(13.4) (1) Z(EndF(V)) is the set of scalar transformations. The image of Z(EndF(V)) under Mx is the set of scalar matrices.

(2) Z(GL(V)) is the set of nonzero scalar transformations. (3) Z(GL(V)) = ker(P).

By 13.4, the projective general linear group PGL(V) is isomorphic to the group of all n by n nonsingular matrices modulo the subgroup of scalar matrices. Often it will be convenient to regard these groups as the same.

Given any FG-representation n : G + GL(V), n can be composed with P to obtain a homomorphism nP : G + PGL(V). Observe that n P is a representation of G on the projective geometry PG(V).

For y E EndF(V) define the determinant of y to be det(x) = det(Mx(y)). That is the determinant of y is the determinant of its associated matrix. Similarly define the trace of y to be Tr(y) = Tr(Mx(y)). So the trace of y is the trace of its associated matrix. If A is a matrix and B is a nonsingular matrix then d e t ( ~ ~ ) = det(A) and T ~ ( A ~ ) = Tr(A), so det(y) and Tr(y) are independent of the choice of basis X by 13.1.2.

Define the special linear group SL(V) to be the set of elements of GL(V) of determinant 1. The determinant map is a homomorphism of GL(V) onto the multiplicative group of F with SL(V) the kernel of this homomorphism, so SL(V) is a normal subgroup of GL(V) and GL(V)/SL(V) Z F'. Also write SL,(F) for SL(V). The image of SL(V) under P is denoted by PSL(V) or PSL,(F). The group PSL(V) is the projective special linear group. Sometimes PSL,(F) is denoted by L,(F).

Prove the next lemma for GL(V) and then use 5.20 to show the result holdsfor SL(V). See section 15 for the definition of 2-transitivity.

(13.5) SL(V) is 2-transitive on the points of PGL(V).

For v in V and a in EndF(V), [v, a] = va - v is the commutator of v with a.This corresponds with the notion of commutator in section 8. Indeed we canform the semidirect product of V by GL(V) with respect to the natural repre-sentation, and in this group the two notions agree. Similarly, for G < GL(V),

[V, G] = ([v, g]: v E V, g E G)

and, for g E G, [V, g] _ [V, (g)].A transvection is an element t of GL(V) such that [V, t] is a point of

PG(V), Cv(t) is a hyperplane of PG(V), and [V, t] <Cv(t). [V, t] and Cv(t)are called the center and axis of t, respectively. Let xn E V - Cv(t). Then[x,,, t] = x1 generates [V, t] and we choose xi E Cv(t), 1 < i < n, so that X =(xi: 1 < i < n) is a basis of V. Then

1 0 0

MX(t)= 0 1 0

1 0 1

so evidently t is of determinant 1 and 13.1.2 implies GL(V) is transitive on itstransvections. Write diag(al, ..., an) for the diagonal matrix whose (i, i)-thentry is ai. If n > 2let

A = {diag(1, a, 1, ... , 1): a E F#}

and ifn=2 let

A = (diag(a, a): a E F#).

Then A < CGL(v)(t) and either det: A F# is a surjection or n = 2 and someelement of F is not a square in F. In the first case GL(V) = ASL(V), so, asGL(V) is transitive on its transvections, so is SL(V) by 5.20. Further if n > 2and s is the transvection with Cv(s) = Cv(t) and [xn, s] =x2, then st is alsoa transvection. So, as SL(V) is transitive on its transvections, t =s-1(st) is acommutator of SL(V). On the other hand, if n = 2, then for b E F# let t (b) be thetransvection with x2t(b) = x2 + bxl and g = diag(a, a-1). Then t(b)9 = t(a2b).Further if F > 3 then a canbe chosen with a2 1. Thus, setting b = (a2_1)-l'

we have [t(b), g] = t, and again t is a commutator of SL(V).


Prove the next lemma for GL(V) and then use 5.20 to show the result holds for SL(V). See section 15 for the definition of 2-transitivity.

(13.5) SL(V) is 2-transitive on the points of PGL(V).

For v in V and a in EndF(V), [v, a ] = va - v is the commutator of v with a. This corresponds with the notion of commutator in section 8. Indeed we can form the semidirect product of V by GL(V) with respect to the natural representation, and in this group the two notions agree. Similarly, for G 5 GL(V),

[V, GI = ([v, gl: v E V, g E G)

and, for g E G, [V, gl = [V, (g)l. A transvection is an element t of GL(V) such that [V, t] is a point of

PG(V), Cv(t) is a hyperplane of PG(V), and [V, t] I Cv(t). [V, t] and Cv(t) are called the center and axis of t, respectively. Let x, E V - Cv(t). Then [x,,, t] = X I generates [V, t] and we choose xi E Cv(t), 1 < i < n, so that X =

(xi: 1 5 i 5 n) is a basis of V. Then

so evidently t is of determinant 1 and 13.1.2 implies GL(V) is transitive on its transvections. Write diag(al, . . . , a,) for the diagonal matrix whose (i, i)-th entry is a i . If n > 2 let

and if n = 2 let

A = {diag(a, a): a E F') .

Then A 5 CGL(~)(t) and either det: A + F' is a surjection or n = 2 and some element of F is not a square in F . In the first case GL(V) = ASL(V), so, as GL(V) is transitive on its transvections, so is SL(V) by 5.20. Further if n > 2 and s is the transvection with Cv(s) = Cv(t) and [x,, s] =x2, then st is also a transvection. So, as SL(V) is transitive on its transvections, t = s-'(st) is a commutator of SL(V). On the other hand, if n = 2, then forb E F' let t(b) be the transvection with x2t(b) = x2 + bxl and g = diag(a, a-I). Then t(b)g = t(a2b). Further if IF I > 3 then a can be chosen with a2 # 1. Thus, setting b = (a2 - I)-', we have [t(b), g] = t, and again t is a commutator of SL(V).

We have shown:

(13.6) (1) Transvections are of determinant 1.(2) The transvections form a conjugacy class of GL(V).(3) Either the transvections form a conjugacy class of SL(V) or n = 2 and

F contains nonsquares.(4) If IF I > 3 or n > 2, then each transvection is in the commutator group

of SL(V).

(13.7) SL(V) is generated by its transvections.

P r o o f . Let 0 be the set of n-tuples w = (x1, ... , x,,-,, (xn)) such that (x l , ... ,xn) is a basis for V. Let T be the subgroup of G = SL(V) generated by thetransvections of G. I'll show T is transitive on 0. Then, by 5.20, G = TG,,,.But GU, = 1, so the lemma holds.

It remains to show T is transitive on 0. Pick a = (yi, ... , Yn-1, (Yn)) E 0such that a 0 wT, yi =xi for i < m, and, subject to these constraints, with mmaximal. Let U = (xi: i < m), x =xn+1> Y = Y,n+i, and W = (U, x, y). Thendim(W/ U) = k = 1 or 2.

Suppose k = 2 and let H be a hyperplane of V containing U and x - y butnot x. Let t be the transvection with axis H and [y, t] = x - y. Then yt = x andxi t = xi for i < m, so at E wT by maximality of m. Then a E wT, contrary tothe choice of a.

So k = 1. Suppose m = n - 1. As k = 1, ax - y E U for some a E F#. Asm = n - 1 and a # w, ax - y # 0. So there is a transvection t with axis Uand [y, t] = ax - y. Now at = w, contradicting a 0 off. So m < n - 1, andhence there is z E V - W. An argument in the last paragraph shows there aretransvections s and t with U < Ca(t) fl Cv(s), ys = z, and zt =x. But nowxist =xi for i < m and yst = x, contradicting the choice of a.

(13.8) If n > 2 then SL,(F) is perfect unless n = 2 and IF I = 2 or 3.

Proof. Let G = SLn(F). By 13.7 it suffices to show transvections are containedin GM, and this follows from 13.6.4.

14 The dual representationIn section 14, V continues to be an n-dimensional vector space over F and7r: G -+ GL(V) is an FG-representation of a group G.

Let (Vi: -oo < i < oo) be a sequence of FG-modules and

...-+ V_1 VoC"') V1-+ ...


We have shown:

(13.6) (1) Transvections are of determinant 1. (2) The transvections form a conjugacy class of GL(V). (3) Either the transvections form a conjugacy class of SL(V) or n = 2 and

F contains nonsquares. (4) If IF I > 3 or n > 2, then each transvection is in the commutator group

of SL(V).

(13.7) SYV) is generated by its transvections.

Proof. Let i-2 be the set of n-tuples o = (XI, . . . , x,-l, (x,)) such that (xl , . . . , x,) is a basis for V. Let T be the subgroup of G = SL(V) generated by the transvections of G. I'll show T is transitive on a. Then, by 5.20, G = TG,. But G , = 1, so the lemma holds.

It remains to show T is transitive on a. Pick a = (yl, . . . , y,-1, (y,)) E i-2 such that a 4 oT, yi =xi for i i m, and, subject to these constraints, with m maximal. Let U=(xi : i i rn),x=x,,+l, y=y,+l, and W= (U,x, y). Then dim(W/ U) = k = 1 or 2.

Suppose k = 2 and let H be a hyperplane of V containing U and x - y but not x. Let t be the transvection with axis H and [y, t] = x - y. Then yt = x and xit =xi for i ( m, so a t E oT by maximality of m. Then a E oT, contrary to the choice of a.

S o k = l . S u p p o s e m = n - l . A s k = l , a x - y E Uforsomea E F#.As rn = n - 1 and a # o, ax - y # 0. So there is a transvection t with axis U and [y, t] =ax - y. Now a t = o, contradicting a 4 oT. So m < n - 1, and hence there is z E V - W. An argument in the last paragraph shows there are transvections s and t with U ( Cv(t) r l Cv(s), ys = z, and zt =x. But now xist =xi for i 5 m and yst = x, contradicting the choice of a .

(13.8) If n 2 2 then SL,(F) is perfect unless n = 2 and IF I = 2 or 3.

Prooj Let G = SL,(F). By 13.7 it suffices to show transvections are contained in G('), and this follows from 13.6.4.

14 The dual representation In section 14, V continues to be an n-dimensional vector space over F and n: G + GL(V) is an FG-representation of a group G.

Let (K: -oo < i < oo) be a sequence of FG-modules and

The dual representation 47

a sequence of FG-homomorphisms. The latter sequence is said to be exact ifker(ai+1) = Viai for each i. A short exact sequence is an exact sequence ofthe form 0 -+ U - V W -+ 0. The maps 0 -+ U and W --> O are forced tobe trivial. Observe that the hypothesis that the sequence is exact is equivalentto requiring that a be an injection, $ a surjection, and Ua = ker(p). HenceW - V/ Ua and the sequence is essentially 0 -+ Ua -+ V -+ V/ Ua -+ 0 withUa -+ V inclusion and V -+ V/Ua the natural map. The sequence is said tosplit if V splits over Ua. As is well known, the sequence splits if and only ifthere is y E HomFG(W, V) with yp =1, and this condition is equivalent inturn to the existence of S E HomFG(V, U) with aS = 1.

Let V * = HomF (V, F) and recall from section 13 that V* is a vector spaceover F. We call V* the dual space of V. It is well known that n = dimF (V * ).If U is an F-space and a EHomF(U, V) define a* EHomF(V*, U*) byxa* =ax, x E V*.

(14.1) Let U, V, and W be finite dimensional F-spaces, a E HomF(U, V), and,B E HomF(V, W). Then

(1) The map y i-+ y* is an F-space isomorphism of HomF(U, V) withHomF(V*, U*).

(2) (afi)*(3) If U4 V - W is exact then so is W* V*-* U*.

If 7r: G -+ GL(V) is an FG-representation then, from 14.1, 7r*: G GL(V*)is also an FG-representation, where 7r*: g H (g-17r)*. The representation 7r*is called the dual of 7r and the representation module V* of 7r* is called thedual of the representation module V of 7r.

Given a basis X = (xi : 1 < i < n) for V, the dual basis X = (zi : 1 < i < n)of X is defined by xi : xj i-+ Sid . Notice T i aixi is the unique member of V*mapping xi to ai for each i.

(14.2) Let 7r: G -+ GL(V) be an FG-representation and X a basis for V. ThenMX(g7r*) = (MX(g7r)-1)T where BT denotes the transpose of a matrix B.

By 14.2, if 7r is viewed as a matrix representation, then 7r* is just the composi-tion of 7r with the transpose-inverse map on GLn(F). As the transpose-inversemap is of order 2, we conclude

(14.3) (7r *)* is equivalent to rr for each finite dimensional FG-representation 7r.

There is a more concrete way to see this.


a sequence of FG-homomorphisms. The latter sequence is said to be exact if ker(ai+l) = Kai for each i. A short exact sequence is an exact sequence of the form 0 + U 3 V 5 W + 0. The maps 0 + U and W -t 0 are forced to be trivial. Observe that the hypothesis that the sequence is exact is equivalent to requiring that a be an injection, /? a surjection, and Ua = ker(/?). Hence W E V/ Ua and the sequence is essentially 0 + Ua + V + V/ Ua + 0 with Ua + V inclusion and V + V/Ua the natural map. The sequence is said to split if V splits over Ua. As is well known, the sequence splits if and only if there is y E HomFG(W, V) with y/? = 1, and this condition is equivalent in turn to the existence of 6 E HomFG(V, U) with a6 = 1.

Let V* = HomF(V, F ) and recall from section 13 that V* is a vector space over F. We call V* the dual space of V. It is well known that n = dimF(V*). If U is an F-space and a €HomF(U, V) define a* €HomF(V*, U*) by XU* = a x , x E V*.

(14.1) Let U, V, and W be finite dimensional F-spaces, a E HomF(U, V), and /? E HomF(V, W). Then

(1) The map y H y* is an F-space isomorphism of HomF(U, V) with HomF(V*, U*).

(2) (a/?)* = /?*a*.

(3) If U 3 V 5 W is exact then so is W* V* % U*.

If n: G + GL(V) is an FG-representation then, from 14.1, n *: G + GL(V*) is also an FG-representation, where n*: g H (g-ln)*. The representation n * is called the dual of n and the representation module V* of n * is called the dual of the representation module V of n .

Given a basis X = (xi: 1 _( i 5 n) for V, the dual basis 2 = (ai : 1 5 i 5 n) of X is defined by ii: x j H Jij. Notice xi a i f i is the unique member of V* mapping xi to ai for each i .

(14.2) Let n : G + GL(V) be an FG-representation and X a basis for V. Then M2 (gn *) = ( M ~ (gn)-l)T, where B~ denotes the transpose of a matrix B.

By 14.2, if n is viewed as a matrix representation, then n * is just the composition of n with the transpose-inverse map on GL,(F). As the transpose-inverse map is of order 2, we conclude

(14.3) (n*)* is equivalent to n for each finite dimensional FG-representation n.

ere is a more concrete way to see this.

(14.4) Let U and V be finite dimensional F-spaces. Then(1) For each v E V there exists a unique element vB E (V*)* with xv9 = vx

for all x E V*.(2) The map 0: v H vB is an F-isomorphism of V with (V*)*.(3) For each a E HomF(U, V), aB = B(a*)*.(4) 0 defines an equivalence of 7r and (7r*)*.

Proof. To prove (1) let X = (xi: 1 <i <n)beabasisfor V, X=(zi: 1 < i < n)its dual basis, and X =(Xi : 1 < i < n) the dual basis of X in (V*)*. Letv= Y'i aixi E V and v = _Yi biii E (V*)*. Then xv = vx for all x E V* if andonly if zi v = v.xi for all i. Further xi v = bi and vii = ai, so v0 = _Yi ai Xi isuniquely determined.

As 0: _Y aixi H 1: aiXi, (2) holds. Part (4) follows directly from (2) and(3). To prove (3) we must show (u9)(a*)* = (ua)B for each u E U. But, forx E V*, x(u9)(a*)* =xa*(u9) = ax(u9) = u(ax) = (ua)x = x((ua)B), com-pleting the proof.

Notice 14.4 gives a constructive proof of 14.3. It will also be useful in the proofof the next lemma.

(14.5) Let G be a group and U, V, and W finite dimensional FG-modules.Then

(1) HomFG(U*, V*) = (a*: a E HomFG(V, U)}.

(2) 0 U 4 V - W 0 is an exact sequence of FG-modules if and onlyif 0 W * -+ V * U * -+ 0 is. The first sequence splits if and only if thesecond splits.

(3) V is irreducible, indecomposable, semisimple, and homogeneous if andonly if V* has the respective property.

Proof. Part (1) follows from 14.1.1 and 14.1.2. The first part of (2) followsfrom 14.1.3 and 14.4. The second part follows from the remark about splittingat the beginning of this section and 14.1.2. Part (3) follows from (2), since theproperties in (3) can be described in terms of exact sequences and the splittingof such sequences.

(14.6) (1) Let a: V --* U be an FG-homomorphism of finite dimensional mod-ules. Then [G, U] < V a if and only if ker(a*) < CU. (G), while [G, V*] <U*a* if and only if ker(a) < Cv(G).

(2) If U is a finite dimensional FG-module then U = [U, G] if and only ifCU.(G) = 0, while U* = [U*, G] if and only if CU(G) = 0.


(14.4) Let U and V be finite dimensional F-spaces. Then (1) For each v E V there exists a unique element v0 E (V*)* with xu0 = vx

for all x E V*. (2) The map 0: v H v0 is an F-isomorphism of V with (V*)*. (3) For each a E HomF(U, V), a 0 = @(a*)*. (4) 0 defines an equivalence of n and (n *)*.

Proof. To prove (1) let X = (xi : 1 5 i 5 n) be a basis for V, 8 = (ai: 1 5 i 5 n) its dual basis, and 2 = (fi : 1 5 i 5 n) the dual basis of 8 in (V*)*. Let v = xi a.x. , , E V and .fr = xi bizi E (V*)*. Thenx8 = vx for allx E V* if and only if ii 8 = vi i for all i. Further i i D = bi and vii = ai, so v0 = xi aiTi is uniquely determined.

As 0: x aixi H x ai.fi, (2) holds. Part (4) follows directly from (2) and (3). To prove (3) we must show (&)(a*)* = (ua)0 for each u E U. But, for x E V*, x(u0)(a*)* =xa*(u0) = ax(u0) = u(ax) = (ua)x = x((ua)0), completing the proof.

Notice 14.4 gives a constructive proof of 14.3. It will also be useful in the proof of the next lemma.

(14.5) Let G be a group and U, V, and W finite dimensional FG-modules. Then

(1) HOmFG(U*, V*) = {a*: a E HomFc(V, U)}. B (2) 0 + U -% V + W + 0 is an exact sequence of FG-modules if and only

B* if 0 + W* -+ V* % U* -+ 0 is. The first sequence splits if and only if the second splits.

(3) V is irreducible, indecomposable, semisimple, and homogeneous if and only if V* has the respective property.

Proof. Part (1) follows from 14.1.1 and 14.1.2. The first part of (2) follows from 14.1.3 and 14.4. The second part follows from the remark about splitting at the beginning of this section and 14.1.2. Part (3) follows from (2), since the properties in (3) can be described in terms of exact sequences and the splitting of such sequences.

(14.6) (1) Let a: V + U be an FG-homomorphism of finite dimensional modules. Then [G, U] _( Va if and only if ker(a*) ( CU*(G), while [G, V*] _(

U*a* if and only if ker(a) _( CV(G). (2) If U is a finite dimensional FG-module then U = [U, GI if and only if

Cu*(G) = 0, while U * = [U *, GI if and only if CU(G) = 0.

Proof. We have the exact sequence

V-0-'> U - U/Va --> 0

so, by 14.1.3,

0--> (U/Va)*-->

is also exact. Let 7r be the representation of G on U/ Va. Then [G, U] < Vaif and only if Gn = 1. This is equivalent to Gn* = 1 which in turn holds if andonly if G centralizes (U/Va)*. As (U/Va)* is FG-isomorphic to ker(a*) bythe exactness of the second series above, the first part of (1) holds, while thesecond follows from the first and 14.4.3.

Let U [U, G] = V and a: V --> Utheinclusion. By(1),ker(a*)<CU.(G).As a is not a surjection, ker(a*) 0 by 14.5.2. Thus 0 Cu. (G). Hence by14.4.3, if U* 0 [U*, G] then 0 Cu(G).

Similarly, if 0 Cu(G), let ,B: U -> U/CU(G) be the natural map. By (1),[G, U*] < (U/CU(G))*,8*. As ,B is not an injection, ,B* is not a surjection by14.5.2. So U* [G, U*]. Applying 14.4.3 we see that if 0 CU.(G) thenU [G, U], completing the proof of (2).

The character of an FG-representation 7r is the map X: G -+ F defined byX (g) =Tr(g7r). Remember Tr(g r) is the trace of the matrix Mx(gn) and isindependent of the choice of the basis X for 'the representation module of 7r.

(14.7) Let 7r be an FG-representation, 7r* the dual of r, and X and X* thecharacters of 7r and 7r*, respectively. Then X*(g) = X(g-1) for each g E G.

Proof. By 14.2, MX(g7r*) = Mx(g-17r )T, so the lemma follows as Tr(A) _Tr(AT) for each n by n matrix A.

Since characters have now been introduced I should probably record two moreproperties of characters which are immediate from 13.1 and the fact that con-jugate matrices have the same trace.

(14.8) (1) Equivalent FG-representations have the same character.(2) If X is the character of an FG-representation then X (gh) = X (g) for each

g, h E G.

Remarks. The stuff in sections 12 and 13 is pretty basic but section 14 is morespecialized. Section 14 is included here to prepare the way for the l-cohomologyin section 17. That section is also specialized. Both can be safely skipped or

Proof. We have the exact sequence

so, by 14.1.3,

is also exact. Let n be the representation of G on U/ Va. Then [G, U] 5 Va if and only if G n = 1. This is equivalent to Gn* = 1 which in turn holds if and only if G centralizes (U/ Va)*. As (U/ Va)* is FG-isomorphic to ker(a*) by the exactness of the second series above, the first part of (1) holds, while the second follows from the first and 14.4.3.

Let U # [U, GI = V anda: V -+ U theinclusion. By (l), ker (a*) i: CU*(G). As a is not a surjection, ker(a*) # 0 by 14.5.2. Thus 0 # CU*(G). Hence by 14.4.3, if U* # [U*, GI then 0 # CU(G).

Similarly, if 0 # CU(G), let 8: U -+ U/CU(G) be the natural map. By (I), [G, U*] 5 (U/CU(G))*B*. As B is not an injection, B* is not a surjection by 14.5.2. So U* # [G, U*]. Applying 14.4.3 we see that if 0 # CU*(G) then U # [G, U], completing the proof of (2).

The character of an FG-representation n is the map X: G -+ F defined by ~ ( g ) =Tr(gn). Remember Tr(gn) is the trace of the matrix Mx(gn) and is independent of the choice of the basis X for'ihe representation module of n .

(14.7) Let n be an FG-representation, n * the dual of n , and x and x * the characters of n and n*, respectively. Then x*(g) = x ( ~ - ' ) for each g E G.

Proof. By 14.2, M2(gn*) = ~ ~ ( ~ - ' n ) ~ , so the lemma follows as Tr(A) = T ~ ( A ~ ) for each n by n matrix A.

Since characters have now been introduced I should probably record two more properties of characters which are immediate from 13.1 and the fact that conjugate matrices have the same trace.

(14.8) (1) Equivalent FG-representations have the same character. (2) If x is the character of an FG-representation then (gh) = x (g) for each

g, h E G.

Remarks. The stuff in sections 12 and 13 is pretty basic but section 14 is more specialized. Section 14 is included here to prepare the way for thel-cohomology in section 17. That section is also specialized. Both can be safely skipped or

postponed by the casual reader. If so, lemma 17.10 must be assumed in provingthe Schur-Zassenhaus Theorem in section 18. But that's no problem.

The reader who is not familiar with the theory of modules over rings mightwant to bone up on modules before beginning section 12.

Exercises for chapter 41. Let G be a finite subgroup of GL(V), where V is a finite dimensional

vector space over a field F with (char (F), IGI) = 1. Prove(1) V = [G, V] ® Cv(G).(2) If G is abelian then V = (Cv(D): D E A), where A is the set of

subgroups D of G with G/D cyclic.(3) If G = Ep", n > 0, and V = [V, G], then V = ®HEr Cv(H), where

r is the set of subgroups of G of index p.2. Let V be a finite dimensional vector space over a field F, g E EndF(V),

and U a g-invariant subspace of V. Prove(1) g centralizes V/U if and only if [V, g] < U.(2) The map v i-+ [v, g] is a surjective linear transformation of V onto

[V, g] with kernel CV(g).(3) dimF(V) = dimF(CV(g)) +dimF([V, g])

3. Let G be a finite group, F a field of prime characteristic p, and .7r anirreducible FG-representation. Prove Op(G7r) = 1.

4. Let F be a field, r and q be primes, X a group of order r acting irreduciblyon a noncyclic elementary abelian q-group Q, and V = [V, Q] a faithfulirreducible FXQ-module. Then dimF(V) = rk where k = dimF(CV(X))dimF(CV(H)) for some hyperplane H of Q.

5. Let a : G --> Sym (I) be a permutation representation of a finite group G ona finite set I, and let F be a field and V an F-space with basis X = (xi : i E I).The FG-representation 7r induced by a is the representation on V withg7r: xi H xiga for each g E G and i E I. V is called the permutation moduleof a. Let X be the character of 7r. Prove(1) X(g) is the number of fixed points of got on I for each g E G.(2) (>gEG X(g))/IGI is the number of fixed points of G on I.

gEG X(g)2)/ I G I is the permutation rank(3) If G is transitive on I then (E

of G on I. (See section 15 for the definition of permutation rank.)6. Assume the hypothesis of the previous exercise with G transitive on I. Let

Z = G-+iEI Xi, Z = (z), and

U= {(xi: =O,a1 EF .

IEI iEI

U is the core of the permutation module V. Prove(1) Z = CV (G) and U = [V, G].


postponed by the casual reader. If so, lemma 17.10 must be assumed in proving the Schur-Zassenhaus Theorem in section 18. But that's no problem.

The reader who is not familiar with the theory of modules over rings might want to bone up on modules before beginning section 12.

Exercises for chapter 4 1. Let G be a finite subgroup of GL(V), where V is a finite dimensional

vector space over a field F with (char (F), \GI) = 1. Prove (1) V = [G, Vl @ Cv(G). (2) If G is abelian then V = (CV(D): D E A), where A is the set of

subgroups D of G with G/D cyclic. (3) If G Z E,., n > 0, and V = [V, GI, then V = @,,, CV(H), where

r is the set of subgroups of G of index p. 2. Let V be a finite dimensional vector space over a field F, g E EndF(V),

and U a g-invariant subspace of V. Prove (1) g centralizes V/ U if and only if [V, g] 5 U . (2) The map v H [v, g] is a surjective linear transformation of V onto

[V, g] with kernel Cv(g). (3) d im~(V) = dim~(Cv(g)> + dim~([V, gl).

3. Let G be a finite group, F a field of prime characteristic p, and n an irreducible FG-representation. Prove O,(Gn) = 1.

4. Let F be a field, r and q be primes, X a group of order r acting irreducibly on a noncyclic elementary abelian q-group Q, and V = [V, Q] a faithful irreducible FXQ-module. Then dimF(V) = rk where k = dimF(Cv(X)) = dimF(Cv(H)) for some hyperplane H of Q.

5. Let a: G + Sym (I) be apermutation representation of a finite group G on afinite set I , and let F be a field and V an F-space with basis X = (xi : i E I). The FG-representation n induced by a is the representation on V with gn : xi H xiga for each g E G and i E I. V is called thepemzutation module of a. Let x be the character of n. Prove (1) ~ ( g ) is the number of fixed points of g a on I for each g E G. (2) (CgsG x(g))/lG I is the number of fixed points of G on I. (3) If G is transitive on I then (C,,, x (~) ' ) / IG I is the permutation rank

of G on I. (See section 15 for the definition of permutation rank.) 6. Assume the hypothesis of the previous exercise with G transitive on I. Let

z = Cis[ xi, Z = (z), and

U is the core of the permutation module V. Prove (1) Z = Cv(G) and U = [V, GI.

(2) If W is an FG-module, i E I, H = Gi is the stabilizer in G of i, w ECw(H), and W = (wG), then there is a surjective homomorphism ofV onto W.

(3) Assume p = char(F) is a prime divisor of 111. Then V does not splitover U, V does not split over Z, and if O' (G) = G then H 1(G, U/Z) #0. (See section 17 for a discussion of the 1-cohomology group H1; inparticular use 17.11.)

7. Let F be a field, U a 2-dimensional vector space over F with basis {x, y},G = GL(U), and V = F[x, y] the polynomial ring in x and y over F. Prove(1) Ir is an FG-representation on V where Ir is defined by f (x, y)g7r =

f (xg, yg) for f E V and g E G.(2) G acts on the (n + 1)-dimensional subspace Vn of homogeneous poly-

nomials of degree n. Let nn be the restriction of 7r to Vn.(3) If char(F) = p > 0, prove 7rn is not irreducible for p < n -1 mod p,

but 7rn is irreducible for 0 < n < p.(4) ker(7rn) is the group of scalar transformations a I of U with a E F and

an=1.(Hint: In (3) let T be the group of transvections in G with center (x) and fori < n let Mi be the subspace of M = Vn generated by y1xn-j, 0 < j < i.Prove [yixn-1 +Mi_2, T] = Mi_1/Mi_2 for all 1 < i < n. Conclude Mo iscontained in any nonzero FG-submodule of M and then, as Mo is conjugateto (yn) under G, conclude Mi is contained in any such submodule forall i.)

8. Let V be a vector space over a field F and 0 = Vo < V1 < Vn = V asequence of subspaces. Let G be a subgroup of GL(V) centralizing Vi+1 / Vifor each i, 0 2, G =

GL(V), X = (xi: 1 <i < n) a basis for V, Vi = (xj :1 < j <i),T={Vi:1< i< n}, and Y= {(xi): 1< i< n}. Prove(1) T is a flag of PG(V) of type I = {0, ... , n - 1}.GT = B is the group

of lower triangular matrices and B is the semidirect product of thesubgroups U and H where U consists of the matrices in B with 1 onthe main diagonal and H is the group of diagonal matrices.

(2) H = Gy is the direct product of n copies of P. U is nilpotent and Z(U)is the root group of a transvection. (The root group of a transvection tconsists of those g E GL(V) with Cv (t) < Cv (G) and [V, g] < [V, t].)

(3) NA(Y) is the semidirect product of H by Sn. If I F I > 2 then NG (Y)NG(H).

The dual representation 5 1

(2) If W is an FG-module, i E I, H = Gi is the stabilizer in G of i , w E Cw (H), and W = (w G) , then there is a surjective homomorphism of V onto W.

(3) Assume p = char(F) is a prime divisor of I I I. Then V does not split over U, V does not split over 2, and if OP(G) = G then H '(G, U/Z) # 0. (See section 17 for a discussion of the 1-cohomology group H'; in particular use 17.1 1 .)

7. Let F be a field, U a 2-dimensional vector space over F with basis {x, y), G = GL(U), and V = F [x, y] the polynomial ring in x and y over F. Prove (1) n is an FG-representation on V where n is defined by f (x, y)gn =

f(xg, yg)for f E V andg E G. (2) G acts on the (n + 1)-dimensional subspace Vn of homogeneous poly-

nomials of degree n. Let nn be the restriction of n to Vn. (3) If char(F) = p > 0, prove n,, is not irreducible for p 5 n $ - 1 mod p,

but nn is irreducible for 0 5 n < p. (4) ker(nn) is the group of scalar transformations a I of U with a E F and

(Hint: In (3) let T be the group of transvections in G with center (x) and for i 5 n let Mi be the subspace of M = Vn generated by yjxn-1, 0 5 j 5 i . Prove [ylxn-l +Mi-2, TI = Mi-l/Mi-z for all 1 i i 5 n. Conclude Mo is contained in any nonzero FG-submodule of M and then, as Mo is conjugate to (yn) under G, conclude MI is contqined in any such submodule for

8. L e t V b e a v e c t o r s p a c e o v e r a f i e l d F a n d O = V o ~ V l ~ ~ ~ 5 Vn = V a sequence of subspaces. Let G be a subgroup of GL(V) centralizing Vi+' / Vl for each i, 0 5 i < n. Prove (1) G is nilpotent of class at most n - 1. (2) If 0 # U is a G-invariant subspace of V then CU (G) # 0 and [U, GI <

9. Let V be an n-dimensional vector space over a field F with n 2 2, G = GL(V) ,X=(xi :15i ~ n ) a b a s i s f o r V , V l = ( x j : l 5 j ( i ) ,T={V, : 1 5 i < n), and Y = {(xl): 1 5 i 5 n). Prove (1) T is a flag of PG(V) of type I = {O, . . . , n - l j .GT = B is the group

of lower triangular matrices and B is the semidirect product of the subgroups U and H where U consists of the matrices in B with 1 on the main diagonal and H is the group of diagonal matrices.

(2) H = G is the direct product of n copies of F'. U is nilpotent and Z(U) is the root group of a transvection. (The root group of a transvection t consists of those g E GL(V) with Cv(t) 5 Cv(G) and [V, g] 5 [V, t].)

(3) NG(Y) is the semidirect product of H by Sn. If IF I > 2 then NG(Y) =


(4) If F is finite of characteristic p then U E Sylp(G).(5) B = NG(U). Indeed VV is the unique object of type i fixed by U.(6) The residue Fs of a flag S of corank 1 is isomorphic to the projective

line over F, and (G5)rs PGL2(F). (See section 3 for the definitionof residue.)

10. Let F be a field and F = F U {oo} the projective line over F. Let G =GL2(F) be the group of invertible 2 by 2 matrices over F, and for

A = (aij) =a1,1 a1,2

(=- GL2(F)a2,1 a2,2

define O(A): F -+ F by

O(A): z Ha1,1z + a2,1

a1,2z + a2,2

where by convention a/oo = 0 for a E F# and ooo(A) = ai, i /a1,2. Pick abasis B = {x1, x2} for a 2-dimensional vector space V over F, and identifyGL(V) with G via the isomorphism MB : GL(V) -+ G. Let Q be the pointsof the projective geometry of V. Prove(1) For A E G, O(A) is a permutation of 1,.(2) G* = {O(A): A E G} is a subgroup of Sym(F), and 0: G -+ G* is a

surjective group homomorphism with kernel Z(G), so 0 induces anisomorphism : G -->. G*, where G = PGL(V).

(3) Define a: Q -+ F by a: Fx1 i-± oo and a: F(a.x1 +x2) H ? for a. E F.Then a is a bijection such that (tog)a = for each Co E 0 andg E G. Hence a is an equivalence of the permutation representationsof O on Q and G* on F.


(4) If F is finite of characteristic p then U E Syl,(G). (5) B = NG(U). Indeed Vi is the unique object of type i fixed by U. (6) The residue rs of a flag S of corank 1 is isomorphic to the projective

line over F , and (Gs )~s 2 PGL2(F). (See section 3 for the definition of residue.)

10. Let F be a field and I- = F U {co} the projective line over F. Let G = GL(F) be the group of invertible 2 by 2 matrices over F, and for

define @(A): I- + I- by

where by convention alto = 0 for a E F# and co@(A) = al,l/al,2. Pick a basis B = {xl , x2} for a 2-dimensional vector space V over F , and identify GL(V) with G via the isomorphism MB : GL(V) + G. Let !2 be the points of the projective geometry of V. Prove (1) For A E G, @(A) is a permutation of r. (2) G* = {@(A): A E G} is a subgroup of Sym(r), and @: G + G* is a

surjective group homomorphism with kernel Z(G), so @ induces an isomorphism 6: c + G*, where G = PGL(V).

(3) Define a: !2 + r by a: Fxl H co and a: F(hxl +x2) H h for h E F. Then a is a bijection such that (wg)a = (oa)$(g) for each o E !2 and g E c. Hence a is an equivalence of the permutation representations of C on !2 and G* on r.

5

Permutation groups

This chapter derives a number of properties of the alternating and symmetricgroups An and S of finite degree n. For example the conjugacy of elements inAn and S,, is determined, and it is shown that An is simple if n > 5. Section15 also contains a brief discussion of multiply transitive permutation groups.Section 16 studies rank 3 permutation groups.

15 The symmetric and alternating groupsLet X be a set and S the symmetric group on X. A permutation group on Xis a subgroup of S. Let G be a permutation group on X. In this section X isassumed to be of finite order n. Thus S is of order n!, so S and G are finite.

Suppose g E S and let H = (g). Then g is of finite order m and H = {gl: 0 <i < m }. Further H has a finite number of orbits (xi H: 1 < i < k), and the orbitxi H is of finite order li. Let Hl = Hx; be the stabilizer in H of xi. By 5.11,li = I H : Hi 1, so, as H = (g) is cyclic, Hi = (gli) and {gJ : 0 < j < li } is a setof coset representatives for Hi in H. Hence, ,by 5.8, xi H = {xi g1: 0 < j < li }.Therefore g acts on xi H as the following cycle:

gIXix = (xl, xig, xig2, .. , xig1 -1).

This notation indicates that g: xig' i-+ xi gJ+1 for 0 < j < li - 1 and g: xi gli -1 Hxi. The last fact holds as gl fixes xi. Further, as the orbits of H partition X, wecan describe the action of g on X with the following notation:

(x1, xlg, ... , xlg1 -1) (X2, X29, .. -X29 12-1) ... (xk, xkg, .. ,xkgit-1).

This is the cycle notation for the permutation g. It describes g, and the de-scription is unique up to a choice of representative xi for the ith orbit and theordering of the orbits. For example, if X is the set of integers { 1 , 2, ... , n}, therepresentative xi could be chosen to be the minimal member of the i th orbit, andthe orbits ordered so that x1 < x2 < < xk. If so, g can be written uniquelyin cycle notation, and conversely each partition of X and each ordering of thepartition and the members of the partition, subject to these constraints, definessome member of S in the cycle notation.

By convention the terms (xi) corresponding to orbits of H of length 1 areomitted. Thus for example if n = 5 we would write g = (1, 2)(3, 4)(5) as

Permutation groups

This chapter derives a number of properties of the alternating and symmetric oups A, and S, of finite degree n. For example the conjugacy of elements in

and S, is determined, and it is shown that A, is simple if n 2 5. Section also contains a brief discussion of multiply transitive permutation groups.

Section 16 studies rank 3 permutation groups.

15 The symmetric and alternating groups Let X be a set and S the symmetric group on X. A permutation group on X is a subgroup of S. Let G be a permutation group on X. In this section X is assumed to be of finite order n. Thus S is of order n!, so S and G are finite.

Suppose g E S and let H = (g). Then g is of finite order m and H = {g' : 0 5 i < m}. Further H has a finite number of orbits (x, H: 1 5 i 5 k), and the orbit x, H is of finite order I,. Let H, = H,, be the stabilizer in H of x,. By 5.1 1, I, = IH:H,I ,so,asH = (g)iscyclic,H, = (gli)and{g-':Os j < I,}isaset of coset representatives for H, in H. Hence, by 5.8, x, H = {x, g-' : 0 5 j < I, }. Therefore g acts on x, H as the following cycle:

g l x , ~ = (XZ, xzg, xlg2, . - . 9 Xzg 1,-1).

This notation indicates that g: x, g-' H x, gj+l for 0 5 j < 1, - 1 and g: x, g l g - l H

x, . The last fact holds as gb fixes xi. Further, as the orbits of H partition X, we can describe the action of g on X with the following notation:

This is the cycle notation for the permutation g. It describes g, and the description is unique up to a choice of representative xi for the ith orbit and the ordering of the orbits. For example, if X is the set of integers {1,2, . . . , n}, the representative xi could be chosen to be the minimal member of the ith orbit, and the orbits ordered so that xl < x2 < . . . < xk. If so, g can be written uniquely in cycle notation, and conversely each partition of X and each ordering of the partition and the members of the partition, subject to these constraints, defines some member of S in the cycle notation.

By convention the terms (xi) corresponding to orbits of H of length 1 are omitted. Thus for example if n = 5 we would write g = (1,2)(3,4)(5) as

54 Permutation groups

g = (1, 2)(3, 4). Notice g is still uniquely described in this modified cyclenotation.

Subject to this convention, g1 = (xi, xt g, ..., x! gt;-l) is a member of S. Theelements gl, ... , gk are called the cycles of g. Also g is said to be a cycle if Hhas at most one orbit of length greater than 1. Notice the two uses of the term`cycle' are compatible.

Given a subset A of X let Mov(A) be the set of points of X moved by A. Here

x in X is moved by A if ax x for some a E A. Notice Mov(A) = Mov((A))and X is the disjoint union of Mov(A) and Fix(A). Cycles c and d in S are saidto be disjoint if Mov(c) fl Mov(d) is empty.

(15.1) Let A, B C S with Mov(A) fl Mov(B) empty. Then ab = ba for allaEAandbEB.

(15.2) Let gi, ..., gr be the nontrivial cycles of g E S. Then

(1) gjgi = gigs for i # j.(2) g = gi . . . g, is the product in S of its nontrivial cycles.(3) If g = cl ... cs with {cl, ... , cs) a set of nontrivial disjoint cycles then

{cl,...,C5) _ (gl,...,g,).(4) The order of g is the least common multiple of the lengths of its cycles.

By 15.2, each member of S# can be written uniquely as the product of non-trivial disjoint cycles, and these cycles commute, so the order of the product isimmaterial.

For g c S define Cycg to be the function from 7L+ into 1L such that Cycg(i)is the number of cycles of g of length i. Permutations g and h are said to havethe same cycle structure if Cycg = Cych.

(15.3) Let g, h E S with

g =(al,...,aa)(bl,...,bp)...

Then(1) g' = (alh,... , aah)(blh, ... , bah)....(2) s E S is conjugate to g in S if and only ifs and g have the same cycle

structure.

A transposition is an element of S moving exactly two points of X. That is atransposition is a cycle of length 2.

(15.4) S is generated by its transpositions.


g = (1,2)(3,4). Notice g is still uniquely described in this modified cycle notation.

Subject to this convention, gi = (xi, xig, . . . , xig''-') is a member of S. The elements gl, . . . , gk are called the cycles of g. Also g is said to be a cycle if H has at most one orbit of length greater than 1. Notice the two uses of the term 'cycle' are compatible.

Given a subset A of X let Mov(A) be the set of points of X moved by A. Here x in X is moved by A if ax # x for some a E A. Notice Mov(A) = Mov((A)) and X is the disjoint union of Mov(A) and Fix(A). Cycles c and d in S are said to be disjoint if Mov(c) fl Mov(d) is empty.

(15.1) Let A, B g S with Mov(A) n Mov(B) empty. Then ab = ba for all a ~ A a n d b ~ B.

(15.2) Let gl , . . . , gr be the nontrivial cycles of g E s'. Then

(1) gigj =gjgi for i # j. (2) g = gl . . . gr is the product in S of its nontrivial cycles. (3) If g = cl . . . cs with {cl, . . . , c,) a set of nontrivial disjoint cycles then

Iclt . . - , c s ) = Igl, - . . ,g r ) . (4) The order of g is the least common multiple of the lengths of its cycles.

By 15.2, each member of S# can be written uniquely as the product of nontrivial disjoint cycles, and these cycles commute, so the order of the product is immaterial.

For g E S define Cyc, to be the function from Z+ into Z such that Cyc,(i) is the number of cycles of g of length i. Permutations g and h are said to have the same cycle structure if Cyc, = Cyc,.

(15.3) Let g, h E S with

Then (1) gh = (alh, . . . , a,h)(blh, . . . , bah). . .. (2) s E S is conjugate to g in S if and only if s and g have the same cycle

structure.

A transposition is an element of S moving exactly two points of X. That is a transposition is a cycle of length 2.

(15.4) S is generated by its transpositions.

The symmetric and alternating groups 55

Proof. By 15.2 it suffices to show each cycle is a product of transpositions.But (1, 2, ..., m) = (1, 2)(1, 3) ... (l, m).

A permutation is said to be an even permutation if it can be written as theproduct of an even number of transpositions, and to be an odd permutation ifit can be written as the product of an odd number of transpositions. Denote byAlt(X) the set of all even permutations of X.

(15.5) (1) Alt(X) is a normal subgroup of Sym(X) of index 2.(2) A permutation is even if and only if it has an even number of cycles of

even length. A permutation is odd if and only if it has an odd number of cyclesof even length.

Proof. Without loss X = (1, 2, ... , n). Consider the polynomial ring R =7L[xl, ... , xn] in n variables xi over the ring 1L of integers. For S E S definesa:R R by f(x1,...,xn)sa = f(x1s,...,xn5).Check that a:S-+ Aut(R)is a representation of S in the category of rings and ring homomorphisms.

Consider the polynomial P(x1, ... , xn) = P E R defined by

P = P(x1, ... , x'n) = H (X j - xi).1<i<j<n

For S E S, sa permutes the factors x j - xi of P up to a change of sign, soPsa = P or -P. Moreover if t is the transposition (1, 2), then Pta = -P.Thus A = { P, - P) is the orbit of P under S. Let A = Sp be the stabilizerof P in S. A is also the kernel of a so A a S and, by 2.11, IS: Al = JAI = 2.By 15.3, the transpositions form a conjugacy class of S, so, as t E S - A andA 4 S, each transposition is in S - A. But, as IS: Al = 2, the product of melements of S - A is in A if and only if m is even. So A = Alt(X) and S - Ais the set of odd permutations. This proves (1), and, since we saw during theproof of 15.4 that a cycle of length m is the product of m - 1 transpositions,(2) also holds.

The group Alt(X) is the alternating group on X. Evidently the isomorphismtype of Sym(X) and Alt(X) depends only on the cardinality of X, so we maywrite Sn and An for Sym(X) and Alt(X), respectively, when IX I = n. Thegroups Sn and An are the symmetric and alternating groups of degree n.

If m is a positive integer, denote by X1 the set product of m copies of X.If G is a permutation group on X, then G is also a permutation group onX' via g: (x1, ... , x,n) H (x1g, ... , xng) for g c G and xi c X. Assume Gis transitive on X. Then the orbits of G on X2 are called the orbitals of G.One orbital is the diagonal orbital {(x, x): x c X}. The permutation rank of atransitive permutation group G is defined to be the number of orbitals of G.

and alternating groups 55

it suffices to show each cycle is a product of transpositions. But (1,2, . . . , m) = (1,2)(1,3). . . (1, m).

A permutation is said to be an even permutation if it can be written as the product of an even number of transpositions, and to be an odd permutation if it can be written as the product of an odd number of transpositions. Denote by Alt(X) the set of all even permutations of X.

(15.5) (1) Alt(X) is a normal subgroup of Sym(X) of index 2. (2) A permutation is even if and only if it has an even number of cycles of

even length. A permutation is odd if and only if it has an odd number of cycles

Proof. Without loss X = {I, 2, . . . , nj. Consider the polynomial ring R = Z[xl, . . . , x,] in n variables xi over the ring Z of integers. For s E S define sa: R + R by f(x1,. . . ,x,)sa = f(xls, . . .,x,,). Checkthat a : S + Aut(R) is a representation of S in the category of rings and ring homomorphisms.

Consider the polynomial P(xl, . . . , x,) = P E R defined by

P = P(xl , . . . , x,) = n (xj -xi). l ' i < j i n

For s E S, s a permutes the factors x j - xi of P up to a change of sign, so Psa = P or -P. Moreover if t is the tran6position (1, 2), then P t a = - P . Thus A = { P , -P) is the orbit of P unde; S. Let A = Sp be the stabilizer of P i n s . AisalsothekernelofasoA<ISand, by2.11, ( S : A ( = (A/ =2 . By 15.3, the transpositions form a conjugacy class of S, so, as t E S - A and A <IS, each transposition is in S - A. But, as IS: A( = 2, the product of m elements of S - A is in A if and only if m is even. So A = Alt(X) and S - A is the set of odd permutations. This proves (I), and, since we saw during the proof of 15.4 that a cycle of length m is the product of m - 1 transpositions, (2) also holds.

The group Alt(X) is the alternating group on X. Evidently the isomorphism type of Sym(X) and Alt(X) depends only on the cardinality of X, so we may write S, and A, for Sym(X) and Alt(X), respectively, when (XI = n. The groups S, and A, are the symmetric and alternating groups of degree n.

If m is a positive integer, denote by Xm the set product of m copies of X. If G is a permutation group on X, then G is also a permutation group on Xm via g : (xl , . . . , x,) H (xlg, . . . , xmg) for g E G and xi E X. Assume G is transitive on X. Then the orbits of G on x2 are called the orbitals of G. One orbital is the diagonal orbital {(x, x): x E X). The permutation rank of a transitive permutation group G is defined to be the number of orbitals of G.

(15.6) Let G be a transitive permutation group on X, X E G, and (xi : 1 < i < r)representatives for the action of H = Gx on X. Then {(x, xi): 1 < i < r} is aset of representatives for the orbitals of G and (x, y) E (x, xi)G if and only ify E xi H. In particular r is the permutation rank of G.

The regular permutation representation of a group H is the representation of Hon itself by right multiplication. A permutation representation n : H -)- Sym(Y)is semiregular if and only if the identity element is the only element of H fixingpoints of Y. Equivalently Hy = 1 for all y in Y.

(15.8) A permutation representation n of finite degree is semiregular if andonly if the transitive constituents of n are regular.

A regular normal subgroup of G is a normal subgroup of G which is regularon X.

(15.9) Let G be transitive on X, x E X, and H < G. Then H is regular on Xif and only if Gx is a complement to H in G.

(15.10) Let H be the split extension of a normal subgroup K by a subgroup Aof H and let jr be the representation of H on the cosets of A. Then K = Kitand Kit is a regular normal subgroup of Hn.

(15.11) Let H be a regular normal subgroup of G and X E X. Then the mapa: H X defined by a: h H xh is an equivalence of the representation ofGx on H via conjugation with the representation of Gx on X.

Given a positive integer m, G is said to be m-transitive on X if G acts transitivelyon the subset of X' consisting of the m-tuples all of whose entries are distinct.Notice G is 2-transitive if and only if it is transitive of permutation rank 2. Alsom-transitivity implies k-transitivity for k < m.

(15.12) (1) Let m > 2 and x E X. Then G is m-transitive on X if and only ifG is transitive and Gx is (m - 1)-transitive on X - {x}.

(2) Sym(X) is n-transitive on X.(3) Alt(X) is (n - 2)-transitive on X.(4) If G is (n - 2)-transitive on X then G = Sym(X) or Alt(X).

Proof. The first three statements are straightforward. Prove the fourth by in-duction on n using (1) and the following observation: If G is transitive on Xand Gx = Sym(X )x or Alt(X )x then G = Sym(X) or Alt(X), respectively. Theobservation follows from 5.20 plus the fact that, if GX < Alt(X) and n > 4,


(15.6) Let G be a transitive permutation group on X, x E G , and (xi : 1 5 i 5 r) representatives for the action of H = G, on X. Then {(x, xi): 1 _( i 5 r } is a set of representatives for the orbitals of G and ( x , y) E (x, xi)G if and only if y E xi H. In particular r is the permutation rank of G.

The regularpermutation representation of a group H is the representation of H on itself by right multiplication. Apermutation representation n: H + Sym(Y) is semiregular if and only if the identity element is the only element of H fixing points of Y. Equivalently H, = 1 for all y in Y.

(15.8) A permutation representation n of finite degree is semiregular if and ,

only if the transitive constituents of n are regular.

A regular normal subgroup of G is a normal subgroup of G which is regular on X.

(15.9) Let G be transitive on X, x E X, and H I_( G. Then H is regular on X if and only if G, is a complement to H in G.

(15.10) Let H be the split extension of a normal subgroup K by a subgroup A of H and let n be the representation of H on the cosets of A. Then K E K n and K n is a regular normal subgroup of H n .

(15.11) Let H be a regular normal subgroup of G and x E X. Then the map a : H + X defined by a : h H xh is an equivalence of the representation of G, on H via conjugation with the representation of G, on X.

Given apositive integer m, G is said to be m-transitive on X if G acts transitively on the subset of Xm consisting of the m-tuples all of whose entries are distinct. Notice G is 2-transitive if and only if it is transitive of permutation rank 2. Also m-transitivity implies k-transitivity for k 5 m.

(15.12) (1) Let m 2 2 and x E X. Then G is m-transitive on X if and only if G is transitive and G, is (m - 1)-transitive on X - {x}.

(2) Sym(X) is n-transitive on X. (3) Alt(X) is (n - 2)-transitive on X. (4) If G is (n - 2)-transitive on X then G = Sym(X) or Alt(X).

Proof. The first three statements are straightforward. Prove the fourth by induction on n using (1) and the following observation: If G is transitive on X and G, = Sym(X), or Alt(X), then G = Sym(X) or Alt(X), respectively. The observation follows from 5.20 plus the fact that, if G, 5 Alt(X) and n 2 4,

The symmetric and alternating groups 57

then G = (Gy: Y E X) < Alt(X), since Alt(X) a Sym(X) and G is 2-transitiveon X.

(15.13) Let G be m-transitive on X and H a regular normal subgroup of G.Then

(1) If m = 2 then n is a power of some prime p and H is an elementaryabelian p-group.

(2) If m = 3 then either n is a power of 2 or n = 3 and G = Sym(X).(3) If m > 4 then m = 4 = n and G = Sym(X).

Proof. We may take m > 2. Let x E X and K = Gx . By 15.12, K is (m - 1)-transitive on X - {x}, and then, by 15.11, K acts (m - 1)-transitively on H#via conjugation. In particular K is transitive on H#.

Let p be a prime divisor of n. As H is regular on X, n = I H 1. So by Cauchy'sTheorem there is h E H of order p. Thus, as K is transitive on H#, everyelement of H# is of order p. We conclude from Cauchy's Theorem that His a p-group. So n = JHJ is a power of p. By 9.8, H is solvable and as Kis transitive on H#, H is a minimal normal subgroup of G. So, by 9.4, H iselementary abelian.

This completes the proof of (1), so we may take m > 3. Let y = xh. By15.2, KY is transitive on X - {x, y} and so, by 15.11, KY = CK(h) is transitiveon H - {1, h} via conjugation. But CK(h) CK((h)), so either (h) = 11, h}or n = 3. In the first case p = 2 and in the second G = Sym(X) by 15.12.4.

This completes the proof of (2), so we may take m > 4. Let g E H - (h).By 15.11 and 15.12, CK(g) n CK(h) = J is transitive on H - {1, g, h) viaconjugation. But J centralizes gh, so n = 4. Hence G = Sym(X) by 15.12.4.

Recall the definition of a primitive representation from section 5.

(15.14) 2-transitive representations are primitive.

(15.15) If G is primitive on X and 1 H a G then H is transitive on X andG = Gx H for each x E X.

Proof. Let X E X. By 5.19, M = GX is a maximal subgroup of G, while asH a G, MH is a subgroup of G containing M by 1.7. Thus MH = M or G.In the latter case H is transitive on X by 5.20. In the former H < M, so H <ker(r) by 5.7, where r is the representation of G on X. But, as G < S, jr is theidentity map on G and in particular is faithful. This is impossible as H 1.

(15.16) The alternating group A is simple if n > 5.

The symmetric and alternating groups

then G = (G,: y E X) 5 Alt(X), since Alt(X) 1 Sym(X) and G is 2-transitive

(15.13) Let G be m-transitive on X and H a regular normal subgroup of G. Then

(1) If m = 2 then n is a power of some prime p and H is an elementary abelian p-group.

(2) If m = 3 then either n is a power of 2 or n = 3 and G = Sym(X). (3) Ifm 2 4 t h e n m = 4 = n a n d G =Sym(X).

Pro05 We may take m 2 2. Let x E X and K = G,. By 15.12, K is (m - 1)- transitive on X - {x}, and then, by 15.11, K acts (m - 1)-transitively on H' via conjugation. In particular K is transitive on H'.

Let p be a prime divisor of n. As H is regular on X, n = I H I . So by Cauchy's Theorem there is h E H of order p. Thus, as K is transitive on H', every element of H' is of order p. We conclude from Cauchy's Theorem that H is a p-group. So n = IHI is a power of p. By 9.8, H is solvable and as K is transitive on H', H is a minimal normal subgroup of G. So, by 9.4, H is

ntary abelian. s completes the proof of (I), so we may take m 2 3. Let y = xh. By

15.2, K , is transitive on X - {x, y } and so, by 15.1 1, K , = CK (h) is transitive on H - (1, h} via conjugation. But CK(h) CK((h)), so either (h) = (1, h} or n = 3. In the first case p = 2 and in the second G = Sym(X) by 15.12.4.

This completes the proof of (2), so we may take m 2 4. Let g E H - (h). By 15.11 and 15.12, CK(g) f? C,y(h) = J is transitive on H - (1, g, h) via conjugation. But J centralizes gh, so n = 4. Hence G = Sym(X) by 15.12.4.

Recall the definition of a primitive representation from section 5.

(15.14) 2-transitive representations are primitive.

(15.15) If G is primitive on X and 1 # H G then H is transitive on X and G = G,H foreachx E X.

Pro05 Let x E X. By 5.19, M = G, is a maximal subgroup of G, while as H 1 G, MH is a subgroup of G containing M by 1.7. Thus MH = M or G. In the latter case H is transitive on X by 5.20. In the former H 5 M , so H 5 ker(n) by 5.7, where n is the representation of G on X. But, as G 5 S, n is the identity map on G and in particular is faithful. This is impossible as H # 1.

(15.16) The alternating group A, is simple if n 2 5.

Proof. Let n > 5, G = Alt(X), and 1 # H a G. We must show H = G. By15.12, G is (n - 2)-transitive on X, so, by 15.14 and 15.15, H is transitive onX and G = K H, where K = Gx and X E X. If 1 = H fl K then, by 15.9, H isregular on X. But this contradicts 15.13 and the fact that G is (n - 2)-transitivewith n > 5.

So 10 H fl K. But K = Alt(Y), where Y = X - (x), so, by induction onn, either K is simple or n = 5. In the former case, as 1 ¢ H fl K 4 K, K =H fl K < H, so G = KH = H. Thus we may take n = 5. Here at least H fl Kis transitive on Y by 15.12, 15.14, and 15.15. So 4 = I Y (= (H fI K : (H fI K)y Ifor y E Y by 5.11. Similarly 5 = (X ( = ( H : H fl K (, so 20 divides the orderof H. But (S( = 5! = 120 while IS: G(= 2, so IGI = 60. Thus, as IHI divides(G(, 1H( = 20 or 60. In the latter case H = G, so we may assume the former.

By Exercise 2.6, H has a unique Sylow 5-group P. Hence P char H 4 G,so, by 8.1, P 4 G. This is impossible as we have just shown that 4 divides theorder of any nontrivial normal subgroup of G.

(15.17) (Jordan) Let G be a primitive permutation group on a finite set X andsuppose Y is a nonempty subset of X such that ( X - Y ( > 1 and Gy is transitiveonX - Y. Then

(1) G is 2-transitive on X, and(2) if Gy is primitive on X - Y then Gx is primitive on X - (x} for x c X.

Proof. Let F = X - Y and induct on (Y(. If (YI = 1 the result is trivial. Soassume I Y ( > 1 and let x and y be distinct points of Y. By Exercise 5.5, there isgEGwith xEYgandyVYg.LetH=(Gy,Gyg)andS2=FUFg.ThenH < Gx and H acts on 0. Suppose I r I > (Y 1. Then r fl rg is nonempty so His transitive on 0. As H < G, r U (y} is contained in an orbit of G. Sincethis holds for each y E Y - (x}, G is 2-transitive on X by 15.12.1. Further ifGy is primitive on F and Q is a Gx-invariant partition of X' = X - {x}, thenf o r Z E Q either (Z fl r ( < 1 or r c_ Z. As 1171 > (Y ( it follows that (Z (= 1or IX'(. Hence Gx is primitive on X'.

So assume Ir( < (Y( and let a, y be distinct points of F. By Exercise 5.5thereish E Gwithy c rhbut a V rh.LetK = (Gy,Gyh),F' = rUrh,and Y' = X - F'. Then K < Gy,, and as y c r fl Fh, K is transitive on F'and ( F ( > IF'-r(.As (rI < (Y( andy c rflFh,wehave Y' # 0.IfGy isprimitive on F then, as ( F ( > ( r' - F 1, K is primitive on F' by an argumentin the last paragraph. So, replacing Y by Y' and applying induction on I Y 1, theresult holds.

Jordan's Theorem is a useful tool for investigating finite alternating groups andsymmetric groups. See for example Exercises 5.6, 5.7, and 16.2.

5 8 Permutation groups

Proof. Let n 2 5, G = Alt(X), and 1 # H _a G. We must show H = G. By 15.12, G is (n - 2)-transitive on X, so, by 15.14 and 15.15, H is transitive on X a n d G = KH,whereK =G,andx c X . I f 1 = HnKthen,by15.9 ,His regular on X. But this contradicts 15.13 and the fact that G is (n - 2)-transitive with n >_ 5.

So 1 # H n K. But K = Alt(Y), where Y = X - (x), so, by induction on n, either K is simple or n = 5. In the former case, as 1 # H f l K _a K, K = H n K 4 H, soG = K H = H.Thuswemaytaken = 5.HereatleastHflK istransitiveony by 15.12, 15.14,and 15.15.So4 = lYl = l H n K :(HnK),I for y E Y by 5.11. Similarly 5 = (XI = ( H : H n KI, so 20 divides the order of H.But IS1 = 5! = 120while IS: GI = 2,so IGI = 60. Thus, as [HI divides .

IGI, I H I = 20 or 60. In the latter case H = G, so we may assume the former. By Exercise 2.6, H has a unique Sylow 5-group P. Hence P char H _a G,

so, by 8.1, P a G. This is impossible as we have just shown that 4 divides the order of any nontrivial normal subgroup of G.

(15.17) (Jordan) Let G be a primitive permutation group on a finite set X and suppose Y is a nonempty subset of X such that IX - Y I > 1 and Gy is transitive on X - Y. Then

(1) G is Z-transitive on X, and (2) if Gy is primitive on X - Y then G, is primitive on X - (x) for x E X.

Proof. Let r = X - Y and induct on I Y I. If 1 Y I = 1 the result is trivial. So assume IY I > 1 and let x and y be distinct points of Y. By Exercise 5.5, there is g E G with x E Yg and y $ Yg. Let H = (Gy, GYg) and C2 = r U r g . Then H _( G, andH actsonC2.Suppose Irl > IYI.Thenrnrgisnonemptyso H is transitive on C2. As H 5 G,, r U (y} is contained in an orbit of G,. Since this holds for each y E Y - (x}, G is Ztransitive on X by 15.12.1. Further if Gy is primitive on r and Q is a G,-invariant partition of X' = X - (x), then for Z E Q either IZ f? r [ 4 1 or r G Z. As Irl 2 ( Y ( it follows that IZ( = 1 or IX'I. Hence G, is primitive on XI.

So assume Irl _( lYI and let a, y be distinct points of r. By Exercise 5.5 thereis h E G with y E r h but a $ r h . Let K = (Gy, GYh), r' = r U r h , and Y' = X - r ' . Then K _( Gy/, and as y E r f? r h , K is transitive on r' and Irl > Ir' - rl. As Irl 4 IYI and y E r n r h , we have Y' # 0. If Gy is primitive on r then, as I r 1 > I r' - r 1, K is primitive on r' by an argument in the last paragraph. So, replacing Y by Y' and applying induction on IY I, the result holds.

Jordan's Theorem is a useful tool for investigating finite alternating groups and symmetric groups. See for example Exercises 5.6,5.7, and 16.2.

Rank 3 permutation groups 59

16 Rank 3 permutation groupsIn this section G is a transitive permutation group on a finite set X of order n.

Recall the definition of an orbital in the preceding section. Given an orbitalA of G, the paired orbital AP of A is

AP = {(y, x): (x, y) E A).

Evidently AP is an orbital of G with (Ap)p = A. The orbital A is said to beself paired if A = AP.

(16.1) (1) A nondiagonal orbital A of G is self paired if and only if (x, y) is acycle in some g E G, for (x, y) E A.

(2) G possesses a nondiagonal self paired orbital if and only if G is of evenorder.

(3) If G is of even order and (permutation) rank 3 then both nondiagonalorbitals of G are self paired.

Recall the definition of a graph from section 3.

(16.2) Let A be a self paired orbital of G. Then A is a symmetric relation onX, so -9 = (X, A) is a graph and G is a group of automorphisms of -9 transitiveon the edges of -9. 61

In the remainder of this section assume G is of even order and of permutationrank 3. Hence G has two nondiagonal orbitals A and F. By 16.1.3 both A and Fare self paired. For x c X, Gx has two orbits A(x) and F(x) on X - {x}, whereA(x) = {y c X: (x, y) c A} and r(x) = {y c X: (x, y) c r); this holds by15.6. By 16.2, ' = (X, A) is a graph and G is a group of automorphisms of -9transitive on the edges of -9. Notice A(x) is the set of vertices adjacent to x inthis graph. Define x1 = {x}UA(x),k = IA(x)l,l = IF(x)I,,1 = IA(x)nA(y)Ifor y E A(x), and µ = I A(x)n A(z)I for Z E F(x). As Gx is transitive on A(x)and F(x),A and µ are well defined. As G is transitive on X these definitionsare independent of the choice of x.

(16.3) (1) n=1+k+l.(2) µl = k(k - A - 1).

Proof. Part (1) is trivial. To prove (2) count IQI in two different ways, where

0 = {(a, b) : b, x c A(a), b c F(x)}

for fixed x c X.


16 Rank 3 permutation groups In this section G is a transitive permutation group on a finite set X of order n.

Recall the definition of an orbital in the preceding section. Given an orbital A of G, the paired orbital AP of A is

Evidently Ap is an orbital of G with (AP)P = A. The orbital A is said to be self paired if A = AP.

(16.1) ( 1 ) A nondiagonal orbital A of G is self paired if and only if ( x , y) is a cycle in some g E G, for ( x , y) E A.

(2) G possesses a nondiagonal self paired orbital if and only if G is of even order.

(3) If G is of even order and (permutation) rank 3 then both nondiagonal orbitals of G are self paired.

Recall the definition of a graph from section 3.

(16.2) Let A be a self paired orbital of G. Then A is a symmetric relation on X, so g = ( X , A ) is a graph and G is a group of automorphisms of g transitive on the edges of g. ,<

In the remainder of this section assume G is of even order and of permutation rank 3. Hence G has two nondiagonal orbitals A and r. By 16.1.3 both A and r are self paired. For x E X, G, has two orbits A(x ) and r ( x ) on X - { x ) , where A(x ) = ( y E X: ( x , y) E A ) and r ( x ) = { y E X: ( x , y) E r); this holds by 15.6. By 16.2, g = ( X , A ) is a graph and G is a group of automorphisms of g transitive on the edges of g. Notice A(x ) is the set of vertices adjacent to x in this gaph. Defined = {x )UA(x ) , k = lA(x)l,l = Ir(x)l, h = IA(x)nA(y)l for y E A(x) , and p = lA (x )n A(z)l for Z E r ( x ) . As G, is transitive on A(x ) and r ( x ) , h and p are well defined. As G is transitive on X these definitions are independent of the choice of x .

Proof. Part ( 1 ) is trivial. To prove (2) count IC2I in two different ways, where

S2 = {(a, b ) : b , x E A(a), b E r ( x ) )

for fixed x E X.

(16.4) If k < 1 then the following are equivalent:

(1) G is imprimitive.(2) a,=k-1.(3) a = 0.(4) x1 = y1 for y c A(x), {(x1)g: g c G} = Q is a system of imprimitivity

for G, and G is 2-transitive on Q.

Proof. If S is a system of imprimitivity for G and x c 9 E S, then, by 5.18,Gx acts on 9. So, as Gx is transitive on A(x) and F(x), either 9 = x1 or9 = (x) U F(x). By 5.18, 101 divides n, so ask < 1, 9 = x1. Now (4) holds. Ifx1 = y1 then (2) and (3) hold. Also (2) is equivalent to (3). Finally if (2) holdsthen x1 = y1, so Q is a system of imprimitivity for G and hence (1) holds.

(16.5) If µ 0 0 or k then G is primitive.

Proof. By 16.4 we may take k > 1. Let µ = l - (k -,X - 1). If µ = 0 then, by16.3.2, k = µ, contrary to hypothesis. Hence, by 16.4 and symmetry betweenA and F, G is primitive.

(16.6) If G is primitive then ' is connected.

Proof. By 5.19, Gx is maximal in G, so G = _ (Gx, Gy) for x y. Now Exercise

5.2 completes the proof.

(16.7) Assume G is primitive. Then either

(1) k=land µ=,k+l=k/2, or(2) d = (X - µ)2 + 4(k - µ) is a square and setting

D = 2k + (A - µ)(k + 1) we have:(a) d1/2 divides D but 2d1j2 does not if n is even, while(b) 2d1/2 divides D if n is odd.

Proof. Let A be the incidence matrix for -9. That is A = (axy) is the n by nmatrix whose rows and columns are indexed by X, and with axy = 1 if (x, y)is an edge of -9' while axy = 0 otherwise. Let B be the incidence matrix for(X, F), J then by n matrix all of whose entries are 1, and I then by n identitymatrix. Observe:

(i) A is symmetric.(ii) I + A + B = J.


(16.4) If k 5 1 then the following are equivalent:

(1) G is imprimitive. (2) h = k - 1 . (3) p = 0. (4) xL = y' for y E A(x), {(x')~: g E G} = C2 is a system of imprimitivity

for G, and G is 2-transitive on C2.

Proof: If S is a system of imprimitivity for G and x E 0 E S, then, by 5.18, G, acts on 8. So, as G, is transitive on A(x) and r(x), either 0 = x' or 0 = {x) U r(x). By 5.18, 101 divides n, so as k 5 1,e = x'. NOW (4) holds. 1f .

x' = y' then (2) and (3) hold. Also (2) is equivalent to (3). Finally if (2) holds then x' = y', so C2 is a system of imprimitivity for G and hence (1) holds.

(16.5) If p # 0 or k then G is primitive.

Proof: By 16.4 we may take k > 1. Let ,2 = 1 - (k - h - 1). If ,2 = 0 then, by 16.3.2, k = p, contrary to hypothesis. Hence, by 16.4 and symmetry between A and r , G is primitive.

(16.6) If G is primitive then is connected.

Proof. By 5.19, G, is maximal in G, so G = (G,, G,) for x # y. Now Exercise 5.2 completes the proof.

(16.7) Assume G is primitive. Then either

(1) k = l a n d p = h + l =k/2,or (2) d = (A - p)2 + 4(k - p) is a square and setting

D = 2k + (A - p)(k + 1) we have: (a) d112 divides D but 2d1I2 does not if n is even, while (b) 2d1f2 divides D if n is odd.

Proof: Let A be the incidence matrix for s?. That is A = (a,,) is the n by n matrix whose rows and columns are indexed by X, and with a,, = 1 if (x, y) is an edge of while a,, = 0 otherwise. Let B be the incidence matrix for (X, r ) , J the n by n matrix all of whose entries are 1, and I the n by n identity matrix. Observe:

(i) A is symmetric. (ii) I + A + B = J .


(iii) AJ = kJ, so (A - kI)J = 0.(iv) A2 = kI + ,kA + AB.

The first two statements are immediate from the definitions. As (A(x)e = k,each row of A has k entries equal to 1. Thus (iii) holds. By (i) the (x, y)-thentry of A2 is the inner product of the xth and yth rows of A. But this innerproduct just counts IA(x) fl A(y)e, so (iv) holds.

Next (ii), (iii), and (iv) imply:

(v)(A-kI)(A2-(,l-A)A-(k-ll)I)=0,

so the minimal polynomial of A divides

p(x) _ (x -k)(x2 -()-A)x -(k-bl)).The roots of p(x) are k, s, and t, where s = ((A - A) + d1/2)/2 and t =((), - A) - d1/2)/2. Let me be the multiplicity of the eigenvalue e.

Claim Mk = 1. Indeed for c = (cl, ..., cA = kc if and only if ci = clfor all i. To prove this we can take I c1 > Ici for each i. As cA = kc, kc1 =Eiciail, so, as exactly k of the ail are 1 and the rest are 0, c1 = ci for eachi c A(1). But now as -9 is connected it follows that cl = ci for all i E X,completing the proof of the claim.

As Mk = 1 it follows that:

(vi)m,.+mt =n - 1 =k+1.

Also, as A is of trace 0:

(vii) k + m,s + mtt = 0.

Now (vi) and (vii) imply:

(viii) m,. = ((k + l)t + k)/(t - s).

Of course t - s = -d1/2, t = ((A - A) - d1/2)/2, and ms is an integer, Thus

(ix) (Dd-1/2 - (k + l))/2 is an integer, where

D =2k+(A-A)(k+1).

In particular either d is a square or D = 0. If D = 0 then A = A + 1 and 1 = k,so 16.7.1 holds by 16.3.2. If d is a square then 16.7.2 holds by (ix).

Remarks. Wielandt [Wi 2] is a good place to find more information aboutpermutation groups. The material in section 16 comes from Higman [Hi].

Section 16 is somewhat technical and can be safely omitted by the novice.On the other hand the results in section 15 are reasonably basic.

(iii) AJ = kJ , so (A - kI ) J = 0. (iv) A2 = kI +AA+pB.

The first two statements are immediate from the definitions. As lA(x)l = k, each row of A has k entries equal to 1. Thus (iii) holds. By (i) the (x, y)-th entry of A2 is the inner product of the xth and yth rows of A. But this inner product just counts lA(x) n A(y)l, so (iv) holds.

Next (ii), (iii), and (iv) imply:

(v) (A - ~I) (A ' - (A - p)A - (k - p)I) = 0,

so the minimal polynomial of A divides

The roots of p(x) are k, s, and t , where s = ((A - p) + d1I2)/2 and t = ((A - p ) - d1I2)/2. Let me be the multiplicity of the eigenvalue e .

Claim mk = 1. Indeed for c = (el, . . . , c,), cA = kc if and only if ci = cl for all i. To prove this we can take Icll >_ lcil for each i. As cA = kc, kcl = Ziciail, so, as exactly k of the ail are 1 and the rest are 0, cl = ci for each i E A(1). But now as is connected it follows that cl = ci for all i E X, completing the proof of the claim.

As mk = 1 it follows that:

Also, as A is of trace 0:

(vii) k + m,s + mtt = 0.

Now (vi) and (vii) imply:

(viii) m, = ((k + l)t + k)/(t - s).

Of course t - s = -dl/', t = ((A - p) - d1I2)/2, and m, is an integer. Thus

(ix) (Dd-'I2 - (k + 1))/2 is an integer, where

D = 2k + (A - p)(k + I). In particular either d is a square or D = 0. If D = 0 then p = h + 1 and 1 = k, so 16.7.1 holds by 16.3.2. If d is a square then 16.7.2 holds by (ix).

Remarks. Wielandt [Wi 21 is a good place to find more information about permutation groups. The material in section 16 comes from Higman [Hi].

Section 16 is somewhat technical and can be safely omitted by the novice. On the other hand the results in section 15 are reasonably basic.

1. (1)

Exercises for chapter 5Prove A5 has no faithful permutation representation of degree lessthan 5.

(2) Prove that, up to equivalence, A5 has unique transitive representationsof degree 5 and 6. Prove both are 2-transitive.

(3) Prove Aut(A5) = S5.(4) Prove there are exactly two conjugacy classes of subgroups of S6 iso-

morphic to A5. Prove the same for A6-2. Let G be a transitive permutation group on X, A a self paired orbital of

G, (x, y) E A, -9 = (X, E) the graph on X determined by A, and H =(G, Gy). Prove(1) xH U yH is the connected component of -9 containing x, and(2) -9 is connected precisely when one of the following holds:

(i) G = H, or(ii) I G : H 4 = 2 and ' is bipartite with partition. {xH, yH} (i.e. {xH, yH}

is a partition of X and A c_ (xH x yH) U (yH x xH)).3. Let S and A be the symmetric and alternating groups of degree n, respec-

tively, and let a E A. Prove aA 0 as precisely when (*) holds(*) Cyca(2m) = 0 and Cyca(2m - 1) < 1 for each positive integer m.in which case as = aA UbAforsomeb E Awith(a) = (b)andlaAI _ (bAl.

4. Let A be the alternating group on a set X of finite order n > 3 and let V bethe set of subsets of X of order 2. Prove(1) A is a rank 3 permutation group on V.(2) A is a maximal subgroup of A for V E V.(3) If G is a primitive rank 3 group on a set 2 of order 10 then G - A5 or

S5 and the representation of G on 0 is equivalent to its representationon V.

5. Let G be a primitive permutation group on a finite set X and let x and ybe distinct points of X. Let Y be a nonempty proper subset of X and defineS(x) = {Yg: g E G, x E Yg} and T(x) = nzEs(x) Z. Prove(1) T(x) = {x), and(2) there exists g E G with x E Yg but y 0 Yg.

6. Let G be a primitive permutation group on a set X of finite order n. Prove(1) If Y C X such that G y is primitive on X -Y and (Y =m with 1 <

m < n - 2, then G is (m + 1)-transitive on X.(2) If G contains a transposition or a cycle of length 3 then G contains the

alternating group on X.(3) If a, b E G with I Mov(a) fl Mov(b)l = 1 then [a, b] is a cycle of

length 3.(4) Let Y C X be of minimal order subject to Gy = 1. Prove either G

contains the alternating group on X or I Y (< n/2.


Exercises for chapter 5 1. (1) Prove A5 has no faithful permutation representation of degree less

than 5. (2) Prove that, up to equivalence, A5 has unique transitive representations

of degree 5 and 6. Prove both are 2-transitive. (3) Prove Aut(A5) = S5. (4) Prove there are exactly two conjugacy classes of subgroups of S6 iso-

morphic to A5. Prove the same for Ag. 2. Let G be a transitive permutation group on X, A a self paired orbital of

G, (x, y) E A, # = (X, A) the graph on X determined by A, and H = (G,, G,). Prove (1) XH U yH is the connected component of 9 containing x, and

(2) is connected precisely when one of the following holds: (i) G = H, or

(ii) IG : HI = 2 andg is bipartite withpartition.{xH, yH) (i.e. {xH, yH] is a partition of X and A 2 (xH x y H) U (y H x x H)).

3. Let S and A be the symmetric and alternating groups of degree n, respectively, and let a E A. Prove a A # as precisely when (*) holds (*) Cyc,(2m) = 0 and Cyc,(2m - 1) 5 1 for each positive integer m. inwhichcaseas = a A U bAforsomeb E Awith(a) = (b) andlaAI = IbAI.

4. Let A be the alternating group on a set X of finite order n > 3 and let V be the set of subsets of X of order 2. Prove (1) A is a rank 3 permutation group on V. (2) A, is a maximal subgroup of A for v E V. (3) If G is a primitive rank 3 group on a set Q of order 10 then G % A5 or

S5 and the representation of G on S2 is equivalent to its representation on V.

5. Let G be a primitive permutation group on a finite set X and let x and y be distinct points of X. Let Y be a nonempty proper subset of X and define S(x) = {Yg: g E G, x E Yg} and T(x) = n ,,,,,, 2. Prove (1) T(x) = {XI, and (2) there exists g E G with x E Yg but y 6 Yg.

6. Let G be a primitive permutation group on a set X of finite order n. Prove (1) If Y E X such that Gy is primitive on X - Y and IYI = m with 1 5

m 5 n - 2, then G is (m + 1)-transitive on X. (2) If G contains a transposition or a cycle of length 3 then G contains the

alternating group on X. (3) If a , b E G with IMov(a) n Mov(b)J = 1 then [a, b] is a cycle of

length 3. (4) Let Y 2 X be of minimal order subject to Gy = 1. Prove either G

contains the alternating group on X or IY I 5 n/2.

(5) If G does not contain the alternating group on X, prove j Sym(X) : G I >

[(n + 1)/2]!7. Let X be a set of finite order n < 5, A the alternating group on X, and G a

proper subgroup of A. Prove one of the following holds:(1) JA: G) > n.(2) (A:GI =nand G=Ax forsome xEX.(3) (A:GI =n=6andG-A5.

8. Prove that either(1) An has a unique conjugacy class of subgroups isomorphic to or

(2) n = 6 and An has exactly two such classes.


(5) If G does not contain the alternating group on X, prove iSym(X) : GI 3

[(n + 1>/21! 7. Let X be a set of finite order n 5 5, A the alternating group on X, and G a

proper subgroup of A. Prove one of the following holds: (1) IA : GI > n. (2) 1A:GI =nand G =A, for somex E X. (3) 1A:Gl = n = 6 a n d G S A 5 .

8. Prove that either (1) A, has a unique conjugacy class of subgroups isomorphic to A,-1, or (2) n = 6 and A, has exactly two such classes.

6

Extensions of groups and modules

Chapter 6 considers various questions about extensions of groups and modules,most particularly the conjugacy of complements to some fixed normal subgroupin a split group extension. Suppose G is represented on an abelian group orF-space V and form the semidirect product GV. Section 17 shows there is abijection between the set of conjugacy classes of complements to V in GVand the 1-cohomology group H1(G, V). If V is an F-space so is H1(G, V).Moreover if Cv(G) = 0 there is a largest member of the class of FG-modulesU such that Cu(G) = 0 and U is the extension of V by a module centralizedby G. Indeed it turns out that if U is the largest member of this class thenU/V = H1(G, V). Further the dual of the statement is also true: that is ifV = [V, G] then there is a largest FG-module U* such that U* = [U*, G] andU* is the extension of an FG-module Z by V with G centralizing Z. In thiscase Z = H1(G, V*).

These results together with Maschke's Theorem are then used to prove theSchur-Zassenhaus Theorem, which gives reasonably complete informationabout extensions of a finite group B by a finite group A when the orders ofA and B are relatively prime. The Schur-Zassenhaus Theorem is then usedto prove Phillip Hall's extended Sylow Theorem for solvable groups. Hall'sTheorem supplies a good illustration of how restrictions on the compositionfactors of a finite group can be used to derive strong information about the group.

I have chosen to discuss 1-cohomology from a group theoretical point ofview. Homological algebra is kept to a minimum. Still the arguments in section17 have a different flavor than most in this book.

17 1-cohomologyIn this section p is a prime, F is a field of characteristic p, V is an abelian groupwritten additively, G is a finite group, and ir: G --> Aut(V) is a representationof G on V.

Form the semidirect product S(G, V, 7r) of V by G with respect to it andidentify G and V with subgroups of S(G, V, 7r) via the injections of 10.1. ThenS(G, V, 7r) = GV with V 4 GV and G is a complement to V to GV.

A cocycle from G into V is a function y: G - V satisfying the cocyclecondition

(gh)y = (gy)p` + by g, h E G.


Chapter 6 considers various questions about extensions of groups and modules, most particularly the conjugacy of complements to some fixed normal subgroup in a split group extension. Suppose G is represented on an abelian group or F-space V and form the semidirect product GV. Section 17 shows there is a .

bijection between the set of conjugacy classes of complements to V in GV and the 1-cohomology group H1(G, V). If V is an F-space so is H'(G, V). Moreover if C"(G) = 0 there is a largest member of the class of FG-modules U such that CU(G) = 0 and U is the extension of V by a module centralized by G. Indeed it turns out that if U is the largest member of this class then U/VE H1(G, V). Further the dual of the statement is also true: that is if V = [V, GI then there is a largest FG-module U* such that U* = [U*, GI and U* is the extension of an FG-module Z by V with G centralizing Z. In this case Z E H1(G, V*).

These results together with Maschke's Theorem are then used to prove the Schur-Zassenhaus Theorem, which gives reasonably complete information about extensions of a finite group B by a finite group A when the orders of A and B are relatively prime. The Schur-Zassenhaus Theorem is then used to prove Phillip Hall's extended Sylow Theorem for solvable groups. Hall's Theorem supplies a good illustration of how restrictions on the composition factors of a finite group can be used to derive strong information about the group.

I have chosen to discuss 1-cohomology from a group theoretical point of view. Homological algebra is kept to a minimum. Still the arguments in section 17 have a different flavor than most in this book.

17 1-cohomology In this section p is a prime, F is a field of characteristic p, V is an abelian group written additively, G is a finite group, and n : G + Aut(V) is a representation of G on V.

Form the semidirect product S(G, V, n ) of V by G with respect to n and identify G and V with subgroups of S(G, V, n ) via the injections of 10.1. Then S(G, V, n ) = GV with V 9 GV and G is a complement to V to GV.

A cocycle from G into V is a function y: G + V satisfying the cocycle condition

(gh)y = (gylh + hy g, h E G.

1-cohomology 65

Notice the cocycle condition forces each cocycle to map the identity of G tothe identity of V.

Let r(G, V) denote the set of cocycles from G to V and make r(G, V) intoa group (again written additively) via:

g(Y+S)=gy+gS y, S E I'(G,V),g EG.

Let A = Aut(G V) be the group of automorphisms of GV, and let U(G, V) _CA(GV/ v) n CA(V).

If V is a vector space over F and 7r is an FG-representation then r(G, V)is also a vector space over F via:

g(ay) = a(gy) y E r(G, V), g E G, a E F.

(17.1) For y E r(G, V) define S(y) = fggy: g E G}. Then the map S: y r-*S(y) is a bijection of r(G, V) with the set of complements to V in GV.

Proof. The cocycle condition says that S(y) is a subgroup of GV. EvidentlyS(y) is a complement to V in GV and S is injective. Conversely if C isa complement then, for g E G, gV n c contains a unique element gv andy: g r-* v is a cocycle with S(y) = C.

(17.2) For y E I' (G, V) define y0: GV --i GV by

YO: gv H g((gY) + v).

Then the map 0: y H yO is a group isomorphism of r(G, V) with U(G, VFor u E U(G, V) and g E G, uO-1: g r-* g-lg"

Proof. The cocycle condition implies yO is a homomorphism. (-y)O is aninverse for yO, so yO E Aut(GV). By definition yO centralizes V and GV/ V,so yO E U(G, V). For U E U define ul/r: g r-* g-lg". An easy check showsui/r is a cocycle and i1r = 0-1, so 0 is an isomorphism.

(17.3) U(G, V) acts regularly on the set of complements to V in G V. IndeedG" = S(uo-1) for each u E U(G, V).

Proof. As U = U(G, V) acts on V it permutes the complements to V in GV.By definition of the maps 0 and S, G" = S(uO-1) for u E U. Hence, as S and0 are bijections, the action of U is regular.

Lemmas 17.1 and 17.3 give descriptions of the complements to V inGV in terms of r(G, V) and U(G, V), while 17.2 and 17.3 give the corres-pondence between the two descriptions. The next few lemmas describe the

Notice the cocycle condition forces each cocycle to map the identity of G to the identity of V.

Let F(G, V) denote the set of cocycles from G to V and make F(G, V) into a group (again written additively) via:

g(y + 6) = gy + g6 y, 6 E UG, V), g E G.

Let A = Aut(G V) be the group of automorphisms of GV, and let U(G, V) =

CA(GV/V) n CA(V). If V is a vector space over F and n is an FG-representation then F(G, V)

is also a vector space over F via:

d a y ) = a(gy) Y E UG, V), g E G, a E F.

(17.1) For y E F(G, V) define S(y) = {gg y : g E G}. Then the map S: y I-+ S(y) is a bijection of F(G, V) with the set of complements to V in GV.

Proof. The cocycle condition says that S(y) is a subgroup of GV. Evidently S(y) is a complement to V in GV and S is injective. Conversely if C is a complement then, for g E G, gV n C contains a unique element gv and y: g I-+ v is a cocycle with S(y) = C.

(17.2) For y E F(G, V) define y 4: G V -+ G V by

Then the map 4: y I-+ y 4 is a group isomorphism of F(G, V) with U(G, V). Foru E U(G, V)andg E G,u&':g I-+ g-'gU.

Proof. The cocycle condition implies y 4 is a homomorphism. (-y)4 is an inverse for y 4, so y 4 E Aut(G V). By definition y 4 centralizes V and G V/ V, so y 4 E U(G, V). For u E U define u+: g I-+ g-'gU. An easy check shows u+ is a cocycle and + = &', so 4 is an isomorphism.

(17.3) U(G, V) acts regularly on the set of complements to V in GV. Indeed GU = S(u4-') for each u E U(G, V).

Proof. As U = U(G, V) acts on V it permutes the complements to V in GV. By dkfinition of the maps 4 and S, GU = S(u4-') for u E U. Hence, as S and 4 are bijections, the action of U is regular.

Lemmas 17.1 and 17.3 give descriptions of the complements to V in GV in terms of F(G, V) and U(G, V), while 17.2 and 17.3 give the correspondence between the two descriptions. The next few lemmas describe the

66 Extensions of groups and modules

representations of G on F(G, V) and U(G, V), and show these representationsare equivalent.

(17.4) For g E G and y E F(G, V) define yg: G - V by h(yg) = [(hg-' )y]g,for h E G. Then Fr is a representation of G on F(G, V), where gfr: y H yg.If tr is an FG-representation so is ft.

(17.5) For v E V define va: G -+ V by g(va) = [g, v] = v - vg. Then themap a: v va is a G-homomorphism of V into F(G, V) with kernel Cv(G).If 7r is an FG-representation then a is an FG-homomorphism.

Proof. Lemma 8.5.4 says vu is a cocycle. The rest is straightforward.

If V is an FG-module then so is F(G, V) and 0 induces an F-space structureon U(G, V) which makes 0 into an F-space isomorphism. That is, for u EU(G, V) and a E F, au = (a(uo-1))O. Equivalently if g" = gv then g°" =g(av). This is the F-space structure on U(G, V) implicit in the remainder ofthe section.

(17.6) Let c: G V - Aut(G V) and d: Aut(G V) f1 N(V) - Aut(U(G, V)) bethe conjugation maps. Then

aV- F(G, V)

U(G, V)

is a commutative diagram, the maps c, a, and 0 are G-homomorphisms, andif V is an FG-module the maps are FG-homomorphisms. Here 0, a, and frare defined in 17.2, 17.5, and 17.4, respectively, and Jr, Fr, and cd are therepresentations of G on V, F(G, V), and U(G, V), respectively.

Proof. Let v, w E V, U E U(G, V), and g, h E G. Then

vc: gw g°w = g(v - vg + w) = (gw)(va)O,

so c = ao and the diagram commutes.By 17.5, a is a G-homomorphism, and even an FG-homomorphism if V is

an FG-module. By 17.2, 0 is a homomorphism.Next agcd: by H (hv)g-'"g = hg-'"gv. Hence (hv)(ygo) = h(hyg + v) _

h(((ha ')y)g + v) = (hg '(ha ')y)gv = ((hg ')(yo))gv = (hv)((yO)gcd) soygo = (yO)gcd, and therefore 0 is a G-homomorphism.


representations of G on r (G, V) and U(G, V), and show these representations are equivalent.

(17.4) For g E G and y E r(G, V) define yg: G + V by h(y8) = [(hg-i)y]g, for h E G. Then it is a representation of G on r(G, V), where git: y H yg. If n is an FG-representation so is 5.

(17.5) For v E V define v a : G + V by g(va) = [g, v] = v - vg. Then the map a: v H va is a G-homomorphism of V into r (G , V) with kernel CV(G) . If n is an FG-representation then a is an FG-homomorphism.

Proof. Lemma 8.5.4 says va is a cocycle. The rest is straightforward.

If V is an FG-module then so is r (G, V) and # induces an F-space structure on U(G, V) which makes # into an F-space isomorphism. That is, for u E U(G, V) and a E F, au = (a(u4-I))#. Equivalently if gu = gv then gaU = g(av). This is the F-space structure on U(G, V) implicit in the remainder of the section.

(17.6) Let c: G V + Aut(G V) and d: Aut(G V) n N(V) + Aut(U(G, V)) be the conjugation maps. Then

is a commutative diagram, the maps c, a , and # are G-homomorphisms, and if V is an FG-module the maps are FG-homomorphisms. Here #, a , and it are defined in 17.2, 17.5, and 17.4, respectively, and n, it, and cd are the representations of G on V, r (G, V), and U(G, V), respectively.

Proof. Let v, w E V, u E U(G, V), and g, h E G. Then

so c = a# and the diagram commutes. By 17.5, a is a G-homomorphism, and even an FG-homomorphism if V is

an FG-module. By 17.2,# is a homomorphism. Next ugcd: hv H (h~)~-"g = hg-'ugv. Hence (hv)(yg#) = h(hy8 + v) =

h(((hg-i)y)g + v) = (hg-l(hg-')y)gv = ((hg-l)(y#))gv = ( h ~ ) ( ( ~ # ) g ~ ~ ) , SO

yg# = (y#)gcd, and therefore # is a G-homomorphism.

1-cohomology 67

Suppose V is an FG-module. For a E F, (au)g`d = ((a(uO-1))c)g`d =((a(uO-1))9)0 = (a((ug`d)0-1))O = a(ugcd), so cd is an FG-representation.

Finally, as c = ao and a and 0 are G-homomorphisms (or FG-homomor-phisms), so is c. Therefore the proof is complete.

By 17.6, F(G, V)/ Va = U(G, V)/ Vc. The first cohomology group of therepresentation 7r is H 1(G, V) = 1'(G, V)/ V a = U (G, V)/ V c. This is anadditive group and, if V is an FG-module, H1 (G, V) is even a vector spaceover F.

The next lemma says that H1 (G, V) is in one to one correspondence withthe set of conjugacy classes of complements to V in GV.

(17.7) H 1 (G, V) acts regularly on the set of conjugacy classes of complementsto V in GV via

(Vc)u: (Gw)v H (Gwu)v U, W E U(G, V).

In particular the number of conjugacy classes of complements to V in GV is

IH1(G,V)I

Proof. This is a consequence of 17.3 and the fact that U(G, V) acts on V.

(17.8) [Gcd, U(G, V)] < Vc.

Proof. For g E G, U E U(G, V), [gcd, u] _ [gc, u] E U(G, V) f1 GVc, asU(G, V) and GVc are normal in Aut(GV). But U(G, V) fl GVc = Vc.

(17.9) Assume either Cv(G) = 0 or G = OP(G). Then CU(G,v)(Gcd) = 0and if Cv(G) = 0 then c: V - U(G, V) is an injection.

Proof. If U E CU(G,v)(Gcd) then 1 = [u, Gcd] = [u, Gc], so [u, G] <ker(c) = Z(GV) = CZ(G)(V) X Cv(G). I'll show [u, G] < G, so that, by17.3, u = 1, and hence CU(G,v)(Gcd) = 0.

If Cv(G) = 0 then [u, G] < CZ(G)(V) < G. If G = OP(G) then G =OP(GZ(GV)) char GZ(GV), so [u, G] < G. So the claim is established.

As Cv(G) is the kernel of c, c is an injection if Cv(G) = 0.

(17.10) If G is a finite p'-group and V an FG-module, then H '(G, V) = 0.Hence G is transitive on the complements to V in GV in this case.

Suppose V is an FG-module. For a E F, (au)gcd = ((a(u@-'))@)gcd = ((a(u@-'))g)@ = (a((ugcd)@-I))@ = a(ugcd), so cd is an FG-representation.

Finally, as c = a@ and a and @ are G-homomorphisms (or FG-homomorphisms), so is c. Therefore the proof is complete.

By 17.6, r(G, V)/ Va S U(G, V)/ Vc. The jirst cohomology group of the representation n is H'(G, V) 2 r (G, V)/Va S U(G, V)/Vc. This is an additive group and, if V is an FG-module, H'(G, V) is even a vector space over F.

The next lemma says that H1(G, V) is in one to one correspondence with the set of conjugacy classes of complements to V in GV.

(17.7) H1(G, V) acts regularly on the set of conjugacy classes of complements to V in GV via

(Vc)u: (G")' H (G"')' u, w E U(G, V).

In particular the number of conjugacy classes of complements to V in GV is

IH1(G, V)I.

Proof. This is a consequence of 17.3 and the fact that U(G, V) acts on V. ii

(17.8) [Gcd, U(G, V)] 5 Vc.

Proof. For g E G, u E U(G, V), [gcd, u] = [gc, u] E U(G, V) n GVc, as U(G, V) and GVc are normal in Aut(GV). But U(G, V) n GVc = Vc.

(17.9) Assume either Cv(G) = 0 or G = Op(G). Then Cu(~,v)(Gcd) = 0 and if CV (G) = 0 then c: V + U(G, V) is an injection.

Proof. If u E CU(G,v)(G~d) then 1 = [u, Gcd] = [u, Gc], so [u, GI < ker(c) = Z(GV) = Cz(~)(V) x Cv(G). I'll show [u, GI ( G, so that, by 17.3, u = 1, and hence CU(~,v)(Gcd) = 0.

If CV(G) = 0 then [u, GI 5 CZ(G)(V) 5 G. If G = OP(G) then G = Op(GZ(GV)) char GZ(GV), so [u, GI 5 G. So the claim is established.

As CV(G) is the kernel of c, c is an injection if Cv(G) = 0.

(17.10) If G is a finite p'-group and V an FG-module, then H1(G, V) = 0. Hence G is transitive on the complements to V in GV in this case.

Proof. By Maschke's Theorem, the FG-module U = U(G, V) splits over V c.Let W be an FG-complement to V c in U. By 17.8, [Gcd, W] < W fl v c = 0.So, by 17.9, W = 0. Thus H 1(G, V) - W = 0, and 17.7 completes the proof.

Lemma 17.10 will be used to prove the Schur-Zassenhaus Theorem in the nextsection.

(17.11) Let V be an FG-module, 0: V - W an injective FG-homomorphism,and assume [G, W] < V18 and Cw(G) = 0. Then there exists an injective FG-homomorphism y: W - U(G, V) making the following diagram commute:

V

W--> U(G, V)Y

In particular dimF(W/ V,B) < dimF(H1(G, V)).

Proof. Let r' be the representation of G on W and consider the semidirectproducts H = S(F# x G, W, 7r') and S = S(F# x G, V, 7r). As ,B is an injec-tive FG-homomorphism it induces by 10.3 an injective group homomorphism,J: S H which is the identity on F# x G. As [G, W] < V,8, (GV),B 4 H, sothe conjugation map e of H on (G V),B composed with (0-1)*: Aut((G V),B) -Aut(GV) = A maps H into A. As Cw(G) = 0, the restriction y of e(,B-1)*to W is an injection of W into U = U(G, V). As y is the composition of(F# x G)-homomorphisms, y is an (F# x G)-homomorphism, and hence is aG-homomorphism preserving the multiplication by F. In particular if the mul-tiplication u i-+ ua, a e F#, U E U of F# on U is that of the F-space structureon U defined earlier, then y is an FG-homomorphism. For this we need to showgU' = gau for g e G. But if gU = gv then gau = gav, while

gUa = ga-'ua =gua = (gv)a = gav as [a, g] = 1 and Va = av.

It remains to show ,By = c. Keeping in mind that ,B: S H is an injec-tive homomorphism trivial on F# x G, we see that for v, w E V and g E G

we have (gw)°Py = (((gw)P)' )P-1 = (gv wP)fi-1 = (g[g, vP]wP)f-1 =g[g, v]w = g"w = (gw)v = (gw)"`. So ,By = c as desired.

Lemma 17.11 says that if Cv(G) = 0 then U(G, V) is the largest extension Wof V such that [W, G] < V and Cw(G) = 0.

Recall the definition of the dual V* of a finite dimensional FG-module Vgiven in section 14, and define U*(G, V) to be (U(G, V*))*.


Proof. By Maschke's Theorem, the FG-module U = U(G, V) splits over Vc. Let W be an FG-complement to Vc in U. By 17.8, [Gcd, W] 5 W n Vc = 0. So, by 17.9, W = 0. Thus H'(G, V) G W = 0, and 17.7 completes the proof.

Lemma 17.10 will be used to prove the Schur-Zassenhaus Theorem in the next section.

(17.11) Let V be an FG-module, B: V + W an injective FG-homomorphism, and assume [G, W] 5 VB and Cw(G) = 0. Then there exists an injective FG- homomorphism y: W + U(G, V) making the following diagram commute:

In particular dimF(W/ VB) 5 d i m F ( ~ ' (G, V)).

Proof. Let n' be the representation of G on W and consider the semidirect products H = S(F# x G, W, n') and S = S(F' x G, V, IT). As B is an injective FG-homomorphism it induces by 10.3 an injective group homomorphism B: S + H which is the identity on F' x G. As [G, W] 5 VB, (GV)B H, so the conjugation map e of H on (G V)B composed with (,!-I)*: Aut((G V)B) + Aut(GV) = A maps H into A. As Cw(G) = 0, the restriction y of e(Bw')* to W is an injection of W into U = U(G, V). As y is the composition of (F' x G)-homomorphisms, y is an (F' x G)-homomorphism, and hence is a G-homomorphism preserving the multiplication by F. In particular if the multiplication u H ua, a E F', u E U of F' on U is that of the F-space structure on U defined earlier, then y is an FG-homomorphism. For this we need to show

- = gaU for g E G. But if gU = gv then gaU = gav, while = ga-'ua -

gun = (guy = gav as [a, g] = 1 and va = av. It remains to show By = c. Keeping in mind that B: S + H is an injec-

tive homomorphism trivial on F# x G, we see that for v, w E V and g E G we have (gw)vB)' = (((gw)~)VB)~- ' = ( g v B w ~ ) ~ - l = (g[g, v ~ l w ~ ) B - ' = g[g, v]w = gVw = (gw)' = ( g ~ ) ~ ' . SO By = c as desired.

Lemma 17.1 1 says that if CV(G) = 0 then U(G, V) is the largest extension W of V such that [W, GI 5 V and Cw(G) = 0.

Recall the definition of the dual V* of a finite dimensional FG-module V given in section 14, and define U*(G, V) to be (U(G, V*))*.

1-cohomology 69

(17.12) Let V be a finite dimensional FG-module, 0: W - V a surjective FG-homomorphism, and assume ker(,B) < CW(G) and W = [W, G]. Then thereexists a surjective FG-homomorphism y: U*(G, V) - W making the follow-ing diagram commute:

Here e* is the dual of the conjugation map e: V* - U(G, V*) and e* is a sur-jective FG-homomorphism with H'(G, V*) = ker(e*) < CU.(G,v)(G). ThusdimF(ker(,B)) < dimF(H'(G, V*)).

Proof. As W = [W, G], CW.(G) = 0 by 14.6. Similarly, as ker(,B) < Cw (G),[G, W *] < V *,B* by 14.6. Let e: V* - U(G, V*) be the conjugation map. By17.11 there is an injective FG-homomorphism S: W* -+ U(G, V*) such thatthe diagram commutes:

W* U(G, V*)

Then applying 14.1.2 and 14.4.3 we conclude the following diagram commutes,

e*

Y

where y = S*. As S is injective, y is surjective, so e* = y,8 is surjective sincefi is surjective by hypothesis. H1 (G, V*) = U(G, V*)/ V*e, so H'(G, V*) =H' (G, V*)* = ker(e*) by 14.5. As [U(G, V*), G] < V*e, G centralizes ker(e*)by 14.6.

Lemma 17.12 says that if V = [V, G] then U*(G, V) is the largest extensionW of a module Z by V with Z < CW(G) and W = [W, G].

(17.12) Let V be a finite dimensional FG-module, B: W + V a surjective FG- homomorphism, and assume ker(B) 5 Cw(G) and W = [W, GI. Then there exists a surjective FG-homomorphism y: U*(G, V) + W making the following diagram commute:

Here e* is the dual of the conjugation map e: V* + U(G, V*) and e* is a surjective FG-homomorphism with H'(G, V*) E ker(e*) 5 Cv*(c,v)(G). Thus dimF(ker(B)) 5 d i m F ( H 1 ( ~ , V*)).

Proof. As W = [W, GI, Cw*(G) = 0 by 14.6. Similarly, as ker(B) 5 Cw(G), [G, W*] 5 V*B* by 14.6. Let e: V* + U(G, V*) be the conjugation map. By 17.11 there is an injective FG-homomorphism S: W* + U(G, V*) such that the diagram commutes:

W* - U(G, V * ) 6

Then applying 14.1.2 and 14.4.3 we conclude the following diagram commutes,

where y = S*. As 6 is injective, y is surjective, so e* = yB is surjective since j3 is surjective by hypothesis. H'(G, V*) = U(G, V*)/V*e, so H'(G, V*) E H'(G, V*)* E ker(e*) by 14.5. As [U(G, V*), GI 5 V*e, G centralizes ker(e*) by 14.6.

Lemma 17.12 says that if V = [V, GI then U*(G, V) is the largest extension W of a module Z by V with Z 5 Cw(G) and W = [W, GI.

18 Coprime action(18.1) (Schur-Zassenhaus Theorem) Let G be a finite group, H < G, andassume

(i) (IHI, IG/HI) = 1, and(ii) either H or G/H is solvable.

Then(1) G splits over H, and(2) G is transitive on the complements to H in G.

Proof. Let G be a minimal counterexample. Assume first that H is solvableand let M be a minimal normal subgroup of G contained in H. By 9.4, M is anelementary abelian p-group for some prime p. Let G = G/M. By minimalityof G, there is a complement X to H in G and G is transitive on the complementsto H. Also if Y is a complement to H in G then F is a complement to H in G,so it suffices to show X splits over M and X is transitive on its complements toM. Hence, by minimality of G, G = X and H = M. Now Gaschutz' Theorem,10.4, says that G splits over H. Let Y be a complement to H in G.

By 12.1 the conjugation map c: Y Aut(M) is a FY-representation, whereF is the field of order p. Hence by 17.10, G is transitive on the complementsto H = M in G.

Assume next that G/H is solvable. Let G* = G/H and K* a minimal normalsubgroup of G*. By 9.4, K* is an elementary abelian p-group for some primep. Let P E Sylp(K) and observe that P is a complement to H in K. By aFrattini Argument, 6.2, G = KNG(P), so as K = HP, also G = HNG(P).

If Y is a complement to H in G then K = K fl G = K fl HY = H(K fl Y)by the Modular Property of Groups, 1.14. Then R = K fl Y is a complementto H in K, so R E Sylp(K). Hence, by Sylow's Theorem, there is k E K withRk=P.AsK<G,R=KfY<Y,soYk < NG(P). Thus, if NG(P) splitsover NH(P) and is transitive on its complements to NH(P), then the theoremholds. So by minimality of G, P < G.

Finally let G = G/P. If Y is a complement to H in G, then Y contains aSylow p-group of G and hence P < Y. Also f is a complement to H in G.Moreover, by minimality of G, there is a complement C to H in G and G istransitive on its complements to H. Now C is a complement to H in G andthere is g E G with Cg = Y, so Cg = Y, completing the proof.

Remarks. The Odd Order Theorem of Feit and Thompson [FT] says thatgroups of odd order are solvable. Notice that if (I A 1, I G 1) = 1 then eitherA I or I G I is odd, so A or G is solvable. Thus the Odd Order Theorem says

hypothesis (ii) of the Schur-Zassenhaus Theorem can be removed.


18 Coprime action (18.1) (Schur-Zassenhaus Theorem) Let G be a finite group, H 9 G, and assume

(9 (IHl, IGIHI) = 1, and (ii) either H or G/H is solvable.

Then

(1) G splits over H , and

(2) G is transitive on the complements to H in G.

Proof. Let G be a minimal counterexample. Assume first that H is solvable and let M be a minimal normal subgroup of G contained in H. By 9.4, M is an elementary abelian p-group for some prime p. Let G = G/M. By minimality of G, there is a complement to H in G and G is transitive on the complements to H. Also if Y is a complement to H in G then Y is a complement to H in G, so it suffices to show X splits over M and X is transitive on its complements to M. Hence, by minimality of G, G = X and H = M. Now Gaschiitz' Theorem, 10.4, says that G splits over H. Let Y be a complement to H in G.

By 12.1 the conjugation map c: Y -+ Aut(M) is a FY-representation, where F is the field of order p. Hence by 17.10, G is transitive on the complements t o H = M i n G .

Assume next that GI H is solvable. Let G* = GI H and K * a minimal normal subgroup of G*. By 9.4, K* is an elementary abelian p-group for some prime p. Let P E Sylp(K) and observe that P is a complement to H in K. By a Frattini Argument, 6.2, G = KNG(P), so as K = HP, also G = HNG(P).

IfYisacomplementtoHinGthenK = K n G = K n H Y = H ( K n Y ) by the Modular Property of Groups, 1.14. Then R = K n Y is a complement to H in K, so R E Sylp(K). Hence, by Sylow's Theorem, there is k E K with R~ = P . As K 9 G, R = K n Y 9 Y, so yk I NG(P). Thus, if NG(P) splits over NH(P) and is transitive on its complements to NH(P), then the theorem holds. So by minimality of G, P 9 G.

Finally let G = G/P. If Y is a complement to H in G, then Y contains a Sylow p-group of G and hence P I Y. Also P is a complement to H in G. Moreover, by minimality of G, there is a complement c to H in G and G is transitive on its complements to H. Now C is a complement to H in G and there is g E G with cg = P, so Cg = Y, completing the proof.

Remarks. The Odd Order Theorem of Feit and Thompson [FT] says that groups of odd order are solvable. Notice that if (IAl, IGJ) = 1 then either IAI or IGI is odd, so A or G is solvable. Thus the Odd Order Theorem says hypothesis (ii) of the Schur-Zassenhaus Theorem can be removed.

Coprime action 71

Let n be a set of primes. The n part of a positive integer n is n, = r[pen pep,where rlp pep = n is the prime factorization of n. Given a finite group G, n (G)denotes the set of prime factors of I G 1. G is a n -group if r (G) is a subset of7r. A Hall 7r -subgroup of G is a subgroup of order I G Jn . n' denotes the set ofprimes not in n.

The following lemma gives a useful characterization of Hall n-subgroups.

(18.2) H is a Hall n-subgroup of the finite group G if and only if H is an-subgroup of G and (G: HI, = 1.

(18.3) Let G be a finite group and H a 7r -subgroup of G. Then(1) If a: G Ga is a homomorphism then Ha is a rr-subgroup of Ga. If

H is a Hall n-subgroup of G and a is surjective then Ha is a Hall n-subgroupof Ga.

(2) If H is a Hall n-subgroup of G and H < K < G then H is a Hall7r-subgroup of K.

(18.4) If p and q are distinct primes and H and K are Hall p'- and q'-subgroupsof a finite group G, respectively, then

(1) G=HK,and(2) H fl K is a Hall (p, q}'-subgroup of G, a Hall p'-subgroup of K, and a

Hall q'-subgroup of H.


(18.5) (Phillip Hall's Theorem) Let G be a finite solvable group and n a setof primes. Then

(1) G possesses a Hall n-subgroup.(2) G acts transitively on its Hall n-subgroups via conjugation.(3) Any n-subgroup of G is contained in some Hall 7r-subgroup of G.

Proof. Let G be a minimal counterexample and M a minimal normal subgroupof G. By 9.4, M is a p-group for some prime p. Let G* = G/M. By minimalityof G, G* satisfies the theorem. In particular G* possesses a Hall n-subgroupH*. Also if X is a n-subgroup of G then so is X*, so X* is contained in someconjugate H*g of H*, and hence X < Hg.

Suppose P E n. Then H is a Hall n-subgroup of G. Further X is containedin the Hall n-subgroup Hg of G. Indeed if X is a Hall n-subgroup of G, thenIXJ _ IH91 so X = Hg is a conjugate of H.

Thus we may assume p n. By the Schur-Zassenhaus Theorem, 18.1,there is a complement K to M in H, and H is transitive on such complements.

Coprime action 7 1

Let n be a set of primes. The n-part of a positive integer n is n, = n,,, pep, where nppep = n is the prime factorization of n. Given a finite group G, n(G) denotes the set of prime factors of I G I . G is a n-group if n(G) is a subset of n . A Hall n-subgroup of G is a subgroup of order /GI,. n ' denotes the set of primes not in n .

The following lemma gives a useful characterization of Hall n-subgroups.

(18.2) H is a Hall n-subgroup of the finite group G if and only if H is a n-subgroup of G and IG: HI, = 1.

(18.3) Let G be a finite group and H a n-subgroup of G. Then (1) If a : G -+ G a is a homomorphism then H a is a n-subgroup of Ga. If

H is a Hall n-subgroup of G and a is surjective then H a is a Hall n-subgroup of Ga.

(2) If H is a Hall n-subgroup of G and H ( K ( G then H is a Hall n-subgroup of K.

(18.4) If p and q are distinct primes and H and K are Hall p'- and 9'-subgroups of a finite group G, respectively, then

(1) G =HK,and (2) H n K is a Hall (p, 9)'-subgroup of G, a Hall p'-subgroup of K, and a

Hall 9'-subgroup of H .


(18.5) (Phillip Hall's Theorem) Let G be a finite solvable group and n a set of primes. Then

(1) G possesses a Hall n-subgroup. (2) G acts transitively on its Hall n-subgroups via conjugation. (3) Any n-subgroup of G is contained in some Hall n-subgroup of G.

Proof. Let G be a minimal counterexample and M a minimal normal subgroup of G. By 9.4, M is a p-group for some prime p. Let G* = GIM. By minimality of G, G* satisfies the theorem. In particular G* possesses a Hall n-subgroup H*. Also if X is a n-subgroup of G then so is X*, so X* is contained in some conjugate H *g of H *, and hence X 5 Hg .

Suppose p E n . Then H is a Hall n-subgroup of G. Further X is contained in the Hall n-subgroup Hg of G. Indeed if X is a Hall n-subgroup of G, then 1x1 = lHgl so X = Hg is a conjugate of H .

Thus we may assume p 6 n . By the Schur-Zassenhaus Theorem, 18.1, there is a complement K to M in H, and H is transitive on such complements.

But as (H* I = I G I,' these complements are precisely the Hall 7r-subgroupsof G contained in H. Therefore G possesses Hall 7r-subgroups and H istransitive on the Hall 7r-subgroups of G contained in H. So, as Xg < H, Xis conjugate to K if X is a Hall 7r-subgroup of G. This shows G is transi-tive on its Hall 7r-subgroups. Finally if H G then, by minimality of G, Xis contained in a Hall 7r-subgroup of Hg, which is also a Hall 7r-subgroupof G.

SoassumeH = G.ThenG =KMsoXM=XMf1G =XMf1KM=(XMf K)M by the Modular Property of Groups, 1.14. Now X and XMn K areHall 7r-subgroups of XM, so, by (2), there is y E XM with XY == XMl K < K.Thus the proof is complete.

A converse of Phillip Hall's Theorem also holds, as we will see soon. In par-ticular the hypothesis of solvability is necessary to insure the validity of thetheorem. Hall's Theorem is a good example of how restrictions on the compo-sition factors of a finite group can lead to significant restrictions of the globalstructure of the group.

Now to a proof of a converse to Phillip Hall's Theorem. The proof willappeal to Burnside's pagb-Theorem that finite groups G with 17r(G)I = 2 aresolvable. The proof of Burnside's Theorem will be postponed until the chapteron character theory.

(18.6) Let G be a finite group possessing a Hall p'-subgroup for each p E 7r(G).Then G is solvable.

Proof. Let G be a minimal counterexample. By 9.8, G is not a p-group. ByBurnside's pogo Theorem, 35.13, I7r(G)) 2. Thus j7r(G)I > 2.

Let p E 7r (G) and H a Hall p'-subgroup of G. By 18.4, H fl K is a Hall q'-subgroup of H for each prime q distinct from p and each Hall q'-subgroup ofK of G. Therefore H satisfies the hypothesis of the lemma, and hence H issolvable by minimality of G. Let M be a minimal normal subgroup of H. By 9.4,M is an r-group for some primer. As 17r(G)I > 2 there is q E 7r(G) - (p, r}.Let K be a Hall q'-subgroup of G. As q 54 r, K contains a Sylow r-subgroupof G. Hence, by Sylow's Theorem, M is contained in some conjugate of K,which we may take to be K. As q p, G = HK by 18.4. So, as M a H, X =(MG) = (MHK) = (MK) < K. Hence, as subgroups of solvable groups aresolvable, X is solvable. Of course X < G and we conclude from 18.3 that G/ Xsatisfies the hypothesis of the lemma, So, by minimality of G, G/ X is solvable.Therefore G is solvable by 9.3.


But as I H*J = IGlnt these complements are precisely the Hall n-subgroups of G contained in H . Therefore G possesses Hall n-subgroups and H is transitive on the Hall n-subgroups of G contained in H. So, as Xg 5 H, X is conjugate to K if X is a Hall n-subgroup of G. This shows G is transitive on its Hall n-subgroups. Finally if H # G then, by minimality of G, X is contained in a Hall n-subgroup of Hg, which is also a Hall n-subgroup of G.

SoassumeH = G.ThenG = K M s o X M = X M n G = X M n K M = (XMn K)M by the Modular Property of Groups, 1.14. Now X and XMn K are Hall n-subgroups of XM, so, by (2), there is y E XM with XY = XM n K 5 K. Thus the proof is complete.

A converse of Phillip Hall's Theorem also holds, as we will see soon. In particular the hypothesis of solvability is necessary to insure the validity of the theorem. Hall's Theorem is a good example of how restrictions on the composition factors of a finite group can lead to significant restrictions of the global structure of the group.

Now to a proof of a converse to Phillip Hall's Theorem. The proof will appeal to Bumside's paqb-~heorem that finite groups G with Jn(G)I = 2 are solvable. The proof of Bumside's Theorem will be postponed until the chapter on character theory.

(18.6) Let G be afinite group possessing aHall pl-subgroup for each p E x(G). Then G is solvable.

Proof. Let G be a minimal counterexample. By 9.8, G is not a p-group. By Burnside's paqa Theorem, 35.13, In(G)J # 2. Thus In(G)I > 2.

Let p E n(G) and H a Hall pl-subgroup of G. By 18.4, H n K is a Hall ql- subgroup of H for each prime q distinct from p and each Hall ql-subgroup of K of G. Therefore H satisfies the hypothesis of the lemma, and hence H is solvable by minimality of G. Let M be a minimal normal subgroup of H. By 9.4, M is an r-group for some prime r. As [n(G)I > 2 there is q E n(G) - (p, r }. Let K be a Hall qf-subgroup of G. As q # r, K contains a Sylow r-subgroup of G. Hence, by Sylow's Theorem, M is contained in some conjugate of K, which wemay take to be K. As q # p, G = HKby 18.4. So, as M 5 H, X = ( M ~ ) = ( M ~ ~ ) = (MK) 5 K. Hence, as subgroups of solvable groups are solvable, X is solvable. Of course X <I G and we conclude from 18.3 that G / X satisfies the hypothesis of the lemma, So, by minimality of G, G/X is solvable. Therefore G is solvable by 9.3.

Coprime action 73

(18.7) (Coprime Action) Let A and H be finite groups with (IA 1, 1HI) = 1.Assume A is represented as a group of automorphisms of H and either A orH is solvable. Let p be a prime. Then

(1) There exists an A-invariant Sylow p-subgroup of H.

(2) CH(A) is transitive on the A-invariant Sylow p-subgroups of G.

(3) Every A-invariant p-subgroup of H is contained in an A-invariantSylow p-subgroup of H.

(4) Let K bean A-invariant normal subgroup of H and H* = H/K. Then

CH*(A) = NH*(A) = CH(A)*.

Proof. Form the semidirect product G of H by A with respect to the represen-tation of A on H, and identify A and H with subgroups of G via the injectionsof 10.1. Then H < G, A is a complement to H in G, and the action of A on His via conjugation.

NG(A) = NG(A) f1AH=ANH(A) by the Modular Property of Groups, 1.14.Also [A, NH(A)] < A n H = 1, so NH(A) = CH(A) and NG(A) = A x CH(A).

Let P E Sylp(H). By a Frattini Argument, 6.3, G = HNG(P). Then bySchur-Zassenhaus there is a complement B to NH(P) in NG(P) and Bg = Afor some g E G. Hence Pg = Q is an A-invariant Sylow p-subgroup of H.NG(Q) is transitive on AG f1 NG(Q) by Schur-Zassenhaus, so, by 5.21, NG(A)is transitive on the A-invariant Sylow p-subgroups of H. As NG (A) = A CH (A)and A < NG(Q), CH(A) is also transitive on the A-invariant Sylow p-sub-groups of H by 5.20.

Let R be a maximal A-invariant p-subgroup of H. If R Sylp(H) then, byExercise 3.2, R Sylp(NH(R)). But NH(R) is A-invariant so by (1) there isan A-invariant Sylow p-subgroup of NH(R), contradicting the maximality ofR. This establishes (3).

Assume the hypothesis of (4) and let X* = CH. (A). We have already shownX * = NH. (A). Now AK<AX and, by Schur-Zassenhaus, AAX = AAK so

by a Frattini Argument, 6.3, AX = KNAX(A). Hence X = KNx(A) by theModular Property of Groups, 1.14. So as CH(A)* < we concludeCH.(A) = X* = Cx (A)* = CH(A)*, completing the proof of (4).

Again the hypothesis that A or G is solvable can be removed from the statementof 18.7, modulo the Odd Order Theorem.

Remarks. The material in section 18 is basic, while that in section 17 is morespecialized. Thus the reader may wish to skip or postpone section 17. If so,17.10 must be assumed in proving the Schur-Zassenhaus Theorem.

Coprime action

(18.7) (Coprime Action) Let A and H be finite groups with (I A 1, I HI) = 1. Assume A is represented as a group of automorphisms of H and either A or H is solvable. Let p be a prime. Then

(1) There exists an A-invariant Sylow p-subgroup of H .

(2) CH(A) is transitive on the A-invariant Sylow p-subgroups of G.

(3) Every A-invariant p-subgroup of H is contained in an A-invariant Sylow p-subgroup of H.

(4) Let K be an A-invariant normal subgroup of H and H* = HIK. Then CH*(A) = NH*(A) = CH(A)*.

Proof. Form the semidirect product G of H by A with respect to the representation of A on H, and identify A and H with subgroups of G via the injections of 10.1. Then H g G, A is a complement to H in G, and the action of A on H is via conjugation.

NG (A) = NG(A) n AH = ANH (A) by the Modular Property of Groups, 1.14. Also [A, NH (A)] 5 A n H = 1, so NH(A) = C,y(A) and NG(A) = A x C,y(A).

Let P E Sylp(H). By a Frattini Argument, 6.3, G = HNG(P). Then by Schur-Zassenhaus there is a complement B to NH(P) in NG(P) and Bg = A for some g E G. Hence Pg = Q is an A-invariant Sylow p-subgroup of H. NG(Q) is transitive on n NG(Q) by Schur-Zassenhaus, so, by 5.21, NG(A) is transitive on the A-invariant Sylow p-subgsoups of H. As NG(A) = AC,y(A) and A 5 NG(Q), CH(A) is also transitive on the A-invariant Sylow p-subgroups of H by 5.20.

Let R be a maximal A-invariant p-subgroup of H . If R 4 Sylp(H) then, by Exercise 3.2, R 4 Syl,(NH(R)). But NH(R) is A-invariant so by (1) there is an A-invariant Sylow p-subgroup of NH(R), contradicting the maximality of R. This establishes (3).

Assume the hypothesis of (4) and let X* = CH* (A). We have already shown X* = NH.(A). Now AK 5 AX and, by Schur-Zassenhaus, = , so by a Frattini Argument, 6.3, AX = KNAx(A). Hence X = KNx(A) by the Modular Property of Groups, 1.14. So as CH(A)* 5 CH*(A), we conclude CH*(A) = X* = Cx(A)* = CH(A)*, completing the proof of (4).

Again the hypothesis that A or G is solvable can be removed from the statement of 18.7, modulo the Odd Order Theorem.

Remarks. The material in section 18 is basic, while that in section 17 is more specialized. Thus the reader may wish to skip or postpone section 17. If so, 17.10 must be assumed in proving the Schur-Zassenhaus Theorem.

Exercises for chapter 61. Let A be a solvable group acting on G = XY with Y a G, X and YA-

invariant, and (IAA, 1GI) = 1. Then CG(A) = CX(A)Cy(A).2. Prove 18.7 with p replaced by a set of primes jr, under the assumption that

H is solvable.3. Let G be the alternating group on a set I of finite order n > 2, let F = GF(2),

let V be the permutation module of the representation on I, and define Zand the core U of V as in Exercise 4.6. Prove(1) 0, Z, U, and V are the only FG-submodules of V. In particular U =

(U + Z)/Z is an irreducible FG-module.(2) If n is odd prove H1(G, U) = 0.(3) If n is even prove V is an indecomposable FG-module, H 1 (G, U) = F,

and V = U(G, U).(Hint: In (2) and (3) let H be the stabilizer of a point x of I and proceedby induction on n. If n is odd prove H centralizes w c U(G, U) - U, andappeal to Exercise 4.6. If n is even prove H centralizes a complement Wto U in U(G, U) and W is a hyperplane of CU(G, U)(Hy) for x 0 y E I. Usethis to conclude dim(W) < 1.)

4. (Alperin-Gorenstein) Let F be a field of characteristic p, G a finite group,V an FG-module, and A a G-invariant collection of p'-subgroups such that:(1) V=[V,X]foreach XEA,and(2) the graph on A obtained by joining X to Y if [X, Y] = 1 is connected.Prove H1(G, V) = 0.

5. Let A be an abelian r-group acting on an r'-group G. Then G = (CG(B):B < A, A/B cyclic).


Exercises for chapter 6 1. Let A be a solvable group acting on G = XY with Y 9 G, X and YA-

invariant, and (I A(, IGI) = 1. Then CG(A) = Cx(A)Cy(A). 2. Prove 18.7 with p replaced by a set of primes n, under the assumption that

H is solvable. 3. Let G be the alternating group on a set I of finite order n > 2, let F = GF(2),

let V be the permutation module of the representation on I, and define Z and the core U of V as in Exercise 4.6. Prove (1) 0, Z, U, and V are the only FG-submodules of V. In particular U =

(U + Z)/Z is an irreducible FG-module. (2) If n is odd prove H'(G, U) = 0. (3) If n is even prove V is an indecomposable FG-module, H'(G, U) 2 F ,

and = U(G, D). (Hint: In (2) and (3) let H be the stabilizer of a point x of I and proceed by induction on n. If n is odd prove H centralizes w E U(G, 0 ) - 0 , and appeal to Exercise 4.6. If n is even prove H centralizes a complement W to U in U(G, 0 ) and W is a hyperplane of CLI(G,~)(Hy) for x # y E I. Use this to conclude dim( W) 5 1 .)

4. (Alperin-Gorenstein) Let F be a field of characteristic p, G a finite group, V an FG-module, and A a G-invariant collection of pf-subgroups such that: (1) V = [V, XI for each X E A, and (2) the graph on A obtained by joining X to Y if [X, Y] = 1 is connected. Prove H'(G, V) = 0.

5. Let A be an abelian r-group acting on an rf-group G. Then G = (CG(B): B ( A, AIBcyclic).

7

Spaces with forms

Chapter 7 considers pairs (V, f) where V is a finite dimensional vector spaceover a field F and f is a nontrivial sesquilinear, bilinear, or quadratic form onV. We'll be primarily interested in the situation where Aut(V, f) is large; inthat event f satisfies one of several symmetry conditions (cf. Exercises 7.9,7.10, 9.1, and 9.9). Under suitable restrictions on F, such pairs are determinedup to isomorphism in sections 19, 20, and 21. For example if F is finite theisomorphism types are listed explicitly in section 21.

It turns out that such spaces satisfy the Witt property: that is, if X and Y aresubobjects of (V, f) and a: X -> Y is an isomorphism, then a extends to anautomorphism of (V, f ). As a result the representation of Aut(V, f) on (V, f )is particularly useful in investigating Aut(V, f).

The groups Aut(V, f), certain normal subgroups of these groups, and theirimages under the projective map of section 13 are called the classical getups.Section 22 derives various properties of the classical groups. For example forsuitable fields they are essentially generated by their transvections or reflec-tions, and are essentially perfect. It will develop much later in section 41 thatif G is a perfect finite classical group then the projective group PG is simple.Conversely the Classification Theorem for finite simple groups says that, bysome measure, most of the finite simple groups are classical groups.

This chapter is one of the longest and most complicated in the book. More-over the material covered here is in some sense specialized and tangential tomuch of the other material in this book. Still, as I've indicated, the classi-cal groups and their representations on the associated spaces (V, Q) are veryimportant, so the effort seems warranted.

19 Bilinear, sesquilinear, and quadratic formsIn this section V is an n-dimensional vector space over a field F and 0 isan automorphism of F. A sesquilinear form on V with respect to 9 is a mapf: V x V->Fsuch that, forallx,y,zeV and all a E F:

.f (x + y, z) = .f (x, z) + .f (y, z) .f (ax, y) = of (x, y)

.f (x, y + z) = .f (x, y) + f (x, z) .f (x, ay) = ae .f (x, y)

The form f is said to be bilinear if 0 = 1. Usually I'll write (x, y) for f (x, y).

Spaces with forms

Chapter 7 considers pairs (V, f ) where V is a finite dimensional vector space over a field F and f is a nontrivial sesquilinear, bilinear, or quadratic form on V. We'll be primarily interested in the situation where Aut(V, f ) is large; in that event f satisfies one of several symmetry conditions (cf. Exercises 7.9, 7.10,9.1, and 9.9). Under suitable restrictions on F , such pairs are determined up to isomorphism in sections 19, 20, and 21. For example if F is finite the isomorphism types are listed explicitly in section 21.

It turns out that such spaces satisfy the Witt property: that is, if X and Y are subobjects of (V, f ) and a: X -+ Y is an isomorphism, then a extends to an automorphism of (V, f ). As a result the representation of Aut(V, f ) on (V, f ) is particularly useful in investigating Aut(V, f ).

The groups Aut(V, f ), certain normal subgroups of these groups, and their images under the projective map of section 13 are called the classical groups. Section 22 derives various properties of the classical groups. For example for suitable fields they are essentially generated by their transvections or reflections, and are essentially perfect. It will develop much later in section 41 that if G is a perfect finite classical group then the projective group PG is simple. Conversely the Classification Theorem for finite simple groups says that, by some measure, most of the finite simple groups are classical groups.

This chapter is one of the longest and most complicated in the book. More- over the material covered here is in some sense specialized and tangential to much of the other material in this book. Still, as I've indicated, the classical groups and their representations on the associated spaces (V, Q ) are very important, so the effort seems warranted.

19 Bilinear, sesquilinear, and quadratic forms In this section V is an n-dimensional vector space over a field F and 8 is an automorphism of F. A sesquilinear form on V with respect to 8 is a map f : V x V -+ F such that, for all x, y, z E V and all a E F :

f ( x +Y,z ) = f (x , Z) + f(y,z) f(ax, y)=af(x, y)

f (x9 Y + z) = f (x, Y) + f (x, z) f (x, ay) = ae f (x, y).

The form f is said to be bilinear if 8 = 1. Usually I'll write (x, y) for f (x, y).

76 Spaces with forms

I'll always assume 0 is of order at most 2. There is little loss of generalityin this assumption since we are interested in forms with big symmetry groups.Exercises 7.10 and 9.9 make this comment more precise.

f is symmetric if f is bilinear and f (x, y) = f (y, x) for all x, y in V. fis skew symmetric if f is bilinear and f (x, y) _ -f (y, x) for all x, y in V.Finally f is hermitian symmetric if 0 is an involution and f (x, y) = f (y, x)8for all x, y in V. I'll always assume that f has one of these three symmetryconditions. One consequence of this assumption is that

(*) For all x, y in V, f(x,y)=0 if and only if f(y,x)=0.

On the other hand if (*) holds then Exercise 7.10 shows that f (essentially)satisfies one of the three symmetry conditions. Further if our form has a biggroup of automorphisms then Exercises 7.9 and 9.1 show we may as well takef to satisfy one of the conditions.

If f (x, y) = 0 I'll write x 1 y and say that x and y are orthogonal. For X C Vdefine

X'={vEV:x±v forallxEX}

and observe that X 1- is a subspace of V and that X 1- = (X )1. Indeed

(19.1) For X E V, x1- = ker(a), where a E HomF(V, F) is defined by ya =(y, x). Hence dim(x1-) > n - 1 with equality precisely when x 0 V1-.

V -L is called the radical of V. Write Rad(V) for V J-. We say f is nondegenerateif Rad(V) = 0.

The form f will be said to be orthogonal if f is nondegenerate and sym-metric, and if in addition, when char(F) = 2, f (x, x) = 0 for all x in V. Theform f is said to be symplectic if f is nondegenerate and skew symmetric, andin addition when char(F) = 2, f (x, x) = 0 for all x in V. Finally f is said tobe unitary if f is nondegenerate and hermitian symmetric.

A few words to motivate these definitions. I've already indicated why thesymmetry assumptions are appropriate. For any space V, V = Rad(V) ® Ufor some subspace U such that the restriction of f to U is nondegenerate.Thus there is little loss in assuming f to be nondegenerate. Besides, fromExercise 9.1, if (V, f) admits an irreducible group of automorphisms then(essentially) f is forced to be nondegenerate. Observe that if char(F) = 2 thensymmetry and skew symmetry are the same. Also, if char(F) 2 and f is skewsymmetric, notice f (x, x) = 0 for all x E V. This motivates the requirement thatf (x, x) = 0 for all x E V when char(F) = 2 and f is orthogonal or symplectic.Indeed Exercise 7.9.3 shows that this assumption leads to little loss of generalityand is always satisfied if (V, f) admits an irreducible group of automorphisms.


I'll always assume 8 is of order at most 2. There is little loss of generality in this assumption since we are interested in forms with big symmetry groups. Exercises 7.10 and 9.9 make this comment more precise.

f is symmetric if f is bilinear and f (x, y) = f (y, x) for all x, y in V. f is skew symmetric if f is bilinear and f (x, y) = - f (y, x) for all x, y in V. Finally f is hermitian symmetric if 8 is an involution and f (x, y) = f (y, x)' for all x , y in V. I'll always assume that f has one of these three symmetry conditions. One consequence of this assumption is that

(* For all x, y in V, f(x, y)=Oif andonly if f(y,x)=O.

On the other hand if (*) holds then Exercise 7.10 shows that f (essentially) .

satisfies one of the three symmetry conditions. Further if our form has a big group of automorphisms then Exercises 7.9 and 9.1 show we may as well take f to satisfy one of the conditions.

Iff (x, y) = 0 I'll write x I y and say that x and y are orthogonal. For X 5 V define

and observe that X' is a subspace of V and that X' = (x)'. Indeed

(19.1) For x E V, x' = ker(a), where a E HomF(V, F ) is defined by ya = (y, x). Hence dim(xL) >_ n - 1 with equality precisely when x $ v'.

VL is called the radical of V. Write Rad(V) for v'. We say f is nondegenerate if Rad(V) = 0.

The form f will be said to be orthogonal if f is nondegenerate and symmetric, and if in addition, when char(F) = 2, f (x, x) = 0 for all x in V. The form f is said to be symplectic i f f is nondegenerate and skew symmetric, and in addition when char(F) = 2, f (x, x) = 0 for all x in V. Finally f is said to be unitary i f f is nondegenerate and hermitian symmetric.

A few words to motivate these definitions. I've already indicated why the symmetry assumptions are appropriate. For any space V, V = Rad(V) @ U for some subspace U such that the restriction of f to U is nondegenerate. Thus there is little loss in assuming f to be nondegenerate. Besides, from Exercise 9.1, if (V, f ) admits an irreducible group of automorphisms then (essentially) f is forced to be nondegenerate. Observe that if char(F) = 2 then symmetry and skew symmetry are the same. Also, if char(F) # 2 and f is skew symmetric, notice f (x, x) = 0 for all x E V. This motivates the requirement that f (x, x) = 0 for all x E V when char(F) = 2 and f is orthogonal or symplectic. Indeed Exercise 7.9.3 shows that this assumption leads to little loss of generality and is always satisfied if (V, f ) admits an irreducible group of automorphisms.

Bilinear, sesquilinear, and quadratic forms 77

(19.2) Let f be nondegenerate and U < V. Then dim(U1) = codim(U).

Proof. The proof is by induction on m = dim(U). The lemma is trivial if m = 0,so take m > 0. Then as f is nondegenerate there exists x E V - U1. By 19.1,W = U fl x1 is a hyperplane of U. By induction on m, dim(W1) = n - m + 1.Let U E U - W; then U1= W1 fl u1. As x E W1 - u1, U1 is a hyperplaneof W1 by 19.1. So dim(U1)=dim(W1) - 1 = n - m, completing the proof.

A vector x E V is isotropic if f (x, x) = 0. I've already observed that, if f isskew symmetric and char(F) is not 2, then every vector is isotropic. Recall thatthis is part of the defining hypothesis of a symplectic or orthogonal form whenchar(F) = 2,

A subspace U of V is totally isotropic if the restriction of f to U is trivial,or equivalently if U < U1. U is nondegenerate if the restriction of f to U isnondegenerate, or equivalently u fl U1= Rad(U) = 0.

(19.3) Let f be nondegenerate and U a subspace of V. Then(1) U is nondegenerate if and only if V = U ® U1.

(2) (U1)1 = U.(3) If U is totally isotropic then each complement to U in U1 is nondegen-

erate.

(4) If U is totally isotropic then dim(U) < n/2.

Proof. Parts (1), (2), and (4) are easy consequences of 19.2, while (2) im-plies (3).

Assume f is symmetric. A quadratic form on V associated to f is a mapQ: V -* F such that, for all x, y E V and a E F, Q(ax) = a2Q(x) and

Q(x + Y) = Q(x) + Q(Y) + f (x, Y).

Observe that if char(F) 2 this definition forces Q(x) = f (x, x)/2, so thequadratic form is uniquely determined by f, and hence adds no new infor-mation. On the other hand, if char(F) = 2 there are many quadratic formsassociated to f. Observe also that f is uniquely determined by Q, since

f(x, Y) = Q(x +Y) - Q(x) - Q(y)

A symplectic space (V, f) is a pair consisting of a vector space V and asymplectic form f on V. A unitary space is a pair (V, f) with f a unitaryform. An orthogonal space is a pair (V, Q) where Q is a quadratic form on Vwith associated bilinear form f.

(19.2) Let f be nondegenerate and U 5 V. Then dim(^') = codim(U).

Proof. The proof is by induction on m = dim(U). The lemma is trivial if m = 0, so take m > 0. Then as f is nondegenerate there exists x E V - u'. By 19.1, W = U n x' is a hyperplane of U. By induction on m, dim(wL) = n - m + 1. Let u E U - W; then U' = W' n u'. As x E W' - u', U' is a hyperplane of W' by 19.1. So dim(^') = dim(wL) - 1 = n - m, completing the proof.

A vector x E V is isotropic if f (x, x) = 0. I've already observed that, if f is skew symmetric and char(F) is not 2, then every vector is isotropic. Recall that this is part of the defining hypothesis of a symplectic or orthogonal form when char(F) = 2,

A subspace U of V is totally isotropic if the restriction of f to U is trivial, or equivalently if U 5 u'. U is nondegenerate if the restriction of f to U is nondegenerate, or equivalently U n U' = Rad(U) = 0.

(19.3) Let f be nondegenerate and U a subspace of V. Then (1) U is nondegenerate if and only if V = U @ u'. (2) (u')' = U. (3) If U is totally isotropic then each complement to U in U' is nondegen-

erate. (4) If U is totally isotropic then dim(U) n/2.

Proof. Parts (I), (2), and (4) are easy consequences of 19.2, while (2) implies (3).

Assume f is symmetric. A quadratic form on V associated to f is a map Q: V + F such that, for all x, y E V and a E F, Q(ax) = a2e (x ) and

Observe that if char(F) # 2 this definition forces Q(x) = f (x, x)/2, so the quadratic form is uniquely determined by f , and hence adds no new information. On the other hand, if char(F) = 2 there are many quadratic forms associated to f . Observe also that f is uniquely determined by Q, since

A symplectic space (V, f ) is a pair consisting of a vector space V and a symplectic form f on V. A unitary space is a pair (V, f ) with f a unitary form. An orthogonal space is a pair (V, Q) where Q is a quadratic form on V with associated bilinear form f .

In the remainder of this section assume (V, f) is a symplectic or unitaryspace, or Q is a quadratic form on V with associated orthogonal form fand (V, Q) is an orthogonal space. The type of V is symplectic, unitary, ororthogonal, respectively.

A vector v in V is singular if v is isotropic and also Q(v) = 0 when V isan orthogonal space. A subspace U of V is totally singular if U is totallyisotropic and also each vector of U is singular. The Witt index of V is themaximum dimension of a totally singular subspace of V. Notice 19.3 says theWitt index of V is at most n/2.

An isometry of spaces (V, f) and (U, g) is a nonsingular linear transfor-mation a: V -* U such that g(xa, ya) = f (x, y) for all x, y E V. A similar-ity is a nonsingular linear transformation a: V -+ U such that g(xa, ya) =.l(a) f (x, y) for all x, y E V and some A(a) E F# independent of x and y. If(V, Q) and (U, P) are orthogonal I'll also require P(xa) = Q(x) or P(xa) =X(a) Q(x), in the respective case. Forms f and g (or P and Q) on V are said tobe equivalent if (V, f) and (V, g) (or (V, Q) and (U, P)) are isometric. f andg (or P and Q) are similar if the corresponding spaces are similar. O(V, f)(or O(V, Q)) denotes the group of isometries of the space, while O(V, f) (orA(V, Q)) denotes the group of similarities. Evidently O(V, f) a L(V, f).

Let X = (xi : 1 < i < n) be a basis of V. Define J = J(X, f) to be the n by nmatrix J = (Jib) with Ji1 = f (xi, xj). Observe that J uniquely determines theform f .

Suppose Y = (yi : 1 j aijxj,aiJ E F, and A = (aid). Set AB = (ar) and let AT be the transpose of A. ObserveAOT = ATO and J(Y, f) = AJATB. Further

(19.4) A form g on V is similar to f if and only if there exists a basisY = (yi : 1 < i < n) of V with J(Y, g) = AJ(X, f) for some X E F#. Equiva-lence holds precisely when A can be chosen to be 1. If Q and P are quadraticforms associated to f and g, respectively, then Q is similar to P precisely whenY can be chosen so that J(Y, g)=).J(X, f) and P(yi)=).Q(xi) for each i,with X = 1 in case of equivalence.

Proof. Let a: (V, f) -* (V, g) be a similarity and let yi = xi a and Y = (yi : 1 <i < n). Then J(Y, g) =.l(a)J(X, f). Conversely if J(Y, g) _ ,kJ(X, f) leta E GL(V) with xia = yi. Then a is a similarity with A(a) =X. Of course thesame arguments extend to quadratic forms.

(19.5) A form g on V is similar to f if and only if J(X, g) _ ,kA J (X, f )AT O forsome nonsingular matrix A and some A E F#, with X = 1 in case of equivalence.


In the remainder of this section assume (V, f ) is a symplectic or unitary space, or Q is a quadratic form on V with associated orthogonal form f and (V, Q) is an orthogonal space. The type of V is symplectic, unitary, or orthogonal, respectively.

A vector v in V is singular if v is isotropic and also Q(v) = 0 when V is an orthogonal space. A subspace U of V is totally singular if U is totally isotropic and also each vector of U is singular. The Witt index of V is the maximum dimension of a totally singular subspace of V. Notice 19.3 says the Witt index of V is at most n/2.

An isometry of spaces (V, f ) and (U, g) is a nonsingular linear transformation a: V + U such that g(xa, ya) = f (x, y) for all x, y E V. A similarity is a nonsingular linear transformation a: V + U such that g(xa, ya) = h(a) f (x, y) for all x, y E V and some h(a) E F' independent of x and y. If (V, Q) and (U, P ) are orthogonal I'll also require P(xa) = Q(x) or P(xa) = h(a)Q(x), in the respective case. Forms f and g (or P and Q) on V are said to be equivalent if (V, f ) and (V, g) (or (V, Q) and (U, P)) are isometric. f and g (or P and Q) are similar if the corresponding spaces are similar. O(V, f ) (or O(V, Q)) denotes the group of isometrics of the space, while A(V, f ) (or A(V, Q)) denotes the group of similarities. Evidently O(V, f ) 9 A(V, f ).

LetX=(xi: l p i p n)beabasisofV.Define J = J(X, f ) tobe thenbyn matrix J = (Jij) with Jij = f (xi, xi). Observe that J uniquely determines the form f .

Suppose Y = (yi: 1 p i p n) is a second basis for V, let yi = xi aijxj, aij E F , and A = (aij). Set Ae = ( a t ) and let AT be the transpose of A. Observe AeT =ATe and J(Y, f ) = A J A ~ ' . Further

(19.4) A form g on V is similar to f if and only if there exists a basis Y = (yi: 1 5 i 5 n) of V with J(Y, g) = h J(X, f ) for some h E F'. Equiva- lence holds precisely when h can be chosen to be 1. If Q and P are quadratic forms associated to f and g, respectively, then Q is similar to P precisely when Y can be chosen so that J(Y, g) = hJ(X, f ) and P(yi) = hQ(xi) for each i , with h = 1 in case of equivalence.

Proof. Leta:(V, f)+(V,g)beasimilarityandletyi=xiaandY=(yi: 1 p i p n). Then J(Y, g) = h(a) J(X, f ). Conversely if J(Y, g) = h J(X, f ) let a E GL(V) with xia = yi. Then a is a similarity with h(a) = h. Of course the same arguments extend to quadratic forms.

(19.5) A form g on V is similar to f if and only if J(X, g) = hA J(X, f )ATe for some nonsingular matrix A and some h E F', with h = 1 in case of equivalence.

Bilinear, sesquilinear, and quadratic forms 79

Proof. This follows from 19.4 and the remark immediately preceding it.

One can see from the preceding discussion that equivalence of forms corre-sponds to equivalence of the associated defining matrices of the forms.

Given a E GL(V) define MX(a) to be the n by n matrix (aij) defined byxia = Ej ajjxj.

(19.6) Let a E GL(V). Then a E A(V, f) if and only if (xia, xja) =.l(a)(x1, xj) for all i and j, and some .l(a) E F#, with a E O(V, f) precisely whenA(a) =1. If (V, Q) is orthogonal then a E A(V, Q) if and only if a E A(V, f)and Q(x;a) = A(a)Q(x;) for each i, with A(a) =1 for equivalence.

(19.7) Let a E GL(V), A = Mx (a), and J = J(X, f ). Then a E O(V, f) ifand only if J = AJATB. U E A(V, f) if and only if AJ = AJATB for someA E P.

(19.8) If V is not a symplectic space then V contains a nonsingular vector.

Proof. Assume otherwise. Letx E V# and Y E V - x1. Then 1 = ((x, y)-lx, y),so without loss (x, y) = 1. Now (ax + by, ax + by) = abe + bae as x and yare singular and (x, y) = 1. If char(F) # 2 take a = b =1 to get ax + by non-singular. If char(F) = 2 and V is unitary take a = 1 and b # be. Finally if V isorthogonal and char(F) = 2 then as x and y are singular, Q(x) = Q(y) = 0, soQ(x + y) _ (x, y) = 1, and hence x + y is nonsingular.

(19.9) Assume V is not symplectic, and if V is orthogonal assume char(F) # 2.Then there exists a basis X = (x1: 1 < i < n) of V such that the members of Xare nonsingular and distinct members are orthogonal.

Proof. By 19.8there is anonisotropicvector x1EV.By 19.3,V=(xl)®(x1)1with (x1)1 nondegenerate. By induction on n there is a corresponding basis(xi:l <i <n)of(xl)1

A basis like the one of 19.9 will be termed an orthogonal basis. An orthonormalbasis for V is a basis X such that J(X, f) = I.

(19.10) If V is unitary then (x, x) is in the fixed field Fix(O) of 0 for eachxEV.

Bilineal; sesquilineal; and quadratic forms 79

Proof. This follows from 19.4 and the remark immediately preceding it.

One can see from the preceding discussion that equivalence of forms corresponds to equivalence of the associated defining matrices of the forms.

Given a E GL(V) define Mx(a) to be the n by n matrix (aij) defined by xia = C aijxj.

(19.6) Let a E GL(V). Then a E A(V, f ) if and only if (xis, xja) = h(a) (xi, xi) for all i and j, and some h(a) E F#, with a E O(V, f ) precisely when h(a) = 1. If (V, Q) is orthogonal then a E A(V, Q) if and only if a E A(V, f ) and Q(xia) = h(a)Q(xi) for each i, with h(a) = 1 for equivalence.

(19.7) Let a E GL(V), A = Mx(a), and J = J(X, f). Then a E O(V, f ) if and only if J = A JA~O. a E A(V, f ) if and only if h J = A J A ~ ' for some h E F#.

(19.8) If V is not a symplectic space then V contains a nonsingular vector.

Proof. Assumeotherwise.Letx E and E V - x L . ~ h e n 1 = ((x, y)-lx, y), so without loss (x, y) = 1. Now (ax + by, ax + by) =abs + bas as x and y are singular and (x, y) = 1. If char(F) # 2 tdke a = b = 1 to get ax + by nonsingular. If char(F) = 2 and V is unitary take a = 1 and b # be. Finally if V is orthogonal and char(F) = 2 then as x and y are singular, Q(x) = Q(y) = 0, so Q(x + y) = (x, y) = 1, and hence x + y is nonsingular.

(19.9) Assume V is not symplectic, andif V is orthogonal assume char(F) # 2. Then there exists a basis X = (xi : 1 5 i n) of V such that the members of X are nonsingular and distinct members are orthogonal.

Proof. By 19.8 thereisanonisotropicvectorxl E V. By 19.3, V = (xl)@(xl)' with (xl) l nondegenerate. By induction on n there is a corresponding basis (xi: 1 < i 5 n) of (x1lL.

A basis like the one of 19.9 will be termed an orthogonal basis. An orthonormal basis for V is a basis X such that J(X, f ) = I.

(19.10) If V is unitary then (x, x) is in the fixed field Fix(8) of 8 for each XEV.

(19.11) Assume V is not symplectic, and if V is orthogonal assume char(F) 0 2.Assume further that the fixed field Fix(9) of 0 satisfies Fix(O) _ {aae: a E F).Then

(1) V possesses an orthonormal basis.(2) All forms on V of each of the prescribed types are equivalent.

Proof. Notice (1) and 19.4 imply (2). To prove (1), choose an orthogonal basisX as in 19.9. Then by hypothesis and Lemma 19.10, (xi, xi) = ail+e> for somea, E F#. Now replacing xi by aixi, we obtain our orthonormal basis.

V is a hyperbolic plane if n = 2 and V possesses a basis X = (x1, x2) such thatxl and x2 are singular and (xl, x2) = 1. Such a basis will be termed a hyperbolicpair.

(19.12) Let x E V be singular and y c V - x1. Then (x, y) = U is a hyper-bolic plane and x is contained in a hyperbolic pair of U.

Proof. Let b = (x, y)-B. Then (x, by) = 1, so without loss (x, y) = 1. ObserveU is nondegenerate, so if y is singular we are done. Thus we may assume eachmember of U - (x) is nonsingular, so in particular V is not symplectic. Thus,unless V is orthogonal and char(F) = 2, 0 0 (ax +y, ax + y) = a +a" + (y, Y).However if char(F) 0 2 we may take a = -(y, y)/2, and use 19.10 to obtaina contradiction.

Thus char(F) = 2. Suppose V is unitary. Let d E F - Fix(O). Then e = d +dB 0 0. Let c = (y, y)/e and a = cd. By 19.10, c c Fix(O), so a + ae = ce =(y, y), and hence ax + y is singular.

This leaves the case V orthogonal. Then choosing a = Q(y), ax + y is sin-gular, completing the proof.

Here's an immediate corollary to 19.12 and 19.4:

(19.13) Let dim(V) = 2. If V# possesses a singular vector, then V is a hyper-bolic plane. In particular, up to equivalence, there is a unique nondegenerateform on V of each type possessing a nontrivial singular vector.

(19.14) Let U be a totally singular subspace of V, R = (ri : 1 < i < m) a ba-sis for U, and W a complement to U in U1. Then there exists S = (si: 1 <i < m) : V such that ri, si is a hyperbolic pair for the hyperbolic planeUi = (ri, si) and W1 is the orthogonal direct sum of the planes (Ui: 1 <i < m).


(19.11) Assume V is not symplectic, and if V is orthogonal assume char(F) # 2. Assume further that the fixed field Fix(@) of @ satisfies Fix(@) = {aas: a E F] . Then

(1) V possesses an orthonormal basis. (2) All forms on V of each of the prescribed types are equivalent.

Proof. Notice (1) and 19.4 imply (2). To prove (I), choose an orthogonal basis X as in 19.9. Then by hypothesis and Lemma 19.10, (xi, xi) =a:('+') for some ai E F#. Now replacing xi by aixi, we obtain our orthonormal basis.

V is a hyperbolic plane if n = 2 and V possesses a basis X = (XI, x2) such that xl and x2 are singular and (xl , x2) = 1. Such a basis will be termed a hyperbolic pair.

(19.12) Let x E V# be singular and y E V - x'. Then ( x , y) = U is a hyperbolic plane and x is contained in a hyperbolic pair of U.

Proof. Let b = (x, y)-s. Then (x, by) = 1, so without loss (x, y) = 1. Observe U is nondegenerate, so if y is singular we are done. Thus we may assume each member of U - (x) is nonsingular, so in particular V is not symplectic. Thus, unless V is orthogonal and char(F) = 2 ,0 # (ax + y, ax + y) = a +as +(y, y). However if char(F) # 2 we may take a = -(y, y)/2, and use 19.10 to obtain a contradiction.

Thus char(F) = 2. Suppose V is unitary. Let d E F - Fix(@). Then e = d + ds # 0. Let c = (y, y)/e and a = cd. By 19.10, c E Fix(@), so a + as = ce = (y, y), and hence ax + y is singular.

This leaves the case V orthogonal. Then choosing a = Q(y), ax + y is singular, completing the proof.

Here's an immediate corollary to 19.12 and 19.4:

(19.13) Let dim(V) = 2. If V# possesses a singular vector, then V is a hyperbolic plane. In particular, up to equivalence, there is a unique nondegenerate form on V of each type possessing a nontrivial singular vector.

(19.14) Let U be a totally singular subspace of V, R = (ri: 1 5 i 5 m) a basis for U, and W a complement to U in u'. Then there exists S = (si: 1 5 i 5 m) V such that ri, si is a hyperbolic pair for the hyperbolic plane Ui = (ri, si) and W' is the orthogonal direct sum of the planes (Ui: 1 5 i 5 m).

Witt's Lemma 81

Proof. By 19.3 we may take W = 0. Thus U = U1. Let Uo = (r,: 1 < i < m).By 19.2, U is a hyperplane of (Uo)'. Then there exists a complement Ul =(r1, Si) to Uo in Uo and by 19.3, U1 is nondegenerate. By 19.12 we may assumerl, s1 is a hyperbolic pair for U1. By 19.3, V = U1 ® Ul L. Finally by inductionon m we may choose (s1: 1 < i < m) in (U1)1 to satisfy the lemma.

Define V to be hyperbolic if V is the orthogonal direct sum of hyperbolic planes.A hyperbolic basis for a hyperbolic space V is a basis X = (x1: 1 < i < m)such that V is the orthogonal sum of the hyperbolic planes (x2i_1, x20 withhyperbolic pair x2i_1, xt. We say V is definite if V possesses no nontrivialsingular vectors.

As a consequence of 19.14 and 19.3 we have:

(19.15) Let U be a maximal hyperbolic subspace of V. Then V = U ® U1 andU1 is definite. Moreover every totally singular subspace of V of dimension mis contained in a hyperbolic subspace of dimension 2m and Witt index m.

(19.16) All symplectic spaces are hyperbolic. In particular all symplecticspaces are of even dimension and, up to equivalence, each space of even di-mension admits a unique symplectic form.

11Proof. This is immediate from 19.15, 19.4, ,and the fact that all vectors in asymplectic space are singular.

If char(F) = 2 and (V, Q) is orthogonal then (V, f) is symplectic, where f isthe bilinear form determined by Q. Hence by 19.16:

(19.17) If V is orthogonal and char(F) = 2, then V is of even dimension.

20 Witt's LemmaThis section is devoted to a proof of Witt's Lemma. I feel Witt's Lemma isprobably the most important result in the theory of spaces with forms. Hereit is:

Witt's Lemma. Let V be an orthogonal, symplectic, or unitary space. LetU and W be subspaces of V and suppose a: U W is an isometry. Then aextends to an isometry of V.

Before proving Witt's Lemma let me interject an aside. Define an object X ina category i' to possess the Witt property if, whenever Y and Z are subobjects

Witt 's Lemma

Proof. B~ 19.3 we may take w = 0. Thus u = u'. ~ e t Uo = (r,: 1 < i 5 m). By 19.2, U is a hyperplane of ( ~ 0 ) ~ . Then there exists a complement U1 = (r l , s l ) to Uo in U: and by 19.3, U1 is nondegenerate. By 19.12 we may assume rl , sl is a hyperbolic pair for U1. By 19.3, V = U1 @ u:. Finally by induction on m we may choose (si: 1 < i _( m) in (~1 ) ' to satisfy the lemma.

Define V to be hyperbolic if V is the orthogonal direct sum of hyperbolic planes. A hyperbolic basis for a hyperbolic space V is a basis X = (xi: 1 5 i 5 m) such that V is the orthogonal sum of the hyperbolic planes (x2i-1, x2i) with hyperbolic pair x2i-1, xi. We say V is dejinite if V possesses no nontrivial singular vectors.

As a consequence of 19.14 and 19.3 we have:

(19.15) Let U be a maximal hyperbolic subspace of V. Then V = U @ U' and U' is definite. Moreover every totally singular subspace of V of dimension m is contained in a hyperbolic subspace of dimension 2m and Witt index m.

(19.16) All symplectic spaces are hyperbolic. In particular all sy mplectic spaces are of even dimension and, up to equivalence, each space of even dimension admits a unique symplectic form.

Proof. This is immediate from 19.15, 19.4,'and the fact that all vectors in a symplectic space are singular.

If char(F) = 2 and (V, Q) is orthogonal then (V, f ) is symplectic, where f is the bilinear form determined by Q. Hence by 19.16:

(19.17) If V is orthogonal and char(F) = 2, then V is of even dimension.

20 Witt's Lemma This section is devoted to a proof of Witt's Lemma. I feel Witt's Lemma is probably the most important result in the theory of spaces with forms. Here it is:

Witt's Lemma. Let V be an orthogonal, symplectic, or unitary space. Let U and W be subspaces of V and suppose a : U + W is an isometry. Then a extends to an isometry of V.

Before proving Witt's Lemma let me interject an aside. Define an object X in a category 6? to possess the Wittproperty if, whenever Y and Z are subobjects

of X and a: Y -+ Z is an isomorphism, then a extends to an automorphism ofX. Witt's Lemma says that orthogonal spaces, symplectic spaces, and unitaryspaces have the Witt property in the category of spaces with forms and isome-tries. All objects in the category of sets and functions have the Witt property.But in most categories few objects have the Witt property; those that do arevery well behaved indeed. If X is an object with the Witt property and G isits group of automorphisms, then the representation of G on X is usually anexcellent tool for studying G.

Now to the proof of Witt's Lemma. Continue the hypothesis and notation ofthe previous section. The proof involves a number of steps. Assume the lemmais false and let V be a counterexample with n minimal.

(20.1) Let H < U and suppose a I H extends to an isometry P of V. Theny = a$-1: U -+ WEB-' is an isometry with YIH = 1, and a extends to anisometry of V if and only if y does.

(20.2) Assume 0: H < U with H nondegenerate. Then(1) If H1 (Ha)1 then a extends to an isometry of V.(2) If Ha = H then a extends to an isometry of V.

Proof. Notice (1) implies (2). As H is nondegenerate, so is Ha, and V =H ®H1 = Ha ®(Ha)1. Let $: H1 -+ (Ha)1 bean isometry. By minimalityof n, (ajunH±)$-1 extends to an isometry y of H1. Then y$: H1 -+ (Ha)1is an isometry extending a I unH± and a I H + y,B is an isometry of V exten-ding a.

(20.3) If H is a totally singular subspace of Rad(U) and K a complement toH in Rad(U), then there exist subspaces U' and W' of V with K = Rad(U')and U < U' such that a extends to an isometry a: U' -+ W'. If U = H1 thenU'=V.

Proof. Let (r1: 1 < i < m) be a basis for H, X a complement to H in H1containing K, and X' a complement to Ha in (Ha)1 containing (X fl U)a.By 19.14 there is (s;: 1 < i < m) and (s': 1 < i < m) such that X 1 and (X')1are the orthogonal sum of hyperbolic planes (ri, s;) and (r; a, s'), respectively.Extend a to

U'= (U,si:1 <i <m)

by defining s;a = s'.


of X and a: Y + Z is an isomorphism, then a extends to an automorphism of X. Witt's Lemma says that orthogonal spaces, symplectic spaces, and unitary spaces have the Witt property in the category of spaces with forms and isometries. All objects in the category of sets and functions have the Witt property. But in most categories few objects have the Witt property; those that do are very well behaved indeed. If X is an object with the Witt property and G is its group of automorphisms, then the representation of G on X is usually an excellent tool for studying G .

Now to the proof of Witt's Lemma. Continue the hypothesis and notation of the previous section. The proof involves a number of steps. Assume the lemma is false and let V be a counterexample with n minimal.

(20.1) Let H 5 U and suppose a / H extends to an isometry /3 of V. Then y = a/3-': U + W/3-' is an isometry with y J H = 1, and a extends to an isometry of V if and only if y does.

(20.2) Assume 0 # H 5 U with H nondegenerate. Then (1) If H' G ( ~ a ) ' then a extends to an isometry of V. (2) If Ha = H then a extends to an isometry of V.

Proof. Notice ( 1 ) implies (2). As H is nondegenerate, so is Ha, and V = H @I H' = Ha @ I ( H ~ ) ' . Let /3: H' + ( ~ a ) ' be an isometry. By minimality of n, (alunH~)/3-' extends to an isometry y of H'. Then yB: HI + ( ~ a ) ' is an isometry extending alunHl and a l ~ + y/3 is an isometry of V extending a .

(20.3) If H is a totally singular subspace of Rad(U) and K a complement to H in Rad(U), then there exist subspaces U' and W' of V with K = Rad(U1) and U 5 U' such that a extends to an isometry a: U' W'. If U = H' then U' = v.

Proof. Let (ri: 1 5 i 5 m) be a basis for H, X a complement to H in H' containing K, and X' a complement to Ha in ( ~ a ) ' containing (X f l U)a. By 19.14 there is (si: 1 5 i 5 m) and (sj: 1 5 i 5 m) such that X' and (x')' are the orthogonal sum of hyperbolic planes (ri, si) and (ria, s,'), respectively. Extend a to

U'= (U,si: 1 i i _( m)

by defining sia = si.

Witt's Lemma 83

(20.4) V is not symplectic.

Proof. By 20.3 we may assume U is nondegenerate. As U = W, dim(U) =dim(W), so dim(U1) = dim(W1-). Hence, by 19.16, U1- = W1-. Then 20.2contradicts the choice of V as a counterexample.

(20.5) If there exists a totally singular subspace 0: H = Ha of Rad(U) thena extends to V.

Proof. Let L = H1 and L = L/H. Then f (or Q) induces a form f of typef (or Q) on L defined by f (x, y) = f (x, y) and the induced map a: U W

is an isometry, so, by minimality of n, it extends to an isometry P of L. LetX be a basis of L with x fl H and x fl U bases for H and U, respectively,and let P e GL(L) be a map with 8ju = a and xj8 = xp for x e X - U. Byconstruction

(x, Y) = (x, Y) = (0, 0) = (xp, YP)

for x, y e X, so ,B is an isometry of L. Now, by 20.3, ,B, and hence also a,extends to an isometry of V.

(20.6) Assume H is a hyperplane of U with aIH = 1. Assume also that H = 0if V is unitary or char(F) # 2. Then a extends to an isometry of V.

Proof. Let u e U - H and set K = U + W. Assume a does not extend.Suppose U = W. Then Rad(U) 0 0 by 20.2, and as a acts on Rad(U),

Rad(U) is not totally singular by 20.5. Thus char(F) = 2 and V is orthogonal.As a does not extend, a I u # 1, so u 0 ua. Now ua = au +h, for some a E F#and h E H. As a IH = 1, a acts on X = (u, h). By 20.2, Rad(X) 0. Hence aseach member of V# is isotropic (because V is orthogonal and char(F) = 2), Xis totally isotropic. Hence as Q(u) = Q(ua), z = u + ua is singular. Thereforeeither X is totally singular or (z) is the unique singular point in X, and henceis a-invariant. By 20.2, H contains no nondegenerate subspaces, so f I H =0by 19.12. Thus h E Rad(U), so, by 20.5, h is nonsingular. So (z) _ (za) issingular and z H. Hence we may assume z = u. Again by 20.5 there is h' EH -z1. Now a acts on X'= (h', z) and, as X' is nondegenerate, 20.2 supplies acontradiction.

So U : W. Let c = 1 if (u, ua) = 0 and c = (u, ua)B/(u, ua) otherwise.Observe we can extend a to an isometry a' of K with (ua)a' = cu, by 19.6.So, by the first argument in the previous paragraph, char(F) = 2 and V isorthogonal. By definition of a', a' fixes z = u + ua. Now H'= (H, z) is a

Witt 's Lemma

(20.4) V is not symplectic.

Proof. By 20.3 we may assume U is nondegenerate. As U g W, dim(U) = dim(W), so dim(^') = dim(wL). Hence, by 19.16, U' E w'. Then 20.2 contradicts the choice of V as a counterexample.

(20.5) If there exists a totally singular subspace 0 # H = H a of Rad(U) then a extends to V.

Proof. Let L = H' and E = LIH. Then f (or Q) induces a form f of type f (or Q) on defined by f ( f , ji) = f (x, y) and the induced map ti: 0 + w is an isometry, so, by minimality of n, ti extends to an isometry fi of E . Let X be a basis of L with X f l H and X n U bases for H and U , respectively, and let /3 E GL(L) be a map with /3lu = a and 3 = ff i for x E X - U. By construction

for x, y E X, so /3 is an isometry of L. Now, by 20.3, /3, and hence also a , extends to an isometry of V.

(20.6) Assume H is a hyperplane of U with a l ~ = 1. Assume also that H = 0 if V is unitary or char(F) # 2. Then a extends to an isometry of V.

Proof. Let u E U - H and set K = U + W. Assume a does not extend. Suppose U = W. Then Rad(U) # O by 20.2, and as a acts on Rad(U),

Rad(U) is not totally singular by 20.5. Thus char(F) = 2 and V is orthogonal. Asadoesnotextend,alu # 1,sou # ua.Nowua = au+h,forsomea E F# andh E H.AsaIH = l , a a c t s o n X = (u,h).By20.2,Rad(X)#O.Henceas each member of V# is isotropic (because V is orthogonal and char(F) = 2), X is totally isotropic. Hence as Q(u) = Q(ua), z = u + ua is singular. Therefore either X is totally singular or (z) is the unique singular point in X, and hence is a-invariant. By 20.2, H contains no nondegenerate subspaces, so f I H = 0 by 19.12. Thus h E Rad(U), so, by 20.5, h is nonsingular. So (z) = (za) is singular and z 4 H. Hence we may assume z = u. Again by 20.5 there is h' E H -z'. Now a acts on X' = (h', z) and, as X' is nondegenerate, 20.2 supplies a contradiction.

So U # W. Let c = 1 if (u, ua) = 0 and c = (u, ua)'/(u, ua) otherwise. Observe we can extend a to an isometry a' of K with (ua)al = cu, by 19.6. So, by the first argument in the previous paragraph, char(F) = 2 and V is orthogonal. By definition of a', a' fixes z = u + ua. Now H' = (H, z) is a


hyperplane of K with a' (H' = 1, so, by the previous paragraph, a', and hencealso a, extends to an isometry of V.

(20.7) V is orthogonal and char(F) = 2.

Proof. Assume not. By 20.3, we may take U to be nondegenerate. By 19.8 and20.4 there is a nonsingular point L in U. By 20.6 applied to L in the role ofU, a I L extends to an isometry of V. Then by 20.1 we may take a I L = 1. Butnow 20.2 supplies a contradiction.

We are now in a position to complete the proof of Witt's Lemma. Choose U ofminimal dimension so that an isometry a: U -* W does not extend to V. LetH be a hyperplane of U. By minimality of U, ajH extends to an isometry ofV, so by 20.1 we may take a I H = 1. Now 20.6 supplies a contradiction andcompletes the proof.

I close this section with some corollaries to Witt's Lemma.

(20.8) (1) The isometry group of V is transitive on the maximal totally singularsubspaces of V, and on the maximal hyperbolic subspaces of V.

(2) V is the orthogonal direct sum of a hyperbolic space H and a definitespace. Moreover H is a maximal hyperbolic space and this decomposition isunique up to an isometry of V.

(3) The dimension of a maximal hyperbolic subspace of V is twice the Wittindex of V.

Proof. These remarks are a consequence of Witt's Lemma and 19.15.

(20.9) (1) If K is a quadratic Galois extension of F and NF : K -* F is thenorm of K over F, then (K, NF) is a 2-dimensional definite orthogonal spaceover F.

(2) Every 2-dimensional definite orthogonal space over F is similar to aspace (K, NF) for some quadratic Galois extension K of F.

Proof. If K is a quadratic Galois extension of F then Gal(K/F) = (a) is oforder 2 and NF (a) = aa' for a c K. It is straightforward to prove (K, NF )is a definite orthogonal space.

Next a proof of (2). Let (V, Q) be a definite orthogonal space and {x, y) abasis for V. If char(F) 0 2 then by 19.9 we can choose (x, y) = 0, while ifchar(F) = 2 choose (x, y) = 1. Replacing Q by a scalar multiple if necessary,we can assume Q(x) = 1. Let Q(y) = b and p(t) = t2+t(x, y)+b, so that P is


hyperplane of K with a'(H' = 1, SO, by the previous paragraph, a', and hence also a, extends to an isometry of V.

(20.7) V is orthogonal and char(F) = 2.

Proof. Assume not. By 20.3, we may take U to be nondegenerate. By 19.8 and 20.4 there is a nonsingular point L in U . By 20.6 applied to L in the role of U , a 1 extends to an isometry of V. Then by 20.1 we may take a 1 = 1. But now 20.2 supplies a contradiction.

We are now in a position to complete the proof of Witt's Lemma. Choose U of minimal dimension so that an isometry a: U -+ W does not extend to V. Let H be a hyperplane of U. By minimality of U , a l ~ extends to an isometry of V, so by 20.1 we may take a 1 H = 1. Now 20.6 supplies a contradiction and completes the proof.

I close this section with some corollaries to Witt's Lemma.

(20.8) (1) The isometry group of V is transitive on the maximal totally singular subspaces of V, and on the maximal hyperbolic subspaces of V.

(2) V is the orthogonal direct sum of a hyperbolic space H and a definite space. Moreover H is a maximal hyperbolic space and this decomposition is unique up to an isometry of V.

(3) The dimension of a maximal hyperbolic subspace of V is twice the Witt index of V.

Proof. These remarks are a consequence of Witt's Lemma and 19.15.

(20.9) (1) If K is a quadratic Galois extension of F and N;: K -+ F is the norm of K over F, then (K, N;) is a Zdimensional definite orthogonal space over F.

(2) Every z-dimensional definite orthogonal space over F is similar to a space (K, N;) for some quadratic Galois extension K of F.

Proof. If K is a quadratic Galois extension of F then Gal(K/F) = (u ) is of order 2 and N;(a) = aaa for a E K. It is straightforward to prove (K, N;) is a definite orthogonal space.

Next a proof of (2). Let (V, Q) be a definite orthogonal space and (x, y) a basis for V. If char(F) # 2 then by 19.9 we can choose (x, y) = 0, while if char(F) = 2 choose (x, y) = 1. Replacing Q by a scalar multiple if necessary, we can assume Q(x) = 1. Let Q(y) = b and P(t) = t2 +t(x, y) +b, so that P is

Spaces over finite fields 85

a quadratic polynomial over F. As V is definite, P is irreducible. Let K be thesplitting field for P over F and c c K a root of P. Then the map x i-+ 1, y i-+ cinduces an isometry of (V, Q) with (K, NF ).

(20.10) Assume F is algebraically closed. Then(1) If char(F) 0 2 then, up to equivalence, V admits a unique nondegenerate

quadratic form. Moreover V has an orthonormal basis with respect to thatform.

(2) If char(F) = 2 then V admits a nondegenerate quadratic form if and onlyif n is even. The form is determined up to equivalence and V is a hyperbolicspace with respect to this form.

Proof. Part (1) follows from 19.11. To prove part (2) it suffices by 19.17 and20.8 to take V an orthogonal space of dimension 2 and prove V is not definite.But as F is algebraically closed it possesses no quadratic extensions, so V isnot definite by 20.9.

(20.11) If V is an orthogonal space of dimension at least 2 then V has a non-degenerate 2-dimensional subspace.

Proof. If char(F) 0 2 this is a consequence of 19.9. If char(F) = 2 thenby 19.16 the underlying symplectic space is hyperbolic and hence possessesa hyperbolic plane, which is a nondegenerate subspace of the orthogonalspace V.

21 Spaces over finite fieldsIn this section the hypothesis and notation of section 19 continue; in particularV is an orthogonal, symplectic, or unitary space over F. In addition assume Fis a finite field of characteristic p.

(21.1) Assume n = 2. Then up to equivalence there is a unique nondegeneratedefinite quadratic form Q on V. Further there is a basis X = {x, y} of Vsuch that:

(1) If p is odd then (x, y) = 0, Q(x) = 1, and -Q(y) is a generator of F#.(2) If p = 2 then (x, y) = 1, Q(x) = 1, Q(y) = b, and P(t) = t2 + t + b is

an irreducible polynomial over F.

Proof. By 20.9 and its proof, Q is at least similar to such a form. It is then aneasy exercise to prove forms similar to Q are even equivalent to Q. As F isfinite, it has a unique quadratic extension, so Q is unique by 20.9.2.

a quadratic polynomial over F. As V is definite, P is irreducible. Let K be the splitting field for P over F and c E K a root of P . Then the map x H 1, y H c induces an isometry of (V, Q) with (K, N:).

(20.10) Assume F is algebraically closed. Then (1) If char(F) # 2 then, up to equivalence, V admits a unique nondegenerate

quadratic form. Moreover V has an orthonormal basis with respect to that form.

(2) If char(F) = 2 then V admits a nondegenerate quadratic form if and only if n is even. The form is determined up to equivalence and V is a hyperbolic space with respect to this form.

Proof. Part (1) follows from 19.1 1. To prove part (2) it suffices by 19.17 and 20.8 to take V an orthogonal space of dimension 2 and prove V is not definite. But as F is algebraically closed it possesses no quadratic extensions, so V is not definite by 20.9.

(20.11) If V is an orthogonal space of dimension at least 2 then V has a nondegenerate Zdimensional subspace.

Proof. If char(F) # 2 this is a consequence of 19.9. If char(F) = 2 then by 19.16 the underlying symplectic space i~ hyperbolic and hence possesses a hyperbolic plane, which is a nondegenerate subspace of the orthogonal space V.

21 Spaces over finite fields In this section the hypothesis and notation of section 19 continue; in particular V is an orthogonal, symplectic, or unitary space over F . In addition assume F is a finite field of characteristic p.

(21.1) Assume n = 2. Then up to equivalence there is a unique nondegenerate definite quadratic form Q on V. Further there is a basis X = (x, y) of V such that:

(1) If p is odd then (x, y) = 0, Q(x) = 1, and -Q(y) is a generator of F'. (2) If p = 2 then (x, y) = 1, Q(x) = 1, Q(y) = b, and P(t) = t2 + t + b is

an irreducible polynomial over F.

Proof. By 20.9 and its proof, Q is at least similar to such a form. It is then an easy exercise to prove forms similar to Q are even equivalent to Q. As F is finite, it has a unique quadratic extension, so Q is unique by 20.9.2.

Denote by D = D+ and Q = D_ the (isometry type of the) hyperbolic planeand the 2-dimensional definite orthogonal space over F, respectively. WriteDm Qk for the orthogonal direct sum of m copies of D with k copies of Q.

(21.2) Let F be a finite field. Then(1) Dm is a hyperbolic space of Witt index m.(2) Dii-1 Q is of Witt index m - 1.(3) D2m is isometric to Q2m.(4) Every 2m-dimensional orthogonal space over F is isometric to exactly

one of Dm or Dm-1 Q.

Proof. By construction Dm is hyperbolic and, by 19.3.4, Dm is of Witt indexm. By construction Q is of Witt index 0. Let V = Dm-'Q, Dm-1 = U <V, and Q = U1= W. Let X be a maximal totally singular subspace of U.As dim(X) =m - 1,X1= X ® W. For W E W#, Q(w) # 0, so, for x E XQ(x + w) = Q(w) # 0. Thus X is a maximal totally singular subspaceof X1, so X is also a maximal totally singular subspace of V. Hence, by 20.8,Dm-1 Q is not isometric to Dm, as they have different Witt indices.

Let V be a 2m-dimensional orthogonal space over F. By 20.11 V has anondegenerate plane U. By 21.1 and induction on m, U = D or Q and U1Dm-1 or Dm-2 Q. So V = Dm, Dm-1Q, or Dm-2 Q2. Thus to complete theproof of 21.2 it remains to show Q2 D2. This will follow from 20.8 if we canshow Q2 has a 2-dimensional totally singular subspace. So take U = U1 = Q.Let {x, y} and {u, v} be bases for U and U1, respectively. If char(F) = 2then by 21.1 we may choose (x, y) = (u, v) = 1, Q(x) = Q(u) = 1, andQ(y) = Q(v). Then (x + u, y + v) is a totally singular plane of V. So takechar(F) to be odd. Then by 21.1 we may take x, y, u, and v to be orthogonalwith Q(u) _ - Q(x) and Q(v) _ Q(y). Then again (x + u, y + v) is a totallysingular plane.

If F is finite and n = 2m is even, then 21.2 says that, up to equivalence, thereare exactly two quadratic forms on V, and that the corresponding orthogonalspaces have Witt index m and m - 1, respectively. Define the sign of thesespaces to be +1 and -1, respectively, and write sgn(Q) or sgn(V) for the signof the space. Thus the isometry type of an even dimensional orthogonal spaceover a finite field is determined by its sign. If V is an orthogonal space of odddimension over F, then, by 19.17, char(F) is odd. Let's. look at such spacesnext.

(21.3) Let V be an orthogonal space of odd dimension over a finite field F.Then V possesses a hyperplane which is hyperbolic.


Denote by D = D+ and Q = D- the (isometry type of the) hyperbolic plane and the 2-dimensional definite orthogonal space over F , respectively. Write Dm ek for the orthogonal direct sum of m copies of D with k copies of Q.

(21.2) Let F be a finite field. Then (1) Dm is a hyperbolic space of Witt index m. (2) Dm-' Q is of Witt index m - 1. (3) DZm is isometric to eZm. (4) Every 2m-dimensional orthogonal space over F is isometric to exactly

one of Dm or Dm-l Q.

Proof. By construction Dm is hyperbolic and, by 19.3.4, Dm is of Witt index m. By construction Q is of Witt index 0. Let V 2 Dm-' Q, Dm-' 2 U 5 V, and Q 2 U' = W. Let X be a maximal totally singular subspace of U. As dim(X) =m - 1, X' = x @ W. For w E W', ~ ( w ) # 0, so, for x E X Q(x + w) = Q(w) # 0. Thus X is a maximal totally singular subspace of x', so X is also a maximal totally singular subspace of V. Hence, by 20.8, g m - 1 Q is not isometric to Dm, as they have different Witt indices.

Let V be a 2m-dimensional orthogonal space over F. By 20.1 1 V has a nondegenerate plane U. By 21.1 and induction on m, U 2 D or Q and U' 2 g m - 1 or Dm-'Q. So V 2 Dm, Dm-'Q, or 0"-'Q2. Thus to complete the

proof of 21.2 it remains to show Q2 2 0'. This will follow from 20.8 if we can show Q2 has a2-dimensional totally singular subspace. So take U 2 U' 2 Q. Let (x, y} and (u, v} be bases for U and u', respectively. If char(F) = 2 then by 21.1 we may choose (x, y) = (u, v) = 1, Q(x) = Q(u) = 1, and Q(y) = Q(v). Then (x + u, y + v) is a totally singular plane of V. So take char(F) to be odd. Then by 21.1 we may take x, y , u, and v to be orthogonal with Q(u) = - Q(x) and Q(v) = - Q(y). Then again (x + u, y + v) is a totally singular plane.

If F is finite and n = 2m is even, then 21.2 says that, up to equivalence, there are exactly two quadratic forms on V, and that the corresponding orthogonal spaces have Witt index m and m - 1, respectively. Define the sign of these spaces to be + 1 and - 1, respectively, and write sgn(Q) or sgn(V) for the sign of the space. Thus the isometry type of an even dimensional orthogonal space over a finite field is determined by its sign. If V is an orthogonal space of odd dimension over F , then, by 19.17, char(F) is odd. Let's look at such spaces next.

(21.3) Let V be an orthogonal space of odd dimension over a finite field F . Then V possesses a hyperplane which is hyperbolic.

Denote by D = D+ and Q = D_ the (isometry type of the) hyperbolic planeand the 2-dimensional definite orthogonal space over F, respectively. WriteDm Qk for the orthogonal direct sum of m copies of D with k copies of Q.

(21.2) Let F be a finite field. Then(1) Dm is a hyperbolic space of Witt index m.(2) Dni-1 Q is of Witt index m -1.(3) Den` is isometric to Q2rn.(4) Every 2m-dimensional orthogonal space over F is isometric to exactly

one of D' or D'-' Q.

Proof. By construction Dm is hyperbolic and, by 19.3.4, Dm is of Witt indexm. By construction Q is of Witt index 0. Let V = Dm- IQ, Dm-1 - U <V, and Q = U1= W. Let X be a maximal totally singular subspace of U.As dim(X) =m - 1, Xl = X ® W. For W E W#, Q(w) 0 0, so, for x E XQ(x + w) = Q(w) 0. Thus X is a maximal totally singular subspaceof X', so X is also a maximal totally singular subspace of V. Hence, by 20.8,Dm-1 Q is not isometric to Dm, as they have different Witt indices.

Let V be a 2m-dimensional orthogonal space over F. By 20.11 V has anondegenerate plane U. By 21.1 and induction on m, U - D or Q and Ul -Drn-1 or Dm_2Q. So V - Dm, Dm-1 Q, or Drn-2 Q2. Thus to complete theproof of 21.2 it remains to show Q2 - D2. This will follow from 20.8 if we canshow Q2 has a 2-dimensional totally singular subspace. So take U = Ul = Q.Let (x, y} and (u, v} be bases for U and Ul, respectively. If char(F) = 2then by 21.1 we may choose (x, y) = (u, v) = 1, Q(x) = Q(u) = 1, andQ(y) = Q(v). Then (x + u, y + v) is a totally singular plane of V. So takechar(F) to be odd. Then by 21.1 we may take x, y, u, and v to be orthogonalwith Q(u) _ - Q(x) and Q(v) _ Q(y). Then again (x + u, y + v) is a totallysingular plane.

If F is finite and n = 2m is even, then 21.2 says that, up to equivalence, thereare exactly two quadratic forms on V, and that the corresponding orthogonalspaces have Witt index m and m - 1, respectively. Define the sign of thesespaces to be +1 and -1, respectively, and write sgn(Q) or sgn(V) for the signof the space. Thus the isometry type of an even dimensional orthogonal spaceover a finite field is determined by its sign. If V is an orthogonal space of odddimension over F, then, by 19.17, char(F) is odd. Let's look at such spacesnext.

(21.3) Let V be an orthogonal space of odd dimension over a finite field F.Then V possesses a hyperplane which is hyperbolic.

Spaces over finite fields 87

Proof. The proof is by induction on n. The remark is trivial if n = 1, so taken > 3. By 19.9, V possesses a nondegenerate subspace U of codimension 2.By induction on n, U possesses a hyperbolic hyperplane K. If n > 3 then,by induction on n, Kl possesses a hyperbolic plane W. Then K ® W is ahyperbolic hyperplane of V.

Son = 3. Choose a basis X = (x;:1 < i < 3) for V as in 19.9. We mayassume V is definite. Thus (x1, x2) is definite, and hence possesses a vectory such that Q(y) has the same quadratic character as -Q(x3). Thus there isa E F with a2Q(y) _ -Q(x3). Now ay + x3 is singular and 19.12 completesthe proof.

If V is an odd dimensional orthogonal space over a finite field then, by 21.3,V possesses a hyperbolic hyperplane H, and, by 20.8, H is determined up toconjugacy under the isometry group of V. Let x be a generator of Hl anddefine the sign of V (or Q) to be +1 if Q(x) is a quadratic residue in F, and-1 if Q(x) is not a quadratic residue. Then evidently when n is odd there areorthogonal spaces of sign s = +1 and -1, and, by the uniqueness of H (up toconjugacy), these spaces are not isometric. On the other hand if c is a generatorfor the multiplicative group F# of F, then c Q is similar to Q under the scalartransformation cl. Moreover (H, Q) is similar to (H, cQ), so (H, cQ) is alsohyperbolic. Hence, as (cQ)(x) = cQ(x) has different quadratic character fromQ(x), sgn(cQ) : sgn(Q). Thus we have shown:

(21.4) Let F be a field of odd order, n an odd integer, and c a generator of themultiplicative group F# of F. Then

(1) Up to equivalence there are exactly two nondegenerate quadratic formsQ and c Q on an n-dimensional vector space V over F.

(2) sgn(Q) = +1 and sgn(cQ) _ -1.(3) Q and cQ are similar via the scalar transformation cl, so O(V, Q) _

O(V, cQ).

(21.5) Let F be a finite field of square order. Then up to equivalence V admitsa unique unitary form f. Further (V, f) possesses an orthonormal basis.

Proof. As F is of square order it possesses a unique automorphism 0 of order2. Moreover Fix(0) = f aae: a E F}, so 19.11 completes the proof.

The final lemma of this section summarizes some of the previous lemmas inthis chapter, and provides a complete description of forms over finite fields.


(21.6) Let V be an n-dimensional space over a finite field F of order q andcharacteristic p. Then

(1) V admits a symplectic form f if and only if n is even, in which case fis unique up to equivalence and (V, f) is hyperbolic.

(2) V admits a unitary form f if and only if q is a square, in which case fis unique up to equivalence and (V, f) has a orthonormal basis.

(3) If n is even then V admits exactly two equivalence classes of nondegen-erate quadratic forms. Two forms are equivalent precisely when they have thesame sign. If P is such a form then (V, P) is isometric to Dn/2 or D(n/2)-1 Q

of sign +1 and -1, respectively.(4) If n is odd then V admits a nondegenerate quadratic form precisely when

p is odd, in which case there are two equivalence classes of forms. All formsare similar.

22 The classical groupsIn section 22, continue to assume the hypothesis and notation of section 19.Section 22 considers the isometry groups O(V, f) and O(V, Q), certain normalsubgroups of these groups, and the images of such groups under the projectivemap P of section 13. Notice that one can also regard the general linear groupas the isometry group O(V, f), where f is the trivial form f (u, v) = 0 for allu, v c V. The groups G and PG, as G ranges over certain normal subgroupsof O(V, f), are called the classical groups (where f is trivial, orthogonal,symplectic, or unitary). We'll be particularly concerned with classical groupsover finite fields.

Observe that if two spaces are isometric then their isometry groups areisomorphic. This is a special case of an observation made in section 2. As amatter of fact the isometry groups are isomorphic if the spaces are only similar,which is relevant because of 21.6.4. The upshot of these observations is that indiscussing the classical groups we need only concern ourselves with forms upto similarity.

Recall, from 19.16 that if n is even there is, up to equivalence, a uniquesymplectic form f on V. Write Sp(V) for the isometry group O(V, f ). Sp(V)is the symplectic group on V. As V is determined by n and F, I'll also writeSpn(F) for Sp(V). Spn(F) is the n-dimensional symplectic group over F.

If f is unitary then O(V, f) is called a unitary group. Similarly if Q is anondegenerate quadratic form then O(V, Q) is an orthogonal group. In generalthere are a number of similarity classes of forms on V and hence more than oneunitary group or orthogonal group on V. Lemma 21.6 gives precise informationwhen F is finite; we will consider that case in a moment. In any event I'll writeGU(V) or O(V) for a unitary or orthogonal group on V, respectively, even

8 8 Spaces with forms

(21.6) Let V be an n-dimensional space over a finite field F of order q and characteristic p. Then

(1) V admits a symplectic form f if and only if n is even, in which case f is unique up to equivalence and (V, f ) is hyperbolic.

(2) V admits a unitary form f if and only if q is a square, in which case f is unique up to equivalence and (V, f ) has a orthonormal basis.

(3) If n is even then V admits exactly two equivalence classes of nondegenerate quadratic forms. Two forms are equivalent precisely when they have the same sign. If P is such a form then (V, P) is isometric to 0"" or 0tnJ2)-l Q of sign + 1 and - 1, respectively.

(4) If n is odd then V admits a nondegenerate quadratic form precisely when p is odd, in which case there are two equivalence classes of forms. All forms are similar.

22 The classical groups In section 22, continue to assume the hypothesis and notation of section 19. Section 22 considers the isometry groups O(V, f ) and O(V, Q), certain normal subgroups of these groups, and the images of such groups under the projective map P of section 13. Notice that one can also regard the general linear group as the isometry group O(V, f), where f is the trivial form f (u, v) = 0 for all u, v E V. The groups G and PG, as G ranges over certain normal subgroups of O(V, f), are called the classical groups (where f is trivial, orthogonal, symplectic, or unitary). We'll be particularly concerned with classical groups over finite fields.

Observe that if two spaces are isometric then their isometry groups are isomorphic. This is a special case of an observation made in section 2. As a matter of fact the isometry groups are isomorphic if the spaces are only similar, which is relevant because of 21.6.4. The upshot of these observations is that in discussing the classical groups we need only concern ourselves with forms up to similarity.

Recall, from 19.16 that if n is even there is, up to equivalence, a unique symplectic form f on V. Write Sp(V) for the isometry group O(V, f). Sp(V) is the symplectic group on V. As V is determined by n and F, I'll also write Spn(F) for Sp(V). Sp,(F) is the n-dimensional symplectic group over F.

If f is unitary then O(V, f ) is called a unitary group. Similarly if Q is a nondegenerate quadratic form then O(V, Q) is an orthogonal group. In general there are a number of similarity classes of forms on V and hence more than one unitary group or orthogonal group on V. Lemma 21.6 gives precise information when F is finite; we will consider that case in a moment. In any event I'll write GU(V) or O(V) for a unitary or orthogonal group on V, respectively, even

The classical groups 89

though there may be more than one such group. GU(V) is (the) general uni-tary group. Write SU(V) and SO(V) for SL(V) fl GU(V) and SL(V) fl O(V),respectively (recall the special linear group SL(V) is defined and discussed insection 13). SU(V) and SO(V) are the special unitary group and special or-thogonal group, respectively. Write Q(V) for the commutator group of O(V).

Suppose for the moment that F = GF(q) is the finite field of order q. Thenwrite Spn (q) for Spn (F). Also, from 21.6, there is a unitary form on V preciselywhen q = r2 is a square, in which case the form is unique, and I write GUn(r)and SU,(r) for GU(V) and SU(V). Notice r (F1, rather r = IF11/2 in theunitary case. If n is odd there is an orthogonal form on V only when q is odd,in which case all such forms are similar and I write On(q), SOn(q), and Qn(q)for O(V), SO(V), and Q(V). Finally if n is even then up to equivalence thereare just two nondegenerate quadratic forms Qe on V, distinguished by the signsgn(QE) = s = +1 or -1 of the form. Write 01(q), SO' (q), and Q '(q) for thecorresponding groups.

For each group G we can restrict the representation P: GL(V) -* PGL(V)of GL(V) on the projective space PG(V) to G and obtain the image PG of Gwhich is a group of automorphisms of the projective space PG(V). Thus forexample we obtain the groups PSpn(q), PGUn(r), POn(q), PS2n(q), etc. It willdevelop much later that the groups PSpn(q), PSUn(r), and PQ (q) are simpleunless n and q are small. In this section we prove these groups are (usually)perfect (i.e. each group is its own commutator group). This fact together withExercise 7.8 is used in 43.11 to establish the simplicity of the groups.

Recall from section 13 that subspaces of V of dimension 1, 2, and n - 1 arecalled points, lines, and hyperplanes, respectively, and in general subspaces ofV are objects of the projective space PG(V). If V has a form for Q, then fromsection 19 we have a notion of totally singular and nondegenerate subspace,and hence totally singular and nondegenerate points, lines, and hyperplanes.

(22.1) If g E O(V, f) then Cv(g) = [V, g]1.

Proof. Let U = Cv(g). For u E U and v E V, (u, v) = (ug, vg) = (u, vg).Thus as

v + u1 = {x E V : (x, u) = (v, u)]

we have vg E v + u1. Hence

[v, g] En u1 = U1.uEU

Therefore [V, g] < U1. But, by Exercise 4.2.3 and 19.2, dim([V, g]) _dim(U1), so the proof is complete.

The classical groups

though there may be more than one such group. GU(V) is (the) general unitary group. Write SU(V) and SO(V) for SL(V) n GU(V) and SL(V) n O(V), respectively (recall the special linear group SL(V) is defined and discussed in section 13). SU(V) and SO(V) are the special unitary group and special orthogonal group, respectively. Write Q(V) for the commutator group of O(V).

Suppose for the moment that F = GF(q) is the finite field of order q. Then write Sp, (q) for Sp,(F). Also, from 21.6, there is a unitary form on V precisely when q = r2 is a square, in which case the form is unique, and I write GU,(r) and SU,(r) for GU(V) and SU(V). Notice r # IF/, rather r = IF^''^ in the unitary case. If n is odd there is an orthogonal form on V only when q is odd, in which case all such forms are similar and I write O,(q), SO,(q), and Q,(q) for O(V), SO(V), and Q(V). Finally if n is even then up to equivalence there are just two nondegenerate quadratic forms Q, on V, distinguished by the sign sgn(Q,) = E = +1 or -1 of the form. Write Oi(q), SOi(q), and Qi(q) for the corresponding groups.

For each group G we can restrict the representation P: GL(V) + PGL(V) of GL(V) on the projective space PG(V) to G and obtain the image P G of G which is a group of automorphisms of the projective space PG(V). Thus for example we obtain the groups PSp,(q), PGU,(r), POi(q), PQi(q), etc. It will develop much later that the groups PSp,(q), PSU,(r), and PQi(q) are simple unless n and q are small. In this section we prove these groups are (usually) perfect (i.e. each group is its own commutator group). This fact together with Exercise 7.8 is used in 43.1 1 to establish thg simplicity of the groups.

Recall from section 13 that subspaces of V of dimension 1,2, and n - 1 are called points, lines, and hyperplanes, respectively, and in general subspaces of V are objects of the projective space PG(V). If V has a form f or Q, then from section 19 we have a notion of totally singular and nondegenerate subspace, and hence totally singular and nondegenerate points, lines, and hyperplanes.

(22.1) If g E O(V, f ) then Cv(g) = [V, glL.

Proof. Let U = Cv(g). For u E U and v E V, (u, v) = (ug, vg) = (u, vg). Thus as

v + u' = (x E V: (x, U) = (v, u)}

we have vg E v + u'. Hence

[a, g] E u' = u'. UEU

Therefore [V, g] 5 u L . But, by Exercise 4.2.3 and 19.2, dim([V, g]) =

dim(uL), so the proof is complete.

Recall the definition of a transvection in section 13. I'll prove the next twolemmas together.

(22.2) O(V, f) (or O(V, Q)) contains a transvection if and only if each of thefollowing holds:

(1) V possesses isotropic points.(2) If (V, Q) is orthogonal then char(F) = 2.

(22.3) Let G = O(V, f) (or O(V, Q)) and assume t is a transvection in G.Then

(1) U = [V, t] is an isotropic point and Cv(t) = U1.(2) Let AU be the set of transvections with center U = (u) and let R =

RU = (AU) be the root group of t. Then R# = Au, for each r E R and Y E Vwe have yr = y + ar(y, u)u for some ar E F, and one of the following holds:

(i) (V, f) is symplectic and the map r i-- ar is an isomorphism of R withthe additive group of F.

(ii) (V, f) is unitary and r i-- are is an isomorphism of R with Fix(9), wheree E F# with ee = -e.

(iii) (V, Q) is orthogonal, char(F) = 2, R = Z2, U is nonsingular, and at =

Q(u)-'

(3) If (V, f) is symplectic or unitary then each singular point is the center ofa transvection and G is transitive on the root groups of transvections. If (V, Q)is orthogonal each nonsingular point is the center of a unique transvection.

(4) Assume either (V, f) is symplectic and H = G or (V, f) is unitaryand H = SU(V). Then one of the following holds:

(i) R < W)(ii) n = 2 and IFix(9)! < 3.(iii) n = 4, (V, f) is symplectic, and BFI = 2.

Now to the proof of 22.2 and 22.3. First 22.3.1 follows from 22.1 and thedefinition of a transvection. In particular if G possesses a transvection then Vpossesses isotropic points, so we may assume U = (u) is an isotropic point ofV. By 19.12 there is an isotropic vector x E V - U1 with x, u a hyperbolicpair in the hyperbolic hyperplane W = (u, x). Let X = (u, x} C Y be a basisfor V with Y - X a basis for W1. Let t = to be the transvection in GL(V) withCv(t) = U1 and xt = x + au, where a is some fixed member of P. Then, by19.6, t E G if and only if (vlt, vet) = (vl, v2) for all v1, v2 E X, and if (V, Q)is orthogonal also Q(vt) = Q(v) for all v E X. By construction if suffices to


Recall the definition of a transvection in section 13. I'll prove the next two lemmas together.

(22.2) O(V, f ) (or O(V, Q)) contains a transvection if and only if each of the following holds:

(1) V possesses isotropic points. (2) If (V, Q) is orthogonal then char(F) = 2.

(22.3) Let G = O(V, f ) (or O(V, Q)) and assume t is a transvection in G. Then

(1) U = [V, t] is an isotropic point and Cv(t) = U L .

(2) Let Au be the set of transvections with center U = (u) and let R = Ru = (Au) be the root group oft. Then R# = Au, for each r E R and y E V we have yr = y + a,(y, u)u for some a, E F , and one of the following holds:

(i) (V, f ) is symplectic and the map r H a, is an isomorphism of R with the additive group of F .

(ii) (V, f ) is unitary and r H are is an isomorphism of R with Fix(@), where e E F# with ee = -e.

(iii) (V, Q) is orthogonal, char(F) = 2, R E Z 2 , U is nonsingular, and a, =

Q(u)-l.

(3) If (V, f ) is symplectic or unitary then each singular point is the center of a transvection and G is transitive on the root groups of transvections. If (V, Q) is orthogonal each nonsingular point is the center of a unique transvection.

(4) Assume either (V, f ) is symplectic and H = G or (V, f ) is unitary and H = SU(V). Then one of the following holds:

(i) R 5 H(' ) . (ii) n = 2 and (Fix(0)I i 3.

(iii) n = 4, (V, f ) is symplectic, and IF I = 2.

Now to the proof of 22.2 and 22.3. First 22.3.1 follows from 22.1 and the definition of a transvection. In particular if G possesses a transvection then V possesses isotropic points, so we may assume U = (u) is an isotropic point of V. By 19.12 there is an isotropic vector x E V - U' with x, u a hyperbolic pair in the hyperbolic hyperplane W = (u, x). Let X = {u, x} E Y be a basis for V with Y - X a basis for W L . Let t = t, be the transvection in GL(V) with Cv(t) = U' and xt = x + au, where a is some fixed member of F#. Then, by 19.6, t E G if and only if (vlt, v2t) = (vl, v2) for all vl, v2 E X, and if (V, Q) is orthogonal also Q(vt) = Q(v) for all v E X. By construction if suffices to

check these equalities when vl = x = V2, and, if (V, Q) is orthogonal, forv = x. The check reduces to a verification that:

(*) a + sae = 0, and if (V, Q) is orthogonal also a = -Q(u)-1 0, where& = -1 if (V, f) is symplectic, and s = +1 otherwise.

If (V, f) is symplectic then(*) holds for each a E F#, so Au = {ta: a E F#}.Also the map a i-- to is an isomorphism of R with the additive group of F.If (V, f) is unitary then (*) has a solution if and only if there exists e E F#with ee = -e, in which case a is a solution to (*) precisely when a = be withb E Fix(B). Observe there is c E F with c co and, setting e = c - co, ee = -e.Finally if (V, Q) is orthogonal then a is a solution to (*) if and only if a =Q(u) 0 and char(F) = 2. Observe in each case ta: y H y + a(y, u)u foreach y E X, and hence also for each y E V.

So 22.2 and the first two parts of 22.3 are established. The transitivity state-ment in 22.3.3 follows from Witt's Lemma, so it remains to establish 22.3.4.Assume the hypothesis of 22.3.4. Let L be the group generated by the transvec-tions with centers in W. W1 < Cv(L), so L is faithful on W and henceL < O(W, f). Now, by Exercise 7.1 and 13.7, L - SL2(Fix(0)). Then, by13.6.4, either R < LM < HM, or IFix(0)I < 3, and we may assume the latterwith n > 2.

If (V, f) is unitary let v be a nonsingular vector in W1 and Z = (W, v),while if (V, f) is symplectic let Z be a nondegenerate subspace containing Wof dimension 4 or 6, for IFI = 3 or 2, respectively. Let K = CH(Z1), so thatK = Sp(Z) or SU(Z). If R < KM then the proof is complete, so without lossV = Z. But now Exercise 7.1 completes the proof.

(22.4) Assume either (V, f) is symplectic and G = O(V, f) = Sp(V), or(V, f) is unitary of dimension at least 2, Fix(B) = {aae: a E F}, and G =SL(V) fl O(V, f) = SU(V). Then either

(1) G is generated by the transvections in O(V, f), or(2) (V, f) is unitary, I F I = 4, and n = 3.

Proof. If (V, f) is symplectic let F = V# and 0 the set of hyperbolic bases ofV. If (V, f) is unitary let

F = {v E V: (v, v) = 1}.

Let T be the group generated by the transvections in O(V, f). I'll show:

(i) T is transitive on F unless 22.4.2 holds, and(ii) T is transitive on 0 if (V, f) is symplectic.

check these equalities when v l = x = v2, and, if (V, Q) is orthogonal, for v = x . The check reduces to a verification that:

(') a + &ae = 0, and if (V, Q) is orthogonal also a = - Q(u)-' # 0, where E = -1 if (V, f ) is symplectic, and E = +1 otherwise.

If (V, f ) is symplectic then (*) holds for each a E F', so A" = {t,: a E F'). Also the map a H t, is an isomorphism of R with the additive group of F . If (V, f ) is unitary then (") has a solution if and only if there exists e E F# with ee = -e, in which case a is a solution to (*) precisely when a = be with b E Fix(@). Observe there is c E F with c # ce and, setting e = c - ce, ee = -e. Finally if (V, Q) is orthogonal then a is a solution to (*) if and only if a = Q(u) # 0 and char(F) = 2. Observe in each case t,: y H y + a(y, u)u for each y E X, and hence also for each y E V.

So 22.2 and the first two parts of 22.3 are established. The transitivity statement in 22.3.3 follows from Witt's Lemma, so it remains to establish 22.3.4. Assume the hypothesis of 22.3.4. Let L be the group generated by the transvections with centers in W. W' 5 Cv(L), so L is faithful on W and hence L 5 O(W, f). Now, by Exercise 7.1 and 13.7, L E SL2(Fix(@)). Then, by 13.6.4, either R 5 L(') 5 H('), or /Fix(@)/ 5 3, and we may assume the latter with n > 2.

If (V, f ) is unitary let v be a nonsingular vector in W' and Z = (W, v), while if (V, f ) is symplectic let Z be a nondegenerate subspace containing W of dimension 4 or 6, for I FI = 3 or 2, resp$ctively. Let K = c~(z'), SO that K = Sp(Z) or SU(Z). If R 5 K(') then the proof is complete, so without loss V = Z. But now Exercise 7.1 completes the proof.

(22.4) Assume either (V, f ) is symplectic and G = O(V, f ) = Sp(V), or (V, f ) is unitary of dimension at least 2, Fix(@) = {aae: a E F), and G = SL(V) n O(V, f ) = SU(V). Then either

(1) G is generated by the transvections in O(V, f ), or (2) (V, f ) is unitary, I FI = 4, and n = 3.

Proof. If (V, f ) is symplectic let r = V' and S-2 the set of hyperbolic bases of V. If (V, f ) is unitary let

Let T be the group generated by the transvections in O(V, f ). I'll show:

(i) T is transitive on r unless 22.4.2 holds, and (ii) T is transitive on S-2 if (V, f ) is symplectic.

Observe that the stabilizer in G of w E 7 is trivial, so (ii) implies T = G.Observe also that by Exercise 7.1 the lemma holds if n = 2, so we may assume

n>2.Suppose (i) holds. If (V, f) is unitary and x E F then GX = SU(xL), and,

as n > 2, induction on n implies GX < T unless n = 4 and IFS = 4, whereExercise 7.3 says the same thing. Hence, by (i) and 5.20, G = T. So take(V, f) symplectic and let X = (x1: 1 < i < n) and Y = (yi: 1 < i < n) bemembers of R. We need to prove Y = Xs for some s e T. By (i) we maytake xl = yl. Claim Y2 E x2(TX,). As (xl, x2) _ (xl, y2) = 1, y2 = X2 + vfor some v E (xl)L. If X2 V vL then there is a transvection t with center (u)and x2t = y2, and as v E (xl)L, t E Tz,. If X2 E vL the same argument showsx2 and y2 are conjugate to xl + x2 in TX,, and hence also to each other. So theclaim holds and hence we may take x2 = y2. But now X is conjugate to Y inSp((x1, X2) -L) fl T by induction on n, so the lemma holds.

So it remains to prove (i). Let x , y E F and U = (x, y). If U is nondegeneratethen x e y(T fl O(U, f )) since (i) holds when n =2. Thus we may take 0U E Rad(U) and it suffices to show there exists a point (z) (such that z isnonsingular if f is unitary) with (x, z) and (y, z) nondegenerate. If n > 3 justchoose z E uL with (x, z) nondegenerate. Thus we can assume n = 3, so that(V, f) is unitary. Further we may assume I F I > 4. Let u, v be a hyperbolic basisfor yL.We may take x=y +u. Then xL=(u,y-v)and if z=au +y-v,then (y, z) fl yL = (au - v). It suffices to choose z so that z and au - v arenonsingular. Equivalently, if T = T ,,(e) is the trace of F over Fix(B), thenT(a) 0 or 1. Hence as IFS > 4 and T: F -* Fix(B) is surjective, we canchoose z as desired.

A field F is perfect if char(F) = 0 or char(F) = p > 0 and the p-power mapis a surjection from F onto F (i.e. F = F"). For example finite fields andalgebraically closed fields are perfect.

(22.5) Let F be a perfect field of characteristic 2 and (V, Q) orthogonal ofdimension at least 4. Let (u) be a nonsingular point of V, t the transvectionwith center (u), and G = O(V, Q). Then

(1) G(u) = CG(t) is represented as Sp(u-/(u)) on uL/(u), with (t) thekernel of this representation.

(2) CG(t) is transitive on the nonsingular points in uL distinct from (u).(3) If I FI > 2 then uL = [u L, Ca(t)].(4) Either G is generated by, and is transitive on, its transvections, or n = 4,

IFS = 2, and sgn(Q) = +1.

Spaces with forms

trivial, so (ii) implies T = G. Observe also that by Exercise 7.1 the lemma holds if n = 2, so we may assume n > 2.

Suppose (i) holds. If (V, f ) is unitary and x E r then G, = su(xL), and, as n > 2, induction on n implies G, i T unless n = 4 and I FI = 4, where Exercise 7.3 says the same thing. Hence, by (i) and 5.20, G = T. So take (V, f ) symplectic and let X = (x,: 1 5 i 5 n) and Y = (y,: 1 5 i I: n) be members of Q. We need to prove Y = Xs for some s E T. By (i) we may take xl = yl. Claim y2 E xz(T,,). As (XI, xz) = (XI, yz) = 1, y2 = x2 + v for some v E (~1)'. If x2 6 vL then there is a transvection t with center (v) and x2t = y2, and as v E (xl)', t E T,, . If x2 E V' the same argument shows x2 and y2 are conjugate to xl + x2 in T,,, and hence also to each other. So the claim holds and hence we may take xz = yz. But now X is conjugate to Y in Sp((xl, x2) l ) n T by induction on n, so the lemma holds.

So it remains to prove (i). Let x , y E r and U = (x , y) . If U is nondegenerate then x E y(T n O(U, f )) since (i) holds when n = 2. Thus we may take 0 # u E Rad(U) and it suffices to show there exists a point (z) (such that z is nonsingular if f is unitary) with (x, z) and (y, z) nondegenerate. If n > 3 just choose z E uL with (x, z) nondegenerate. Thus we can assume n = 3, so that (V, f ) is unitary. Further we may assume IF I > 4. Let u, v be a hyperbolic basis f o r y L . ~ e m a y t a k e x = y + u . ~ h e n n ' = (u,y - v ) a n d i f z = a u + y - v , then (y, z) n yL = (au - v). It suffices to choose z so that z and au - v are nonsingular. Equivalently, if T = TL~(~) is the trace of F over Fix(@), then T(a) # 0 or 1. Hence as IF1 > 4 and T: F + Fix(@) is surjective, we can choose z as desired.

A field F is perjfect if char(F) = 0 or char(F) = p > 0 and the p-power map is a surjection from F onto F (i.e. F = FP). For example finite fields and algebraically closed fields are perfect.

(22.5) Let F be a perfect field of characteristic 2 and (V, Q) orthogonal of dimension at least 4. Let (u) be a nonsingular point of V, t the transvection with center (u), and G = O(V, Q). Then

(1) G(,, = CG(t) is represented as Sp(uL/(u)) on uL/(u), with (t) the kernel of this representation.

(2) CG(t) is transitive on the nonsingular points in uL distinct from (u). (3) If IFI > 2 then uL = [uL, CG(t)]. (4) Either G is generated by, and is transitive on, its transvections, or n = 4,

IF1 = 2, and sgn(Q) = +l.

Proof. Let U = (u) and H = Gu. As t is the unique transvection with centerU, H = CG(t) and t is the kernel of the representation Tr of H on UL/ U = M.Observe that, if f is the bilinear form on V defined by Q, then (M, f) is asymplectic space, where 1(x, y) = f (x, y). Let. E M# and W = (x, u). AsF is perfect we can choose u with Q(u) = 1, and W contains a unique singularpoint (w). For a E F#, let ta be the transvection with center (aw + u). Then ta7ris a transvection on M with center W and, for y E M, y(ta7r) = y +a2(y, w)wby 22.3.2. Thus, as F = F2, Rn = (tan: a E F#) is the full root group of tan inSp(M) by 22.3.2. So, by 22.4, H7r = Sp(M). Therefore (1) is established. Also[y, ta] = a(y, w)(aw + u) by 22.3.2, so, if IFI > 2, then, choosing y V wL,we have u E [y, R], so, as M = [M, H7r], (3) holds. To prove (2) observe thatH7r is transitive on M# by (1), so it suffices to show NH(W) is transitive on theset F of points of W distinct from (u) and (w). Let w, v be a hyperbolic pairin M. By (1), for a E F# there is ga E H with wga = aw and vga = a-1v.Hence (ga: a E F#) is transitive on F.

It remains to prove (4). Let T be the group generated by the transvectionsin G. We've seen that H < T, so to prove T = G it suffices by 5.20 to showT is transitive on the set A of nonsingular points of V. This will also showG is transitive on its transvections. Let Z be a second nonsingular point. Wemust show Z E UT T. This follows from (2) if (Z + U)L contains a nonsingularpoint distinct from U and Z, so assume otherwise. Then U + Z = (U + Z)Land U and Z are the only nonsingular point§ in U + Z. This forces n = 4 andIF) = 2. Let A = Rad(U + Z). If B is a nonsingular point in AL - (U + Z)then U + B and Z + B are nondegenerate, so U, Z E BT and hence Z E UT T.Thus no such point exists, which forces sgn(Q) = +1.

If char(F) ; 2 and (V, Q) is orthogonal, then a reflection on V is an elementr in O(V, Q) such that [V, r] is a point of V. [V, r] is called the center of r.

(22.6) Let char(F) 2 and (V, Q) orthogonal. Then(1) If r is a reflection on V then r is an involution, [V, r] is nonsingular,

and Cv(r) [V, r]-L.

(2) If U = (u) is a nonsingular point of V then there exists a unique reflec-tion r,, on V with center U. Indeed xra = x - (x, u)u/Q(u) for each x E V.

Proof. Let r be a reflection on V. Then [V, r] = (v) is a point. By 22.1,Cv (r) = vL, so vL is a hyperplane by 19.2. By 22.2, r is not a transvection, sov V vL and hence v is nonsingular. Next yr = av for some 10 a E F# andQ(v) = Q(vr) = Q(av) = a2Q(v), so a = -1. Hence r is an involution.


Proof. Let U = (u) and H = GU. As t is the unique transvection with center U, H = Cc(t) and t is the kernel of therepresentation 7t of H on u'/ U = M. Observe that, if f is the bilinear form on V defined by Q, then (M, 7) is a symplectic space, where f(Z, 5,) = f (x, y). Let Z E M# and W = (x, u). As F is perfect we can choose u with Q(u) = 1, and W contains a unique singular point (w ) . For a E F#, let t, be the transvection with center (aw + u) . Then tan is a transvection on M with center w and, for y E M, ?(tan) = 5, + a2(ji, #)a by 22.3.2. Thus, as F = F2, R7t = (tan: a E F#) is the full root group of tan in s ~ ( M ) by 22.3.2. So, by 22.4, H7t = Sp(M). Therefore (1) is established. Also [y, t,] = a(y, w)(aw + u) by 22.3.2, so, if I FI > 2, then, choosing y 6 wL, we have u E [y, R], so, as M = [M, Hn], (3) holds. To prove (2) observe that H7t is transitive on M# by (I), so it suffices to show NH(W) is transitive on the set r of points of W distinct from (u) and (w ). Let ii, , 6 be a hyperbolic pair in M. By (I), for a E F# there is g, E H with wg, = aw and vg, = a-'v. Hence (g,: a E F#) is transitive on r .

It remains to prove (4). Let T be the group generated by the transvections in G. We've seen that H ( T, so to prove T = G it suffices by 5.20 to show T is transitive on the set A of nonsingular points of V. This wiU also show G is transitive on its transvections. Let Z be a second nonsingular point. We must show Z E U T. This follows from (2) if (Z + u ) ~ contains a nonsingular point distinct from U and Z, so assume otherwise. Then U + Z = (U + z)' and U and Z are the only nonsingular point? in U + Z. This forces n = 4 and JFI = 2. Let A = Rad(U + Z). If B is a nonsingular point in A' - (U + Z) then U + B and Z + B are nondegenerate, so U, Z E BT and hence Z E U T. Thus no such point exists, which forces sgn(Q) = +l.

If char(F) # 2 and (V, Q) is orthogonal, then a rejection on V is an element r in O(V, Q ) such that [V, r] is a point of V. [V, r ] is called the center of r.

(22.6) Let char(F) # 2 and (V, Q) orthogonal. Then (1) If r is a reflection on V then r is an involution, [V, r] is nonsingular,

and Cv(r) = [V, r]'. (2) If U = (u) is a nonsingular point of V then there exists a unique reflec-

tion r, on V with center U. Indeed xr, = x - (x, u)u/Q(u) for each x E V.

Proof. Let r be a reflection on V. Then [V, r] = (v) is a point. By 22.1, Cv(r) = v', so v' is a hyperplane by 19.2. By 22.2, r is not a transvection, so v 6 v' and hence v is nonsingular. Next vr = av for some 1 # a E F# and Q(v) = Q(vr) = Q(av) = a2e(v), so a = -1. Hence r is an involution.

Conversely let U = (u) be a nonsingular point. By (1) there is at most onereflection with center U, while a straightforward calculation shows the maplisted in (2) is such a reflection.

The proof of the following lemma comes essentially from 1.5.1 on page 19 ofChevalley [Ch 2].

(22.7) Let (V, Q) be an orthogonal space. Then either

(1) O(V, Q) is generated by its transvections or reflections, or(2) 1 F I = 2, n = 4, and sgn(Q) = +1.

Proof. If n = 2 or IFS = 2 the result follows from Exercise 7.2 or 22.5.4,respectively. If n = 1, then char(F) # 2 and O(V, Q) is generated by thereflection -I. Thus we may taken > 2 and IF) > 2.

Let T be the group generated by all transvections or reflections in G =O(V, Q) and suppose h E G - T. Pick h so that dim(Cv(h)) is maximal in thecoset hT.

Suppose Y E V with z = [y, h] nonsingular. yh = y +z and Q(yh) = Q(y),so Q(z)+(y, z)=O. Inparticulary 0 z1 andy+z = y-(y, z)z/Q(z) = yrz,where rZ is the transvection or reflection with center (z). Thus yh = yrz,so y E Cv(hrz). By 22.1, Cv(h) = [V, h]L C z1 = Cv(rz), so Cv(h) <Cv(hrz). Hence dim(Cv(hrr)) > dim((Cv(h), y)) > dim(Cv(h)), contrary tothe choice of h.

Therefore [V, h] is totally singular. I claim next that T is transitive on themaximal totally singular subspaces of V. Assume not and pick two such spacesM and N such that M 0 NT and, subject to this constraint, with dim(M fl N)maximal. Then M # N so M < M + N, and hence by maximality of M thereis a nonsingular vector x = m + n E M + N. As x is nonsingular, (m, n) 54 0.Also Q(x) = (m, n), so mrx = -n E N, while M n N < x1 = Cv(rx). ThusM n N < (M n N, n) < Mrx n N, and then, by maximality of M n N, Mrx ENR. But now M E NR, contrary to the choice of M and N.

So the claim is established. Next there is a maximal totally singular subspaceM with [V, h] < M. Then M1 < [V, h]1 = Cv(h). Let H = Cc(M1) nCc(V/M1), so that h E H. As T is transitive on maximal totally singularsubspaces, it follows that G = NG(M)T and each member of G has a T-cosetrepresentative in H9 for some g E G. Then as H < NG(M), HT < G, so eachmember of G has a T-coset representative in HT, and hence G = HT.

By Exercise 4.8, H is abelian, so G/T is abelian. Thus [h, g] E T for eachg E G, so [h, g] 0 ho. If h acts on a proper nondegenerate subspace U of


Conversely let U = (u) be a nonsingular point. By (1) there is at most one reflection with center U, while a straightforward calculation shows the map listed in (2) is such a reflection.

The proof of the following lemma comes essentially from 1.5.1 on page 19 of Chevalley [Ch 21.

(22.7) Let ( V , Q ) be an orthogonal space. Then either

( 1 ) O ( V , Q ) is generated by its transvections or reflections, or (2) IF I = 2, n = 4, and sgn(Q) = + 1.

Proof. If n = 2 or 1 FI = 2 the result follows from Exercise 7.2 or 22.5.4, respectively. If n = 1, then char(F) # 2 and O(V, Q ) is generated by the reflection - I . Thus we may take n > 2 and IF I > 2.

Let T be the group generated by all transvections or reflections in G = O(V, Q ) and suppose h E G - T. Pick h so that dim(Cv(h)) is maximal in the coset h T.

Suppose y E V with z = [ y , h ] nonsingular. yh = y +z and Q(yh) = Q(y) , so Q(z) + ( y , z)=O. In particular y $ z' and y +z = y - ( y , z ) z / Q ( z ) = yr,, where r, is the transvection or reflection with center ( 2 ) . Thus yh = yr,, so y E Cv(hr,). By 22.1, C v ( h ) = [ V , h]' E z' = Cv(r,), so C v ( h ) 5 Cv (hr,). Hence dim(Cv (hr,)) > dim((Cv (h ) , y ) ) > dim(Cv (h)) , contrary to the choice of h.

Therefore [ V , h ] is totally singular. I claim next that T is transitive on the maximal totally singular subspaces of V . Assume not and pick two such spaces M and N such that M $ NT and, subject to this constraint, with dim(M n N) maximal. Then M # N so M < M + N , and hence by maximality of M there is a nonsingular vector x = m + n E M + N . As x is nonsingular, ( m , n ) # 0. Also Q ( x ) = ( m , n ) , so mr, = -n E N , while M fl N 5 x L = Cv(r,). Thus M n N < ( M n N , n ) 5 Mr, n N , and then, by maximality of M n N , Mr, E

NR. But now M E NR, contrary to the choice of M and N . So the claim is established. Next there is a maximal totally singular subspace

M with [ V , h] 5 M. Then M' ( [ V , hlL = Cv(h). Let H = C G ( M L ) n C G ( v / M L ) , SO that h E H . As T is transitive on maximal totally singular subspaces, it follows that G = NG(M)T and each member of G has a T-coset representative in Hg for some g E G . Then as H I! NG(M), H T a_ G , so each member of G has a T-coset representative in HT, and hence G = HT.

By Exercise 4.8, H is abelian, so G / T is abelian. Thus [h , g] E T for each g E G , so [h, g] $ hG. If h acts on a proper nondegenerate subspace U of

V then h = h1h2 with h1 E O(U, Q) and h2 E O(U1-, Q). By induction onn, h, E T, so h E T. Hence h acts on no such subspace. In particular C1, (h)is totally isotropic, so, as [V, h] is totally singular, 22.1 and 19.3.2 implyCv(h) = [V, h] = M, and V is hyperbolic. Further, for V E V, [v, h] E v1-,since (v, [v, h]) is not nondegenerate.

By 19.14 there is a totally singular subspace N of V with V = M ® N.Let x1 E M#. As M = [V, h] _ [N, h], ¢: n r+ [n, h] is a vector spaceisomorphism of N and M. Hence there is y2 E N with x1 = [y2, h]. By thelast remark of the previous paragraph, (X1, y2) = 0. As ¢ is an isomorphism,(N - x1 )¢ ¢ y2 L, so there is y1 E N - xi with ¢(y1) = X2 0 y2 L. LetX = (x1, x2, Y1, Y2). Then V1 = (X) is nondegenerate and h-invariant, soV V1.

For a E F# define ah E GL(V) by v(ah) = v + a[v, h] for each v E V.Notice Q(v(ah)) = Q(v) as [v, h] E v1 fl M. Therefore ah E H. Indeedsetting Xa = (axi, x2, a-1y1, y2) we have MX(h) = MXa(ah) and J(X, f) _J(Xa, f). So the element g E GL(V) with Xg = Xa is in G by 19.6 andhg = ah. Now, as IFS > 2, we can choose a with a - 1 # 0. Then [h, g] _h-'hg = ((-1)h)(ah) = (a - 1)h E hG, contrary to an earlier remark. Theproof is complete.

Let (V, Q) be an orthogonal space. We next construct the Clifford algebra C =C(Q) of (V, Q). The treatment here will belabbreviated. For a more completediscussion see chapter 2 in Chevalley [Ch 1]. C is the tensor algebra (cf. Lang[La], chapter 16, section 5) of V, modulo the relations x ® x - Q(x)1 = 0, forx E V. For our purposes it will suffice to know the following:

(22.8) Let (V, Q) be an orthogonal space with ordered basis X. Choose X tobe orthogonal if char(F) # 2 and choose X to be a hyperbolic basis of the un-derlying symplectic space (V, f) if char(F) = 2. Let C = C(Q) be the Cliffordalgebra of (V, Q). Then C is an F-algebra with the following properties:

(1) There is an injective F-linear map p: V -- C such that C is generated asan F-algebra by Vp. Write ex for xp if x E X.

(2) For S = {x1, ... , x,"} C X with x1 < < write es = ex, ... exm.Then (es: S C X) is a basis for C over F. In particular eH = 1 = 1, anddimF(C) = 2".

(3) For u, v E V, (up)2 = Q(u) 1 and upvp + vpup = (u, v) 1.

(4) Let Cl be the subspace of C spanned by the vectors es, S C X, ISIimod 2, i = 0, 1. Then {C0, C1 } is a grading of C. That is C = Co ® C1and C;Cj C Ci+j, for i, j E {0, 11, where i + j is read mod 2.


V then h = hlh2 with h l E O(U, Q ) and h2 E O ( u L , Q) . By induction on n , hi E T , so h E T. Hence h acts on no such subspace. In particular C v ( h ) is totally isotropic, so, as [ V , h ] is totally singular, 22.1 and 19.3.2 imply C v ( h ) = [ V , h ] = M , and V is hyperbolic. Further, for v E V , [ v , h ] E vL , since ( v , [v , h ] ) is not nondegenerate.

By 19.14 there is a totally singular subspace N of V with V = M @ N . Let xl E M'. As M = [ V , h ] = [ N , h ] , 4: n I-+ [n, h ] is a vector space isomorphism of N and M . Hence there is y2 E N with xl = [y2, h] . By the last remark of the previous paragraph, ( x l , y2) = 0. As @ is an isomorphism, ( N - x:)@ g y$, SO there is yl E N - x: with @(yl) = x2 $ yj!-. Let X = ( x l , x2, yl , y2). Then Vl = ( X ) is nondegenerate and h-invariant, so v = vl.

For a E F# define ah E G L ( V ) by v(ah) = v + a [ v , h ] for each v E V . Notice Q(v(ah)) = Q(v) as [v , h ] E v' n M . Therefore a h E H. Indeed setting X, = ( a x l , xp, a - l y l , y2) we have Mx(h) = Mx,(ah) and J ( X , f ) = J(X,, f ) . So the element g E G L ( V ) with Xg = X, is in G by 19.6 and hg = ah. Now, as I FI > 2, we can choose a with a - 1 # 0. Then [h , g] = h-lhg = ( ( - l )h)(ah) = (a - l ) h E hG, contrary to an earlier remark. The proof is complete.

Let ( V , Q ) be an orthogonal space. We next construct the ClifSord algebra C = C ( Q ) of ( V , Q) . The treatment here will be4abbreviated. For a more complete discussion see chapter 2 in Chevalley [Ch 11. C is the tensor algebra (cf. Lang [La], chapter 16, section 5) of V , modulo the relations x @ x - Q(x) l = 0 , for x E V . For our purposes it will suffice to know the following:

(22.8) Let ( V , Q ) be an orthogonal space with ordered basis X. Choose X to be orthogonal if char(F) # 2 and choose X to be a hyperbolic basis of the underlying symplectic space ( V , f ) if char(F) = 2. Let C = C ( Q ) be the Clifford algebra of ( V , Q). Then C is an F-algebra with the following properties:

(1) There is an injective F-linear map p: V + C such that C is generated as an F-algebra by Vp. Write ex for xp if x E X.

(2) For S = { x l , . . . , x,} E X with xl < . - . < x, write es = ex, . . . ex,,,. Then (es: S E X ) is a basis for C over F . In particular e, = 1 = 1, and dimF(C) = 2".

(3) For u , v E V , = Q(u) 1 and upvp + vpup = ( u , v ) . 1. (4) Let Ci be the subspace of C spanned by the vectors es, S E X , IS1 =

imod 2, i = 0 , l . Then {Co, C 1 } is a grading of C. That is C = Co @ C1 and C i C j Ci+j, for i, j E {0, 11, where i + j is read mod 2.

(5) If u is a nonsingular vector of V then up is a unit in C with inverseQ(u)-lup, and, for v c V, (vp)uP = -(vru)p, where ru is the reflectionor transvection in O(V, Q) with center (u).

(6) The Clifford group G of C is the subgroup of units in C which permuteVp via conjugation. The representation 7r of G on V (subject to the iden-tification of V with Vp via p) is a representation of G on (V, Q) withGir =O(V, Q) if n is even and Gir = SO(V, Q) if n is odd. If u is anonsingular vector in V then up7r = -ru.

(7) ker(,-r) is the set of units in Z(C) and Z(C) = F 1 or F 1 + Fex for n even orodd, respectively. If n is odd, no unit in C induces -I on C by conjugation.

(8) There is an involutory algebra antiisomorphism t of C such that est =ex,, ... ex, for each S = {x1, ..., xm } C X.

Proof. I'll sketch a proof. If char(F) 2 the multilinear algebra can be avoidedas in chapter 5, section 4 of Artin [Ar]. A full treatment can be found inChevalley [Ch 1].

By definition, C = T/K where T = ®°°o Ti(V) is the tensor algebra andK = (x ®x - Q(u)1: X E V). In particular To(V) = F1 and there is a naturalisomorphism po: V -+ T1(V). Then p: V -+ C is the map v H vpo + Kinduced by po, and (1) will follow from (2), once that part is established.es = ex, ... ex,n = x1 ®. ®xm + K, so Ci is the image of ®j_i Tj(V)in C. Hence (4) follows from (2) and the definition of multiplication in T.The universal property of T implies there is an involutory antiisomorphism toof T with (xl ® . . (9 xm)to = xm ® ® x1. As to preserves K it inducest on C. Thus (8) holds. Part (3) is a direct consequence of the definition ofC, since x ® x - Q(u)1 E K. An easy induction argument using (3) showseseT is a linear combination of the elements eR, R C X, for each S, T C X,so (eR: R C X) spans C. Using the universal property of the tensor algebra,Chevalley shows on page 39 of [Ch 1] that there is a homomorphic image ofC of dimension 2", completing the proof of (2), and hence also of (1) and (4).I omit this demonstration.

Part (5) is a straightforward consequence of (3). If char(F) 0 2 then (3)shows e' = (-1)mes for x E X and S C X, where m = BSI if x SandM = IS I -1 if x E S. Since Xp generates C as an F-algebra, (7) follows in thiscase. If char(F) = 2 then choose X so that each of its members is nonsingular.Then (3) shows [ex, es] = 0 if S C_ x1, while [ex, es] = Q(x)-les+y if Scontains the unique y in X - x1, where S + y is the symmetric difference ofS with {y}. It follows that [ex, C] = (es: S c x1) is of dimension 2a-1. So, as

2n-1 = dim(C)/2 and ex is an involution, [ex, C] = Cc(ex). Thus (7) holds inthis case too.

Let G and 7r be as in (6). For g E G and V E V, Q(vgir)1 = ((vp)g)2 =((vp)2)g = Q(v)1, so Q(vgir) = Q(v). Hence Gir < O(V, Q). Let Go be the


(5) If u is a nonsingular vector of V then up is a unit in C with inverse Q(u)-'up, and, for v E V, = -(vru)p, where r, is the reflection or transvection in O(V, Q) with center (u).

(6) The Clifford group G of C is the subgroup of units in C which permute Vp via conjugation. The representation n of G on V (subject to the iden- tification of V with Vp via p) is a representation of G on (V, Q) with Gn =O(V, Q) if n is even and G n = SO(V, Q) if n is odd. If u is a nonsingular vector in V then upn = -r,.

(7) ker(n) is the set of units in Z(C) and Z(C) = F 1 or F 1 + Fex for n even or odd, respectively. If n is odd, no unit in C induces -I on C by conjugation.

(8) There is an involutory algebra antiisomorphism t of C such that est = e,,, . . . ex, for each S = {XI, . . . , x,} G X.

Proof. I'll sketch aproof. If char(F) # 2 the multilinear algebracan be avoided as in chapter 5, section 4 of Artin [Ar]. A full treatment can be found in Chevalley [Ch 11.

By definition, C = T/K where T = $Eo z (V) is the tensor algebra and K = (x @ x - Q(u)l: x E V). In particular To(V) = F 1 and there is a natural isomorphism po: V + Tl(V). Then p: V + C is the map v H vpo + K induced by po, and (1) will follow from (2), once that part is established. es = ex, . . .ex,,, = xl @ . . - 8 x, + K, so Ci is the image of ej-, Tj(V) in C. Hence (4) follows from (2) and the definition of multiplication in T. The universal property of T implies there is an involutory antiisomorphism to of T with (xl 8 . . . 8 xm)to = x, 8 . - . @3 xl. As to preserves K it induces t on C. Thus (8) holds. Part (3) is a direct consequence of the definition of C, since x 8 x - Q(u)l E K. An easy induction argument using (3) shows e s e ~ is a linear combination of the elements e ~ , R X, for each S, T X, so (eR: R X) spans C. Using the universal property of the tensor algebra, Chevalley shows on page 39 of [Ch 11 that there is a homomorphic image of C of dimension 2", completing the proof of (2), and hence also of (1) and (4). I omit this demonstration.

Part (5) is a straightforward consequence of (3). If char(F) # 2 then (3) shows e: = (-l),es for x E X and S G X, where m = IS] if x $ S and m = IS1 - 1 if x E S. Since Xp generates C as an F-algebra, (7) follows in this case. If char(F) = 2 then choose X so that each of its members is nonsingular. Then (3) shows [e,, es] = 0 if S E xL, while [ex, es] = Q(x)-'es+, if S contains the unique y in X - xL, where S + y is the symmetric difference of S with {y}. It follows that [ex, C] = (es: S G xL) is of dimension 2'+'. So, as 2"-1 = dim(C)/2 and ex is an involution, [ex, C] = Cc(ex). Thus (7) holds in

this case too. Let G and n be as in (6). For g E G and v E V, Q(vgn)l = ((vp)s)' =

((vp)')g = Q(v)l, so Q(vgn) = Q(v). Hence G n 5 O(V, Q). Let Go be the

subgroup of G generated by the elements up as u varies over the nonsingularvectors of V. By (5), up,-r = -ru, so, by 22.7, G,-r = Gon = O(V, Q) ifchar(F) = 2. Further if char(F) 2 and n is even then -1 is a product ofelements -ru, so ru c Go7r and, again by 22.7, G7r = Goir = O(V, Q). Finallyif n is odd then det(-ru) = +1, so Goir < SO(V, Q), and then, as O(V, Q) =(-I) x SO(V, Q), 22.7 says G07r = SO(V, Q). Then to complete the proofof (6) it remains only to observe that, by (7), -1 G7r, so G0ir = G7r.

(22.9) Let (V, Q) be an orthogonal space, C = C(Q) its Clifford algebra, Gthe Clifford group of (V, Q), and G+ = G fl Co the special Clifford group. Let7r be the representation of 22.8.6. Then

(1) G+ is a subgroup of G of index 2.(2) If char(F) 2 then G+ 7r = SO(V, Q).(3) If char(F) = 2 then G+ 7r is of index 2 in O(V, Q).(4) G+n contains no transvections or reflections.

Proof. Part (1) is a consequence of 22.8.4. If n is even then, by 22.8.7, ker(7r) <G+ and, by 22.8.6, G,-r = O(V, Q), so G+ 7r is of index 2 in O(V, Q). Also-1 E G+n by the proof of 22.8, so by 22.8.6 each transvection or reflectionru is not in G+,-r. Thus the lemma holds in this case as reflections are notin SO(V, Q). If n is odd then Gir = SO(V, Q) by 22.8.6, while, by 22.8.7,G = G+ker(,-r). So again the lemma holds.

(22.10) Let v1, ... , vm be nonsingular vectors in the orthogonal space (V, Q)such that r,,, ... r,,, = 1. Then the product Q(vl) ... Q(vm) is a square in F.

Proof. Let c = vl p ... vm p. As r,,, ... r,,,, = 1, m is even, because by 22.9 thereis a subgroup of O(V, Q) of index 2 containing no transvection or reflection.Hence c7r = (-1)m r,,, . . . r,,, =1 by 22.8.6. So C E ker(7r ). Indeed, as m is even,

c E G+ fl ker(7r) = F# 1 by 22.8.7. So c = a 1 for some a c F#.Let t be the antiisomorphism of 22.8.8. It follows that c(ct) = c2 = a2 1.

On the other hand c(ct) = vi p ... vm pvm p ... v1 p as t is an antiisomorphism.Further (vi p)2 = Q (vi) 1, so c(ct) _ (Q(v1)... Q (vm)) 1, completing the proof.

Let F2 = {a2: a E F#} be the subgroup of squares in F#, and consider thefactor group F#/F2. For example if F is a finite field of odd order then F#/F2is of order 2, while if F is perfect of characteristic 2 then F# = F2. Definea map 9: O(V, Q) -+ F#/F2 as follows. For g E O(V, Q), g = rx, . . . r,,,,, forsuitable transvections or reflections rX, with center (xi). (Except in the ex-ceptional case of 22.7, which I'll ignore.) Define 9(g) = Q(xl) ... Q(xm)F2.Observe first that Q(axi) =a 2 Q(xi) E Q(xi )F2, so the definition of 9 is in-dependent of the choice of generator xi of (xi). Also if g = ry, . . . ryk then

subgroup of G generated by the elements up as u varies over the nonsingular vectors of V. By (5 ) , upn = -r,, so, by 22.7, Gn = Gon = O(V, Q) if char(F) = 2. Further if char(F) # 2 and n is even then -I is a product of elements -r,, sor, E Gon and, again by 22.7, G n = Gon = O(V, Q). Finally if n is odd then det(-r,) = +1, so Gon 5 SO(V, Q), and then, as O(V, Q) =

(-I) x SO(V, Q), 22.7 says Gon = SO(V, Q). Then to complete the proof of (6) it remains only to observe that, by (7), -I $ Gn, so Gon = Gn.

(22.9) Let (V, Q) be an orthogonal space, C = C(Q) its Clifford algebra, G the Clifford group of (V, Q), and G+ = G n Co the special Clifford group. Let n be the representation of 22.8.6. Then

(1) Gf is a subgroup of G of index 2. (2) If char(F) # 2 then G+n = SO(V, Q). (3) If char(F) = 2 then G+n is of index 2 in O(V, Q). (4) Gf n contains no transvections or reflections.

Proof. Part (1) is aconsequence of 22.8.4. If n is even then, by 22.8.7, ker(n) 5 G+ and, by 22.8.6, G n = O(V, Q), so G+n is of index 2 in O(V, Q). Also -I E Gf n by the proof of 22.8, so by 22.8.6 each transvection or reflection r, is not in G+n. Thus the lemma holds in this case as reflections are not in SO(V, Q). If n is odd then G n = SO(V, Q) by 22.8.6, while, by 22.8.7, G = G+ker(n). So again the lemma holds. 9

(22.10) Let vl, . . . , urn be nonsingular vectors in the orthogonal space (V, Q) such that r,, . . . r,,,, = 1. Then the product Q(vl) . . . Q(vrn) is a square in F.

Proof. Let c = v l p . . . vmp. As r,, . . . r, = 1, m is even, because by 22.9 there is a subgroup of O(V, Q) of index 2 containing no transvection or reflection. Hence cn = (- l)"r,, . . . r,,, = 1 by 22.8.6. So c E ker(n). Indeed, as m is even, c E G+ n ker(n) = F# . 1 by 22.8.7. So c = a . 1 for some a E F#.

Let t be the antiisomorphism of 22.8.8. It follows that c(ct) = c2 = a2 . 1. On the other hand c(ct) = vlp . . . v,pvrnp .. . vlp as t is an antiisomorphism. Further (vip)' = Q(vi)l, so c(ct) = (Q(v1) . . . Q(vrn))l, completing the proof.

Let F2 = {a2: a E F#} be the subgroup of squares in F#, and consider the factor group F#/F2. For example if F is a finite field of odd order then F#/F2 is of order 2, while if F is perfect of characteristic 2 then F# = F2. Define a map 8: O(V, Q) + F'#/F2 as follows. For g E O(V, Q), g = r,, . . . r,,, for suitable transvections or reflections r,, with center (xi). (Except in the ex- ceptional case of 22.7, which I'll ignore.) Define B(g) = Q(xl) . . . Q(x,)F'. Observe first that ~ ( a x i ) =a2Q(xi) E Q(x~)F', SO the definition of 8 is independent of the choice of generator xi of (xi). Also if g = r,, . . . r,, then


1 = r,,, ... r,,,,, rYk ... ry, , so, by 22. 10, Q (xi) ... Q (xm) F2 = Q (yi) ... Q (Yk)F2Thus 0 is independent of the choice of transvections and reflections too. 0 iscalled the spinor norm of O(V, Q). From the preceding discussion it is evidentthat.

(22.11) The spinor norm 0 is a group homomorphism of O(V, Q) into F#/F2.

(22.12) If Q is not definite then the spinor norm maps G+n surjectively ontoF#/F2.

Proof. V contains a hyperbolic plane U and for each a E F# there exist u,v c U with Q(v) = 1 and Q(v) = a. Now a.

(22.13) Let (V, Q) be a hyperbolic orthogonal space, let r = F(V) be the setof maximal totally singular subspaces of V, and define a relation - on r byA - B if dim(A/(A fl B)) is even. Then - is an equivalence relation withexactly two equivalence classes.

Proof. Given a triple A, B, C of members of r define

S(A, B, C) = dim(A/(A fl B)) + dim(A/(A fl C)) + dim(B/(B fl c)).

Observe that the lemma is equivalent to the assertion that S(A, B, C) is evenfor each triple A, B, C from F.

Assume the lemma is false and pick a counterexample V with n minimal.As V is hyperbolic, n = 2m is even. If m = 1 then IF I = 2, so the result holds.Thus n > 1. Let A, B, C E F with S(A, B, C) odd.

Let D = A fl B fl C and suppose D 0. Let U = D1 and U = U/ D. As wehave seen several times already, Q induces a quadratic form Q on U. FurtherU is hyperbolic with A, B, C E I'(U), so, by minimality of V, 8(A, B, C) iseven. As S(A, B, C) = S(A, B, C) we have a contradiction to the choice ofA, B, C. Hence D=0.

Suppose next E = A fl B 0 and let Co = (C fl E1) + E. By 19.2,Co E F. 0 E = A fl B fl Co, so, by a previous case, S(A, B, Co) is even.But X fl Co = (X fl C) + E and X fl C fl E = 0 for X = A and B, soS(A, B, C) - S(A, B, Co) mod 2, again a contradiction.

Thus S(A, B, C) = 3m. Hence m is odd. Therefore if T, S E F with T tiA S then A fl T 0, so, by the last case, T - S. Hence - is an equivalencerelation. Finally let R E F with A fl R a hyperplane of A. Then 0 A fl R,so, by the last case, 8(A, B, R) is even, and hence B R. This shows - hastwo classes and completes the proof.


1 = r,, . . . r,,,,r, . . . ry,,so,by22.10, Q(xl). . . Q(xm>F2 = Q(y1). . . e(yk)F2. Thus 0 is independent of the choice of transvections and reflections too. 0 is called the spinor norm of O(V, Q). From the preceding discussion it is evident that.

(22.11) The spinor norm 0 is a group homomorphism of O(V, Q) into F#/F2.

(22.12) If Q is not definite then the spinor norm maps G+n surjectively onto F#/F2.

Proof. V contains a hyperbolic plane U and for each a E F# there exist u ,

v E U with Q(v) = 1 and Q(v) = a. Now O(r,ru) = a.

(22.13) Let (V, Q) be a hyperbolic orthogonal space, let r = r (V) be the set of maximal totally singular subspaces of V, and define a relation -- on r by A -- B if dim(A/(A n B)) is even. Then -- is an equivalence relation with exactly two equivalence classes.

Proof. Given a triple A, B, C of members of r define

6(A, B, C) = dim(A/(A n B)) + dim(A/(A n C)) + dim(B/(B n C)).

Observe that the lemma is equivalent to the assertion that 6(A, B, C) is even for each triple A, B, C from r.

Assume the lemma is false and pick a counterexample V with n minimal. As V is hyperbolic, n = 2m is even. If m = 1 then )r I = 2, so the result holds. Thus n > 1. Let A, B, C E r with 6(A, B, C) odd.

let^ = A n ~ n C a n d s u p p o s e ~ # 0 . ~ e t U = ~ ' a n d u = U/D.Aswe have seen several times already, Q induces a quadratic form Q on 0 . Further 0 is hyperbolic with A, B , C E r(U), so, by minimality of V, 6(A, B , C) is even. As 6(A, B, C) = 6(A, B, C) we have a contradiction to the choice of A, B, C.Hence D = 0.

Suppose next E = A n B # 0 and let Co = (C n E') + E. By 19.2, Co E r. 0 # E = A n B n Co, so, by a previous case, 6(A, B, Co) is even. But X n Co = (X n C) + E and X n C n E = 0 for X = A and B, so

6(A, B, C) 6(A, B, Co) mod 2, again a contradiction. Thus 6(A, B, C) = 3m. Hence m is odd. Therefore if T, S E r with T --

A -- S then A n T # 0, so, by the last case, T -- S. Hence -- is an equivalence relation. Finally let R E r with A n R a hyperplane of A. Then 0 # A n R, so, by the last case, 6(A, B, R) is even, and hence B -- R. This shows -- has two classes and completes the proof.

(22.14) Let (V, Q) be a hyperbolic orthogonal space, let G = O(V, Q), andlet H be the subgroup of G preserving the equivalence relation of 22.13. Then

(1) IG: HI = 2.(2) H is the image of the special Clifford group under the map of 22.8.6.(3) Reflections and transvections are in G - H.

Proof. By Witt's Lemma, G is transitive on the set r of maximal totally singularsubspaces of V, so IG: HI is the number of equivalence classes of r. That isI G: H I = 2 by 22.13. It's easy to check that (3) holds. Then, by (1), (3), and22.7, H is the subgroup of G consisting of the elements which are the productof an even number of transvections or reflections, while by 22.9 this subgroupis the image of the special Clifford group under the map of 22.8.6.

I close this section with a brief discussion of some geometries associated to theclassical groups. A few properties of these geometries are derived in Exercise7.8, while chapter 14 investigates these geometries in great detail.

Assume the Witt index m of (V, f) or (V, Q) is positive. In this case thereare some interesting geometries associated to the space and preserved by itsisometry group. The reader may wish to refer to the discussion in section 3 ongeometries.

The polar geometry r of (V, f) or (V, Q) is the geometry over I =10, 1, ... , m - 1 } whose objects of type i are the totally singular subspaces ofV of projective dimension i, with incidence defined by inclusion. EvidentlyO(V, f) is represented as a group of automorphisms of F. Indeed the similaritygroup O(V, f) is also so represented.

If (V, Q) is a hyperbolic orthogonal space there is another geometry associ-ated to (V, Q) which is in many ways nicer than the polar geometry. Assume thedimension of V is at least 6, so that the Witt index m of (V, Q) is at least 3. Theoriflamme geometry r of (V, Q) is the geometry over I = 10, 1, ... , m - l}whose objects of type i < m - 2 are the totally singular subspaces of projectivedimension i, and whose objects of types m - 1 and m - 2 are the two equiv-alence classes of maximal totally singular subspaces of (V, Q) defined by theequivalence relation of 22.13. Incidence is inclusion, except between objectsU and W of type m - 1 and m - 2, which are incident if U fl W is a hyper-plane of U and W. In this case the subgroup of O(V, Q) of index 2 preservingthe equivalence relation of 22.13 is represented as a group of automorphismsof F.

Remarks. The standard reference for much of the material in this chapter isDieudonne [Di]. In particular this is a good place to find out who first provedwhat in the subject.

(22.14) Let (V, Q) be a hyperbolic orthogonal space, let G = O(V, Q), and let H be the subgroup of G preserving the equivalence relation of 22.13. Then

(1) IG: HI = 2. (2) H is the image of the special Clifford group under the map of 22.8.6. (3) Reflections and transvections are in G - H .

Proof. By Witt's Lemma, G is transitive on the set r of maximal totally singular subspaces of V, so IG: HI is the number of equivalence classes of r. That is IG: HI = 2 by 22.13. It's easy to check that (3) holds. Then, by (I), (3), and 22.7, H is the subgroup of G consisting of the elements which are the product of an even number of transvections or reflections, while by 22.9 this subgroup is the image of the special Clifford group under the map of 22.8.6.

I close this section with a brief discussion of some geometries associated to the classical groups. A few properties of these geometries are derived in Exercise 7.8, while chapter 14 investigates these geometries in great detail.

Assume the Witt index m of (V, f ) or (V, Q) is positive. In this case there are some interesting geometries associated to the space and preserved by its isometry group. The reader may wish to refer to the discussion in section 3 on geometries.

The polar geometry r of (V, f ) or (V, Q) is the geometry over I = (0, 1, . . . , m - 1) whose objects of type i arq the totally singular subspaces of V of projective dimension i, with incidence defined by inclusion. Evidently O(V, f ) is represented as a group of automorphisms of r. Indeed the similarity group A(V, f ) is also so represented.

If (V, Q ) is a hyperbolic orthogonal space there is another geometry associated to (V, Q) which is in many ways nicer than the polar geometry. Assume the dimension of V is at least 6, so that the Witt index m of (V, Q) is at least 3. The orijlamme geometry r of (V, Q) is the geometry over I = (0, 1, . . . , m - 1) whose objects of type i < m - 2 are the totally singular subspaces of projective dimension i, and whose objects of types m - 1 and m - 2 are the two equivalence classes of maximal totally singular subspaces of (V, Q) defined by the equivalence relation of 22.13. Incidence is inclusion, except between objects U and W of type m - 1 and m - 2, which are incident if U fl W is a hyperplane of U and W. In this case the subgroup of A(V, Q) of index 2 preserving the equivalence relation of 22.13 is represented as a group of automorphisms of r.

Remarks. The standard reference for much of the material in this chapter is DieudonnC [Di]. In particular this is a good place to find out who first proved what in the subject.

We will encounter groups generated by reflections again in sections 29and 30.

Observe that, by 13.8, 22.4.4, 22.4, and Exercise 7.6, almost all the finiteclassical groups SLn(q), Spn(q), Q '(q), and SUn(q) are perfect. This fact isused to prove in 43.12 that the projective groups PSL,,(q), PSpn(q), PQ (q),and PSUn(q) are simple unless n and q are small.

Since by some measure most finite simple groups are classical, the studyof the classical groups is certainly important. Moreover along with Lie theory(cf. chapter 14 and section 47) the representations of the classical groups ontheir associated spaces is the best tool for studying the classical groups. On theother hand the study of the classical groups is a special topic and the materialin this chapter is technical. Thus the casual reader may wish to skip, or at leastpostpone, this chapter.

Exercises for chapter 71. Let V be an n-dimensional vector space over F. Prove:

(1) If n = 2 then SL(V) = Sp(V).(2) If 9 is an automorphism of F of order 2, n = 2, and (V, f) is a

hyperbolic unitary space, then SL(V) n O(V, f) = SL2(Fix(9)).(3) Let IF I = q2 < oo and (V, f) a 3-dimensional unitary space over F.

Then there exists a basis X of V such that

J(X, f) _ 0 1 0

0 0

'011 0

1 a c

Mx(g) = 0 1 b

0 0 1

then P 1 = q3 and [P, P] is a root group of SU(V).(4) Let IF! = q < oo and (V, f) a 2m-dimensional symplectic space

over F with m > 1. Then there exists a basis X = (xi: 1 < i < 2m)such that

J(X, f) _0 Im

-IM 0 /where In is the m by m identity matrix. Moreover if P consists of


We will encounter groups generated by reflections again in sections 29 and 30.

Observe that, by 13.8, 22.4.4, 22.4, and Exercise 7.6, almost all the finite classical groups SL,(q), Sp,(q), Qi(q), and SU,(q) are perfect. This fact is used to prove in 43.12 that the projective groups PSL,(q), PSp,(q), PQ:(q), and PSU,(q) are simple unless n and q are small.

Since by some measure most finite simple groups are classical, the study of the classical groups is certainly important. Moreover along with Lie theory (cf. chapter 14 and section 47) the representations of the classical groups on their associated spaces is the best tool for studying the classical groups. On the other hand the study of the classical groups is a special topic and the material in this chapter is technical. Thus the casual reader may wish to skip, or at least postpone, this chapter.

Exercises for chapter 7 1. Let V be an n-dimensional vector space over F . Prove:

(1) If n = 2 then SL(V) = Sp(V). (2) If 8 is an automorphism of F of order 2, n = 2, and (V, f ) is a

hyperbolic unitary space, then SL(V) n O(V, f ) E SLz(Fix(8)). (3) Let I FI = q2 < oo and (V, f ) a 3-dimensional unitary space over F .

Then there exists a basis X of V such that

Moreover if P consists of those g E SU(V) with

then I PI = q3 and [ P , PI is a root group of SU(V). (4) Let IF] = q < oo and (V, f ) a 2m-dimensional symplectic space

over F with m > 1. Then there exists a basis X = (xi: 1 F: i 5 2m) such that

where I, is the m by m identity matrix. Moreover if P consists of

those g c Sp(V) with

Mx(g) =1 0 0

a 12._2 0b c I

where a and c are column and row vectors, respectively, then I P I =q2'-1. If q is odd, [P, P] is a root group of Sp(V). If q = 2 an d m = 3then P contains a transvection and P = [P, H] where H = Sp(U)and U = (XI, x2m)1

2. Let (V, Q) be a 2-dimensional orthogonal space over F. Prove:(1) O(V, Q) is the semidirect product of a subgroup H by (r) = Z2, where

r inverts H.(2) Each element of O(V, Q) - H is a transvection or reflection. In par-

ticular O(V, Q) is generated by such elements.(3) If O(V, Q) is hyperbolic then H is isomorphic to the multiplicative

group F# of F.(4) If (V, Q) is definite then there exists a quadratic Galois extension K of

F such that (V, Q) is similar to (K, NF) and H - {a E K: aae = I),where (0) = Gal(K/F).

3. Let (V, f) be a 4-dimensional unitary space over a field F of order 4, Xan orthonormal basis for V, A _ {(x):x c X}, and G = SU(V). ProveNG(A)° - S4, Go = E27, and NG(A) s generated by transvections. LetD E A, T the subgroup of G generated by the transvections in G, and F theset of conjugates of A under NG(D). Prove NG(D)r - A4 and I Gr = 54.Prove NG(D) < T.

4. Let q be a prime power. Prove(1) Z(GU,,(q)) and are cyclic of order q + 1.(2) Z(SU,(q)) and PGU,(q)/ U,,(q) are cyclic of order (q + 1, n).

5. Assume the hypothesis and notation of Exercise 4.7 with char(F) 2. LetW = V3, a = 7r3, and define Q: W --> F by

Q(ax2 + bxy + cy2) = b2 - 4ac.

Prove(1) Q is a nondegenerate quadratic form on W with bilinear form (ax 2 +

bxy + cy2, rx2 + sxy + ty2) = 2bs - 4(at + cr).(2) For each g E G, ga is a similarity of (W, Q) with A(ga) = det(g)2.(3) (Ga)S is the group A(W, Q) of all similarities of (W, Q), where S is

the group of scalar maps on W.(4) Up to similarity, (W, Q) is the unique 3-dimensional nondefinite or-

thogonal space over F.

those g E Sp(V) with

where a and c are column and row vectors, respectively, then I P I = qZm-'. If q is odd, [P, PI is a root group of Sp(V). If q = 2 and m = 3 then P contains a transvection and P = [P, HI where H = Sp(U) and U = (xr, ~2,)'.

2. Let (V, Q) be a 2-dimensional orthogonal space over F. Prove: (1) O(V, Q) is the semidirect product of a subgroup H by ( r ) E Z2, where

r inverts H. (2) Each element of O(V, Q) - H is a transvection or reflection. In par-

ticular O(V, Q) is generated by such elements. (3) If O(V, Q) is hyperbolic then H is isomorphic to the multiplicative

group F# of F . (4) If (V, Q) is definite then there exists a quadratic Galois extension K of

F such that (V, Q) is similar to (K, N;) and H E {a E K: aae = 11, where (0) = Gal(K/F).

3. Let (V, f ) be a 4-dimensional unitary space over a field F of order 4, X an orthonormal basis for V, A = {(x): x E X I , and G = SU(V). Prove N ~ ( A ) ~ E S4, GA E EZ7, and NG(A) is generated by transvections. Let D E A, T the subgroup of G generated b i the transvections in G, and I? the set of conjugates of A under NG(D). Prove N ~ ( D ) ~ 2 A4 and )Gr 1 = 54. Prove Nc(D) 5 T.

4. Let q be a prime power. Prove (1) Z(GU,(q)) and GU,(q)/SU,(q) are cyclic of order q + 1. (2) Z(SU,(q)) and PGU,(q)/ U,(q) are cyclic of order (q + 1, n).

5. Assume the hypothesis and notation of Exercise 4.7 with char(F) # 2. Let W = V3, a = n3, and define Q: W + F by

e(ax2 + bxy + cy2) = b2 - 4ac.

Prove (1) Q is a nondegenerate quadratic form on W with bilinear form (ax2 +

bxy + cy2, rx2 + sxy + ty2) = 2bs - 4(at + cr). (2) For each g E G, g a is a similarity of (W, Q) with h(ga) = det(g)2. (3) (Ga)S is the group A(W, Q) of all similarities of (W, Q), where S is

the group of scalar maps on W. (4) Up to similarity, (W, Q) is the unique 3-dimensional nondefinite or-

thogonal space over F .

(5) If F is finite or algebraically closed every 3-dimensional orthogonalspace over F is similar to (W, Q).

(6) (rrh: r E R, h E Get) = O(W, Q)(l) = L2(F), where R is the set ofreflections in O(W, Q).

6. Let (V, Q) be an n-dimensional orthogonal space over a field F with n > 3.(1) Assume (V, Q) is not definite and if IF I < 3 and n < 4 assume n = 4

and sgn(Q) _ -1. Prove the following subgroups are equal.

(i) n(V, Q).(ii) The kernel in G+ 7r of the spinor norm, where G+ is the special

Clifford group.(iii) (rrg: r reflection or transvection, g E O(V, Q)),

(2) If char(F) 0 2 prove Q(V, Q) is perfect unless IFI = 3 and eithern=3orn =4andsgn(Q) _+1.IfFisfinite prove O(V, Q)/Q(V, Q)

E4, and -I E Q(V, Q) if and only if n is even and sgn(Q) - IFI"/zmod 4.

(3) If char(F) = 2 and F is perfect, prove either Q(V, Q) is perfect andIO(V, Q): SZ(V, Q)j = 2 or IFI = 2, n = 4, and sgn(Q) = +1.

(Hint: To prove Q(V, Q) perfect use (1) and show rrg is contained in aperfect subgroup of O(V, Q) for each reflection or transvection r and eachg E G. Toward that end use Exercise 7.5 in (2) and 22.5 in (3).)

7. Let G be a permutation group on a set I of finite order n and V thepermutation module for G over F with G-invariant basis X = (x1: i E I).Define a bilinear form f on V by f (x1, xj) = Sjj (the Kronecker delta)for i, j e I. Let z = F_ZEI xi, U the core of the permutation moduleV, and V = V/(z). If char(F) = 2 define a quadratic form Q on U byQ(F_ a;x1) _ E a? + Ei<j a1ap Prove:(1) U=z1.(2) G < O(V, f).(3) If char(F) 54 2 then (V, f) is an orthogonal space. If char(F) = 2

then (U, f) is a symplectic space preserved by G and if further b 2

mod 4 then (U, Q) is an orthogonal space preserved by G, wheref (u, v) = f (u, v) and Q(u) = Q(u) for u, v E U.

(4) If G = S6, n = 6, and IF) = 3 then (U, f) is of sign -1, O(U, f) _(-I) x G, and A6 - Q4 -Q).

(5) If G = S5, n = 5, and BFI = 2, then (U, Q) is of sign -1 and0 ( 0 ,G0 4- ( 2 ) .

(6) IfG=S6,n=6, and IF I =2,then G=O(U, f) =Sp4(2).(7) If G = S8, n = 8, and IFI = 2, then G = O(U, Q) - O6+(2).

8. Let (V, f) be a symplectic or unitary space over F, or (V, Q) an orthogonalspace over F. Assume the Witt index m of the space is positive. Let r be thepolar geometry of the space over I = {0, 1, ... , m - 1) and G the isometry


(5) If F is finite or algebraically closed every 3-dimensional orthogonal space over F is similar to (W, Q).

(6) (rrh: r E R, h E Ga) = A(W, Q)") S L2(F), where R is the set of reflections in O(W, Q).

6. Let (V, Q ) be an n-dimensional orthogonal space over a field F with n 2 3. (1) Assume (V, Q) is not definite and if I FI 5 3 and n 5 4 assume n = 4

and sgn(Q) = - 1. Prove the following subgroups are equal.

(9 Q(V, Q). (ii) The kernel in G+n of the spinor norm, where G+ is the special

Clifford group. (iii) (rrg: r reflection or transvection, g E O(V, Q)).

(2) If char(F) # 2 prove Q( V, Q) is perfect unless I F I = 3 and either n = 3 orn = 4andsgn(Q) = +l.If F is finiteproveO(V, Q)/ Q(V, Q) S E4, and -I E Q(V, Q) if and only if n is even and sgn(Q) r 1 F In/' mod 4.

(3) If char(F) = 2 and F is perfect, prove either Q(V, Q) is perfect and IO(V, Q): Q(V, Q)i = 2 or IF1 = 2, n = 4, and sgn(Q) = +l .

(Hint: To prove Q(V, Q) perfect use (1) and show rrg is contained in a perfect subgroup of O(V, Q) for each reflection or transvection r and each g E G. Toward that end use Exercise 7.5 in (2) and 22.5 in (3).)

7. Let G be a permutation group on a set I of finite order n and V the permutation module for G over F with G-invariant basis X = (xi: i E I). Define a bilinear form f on V by f (xi, xi) = Sij (the Kronecker delta) for i, j E I . Let z = CiEI xi, U the core of the permutation module V, and ii = V/(z). If char(F) = 2 define a quadratic form Q on U by

Q ( 1 aixi) = 1 a: + aiaj. Prove: (1) u = zL.

(2) G 5 OW, f 1. (3) If char(F) # 2 then (V, f ) is an orthogonal space. If char(F) = 2

then (U, f ) is a symplectic space preserved by G and if further b f 2 mod 4 then (U, (2) is an orthogonal space preserved by G, where f(ii, fi) = f (u, v) and ~ ( i i ) = Q(u) for u , v E U .

(4) If G = S6, n = 6, and IF1 = 3 then ( 0 , f ) is of sign -1,O(o, f ) = ( - I ) x G, and A6 QT(3).

(5) If G = S5, n = 5, and (F I = 2, then ( 0 , (2) is of sign - 1 and O(0, Q) = G 2 0,(2).

(6) If G = S6, n = 6, and IF1 = 2, then G = 0(0, f ) 2 Sp4(2). (7) If G = S8, n = 8, and IF1 = 2, then G = 0(0, Q) S 0:(2).

8. Let (V, f ) be a symplectic or unitary space over F , or (V, Q) an orthogonal space over F. Assume the Witt index m of the space is positive. Let r be the polar geometry of the space over I = {0, 1, . . . , m - 1) and G the isometry

group of the space or (V, Q) a hyperbolic orthogonal space with m >3, 1' the oriflamme geometry of (V, Q), and G the subgroup of O(V, Q)preserving the equivalence relation of 22.13. Let Z be a maximal hyperbolicsubspace of V, X = (xi: 1 < i < 2m) a hyperbolic basis for Z, Vi =(x2j_1: 1 < j < i), and Y = {(x):x E X}. Let T = {Vi: 1 < i < m) if ris the polar geometry and T = {Vi, 1 1

then G is rank 3 on these points.(2) G is flag transitive on r.(3) T is a flag of r of type I.(4) B = GT is the semidirect product of U with H = Gy, where U is

the subgroup of G centralizing V1, Vi+1 / Vi, 1 < i < m - 1, and(a) Vm/Vm_i and (V.)'/ V,, if r is a polar space, or(b) (Vm_i n Vm)/Vm_2 and (Vm_1 + V,,,)/(Vm_i n vif r is an

oriflamme geometry.(5) U is nilpotent and H is the direct product of m copies of F# with

O(Z1-, f) (or O(Z1, Q)).(6) Let i E I. Then either U fixes a unique object of type i in r or r is

a polar geometry and V is a hyperbolic orthogonal space.(7) B = NG(U).(8) Assume F is finite of characteristic p and r is oriflamme if V is

hyperbolic orthogonal. Then U E Sy1p(G).(9) NG (Y) Y is 7L2wrSm or of index 2 in that group, for r a polar space or

oriflamme geometry, respectively.(10) Let S be a flag of corank 1 in T. Then either the residue I's of S is

isomorphic to the projective line over F and (Gs)r, - PGL2(F) orL2(F), or r is a polar geometry, S is of type {0, ... , m - 2}, I's isisomorphic to the set of singular points of W = (V,,-,)-L/ V,,-,, andeither (Gs)1'S PO(W, f) (or PO(W, Q)) or V is hyperbolic orthog-onal and I I's I = 2.

9. Let V be a finite dimensional vector space over a field F and f a nontrivialsesquilinear form on V. Then(1) If char(F) # 2 and f is bilinear then f = g + h where g and h are sym-

metric and skew symmetric forms on V, respectively, and O(V, f) <O(V, g) n O(V, h).

(2) If char(F) = 2 and f is bilinear there exists a nontrivial symmetricbilinear form g on V with O(V, f) < O(V, g).

(3) Let char(F) = 2 and assume f is bilinear and symmetric. Let U ={x E V : f (x, x) = 0). Prove U is a subspace of V which is of codi-mension at most 1 if F is perfect.

group of the space or (V, Q) a hyperbolic orthogonal space with m 2 3, r the oriflamme geometry of (V, Q), and G the subgroup of O(V, Q) preserving the equivalence relation of 22.13. Let Z be a maximal hyperbolic subspace of V , X = (xi: 1 5 i 5 2m) a hyperbolic basis for Z, Vi = (xzj-l: 1 5 j 5 i) , and Y = {(x):x E X}. Let T = {K: 1 5 i 5 m) if r is the polar geometry and T = {Vi, Vh-l: 1 5 i 5 m, i # m - 1) if r is the oriflamme geometry, where VhW1 = (V,-2, X~rn-3, ~2,). Prove

(1) If m = 1 then G is 2-transitive on the points of r , while if m > 1 then G is rank 3 on these points.

(2) G is flag transitive on r . (3) T is a flag of r of type I. (4) B = GT is the semidirect product of U with H = Gy, where U is

the subgroup of G centralizing Vl , Vi+1 / Vi , 1 5 i < m - 1, and (a) V,/ Vm-l and (V,)'/ V, if r is a polar space, or (b) (V,-l n V,)/V,-z and (V,-1 + V,)/(V,-l f l V,) if r is an

oriflamme geometry. (5) U is nilpotent and H is the direct product of m copies of F' with

O(Z', f > (or O(Z'? Q)). (6) Let i E I. Then either U fixes a unique object of type i in r or r is

a polar geometry and V is a hyperbolic orthogonal space. (7) B = NG(U). (8) Assume F is finite of characteristi9 p and r is oriflamme if V is

hyperbolic orthogonal. Then U E Syl,(G). (9) NG(y)' is Z2wrS, or of index 2 in that group, for r a polar space or

oriflamme geometry, respectively. (10) Let S be a flag of corank 1 in T. Then either the residue rs of S is

isomorphic to the projective line over F and (Gs)~. S PGLz(F) or L2(F), or r is a polar geometry, S is of type {O, . . . , m - 21, rs is isomorphic to the set of singular points of W = (v,-~)'/ V,-l, and either ( G S ) ~ ~ S PO(W, f ) (or PO(W, Q)) or V is hyperbolic orthogonal and 1 rs 1 = 2.

9. Let V be a finite dimensional vector space over a field F and f a nontrivial sesquilinear form on V. Then (1) If char(F) # 2 and f is bilinear then f = g + h where g and h are sym-

metric and skew symmetric forms on V, respectively, and O(V, f ) 5

OW, g) n OW, h). (2) If char(F) = 2 and f is bilinear there exists a nontrivial symmetric

bilinear form g on V with O(V, f ) i O(V, g). (3) Let char(F) = 2 and assume f is bilinear and symmetric. Let U =

{x E V: f (x, x) = O}. Prove U is a subspace of V which is of codimension at most 1 if F is perfect.

(4) Assume f is sesquilinear with respect to the involution 0 and f is skewhermitian; that is f (x, y) = -f (y, x)° for all x, y E V. Prove f issimilar to a hermitian form.

(5) If f is sesquilinear with respect to an involution 0, then there exists anontrivial hermitian symmetric form g on V with O(V, f) < O(V, g).

10. Let F be a field and f a sesquilinear form on V with respect to the auto-morphism 0 of F, such that for all x, y e V, f (x, y) = 0 if and only iff (y, x) = 0. Prove that either(1) f (x, x) = 0 for all x E V, 0 = 1, and f is skew symmetric, or(2) there exists x E V with f (x, x) 0 0 and one of the following holds:

(a) 0 = 1 and f is symmetric.(b) 191 = 2 and f is similar to a hermitian symmetric form.(c) X01 > 2 and Rad(V) is of codimension 1 in V.


(4) Assume f is sesquilinear with respect to the involution 8 and f is skew hermitian; that is f (x, y) = - f (y, x)' for all x, y E V. Prove f is similar to a hermitian form.

(5) I f f is sesquilinear with respect to an involution 8, then there exists a nontrivial hermitian symmetric form g on V with O(V, f ) 5 O(V, g).

10. Let F be a field and f a sesquilinear form on V with respect to the automorphism 8 of F, such that for all x, y E V, f (x, y ) = 0 if and only if f (y, x) = 0. Prove that either (1) f (x, x) = 0 for all x E V, 8 = 1, and f is skew symmetric, or (2) there exists x E V with f (x, x) # 0 and one of the following holds:

(a) 8 = 1 and f is symmetric. (b) 181 = 2 and f is similar to a hermitian symmetric form. (c) 18 1 2 and Rad(V) is of codimension 1 in V.

p-groups

Chapter 8 investigates p-groups from two points of view: first through a study ofp-groups which are extremal with respect to one of several parameters (usuallyconnected with p-rank) and second through a study of the automorphism groupof the p-group.

Recall that if p is a prime then the p-rank of a finite group is the maximumdimension of an elementary abelian p-subgroup, regarded as a vector spaceover GF(p). Section 23 determines p-groups of p-rank 1, p-groups in whicheach normal abelian subgroup is cyclic, and, for p odd, p-groups in which eachnormal abelian subgroup is of p-rank at most 2. Perhaps most important, thep-groups of symplectic type are determined (a p-group is of symplectic typeif each of its characteristic abelian subgroups is cyclic).

The Frattini subgroup is introduced to study p-groups and their automor-phisms. Most attention is focused on p'-groups of automorphisms of p-groups;a variety of results on the action of p'-groups on p-groups appear in section24. One very useful result is the Thompson A,x B Lemma. Also of importanceis the concept of a critical subgroup.

23 Extremal p-groupsIn this section p is a prime and G is a p-group.

The Frattini subgroup of a group H is defined to be the intersection of allmaximal subgroups of H. (D(H) denotes the Frattini subgroup of H.

(23.1) (1) '1(H) char H.(2) If X C H with H = (X, (D(H)), then H = (X).(3) If H/(D(H) is cyclic, then H is cyclic.

(23.2) If G is a p-group then (D (G) is the smallest normal subgroup H of Gsuch that G/H is elementary abelian.

Proof. If M is a maximal subgroup of G then, by Exercise 3.2, M 4 G andI G : MI = p, so, by 8.8, G(1 < M. Hence G(1) < t (G), so, by 8.8, GI (D (G)is abelian. Also, as G/M = Zp, gP E M for each g e G. so 9° E (D (G). HenceG/(D(G) is elementary abelian.

p-groups

Chapter 8 investigates p-groups from two points of view: first through a study of p-groups which are extremal with respect to one of several parameters (usually connected with p-rank) and second through a study of the automorphism group of the p-group.

Recall that if p is a prime then the p-rank of a finite group is the maximum dimension of an elementary abelian p-subgroup, regarded as a vector space over GF(p). Section 23 determines p-groups of p-rank 1, p-groups in which each normal abelian subgroup is cyclic, and, for p odd, p-groups in which each normal abelian subgroup is of p-rank at most 2. Perhaps most important, the p-groups of symplectic type are determined (a p-group is of symplectic type if each of its characteristic abelian subgroups is cyclic).

The Frattini subgroup is introduced to study p-groups and their automorphisms. Most attention is focused on p'-groups of automorphisms of p-groups; a variety of results on the action of p'-groups on p-groups appear in section 24. One very useful result is the Thompson A,x B Lemma. Also of importance is the concept of a critical subgroup.

23 Extremal p-groups In this section p is a prime and G is a p-group.

The Frattini subgroup of a group H is defined to be the intersection of all maximal subgroups of H. @(H) denotes the Frattini subgroup of H.

(23.1) (1) @(H) char H . (2) If X C H with H = (X, @ ( H ) ) , then H = (X). (3) If H/@(H) is cyclic, then H is cyclic.

(23.2) If G is a p-group then @(G) is the smallest normal subgroup H of G such that GIH is elementary abelian.

Proof. If M is a maximal subgroup of G then, by Exercise 3.2, M L] G and lG : MI = p, so, by 8.8, G(') 5 M. Hence G(') 5 @(G), so, by 8.8, G/@(G) is abelian. Also, as GIM Z Zp, gp E M for each g E G. so gp E @(G). Hence G/@(G) is elementary abelian.

106 p-groups

Conversely let H a G with G/H = G* elementary abelian. Then G* _G *x x .. x G* with Gi = l,, so setting H, = (G1: j i), I G : H, I = p andH = n Hl. Thus H, is maximal in G so H = n H1 > 4)(G).

Observe that, as a consequence of 23.2, a p-group G is elementary abelian ifand only if c(G) = 1.

Recall that if n is a positive integer then Qn (G) is the subgroup of G generatedby all elements of order at most pn.

(23.3) Let G = (x) be cyclic of order q = pn > 1 and let A = Aut(G), Then(1) The map a i-+ m(a) is an isomorphism of A with the group U(q) of

units of the integers modulo q, where m(a) is defined by xa = xm(a) for a E A.In particular A is abelian of order O(q) = pn-1(p - 1).

(2) The subgroup of A of order p - 1 is cyclic and faithful on Stl(G).(3) If p is odd then a Sylow p-group of A is cyclic and generated by the

element b with m(b) = p + 1. In particular the subgroup of A of order p isgenerated by the element b0 with m(bo) = pn-1 + 1.

(4) If q = 2 then A = l while if q = 4 then A = (c) = LL2, where m (c) _ -1.(5) If p = 2 and q > 4 then A = (b) x (c) where b is of order 2n-2 with

m(b) = 5, and c is of order 2 with m(c) = -1. The involution b0 in (b) satisfiesm(bo) = 2n-1 + 1 and m(cb0) = 2n-1 - 1.

Proof. I leave part (1) as an exercise and observe also that a: a i-+ m(a) modp is a surjective homomorphism of A onto U(p) with kernel CA(QI(G)). So,as IU(p)I p = p - 1= IU(q)Ip,, the subgroup of A of order p - 1 is isomor-phic to U(p) and faithful on cZ 1(G), while ker(a) _ {a E A: m (a) -1 mod p} ESylp(A). Next U(p) is the multiplicative group of the field of order p, andhence cyclic, so (2) holds. Thus we may take q > p. Evidently if m(c) = -1then c is of order 2. So, as J A I =2 if q =4, (4) holds. Thus we can assumen > 1, and n > 2 if p = 2. Choose b as in (3) or (5). Then b E ker(a), sobP"-' = 1. Thus if p is odd it remains to show by"-2 = bo and if p=2 showb2" = b0.

Observe:

(kpm + 1)P (1 + kpm+1 + k2 p2-+1(p _ 1)/2) mod pm+2

1 + kpm+1

if m> 1 or p is odd. Hence as m(b) =1 +s with s = p if p is odd ands =4 if p =2, it follows that m(bpi-z) =1 + pn-1=m(bo) if p is odd, while

m(b2"-3) = 2n-1 + 1 = m(bo) if p = 2. So the proof is complete.

106 p-groups

Conversely let H G with G / H = G* elementary abelian. Then G* = GTx . . . x G; with GfSZ , , so setting Hi=(Gj: j # i ) , IG:HiI=p and H = n Hi. Thus Hi is maximal in G so H = n Hi 2 @(G).

Observe that, as a consequence of 23.2, a p-group G is elementary abelian if and only if @(G) = 1.

Recall that if n is a positive integer then Qn (G) is the subgroup of G generated by all elements of order at most pn.

(23.3) Let G = (x) be cyclic of order q = pn > 1 and let A = Aut(G), Then

(1) The map a H m(a) is an isomorphism of A with the group U(q) of units of the integers modulo q, where m(a) is defined by xu = xm(a) for a E A. In particular A is abelian of order @(q) = pn-'(p - 1).

(2) The subgroup of A of order p - 1 is cyclic and faithful on nl(G). (3) If p is odd then a Sylow p-group of A is cyclic and generated by the

element b with m(b) = p + 1. In particular the subgroup of A of order p is generated by the element bo with m(bo) = pn-' + 1.

(4) I fq=2thenA=1whi le i fq=4thenA=(c)~Z2,wherem(c)=-1 . (5) If p = 2 and q > 4 then A = (b) x (c) where b is of order 2n-2 with

m(b) = 5, and c is of order 2 with m(c) = - 1. The involution bo in (b) satisfies m(bo) = 2"-' + 1 and m(cbo) = 2"-' - 1.

ProoJ: I leave part (1) as an exercise and observe also that a : a H m(a) mod p is a surjective homomorphism of A onto U(p) with kernel CA(n1(G)). SO, as I U(p) 1 ,I = p - 1 = I U (q) 1 ,I, the subgroup of A of order p - 1 is isomorphic to U(p) and faithful on a1 (G), while ker(a) = (a E A: m(a) - 1 mod p) E

Syl,(A). Next U(p) is the multiplicative group of the field of order p, and hence cyclic, so (2) holds. Thus we may take q > p. Evidently if m(c) = -1 then c is of order 2. So, as IAl= 2 if q =4, (4) holds. Thus we can assume n > 1, and n > 2 if p =2. Choose b as in (3) or (5). Then b E ker(a), so ,

bpn-' = 1. Thus if p is odd it remains to show bpn-' = bo and if p = 2 show b2"-3 = bo.

Observe:

(kpm + l)P = (1 + kpm+l + k2p2m+1(p - 1)/2) mod pm+2

if m > 1 or p is odd. Hence as m(b)=l + s with s = p if p is odd and s = 4 if p = 2, it follows that m(b~"-') = 1 + pn-' = m(bo) if p is odd, while

(b2"-3 ) = 2"-' + 1 = m(bo) if p = 2. So the proof is complete.

Extremal p-groups 107

Next the definition of four extremal classes of p-groups. The modular p-groupModp" of order p' is the split extension of a cyclic group X = (x) of order pii-1by a subgroup Y = (y) of order p with xy

=xPi-2+1.

Modpn is defined onlywhen n > 3, where, by 23.3 and 10.3, Mode- is well defined and determinedup to isomorphism. Similar comments hold for the other classes. If p = 2 andn > 2 the dihedral group D2 is the split extension of X by Y withxy = x-1 andif n > 4 the semidihedral group SD2 is the split extension with xy =

x2^-2-1

The fourth class is a class of nonsplit extensions. Let G be the split extensionof X = (x) of order 21-1 > 4 by Y = (y) of order 4 with xy =x-1. Notice

(x2"2, y2) = Z(G). Define the quaternion group Q2n of order 2" to be thegroup G/(x2ii-2y2).

The modular, dihedral, semidihedral, and quaternion groups are discussedin Exercises 8.2 and 8.3. Observe Mod8 = D8.

(23.4) Let G be a nonabelian group of order p'° with a cyclic subgroup of indexp. Then G - Modp,,, Den, SD2-, or Q2'.

Proof. Notice that, as G is nonabelian, n > 3 by Exercise 2.4. Let X = (x) be ofindex pin G. By Exercise 3.2, X 4 G. As X is abelian but G is not, X = CG(X)by Exercise 2.4. So y E G - X acts nontrivially on X. As yP E X, y inducesan automorphism of X of order p. By 23.3, Aut(X) has a unique subgroup oforder p unless p = 2 and n > 4, where Aut(X) has three involutions. In the firstcase by 23.3, xy = xz for some z of order p in X. In the remaining case p = 2and xy =x-1z', where s = 1 or 0 and z is the involution in X.

Now if the extension splits we may choose y of order p and by definitionGModP,,, D2,,, or SD2,,. So assume the extension does not split. ObserveCX(y) = (x") ifxy =xz, while CX(y) = (z) otherwise. Also yP E Cx(y). As Gdoes not split over X, (y, Cx(y)) does not split over Cx(y), so, as (y, Cx(y)) isabelian, it is cyclic. Thus Cx(y) = (yP). Hence we may take yP = xP if xy = xz

and y2 = z otherwise.Suppose xy = xz. Then z = [x, y] centralizes x and y, so, by 8.6, (yx-1)P =

y px-pZp(p-1)/2 = Zp(p-1)/2 while zP(P-1)/2 = 1 unless p = 2. So, as G does not

split, p = 2. Here z =x2n_2

and if n > 4 then, setting i = 2i-3 - 1, (yx` )2 = 1.If n = 3 then xy = x-1, which we handle below.

So p = 2, xy = x-1zE, and y2 = z. Ifs = 0, then by definition G = Q2', sotake s = 1. Then, as z c Z(G), (yx)2 = y2xyx = zx-1zx = 1, so the extensiondoes indeed split.

(23.5) Let G be a nonabelian p-group containing a cyclic normal subgroup Uof order p" with CG(U) = U. Then either


Next the definition of four extremal classes of p-groups. The modularp-group Mod,. of order pn is the split extension of a cyclic group X = (x) of order pn-' by a subgroup Y = (y) of order p with xY = XP"-'+I. Mod,. is defined only when n 2 3, where, by 23.3 and 10.3, Mod,. is well defined and determined up to isomorphism. Similar comments hold for the other classes. If p = 2 and n 2 2 the dihedral group DT is the split extension of X by Y withxy = x-' and if n 2 4 the semidihedral group SD2?, is the split extension with xy = x~~-'- ' .

The fourth class is a class of nonsplit extensions. Let G be the split extension of X = (x) of order 2"-' 2 4 by Y = (y) of order 4 with xY =x-'. Notice (x2"-', Y2) = Z(G). Define the quaternion group Q2. of order 2" to be the group G / ( X ~ ~ - ' y2).

The modular, dihedral, semidihedral, and quaternion groups are discussed in Exercises 8.2 and 8.3. Observe Mods = D8.

(23.4) Let G be a nonabelian group of order pn with a cyclic subgroup of index p. Then G S Mod,. , D2", SD2., or 122".

ProoJ: Notice that, as G is nonabelian, n 2 3 by Exercise 2.4. Let X = (x) be of index p in G. By Exercise 3.2, X G. As X is abelian but G is not, X = CG(X) by Exercise 2.4. So y E G - X acts nontrivially on X. As yp E X, y induces an automorphism of X of order p. By 23.3, Aut(X) has a unique subgroup of

4 order p unless p = 2 and n 2 4 , where Aut(Xj has three involutions. In the first case by 23.3, xY = xz for some z of order p in X. In the remaining case p = 2 and X Y = X-'zE, where E = 1 or 0 and z is the involution in X.

Now if the extension splits we may choose y of order p and by definition G S Mod,. , D2", or SD2.. So assume the extension does not split. Observe Cx(y) = ( x p ) if xY =xz, while Cx(y) = (z) otherwise. Also yp E Cx(y). As G does not split over X, (y, Cx(y)) does not split over Cx(y), so, as (y, CX(Y)) is abelian, it is cyclic. Thus Cx(y) = (yp). Hence we may take yp = xP if xY = xz and y2 = z otherwise.

Suppose xY =xz. Then z = [x, y] centralizes x and y, so, by 8.6, (YX- ' )~ = y ~ ~ - ~ z ~ ( ~ - 1 ) 1 2 = Z~(~-1)12 , while zp(p-1)/2 = 1 unless p = 2. So, as G does not

2"-2 split, p = 2 . Here z = x and if n 2 4 then, setting i = 2n-3 - 1 , (YX')~ = 1 . If n = 3 then X Y = X-', which we handle below.

SO p = 2 , X Y =x-'zE, and y2 = z. If E = 0 , then by definition G S Q2., so take E = 1. Then, as z E Z(G), (yx)2 = y 2 ~ y ~ = ZX-'ZX = 1, SO the extension does indeed split.

(23.5) Let G be a nonabelian p-group containing a cyclic normal subgroup U of order pn with CG(U) = U. Then either

108 p-groups

(1) G Den+I, Q2""+', or SD2n+I, or

(2) M = Cc (Z51(U)) = Mod p»+1 and Ep2 - S21(M) char G.

Proof. Let G* = G/U. As U = CG(U), G* = AutG(U) < Aut(U). As G is non-abelian, G* 0 1 and n > 2. If G* is of order p then the lemma holds by 23.4 andExercise 8.2, so assume IG*I > p. Then by 23.3 there exists y* E G* of orderp with uY = up"-'+', where U = (u). Let M = (y, U). By 23.4, M - Modpn+1and, by Exercise 8.2, E = S21(M) = Epz. It remains to show E char G. By23.3, G* is abelian and either G* is cyclic, or p = 2 and there exists g* E G*with ug = u-1. In the first case Q, (G*) = M*, so E = Q1(M) = Q1(G) char G.In the second Z51(U) = (u2) = ([u, g]) and as G* is abelian, G(1) < U. HenceG(1)=ZS1(U) or U, and in either case Z51(U) char G(1), so Z31 (U) char G.Therefore E = S21(CG (S2' (U))) char G.

A critical subgroup of G is a characteristic subgroup H of G such that1)(H) < Z(H) > [G, H] and CG(H) = Z(H). Observe that in particular acritical subgroup is of class at most 2.

(23.6) Each p-group possesses a critical subgroup.

Proof. Let S be the set of characteristic subgroups H of G with (D(H) < Z(H) >[G, H]. Let H be a maximal member of S; I claim H is a critical sub-group of G. Assume not and let K = CG(H), Z = Z(H), and define X byX/Z = S21(Z(G/Z)) fl K/Z. Then K H and Z = H fl K, so, as K a G,X 0 Z by 5.15. But notice XH E S, contradicting the maximality of H.

A p-group G is special if t(G) = Z(G) = G(1). A special p-group is said tobe extraspecial if its center is cyclic.

(23.7) The center of a special p-group is elementary abelian.

Proof. Let G be special and g, h e G. Then gP e 1(G) = Z(G), so, by 8.6.1,1= [gP, h] _ [g, h]p. Hence G(') is elementary, so, as Z(G) = GM, the lemmaholds.

(23.8) Let E be an extraspecial subgroup of G with [G, E] < Z(E). ThenG = ECG(E).

Proof. Let Z = (z) = Z(E). As E/Z < AutG(E) < C = it sufficesto show E/Z= C. Let a E C and (x; Z: 1 < i < n) a basis for E/Z. Then

108 p-groups

(1) G Z Dz.+l, Q2,,+1, or SDZn+l, or (2) M = C G ( u l (u)) S Modpgt+1 and Ep2 2 Q1 ( M ) char G.

Proof. Let G* = G/U. As U = CG(U), G* = AutG(U) ( Aut(U). As G is nonabelian, G* # 1 and n > 2. If G* is of order p then the lemma holds by 23.4 and Exercise 8.2, so assume IG*l > p. Then by 23.3 there exists y* E G* of order p with U Y = upn-'+', where U = (u). Let M = ( y , U ) . By 23.4, M S Modp.+l and, by Exercise 8.2, E = Q 1 ( M ) S Ep2. It remains to show E char G . By 23.3, G* is abelian and either G* is cyclic, or p = 2 and there exists g* E G* with~~=~-'.InthefirstcaseQ~(G*)=M*,soE=Q~(M)=Q~(G)charG. In the second U ' ( U ) = (u2) = ([u, g]) and as G* is abelian, G(') ( U . Hence G(')=u'(u) or U , and in either case u l ( U ) char G('), so U 1 ( U ) char G . Therefore E = Q (CG (Q ' (u))) char G .

A critical subgroup of G is a characteristic subgroup H of G such that @ ( H ) ( Z ( H ) 2 [G, HI and C G ( H ) = Z ( H ) . Observe that in particular a critical subgroup is of class at most 2.

(23.6) Each p-group possesses a critical subgroup.

Proof. Let S be the set of characteristic subgroups H of G with @ ( H ) i Z ( H ) 2 [G, HI. Let H be a maximal member of S; I claim H is a critical subgroup of G. Assume not and let K = Cc(H) , Z = Z ( H ) , and define X by X / Z = f i l ( Z ( G / Z ) ) fl K / Z . Then K -& H and Z = H fl K , so, as K G , X # Z by 5.15. But notice XH E S, contradicting the maximality of H .

A p-group G is special if @(G) = Z ( G ) = ~ ( ' 1 . A special p-group is said to be extraspecial if its center is cyclic.

(23.7) The center of a special p-group is elementary abelian.

Proof. Let G be special and g, h 6 G. Then gp E @(G) = Z(G) , so, by 8.6.1, 1 = [gp, h] = [g, h]P. Hence G(') is elementary, so, as Z ( G ) = G(') , the lemma holds.

(23.8) Let E be an extraspecial subgroup of G with [G, El ( Z(E) . Then G = ECG(E).

Proof. Let Z = (z) = Z ( E ) . As E/Z( Autc(E) 5 C = CA"~(E)(E/Z), it suffices to show E/Z= C. Let a E C and (x iZ: 1 5 i i n ) a basis for E/Z. Then

[xi, a] = z'i for some 0 < mi < p, and, as E = (xi: 1 < i < n) by 23.1, a is de-termined by the integers (m i : I< i <n). Thus I CI < p" = J E/ZI , and the lemmaholds.

A p-group is said to be of symplectic type if it has no noncyclic characteristicabelian subgroups.

(23.9) If G is of symplectic type then G = E * R where

(1) Either E is extraspecial or E = 1, and(2) Either R is cyclic, or R is dihedral, semidihedral, or quaternion, and of

order at least 16.

Proof. By 23.6, G possesses a critical subgroup H. Let U = Z(H). By hy-pothesis U is cyclic. Let Z be the subgroup of U of order p and G* = G/Z.For h, k H,hneU,so[h,k]p=[hp,k]=1,by8.6.Thus H(1) <ZsoH*isabelian. Let K* = Qi(H*) and E* a complement to Z(K)* in K*. K char G,so Z(K) is cyclic. Hence if K is abelian then H* is cyclic, so, by Exercise 2.4,U = H. Now 23.5 applies and says the lemma holds with E = 1 and R = G orE=G=D8 or Q8 and R=1.

So K is nonabelian and then E is extraspecial. For g e G and e E E, eg = eufor some u E U fl K, and ep e Z < Z(G), soeP = eP9 = (eu)P = enup. Henceup = 1, so that [G, E] = Z. Therefore, by 23.8, G = E * R, where R = CG(E).Thus H = E * CH(E). Recall H* is abelian and S21(H*) = K* = E* x Z(K)*with Z(K) cyclic. We conclude CH(E)* is cyclic and hence CH(E) is abelianby Exercise 2.4. Thus as H = ECH(E), CH(E) = U. Also CR(U) = CR(H) <R fl H = U, and we may assume R 0 U. So J U I > p2 and 23.5 applies to R.If IU I = p2 then Z51(U) = Z < Z(G). So, by 23.5, R is dihedral, quaternion ormodular of order p3, and in particular R is extraspecial. But then G = E*R isalso extraspecial, so the lemma holds. Thus we may take I U I > p2 and assumeR satisfies 23.5.2. Let M = CR(Z51(U)). Then N = CG (Z51(U)) char G andN = EM. By Exercise 8.2, Q2(N) = EQ2(M) =7Lp x (E * 7Lp2) SO Z(Q2(N))is noncyclic, a contradiction.

(23.10) Let E be an extraspecial p-group, Z = Z(E), and E = E/Z.(1) Regard Z as the field of integers modulo p and t as a vector space over

Z. Define f : E x k -+ Z by f (x, y) = [x, y]. Then f is a symplectic form onE, so (E, f) is a symplectic space over Z.

(2) m(E) = 2n is even.(3) If p=2 define Q: E - Z by Q(x)=x 2. Then Q is a quadratic form

on t associated to f, so (E, Q) is an orthogonal space over Z.

Extremal p-groups

[xi , a ] = zmi for some 0 (mi < p, and, as E = (x i : 1 5 i ( n ) by 23.1, a is determined by the integers (mi: 1 ( i ( n). Thus ( C ( 5 pn = (E/Z\ , and the lemma holds.

A p-group is said to be of symplectic type if it has no noncyclic characteristic abelian subgroups.

(23.9) If G is of symplectic type then G = E * R where

(1) Either E is extraspecial or E = 1, and (2) Either R is cyclic, or R is dihedral, semidihedral, or quaternion, and of

order at least 16.

Proof. By 23.6, G possesses a critical subgroup H. Let U = Z ( H ) . By hypothesis U is cyclic. Let Z be the subgroup of U of order p and G* = G/Z. For h, k E H , hP E U , SO [h, k]P = [hp, k] = 1, by 8.6. Thus H(') 5 Z so H* is abelian. Let K* = Q1(H*) and E* a complement to Z(K)* in K". K char G , so Z ( K ) is cyclic. Hence if K is abelian then H* is cyclic, so, by Exercise 2.4, U = H . Now 23.5 applies and says the lemma holds with E = 1 and R = G or E = G Z D s or Qs and R=1.

So K is nonabelian and then E is extraspecial. For g E G and e E E , eg = eu for some u E U n K , and ep E Z ( Z ( G ) , so,@' = epg = (eu)p = eJ'uP. Hence UP = 1, SO that [G, El = Z . Therefore, by 23.8, G = E * R, where R = CG(E). Thus H = E * C H ( E ) . Recall H* is abelian and Ql(H*) = K* = E* x Z(K)* with Z ( K ) cyclic. We conclude CH(E)* is cyclic and hence C H ( E ) is abelian by Exercise 2.4. Thus as H = ECH ( E ) , CH ( E ) = U . Also C R ( U ) = C R ( H ) ( R n H = U , and we may assume R # U . So IUI 2 p2 and 23.5 applies to R. If IUI = p2 then 7J1(u) = Z ( Z(G). So, by 23.5, R is dihedral, quatemion or modular of order p3, and in particular R is extraspecial. But then G = E* R is also extraspecial, so the lemma holds. Thus we may take IUI > p2 and assume R satisfies 23.5.2. Let M = C R ( u l ( u ) ) . Then N = CG(U'(U)) char G and N =EM. By Exercise 8.2, Q 2 ( N ) = EQ2(M) 2 Zp x ( E * Zp2) SO Z(Q2(N)) is noncyclic, a contradiction.

(23.10) Let E be an extraspecial p-group, Z = Z ( E ) , and ,!? = E I Z . (1) Regard Z as the field of integers modulo p and ,!? as a vector space over

Z . Define f : ,!? x ,!? + Z by f (2, y) = [ x , y]. Then f is a symplectic form on E , so (,!?, f ) is a symplectic space over Z .

(2) m(E) = 2n is even. (3) If p = 2 define Q : ,!? + Z by Q(2) = x2. Then Q is a quadratic form

on ,!? associated to f , so (,!?, Q ) is an orthogonal space over Z .

110 p-groups

(4) Let Z < U < E. Then U is extraspecial or abelian if and only if U isnondegenerate or totally isotropic, respectively. If p = 2 then U is elementaryabelian if and only if U is totally singular.

Proof. As Z = c(E), k is elementary abelian, so by 12.1 we can regard k as avector space over Z in a natural way. Notice that under this convention the groupoperations on k and Z are written additively. By 8.5.4, [xy, z] = [x, z]Y[y, z] =[x, z][y, z], with the latter equality holding as E is of class 2. This says f islinear in its first variable and a similar argument gives linearity in the secondvariable. As Z = Z(E), f is nondegenerate. [x, y] _ [y, x]-1, or, in additivenotation, f (z, y) = - f (y, z). Thus (1) holds.

Notice (1) and 19.16 imply (2). Let p =2. By 8.6, (xy)2 =x2y2[x, y], or,in additive notation, Q(. + y) = Q(z) + Q(y) + f (i, y). Thus (3) holds. Theproof of (4) is straightforward.

(23.11) Assume p is odd and G is of class at most 2. Then S21(G) is of expo-nent p.

Proof. Let x and y be elements of G of order p. Then [x, y] = z E Z(G). By8.6.1, zp = [x", y] = 1, so, by 8.6.2, (xy)p =xpypZp(p-1)12 = 1.

(23.12) Let p be odd and E an extraspecial p-group. Then 01(E) is of exponentp and index at most p in E. If 01(E) # E then S21(E) = X x Eo where X = 7Lpand Eo is of order p or extraspecial, and E = El * Eo with El = Modp3.

Proof. Let Y = 01(E). By 23.11, Y is of exponent p. Suppose I E : Y) = p.Then, in the notation of 23.10, f is a hyperplane of E, and hence of odddimension, so, as all symplectic spaces are of even dimension, Y is degenerate.Let R be a point in Rad(Y). As RI is a hyperplane of E, Rl = Y. Hence, by23.10.4, Y = CE(R). As Y is of exponent p, R = X x Z for some X of orderp. Let Eo be a complement to R in Y. By 19.3 and 23.10.4, Eo is extraspecialor Eo = Z; of course Y = X x E0. Let El = CE(EO). By 19.3 and 23.10, El isextraspecial. As Y # E, El > 01(E1) so, by 23.4, E1= Modp3.

It remains to show I E : YI < p. Let u, v E E and U = (u, v). It suffices toshow IU: 01(U)I < p. If U is abelian this holds because Z = (D (E). If U isnonabelian appeal to 23.4.

By 23.10 an extraspecial p-group is of order p1+2n for some positive integerp. If p is odd, denote by p1+2n an extraspecial p-group of exponent p andorder p 1+2n .21 +2denotes any extraspecial 2-group of order 21+2n; by 1.13

110 p-groups

(4) Let Z ( U ( E. Then U is extraspecial or abelian if and only if ~ is nondegenerate or totally isotropic, respectively. If p = 2 then U is elementary abelian if and only if 0 is totally singular.

ProoJ As Z = (P(E), E is elementary abelian, so by 12.1 we can regard E as a vector space over Z in a natural way. Notice that under this convention the group operations on E and Z are written additively. By 8.5.4, [xy , z ] = [ x , zIY [ y , zl = [ x , z ] [ y , z] , with the latter equality holding as E is of class 2. This says f is ,

linear in its first variable and a similar argument gives linearity in the second variable. As Z = Z ( E ) , f is nondegenerate. [x , y] = [ y , X I - ' , or, in additive notation, f (K, y") = - f (y", 2) . Thus (1) holds.

Notice (1) and 19.16 imply (2). Let p =2. By 8.6, ( x ~ ) ~ = X ~ ~ ~ [ X , y] , or, in additive notation, Q(K + y") = Q(K) + Q(y") + f ( 2 , 7). Thus (3) holds. The proof of (4) is straightforward.

(23.11) Assume p is odd and G is of class at most 2. Then Q 1 ( G ) is of exponent p.

ProoJ Let x and y be elements of G of order p. Then [ x , y] = z E Z(G). By 8.6.1, zP = [xp, y] = 1, so, by 8.6.2, (xy)P = xpypzp(p-')I2 = 1.

(23.12) Let p be odd and E an extraspecial p-group. Then Q1 ( E ) is of exponent p and index at most p in E. If Q1 ( E ) # E then Q1 ( E ) = X x EO where X Z Z p and Eo is of order p or extraspecial, and E = El * Eo with El Z Modp3.

ProoJ Let Y = Q 1 ( E ) . By 23.11, Y is of exponent p. Suppose I E : Y I = p . Then, in the notation of 23.10, f is a hyperplane of E, and hence of odd dimension, so, as all symplectic spaces are of even dimension, P is degenerate. Let I? be a point in ~ad(Y) . As fiL is a hyperplane of E , I?' = P. Hence, by 23.10.4, Y = CE(R) . As Y is of exponent p , R = X x Z for some X of order p. Let Eo be a complement to fi in E. By 19.3 and 23.10.4, Eo is extraspecial or Eo = Z ; of course Y = X x Eo. Let El = CE(Eo). By 19.3 and 23.10, El is extraspecial. As Y # E , El > Ql(E1) so, by 23.4, El Z Modp3.

It remains to show IE : Y I ( p. Let u , v E E and U = (u , v). It suffices to show IU: Q1(U)I ( p. If U is abelian this holds because Z = (P(E). If U is nonabelian appeal to 23.4.

By 23.10 an extraspecial p-group is of order for some positive integer p. If p is odd, denote by an extraspecial p-group of exponent p and order 21+2n denotes any extraspecial 2-group of order 21+2"; by 1.13

there are no such groups of exponent 2. Write D8 Q8 for a central product ofn copies of D8 with m copies of Q8, and all centers identified.

(23.13) Let p be an odd prime and n a positive integer. Then up to isomorphismthere is a unique extraspecial p-group E of order p1+2n and exponent p. E isthe central product of n copies of pl+z

Proof. By 23.10 and 19.16, E is a central product of n extraspecial subgroupsE1, 1 < i < n, of order p3 and center Z = Z(E). Now Exercise 8.7 completesthe proof.

(23.14) Let n be a positive integer. Then up to isomorphism D8 and D8-1 Q8are the unique extraspecial groups of order 22n+1 D8 has 2-rank n + 1 whileD8-1 Q8 has 2-rank n, so the groups are not isomorphic.

Proof. By 23.10 and 21.2, E is a central product of n extraspecial groupsE1, 1 <i <n,oforder 8with Z(E;)=Z(E)=Z. Z is of order 2 so Aut(Z) = 1.Hence 11.2 says E is determined up to isomorphism by the groups Ej. Againby 23.10 and 21.2, we can choose (E,, Q) hyperbolic for i <n and (En, Q)either hyperbolic or definite. By 1.13, EZ is not of exponent 2, so, by 23.4EI - D8 or Q8. By 23.10.4, (Ei, Q) is hyperbolic in the first case and definitein the second. Finally 23.10 and 21.2 imply the remark about the 2-rank andcomplete the proof.

(23.15) Let A be a maximal abelian normal subgroup of G and Z = S2(A).Then

(1) A=CA(A).(2) (CG(A/Z) n C(Z)(1) <A.(3) If p is odd then Qj(CG(Z)) < CG(A/Z).

Proof. Let C = Cc (A). A < C < G, so if C # A there is D/A of order p inZ(G/A) n C/A. Then D < G and D is abelian by Exercise 2.4, contradict-ing the maximality of A. A straightforward calculation shows (CG(A/Z) nC(Z))(' < C(A), so (1) implies (2). Let p be odd, x of order p in CG(Z)and X = (x, A). Let Y = (x, CA ((X, Z)/Z)). Then Y is of class at most 2, so,by 23.11, W = 01(Y) is of exponent p. Thus W = (x, Z). But W char Y soNx (Y) < Nx (W) = Y, so Y = X and (3) holds.

(23.16) Let p be an odd prime and Z a maximal elementary abelian normalsubgroup of G. Then Z = 01(CG (Z)).

Extremal p-groups 1 1 1

there are no such groups of exponent 2. Write D i Q r for a central product of n copies of Ds with m copies of Q8, and all centers identified.

(23.13) Let p be an odd prime and n a positive integer. Then up to isomorphism there is a unique extraspecial p-group E of order and exponent p. E is the central product of n copies of

Proof. By 23.10 and 19.16, E is a central product of n extraspecial subgroups Ei , 1 ( i ( n, of order p3 and center Z = Z(E) . Now Exercise 8.7 completes the proof.

(23.14) Let n be a positive integer. Then up to isomorphism D: and D,"-' Q8 are the unique extraspecial groups of order 22"+'. Di has 2-rank n + 1 while D:-' Q8 has 2-rank n, so the groups are not isomorphic.

Proof. By 23.10 and 21.2, E is a central product of n extraspecial groups Ei, 1 ( i 5 n, of order 8 with Z ( E i ) = Z ( E ) = Z . Z is of order 2 so Aut(Z) = 1. Hence 11.2 says E is determined up to isomorphism by the groups Ei. Again by 23.10 and 21.2, we can choose ( E , , Q ) hyperbolic for i < n and ( E n , Q ) either hyperbolic or definite. By 1.13, Ei is not of exponent 2, so, by 23.4 Ei G D8 or Q 8 . By 23.10.4, ( E i , Q ) is hyperbolic in the first case and definite in the second. Finally 23.10 and 21.2 imply the remark about the Zrank and complete the proof.

(23.15) Let A be a maximal abelian normal subgroup of G and Z = Q 1 ( A ) . Then

(1) A = CG(A) . (2) (cG ( A / z ) n C(Z)( ' ) 5 A. (3) If p is odd then Ql(CG(Z) ) 5 C G ( A / Z ) .

ProoJ Let C = CG (A) . A 5 C L! G , so if C # A there is D / A of order p in Z(G/A) n C/A. Then D i? G and D is abelian by Exercise 2.4, contradicting the maximality of A. A straightforward calculation shows (CG(A/z> n C(Z) ) ( ' ) 5 C ( A ) , so (1) implies (2). Let p be odd, x of order p in C G ( Z ) and X = ( x , A ) . Let Y = ( x , C A ( ( x , Z ) / Z ) ) . Then Y is of class at most 2, so, by 23.1 1, W = Q 1 ( Y ) is of exponent p. Thus W = ( x , Z ) . But W char Y so Nx(Y) ( N x ( W ) = Y , so Y = X and (3) holds.

(23.16) Let p be an odd prime and Z a maximal elementary abelian normal subgroup of G . Then Z = Q1(CG(Z) ) .

112 p-groups

Proof. Let X = S21(CG(Z)). I'll show X is of exponent p. Hence if X Zthen there is D/Z of order p in Z(G/Z) n X/Z and, by Exercise 2.4, D iselementary abelian, contradicting the maximality of Z.

Let A be a maximal abelian normal subgroup of G containing Z. ThenZ = S21(A) by maximality of Z. By 23.15.3, [X, A] < Z, so by 23.15.2,XM < A. Choose U < X of minimal order subject to U = Q, (U) and U notof exponent p. Then there exist x and y in U of order p with xy not of orderp. By minimality of U, U = (x, y). By 7.2, V = (xU) 0 U, so V is of expo-nent p. Hence [x, y] E V is of order at most p, so, as X" < A, [x, y] E Z. AsX < C(Z), U is of exponent p by 23.11, contrary to the choice of U.

(23.17) Let p be an odd prime and assume G contains no normal abeliansubgroup of rank 3. Then G is of p-rank at most 2.

Proof. By Exercise 8.4 we may assume Ep2 = Z a G. Let H = CG(Z)and Ep3 = A < G. Then IA: A n HI 3. Thusm(H) > 3. However by hypothesis Z is a maximal elementary abelian normalsubgroup of G, so Z = c21(H) by 23.16.

24 Coprime action on p-groupsIn this section p is a prime, G is a p-group, and A is a p'-group of automor-phisms of G, unless the conditions are explicitly relaxed as in the ThompsonA x B Lemma.

(24.1) A is faithful on GI (D (G).

Proof. Suppose b c A centralizes GI (D (G). We wish to show b = 1. If not thereis a prime q and a nontrivial power of b which is a q-element and centralizesG/(D(G), so without loss b is a q-element. Let B = (b) and g c G. Then B actson the coset X =g(D(G). By 5.14, m = JXJmod q, where m is the number offixed points of B on X, and, as X j = J(D(G) l is a power of p, X i 0 0 mod q, soB centralizes some x E X. Hence B centralizes a set Y of coset representativesfor c(G) in G, so, by 23.1.2, G = (Y) < CG(B). Hence B = 1, completing theproof.

(24.2) (Thompson A x B Lemma). Let AB be a finite group represented as agroup of automorphisms of a p-group G, with [A, B] = 1= [A, CG(B)], B ap-group, and A = OP(A). Then [A, G] = 1.

Proof. Form the semidirect product H of G by AB and identify AB and G withsubgroups of H. We may assume [A, G] 0 1 so, as A = OP(A), [X, G] 0 1

112 p-groups

Proof. Let X = Q1(CG(Z)) . I'll show X is of exponent p. Hence if X # Z then there is D / Z of order p in Z ( G / Z ) n X / Z and, by Exercise 2.4, D is elementary abelian, contradicting the maximality of Z.

Let A be a maximal abelian normal subgroup of G containing Z. Then Z = Q1(A) by maximality of Z. By 23.15.3, [ X , A] 5 Z , so by 23.15.2, x(') 5 A. Choose U X of minimal order subject to U = Q1(U) and U not of exponent p. Then there exist x and y in U of order p with xy not of order p. By minimality of U , U = ( x , y). By 7.2, V = ( x u ) # U , so V is of exponent p. Hence [ x , y] E V is of order at most p, so, as x(') ( A , [ x , y] E Z. As X I. C(Z) , U is of exponent p by 23.1 1, contrary to the choice of U.

(23.17) Let p be an odd prime and assume G contains no normal abelian subgroup of rank 3. Then G is of p-rank at most 2.

Proof. By Exercise 8.4 we may assume Epz Z Z 9 G. Let H = CG(Z) and Ep3 S A 5 G. Then ]A: A n HI 5 p and hence m((A n H ) Z ) 2 3. Thus m ( H ) 2 3. However by hypothesis Z is a maximal elementary abelian normal subgroup of G , so Z = Q1(H) by 23.16.

24 Coprime action on p-groups In this section p is a prime, G is a p-group, and A is a p'-group of automorphisms of G, unless the conditions are explicitly relaxed as in the Thompson A x B Lemma.

(24.1) A is faithful on G/@(G).

Proof. Suppose b E A centralizes G/@(G). We wish to show b = 1. If not there is a prime q and a nontrivial power of b which is a q-element and centralizes G/@(G) , so without loss b is a q-element. Let B = (b) and g E G. Then B acts on the coset X =g@(G). By 5.14, m = IXlmod q, where rn is the number of fixedpointsofBonX,and,as/XI=~@(G)~isapowerofp,~X~ + Omod q,so B centralizes some x E X. Hence B centralizes a set Y of coset representatives for @(G) in G , so, by 23.1.2, G = ( Y ) 5 CG(B). Hence B = 1, completing the proof.

(24.2) (Thompson A x B Lemma). Let AB be a finite group represented as a group of automorphisms of a p-group G , with [ A , B] = 1 = [ A , CG(B)] , B a p-group, and A = Op(A). Then [ A , GI = 1.

Proof. Form the semidirect product H of G by AB and identify AB and G with subgroups of H. We may assume [ A , GI # 1 so, as A = OP(A), [ X , GI # 1

Coprime action on p-groups 113

for some p'-subgroup X of A, and replacing A by X we may assume A is ap'-group.

G < H = GBA with A < NH(B), so GB is a normal p-subgroup of H,and, replacing G by GB, we may assume B < G. Then B < Q = CG (A), soCG(Q) < CG(B). Also CG(B) < Q by hypothesis, so CG(Q) < Q.

As [A, G] 0 1, Q 0 G, so, by Exercise 2.2, Q is properly contained inNG(Q). So, by definition of Q, [A, NG(Q)] 0 1, and hence we may assumeQ <G.

Let G* = G/Q. As A is a p'-group, CG(A) = Q* = 1, by 18.7.4.Thus [G, A] Q.

As Q< G, [G, Q, A] < [Q, A] =1, so [G, Q, A] = [Q, A, G] =1. Henceby the Three-Subgroup Lemma, 8.7, [A, G, Q] =1. Thus [A, G] < CG (Q) < Q,by paragraph 2, contrary to the last paragraph.

(24.3) If G is abelian then A is faithful on c21(G).

Proof. Without loss, A centralizes S21(G). Let X be of order p in G andG* = G/X. By Exercise 3.1, A is faithful on G*, so, by induction on the orderof G, A is faithful on Q, (G*), and hence without loss G* = Q, (G*). Now, by12.1 and Exercise 4.1.1, we may take CG. (A) =1. Thus, by 18.7.4, X = CG (A),so X = S21(G). Hence, as G is abelian, 1.11 implies G is cyclic. Now 23.3supplies a contradiction.

(24.4) G = [G, A]CG(A).

Proof. Let G* =G/4(G). By 23.2, G* is an elementary abelian p-group, so,by Exercise 4.1, G* = [G*, A] X CG.(A). By 8.5.3, [G*, A] = [G, A]* and,by 18.7.4, CG.(A) = CG(A)*. Hence G = ([G, A], CG(A), (D (G)), so, by 23.1,G = ([G, A], CG(A)). Finally, by 8.5.6, [G, Al <G, so ([G, A], CG(A)) _[G, A]CG(A) by 1.7.2.

(24.5) [G, A] = [G, A, Al

Proof. Let H = [G, A]. By 8.5.6, H < G and [H, Al :< H. Thus CG (A) acts on[H, A], so [H, A] < HCG (A) = G. Next H = [H, A]CH (A) so [G, A] < [H, A]by 8.5. But of course [H, A] < [G, A] as H < G.

(24.6) If G is abelian then G = [G, A] X CG (A).


for some pl-subgroup X of A, and replacing A by X we may assume A is a pf-group.

G 9 H = GBA with A 5 NH(B), so GB is a normal p-subgroup of H, and, replacing G by GB, we may assume B ( G. Then B ( Q = CG(A), so CG(Q) ( CG(B). Also CG(B) ( Q by hypothesis, so Cc(Q) ( Q.

As [A, G I # 1, Q # G, so, by Exercise 2.2, Q is properly contained in NG(Q). SO, by definition of Q, [A, NG(Q)] # 1, and hence we may assume

Q g G . Let G* = G/Q. As A is a pl-group, CG*(A) = CG(A)* = Q* = 1, by 18.7.4.

Thus [G, A] -$ Q. As Q 9 G, [G, Q, A] ( [Q, A] = 1, so [G, Q, A] = [Q, A, GI = 1. Hence

by theThree-SubgroupLemma, 8.7, [A, G, Q] = 1. Thus [A, GI ( CG(Q) ( Q, by paragraph 2, contrary to the last paragraph.

(24.3) If G is abelian then A is faithful on Ql(G).

Proof. Without loss, A centralizes Q1(G). Let X be of order p in G and G* = G/X. By Exercise 3.1, A is faithful on G*, so, by induction on the order of G, A is faithful on Q1(G*), and hence without loss G* = Q1(G*). Now, by 12.1 and Exercise 4.1.1, we may take CG* (A) = 1. Thus, by 18.7.4, X = Cc(A), so X = Q1(G). Hence, as G is abelian, 1.11 implies G is cyclic. Now 23.3 supplies a contradiction.

! (24.4) G = [G, A]CG (A).

Proof. Let G* = G/Q(G). By 23.2, G* is an elementary abelian p-group, so, by Exercise 4.1, G* = [G*, A] x CG*(A). By 8.5.3, [G*, A] = [G, A]* and, by 18.7.4, CG* (A) = CG (A)*. Hence G = ([G, A], CG(A), Q(G)), so, by 23.1, G = ([G, A], CG(A)). Finally, by 8.5.6, [G, A] 9 G, so ([G, A], CG(A)) = [G, A]CG(A) by 1.7.2.

(24.5) [G, A] = [G, A, A].

Proof. Let H = [G, A]. By 8.5.6, H 9 G and [H, A] 9 H. Thus CG(A) acts on [H, A],so[H, A] 9 HCG(A)= G.Next H = [H, A]CH(A)so[G, A] ( [H, A] by 8.5. But of course [H, A] ( [G, A] as H ( G.

(24.6) If G is abelian then G = [G, A] x Cc(A).

114 p-groups

Proof. Let G be a minimal counterexample and X = [S21(G), A]. By 24.3,X 0 1 and, by 12.1 and Exercise 4.1.1, Cx(A) = 1. By minimality of G,C([G,AI/x)(A) =1, so C[G,AI(A) =1. Now 24.4 completes the proof.

(24.7) If G = [G, A] and A centralizes every characteristic abelian subgroupof G, then G is special and Z(G) = CG (A).

Proof. As A centralizes each characteristic abelian subgroup of G, so doesG = [G, A] by Exercise 3.6. Thus Z = Z(G) is the unique maximal char-acteristic abelian subgroup. [Z2(G), G, G] = 1, so, by the Three-SubgroupLemma, Z2(G) centralizes GM. Hence Z2(G) f1 GM is abelian, and there-fore contained in Z, so Gf' < Z. By 24.6, G/G1> = (Z/Gf1) x [GIG('), A]so, as G = [G, A], Z = GO). Finally suppose G has exponent p" > p. Letg, h E G. By 8.6, [gp"-' , V_'] _ [gp°, hp"-Z] =

1,so 73' 1 (G) is abelian and

hence ?5n-1(G) < Z. But then G/Z is of exponent p. So Z = I(G).

(24.8) If p is odd then A is faithful on S21(G).

Proof. Choose G to be a minimal counterexample and let a E A# centralizeS21(G). By 24.5 and minimality of G, G = [G, a]. By 24.3, a centralizeseach characteristic abelian subgroup of G, so, by 24.7, G is special withZ = Z(G) = CG(a). By 23.7, Z = S21(G). Let g E G - Z, z = gp and v =[g, g-°]. Then z, v E Z = S21(G), so vp =1. Notice that, as Z = CG (a), (g-Q)p =

Z-1, andgg_a = h 0 Z by 18.7.4. Now, by 8.6, hp = zz-1 vp(p-1)/2 = 1, contra-

dicting Z = SZ 1(G).

(24.9) Let H be a critical subgroup of G. Then(1) A is faithful on H.(2) If p is odd then A is faithful on S21(H), and there exists a critical sub-

group H of G such that SZ1(H) contains each element of order pin CG(SZ 1(H)).

Proof. By definition of H, CG(H) < H, so by the Thompson A x B Lemma(applied to `A' = CA(H) and `B' = H), CA (H) = 1. Thus (1) holds. Part (1)and 24.8 imply the first statement in (2). To prove the second, choose H withL = S21(H) maximal. It suffices to show Y = 01(CG (L)) < L. Assume not andlet V be a maximal elementary abelian normal subgroup of Y. By 23.16,V = S21(Cy(V)), so, as Y L, V L. Thus v f1 Z2(Y) L, so S21(Z2(Y)) =K L. By 23.11, K is of exponent p, so X L, whereX/Z(L) = Z(G/Z(L))l(K/Z(L)). Now define S as in the proof of 23.6. Then XL E S so, by the proofof 23.6, XL is contained in a critical subgroup C of G. But L < XL < SZ 1(C),contradicting the choice of H.

114 p-groups

Proof. Let G be a minimal counterexample and X = [Q1(G), A]. By 24.3, X # 1 and, by 12.1 and Exercise 4.1.1, Cx (A) = 1. By minimality of G, C([G,Allx)(A) = 1, so CIG,Al(A) = 1. Now 24.4 completes the proof.

(24.7) If G = [G, A] and A centralizes every characteristic abelian subgroup of G, then G is special and Z(G) = CG(A).

Proof. As A centralizes each characteristic abelian subgroup of G, so does G = [G, A] by Exercise 3.6. Thus Z = Z(G) is the unique maximal characteristic abelian subgroup. [Z2(G), G, GI = 1, so, by the Three-Subgroup Lemma, Z2(G) centralizes ~ ( ' 1 . Hence Z2(G) f l G(') is abelian, and therefore contained in Z, so G(') ( Z. By 24.6, GIG(') = (z/G(')) x [GIG('), A] so, as G = [G, A], Z = G('). Finally suppose G has exponent pn > p. Let g, h E G. By 8.6, [gp"-', hpn-I] = tgp", hpn-'1 = 1, SO W-'(G) is abelian and hence U""(G) 5 Z. But then G/Zis of exponent p. So Z = @(G).

(24.8) If p is odd then A is faithful on Q1(G).

Proof. Choose G to be a minimal counterexample and let a E A' centralize Q1(G). By 24.5 and minimality of G, G = [G, a]. By 24.3, a centralizes each characteristic abelian subgroup of G, so, by 24.7, G is special with Z = Z(G) = CG(a). By 23.7, Z = Q1(G). Let g E G - Z, z = gp and v = [g, gVa]. Thenz, v E Z = Ql(G), so v p = 1. Notice that, as Z = CG(a), (g-')P = z-', and gg-a = h 4 Z by 18.7.4. Now, by 8.6, hP = ~ z - ' v p ( p - ' ) / ~ = 1, contradicting Z = Q ' (G).

(24.9) Let H be a critical subgroup of G. Then

(1) A is faithful on H.

(2) If p is odd then A is faithful on Ql(H), and there exists a critical subgroup H of G such that Q1 (H) contains each element of order p in CG(Q1 (H)).

Proof. By definition of H, CG(H) 5 H, so by the Thompson A x B Lemma (applied to 'A' = CA(H) and 'B' = H), CA(H) = 1. Thus (1) holds. Part (1) and 24.8 imply the first statement in (2). To prove the second, choose H with L = Ql(H) maximal. It suffices to show Y = Q~(CG(L)) I L. Assume not and let V be a maximal elementary abelian normal subgroup of Y. By 23.16, V = Q1(CY(V)), SO, as Y -$ L, V $ L. Thus V f l Zz(Y) $ L, so Q1(Z2(Y)) = K $ L. By 23.1 1, K is of exponent p, so X $ L, whereX/Z(L) = Z(G/Z(L)) f l

(K/Z(L)). Now define S as in the proof of 23.6. Then XL E S so, by the proof of 23.6, XL is contained in a critical subgroup C of G. But L < XL ( QI (C), contradicting the choice of H.

Remarks. The discussion of p-groups in this chapter is essentially the sameas Gorenstein's treatment of p-groups [Gor 4], which was influenced in turnby lecture notes of Phillip Hall.

P. Hall originally classified the p-groups of symplectic type. The notion of a`critical subgroup' is due to J. Thompson as is of course the Thompson A x BLemma.

Almost all of the material in this chapter is basic and belongs in the repertoire

of any finite group theorist. For the simple group theorist it represents animportant part of the foundation of the local group theory involved in theclassification. For example the importance of p-groups of symplectic type isreflected in the second case of Theorem 48.3. More generally the results of thischapter will be used repeatedly in chapters 10 through 16.

Exercises for chapter 81. Let q be a prime and A an elementary abelian q-group acting on a q'-group

G. Prove G = (CG(B): IA: BI =q). (Hint: Use 18.7 to reduce to the caseG a p-group. Then use Exercise 4.1 and 23.1.)

2. Let G = Model. , n > 3, with 7Lp"-1 = X = (x) 4 G, y of order p in G - X,and xy =xp-1. Prove(1) G is of class 2 with Z(G) _ t(G) = (xP) -l pf-2.(2) G(1) = (xpn 2) =7L p.(3) SZm(G) = (xp° m, y) -7Lpm x 7Lp for.p < m <n - 1, unless p° = 8.

3. Let G = Dzn , n > 2, Qzn , n > 3, or SDzn , n > 4. Let 7L2--1 X = (x) 4 Gand y (=- G - X with y an involution if G is dihedral or semidihedral andy of order 4 if G is quaternion. Prove

(1) GO) = (D(G) = (x2) =7L2 -2.(2) Either G is dihedral of order 4 or Z(G) = (x2 -2) is of order 2.(3) G is of class n - 1.(4) X is the unique cyclic subgroup of G of index p, unless G is dihedral

of order 4 or quaternion of order 8.(5) G - X is the union of two conjugacy classes of G with representatives

y and yx. Each member of G - X is an involution if G is dihedral,each is of order 4 if G is quaternion, while if G is semidihedral theny is of order 2 and xy of order 4.

(6) G has two maximal subgroups distinct from X. If G is dihedral oforder at least 8, both are dihedral. If G is quaternion of order at least16, both are quaternion. If G is semidihedral then one is dihedral andthe other quaternion.

(7) Quaternion groups have a unique involution.4. Let G be a p-group with no noncyclic normal abelian subgroups. Prove G

is cyclic, quaternion, semidihedral, or dihedral, and in the last case I G I > 8.

Coprime action on p-groups

Remarks. The discussion of p-groups in this chapter is essentially the same as Gorenstein's treatment of p-groups [Gor 41, which was influenced in turn by lecture notes of Phillip Hall.

P. Hall originally classified the p-groups of symplectic type. The notion of a 'critical subgroup' is due to J. Thompson as is of course the Thompson A x B Lemma.

Almost all of the material in this chapter is basic and belongs in the repertoire of any finite group theorist. For the simple group theorist it represents an important part of the foundation of the local group theory involved in the classification. For example the importance of p-groups of symplectic type is reflected in the second case of Theorem 48.3. More generally the results of this chapter will be used repeatedly in chapters 10 through 16.

Exercises for chapter 8 1. Let q be a prime and A an elementary abelian q-group acting on a q'-group

G. Prove G = (CG(B): !A: BI =q) . (Hint: Use 18.7 to reduce to the case G a p-group. Then use Exercise 4.1 and 23.1 .)

2. Let G S Mod,. , n 2 3, with Z,.-I S X = (x) <I G, y of order p in G - X, and X Y = X P - ~ . Prove (1) G is of class 2 with Z(G) = @(G) = (xp) Z Zpn-2. (2) G(" = (xpn-') Z Z,. (3) Q,(G) = (XP"-~, y) Z Zpm x Zp for &I < m c n - 1, unless pn = 8.

3. Let G 2 D 2 " , n > 2 , Q2.,n23, O ~ S D ~ ~ , ~ > ~ . L ~ ~ Z ~ ~ - L Z X = ( X ) < ] G and y E G - X with y an involution if G is dihedral or semidihedral and y of order 4 if G is quaternion. Prove (1) G(') = @(G) = (x2) S Z2n-2. (2) Either G is dihedral of order 4 or Z(G) = (x2'-') is of order 2. (3) G isof classn - 1. (4) X is the unique cyclic subgroup of G of index p, unless G is dihedral

of order 4 or quaternion of order 8. (5) G - X is the union of two conjugacy classes of G with representatives

y and yx. Each member of G - X is an involution if G is dihedral, each is of order 4 if G is quaternion, while if G is semidihedral then y is of order 2 and xy of order 4.

(6) G has two maximal subgroups distinct from X. If G is dihedral of order at least 8, both are dihedral. If G is quaternion of order at least 16, both are quaternion. If G is semidihedral then one is dihedral and the other quaternion.

(7) Quaternion groups have a unique involution. 4. Let G be a p-group with no noncyclic normal abelian subgroups. Prove G

is cyclic, quaternion, semidihedral, or dihedral, and in the last case I GI > 8.

116 p-groups

If H is a p-group with just one subgroup of order p, prove H is cyclic orquaternion.

5. Let G be a nonabelian p-group of symplectic type and exponent p or 4.Set Z = Z(G), G, = G/Z, A = Aut(G), and A* = Out(G). Prove(1) Inn(G) = CA((3).(2) CA(Z)* = Sp(O) and A* is the group of all similarities of some sym-

plectic form on G if p is odd.(3) If p = 2 then G - D', Di-1 Q, or Z4 * D', and A* - OZ (2), OZn(2),

or SP2n(2), respectively.6. Let G be a 2-group containing an involution x with CG(x) - E. Then G

is dihedral or semidihedral.7. Let p be an odd prime. Prove

(1) Up to isomorphism there is a unique extraspecial group E of order p3and exponent p.

(2)(3) Up to isomorphism there is a unique central product of n copies of E

with identified centers.8. Let A be a 7r'-group acting on a 7r-group G. Prove

(1) G = [G, A]CG(A), and(2) [G, A] _ [G, A, A].

9. Let r be a prime, A an elementary abelian r-group acting on a solvabler'-group G, D < A, and B a noncyclic subgroup of A. Prove [G, D]([CG(b), D]: b E B#).

10. Let A be a p'-group with a unique minimal normal subgroup B, assumeA acts faithfully on a nontrivial p-group P, and assume A is faithful onno proper subgroup of P. Prove that either(1) P is elementary abelian and A is irreducible on P, or(2) P = [P, B] is special, [B, Z(P)] = 1, and A is irreducible on P/Z(P).

If [A, Z(P)] = 1 and AP possesses a faithful irreducible representationover some field, then P is extraspecial.

11. Let p be an odd prime and G a p-group with m(G) > 3. Prove G has anormal abelian subgroup of p-rank at least 4. (Hint: Let G be a coun-terexample, V an elementary abelian normal subgroup of G of max-imal rank, H = CG(V), and Ep4 - A < G. Show V = S21(H) _2-- EP 3,m(Af H) = 2, and A is the unique Ep4-subgroup ofAV. Let K = NG (A) andg E NG(K) - K. Show AA9 is of class at most 2 and AA9 <AV.)

12. Let G be a p-group and (Gi : 1 < i < n) a family of subgroups of G whichgenerates G. Then, for any family (xi : 1 < i < n) of elements of G, G isgenerated by ((G): 1 < i < n).

116 p-groups

If H is a p-group with just one subgroup of order p, prove H is cyclic or quaternion.

5. Let G be a nonabelian p-group of symplectic type and exponent p or 4. Set Z = Z(G), G, = G/Z, A = Aut(G), and A* = Out(G). Prove (1) Inn(G) = cA(G). (2) CA(Z)* = sp(G) and A* is the group of all similarities of some sym-

plectic form on G if p is odd. (3) If p = 2 then G E Dn, D"-' Q, or Z4 * Dn, and A* E 0L(2), 0,(2),

or Sp2,(2), respectively. 6. Let G be a 2-group containing an involution x with CG(x) E E4. Then G

is dihedral or semidihedral. 7. Let p be an odd prime. Prove

(1) Up to isomorphism there is a unique extraspecial group E of order p3

and exponent p . (2) Au~A~~(E)(Z(E)) E Aut(Z(E)). (3) Up to isomorphism there is a unique central product of n copies of E

with identified centers. 8. Let A be a nl-group acting on a n-group G. Prove

(1) G = [G, A]CG(A), and (2) [G, A1 = [G, A, A].

9. Let r be a prime, A an elementary abelian r-group acting on a solvable rl-group G, D 5 A, and B a noncyclic subgroup of A. Prove [G, Dl = ([CG (b), Dl: b E B') .

10. Let A be a p'-group with a unique minimal normal subgroup B, assume A acts faithfully on a nontrivial p-group P , and assume A is faithful on no proper subgroup of P . Prove that either (1) P is elementary abelian and A is irreducible on P , or (2) P = [P, B] is special, [B, Z(P)] = 1, and A is irreducible on P/Z(P).

If [A, Z(P)] = 1 and AP possesses a faithful irreducible representation over some field, then P is extraspecial.

11. Let p be an odd prime and G a p-group with m(G) > 3. Prove G has a normal abelian subgroup of p-rank at least 4. (Hint: Let G be a counterexample, V an elementary abelian normal subgroup of G of maximal rank, H = CG(V), and EP4 E A ( G. Show V = a l ( H ) E Eps, m(An H) = 2, and A is the unique Ep4-subgroup ofAV. Let K = NG(A) and g E NG(K) - K. Show A A g is of class at most 2 and AAg (AV.)

12. Let G be a p-group and (Gi: 1 5 i 5 n) a family of subgroups of G which generates G. Then, for any family (xi: 1 5 i 5 n) of elements of G, G is generated by ((Gi)X1 : 1 5 i 5 n).

9

Change of field of a linear representation

Let jr: G -> GL(V, F) be an FG-representation, E a subfield of F, and K anextension field of F. Then V is also a vector space over E with GL(V, F) <GL(V, E), so 7r also defines an EG-representation. Further, by a tensoringprocess discussed in section 25, n induces a KG-representation 7rK on a K-space VK. This chapter investigates the relationship among these represen-tations. It will often be very useful to extend F to K by passing from 7r to7r K. For example several results at the end of chapter 9 are established in thisway.

Tr is said to be absolutely irreducible if 7r K is irreducible for each extensionK of F, and F is said to be a splitting field for G if every irreducible FG-representation is absolutely irreducible. It develops in section 25 that 7r isabsolutely irreducible precisely when F = EndFG(V) and in section 27 that ifG is finite then a splitting field is obtained by adjoining a suitable root of unity toF. It's particularly nice to work over a splitting field. For example in section 27it is shown that, over a splitting field, the irreducible representations of the directproduct of groups are just the tensor products of irreducible representations ofthe factors.

Section 26 investigates representations over finite fields, where change offield goes very smoothly. Lemma 26.6 summarizes many of the relationshipsinvolved.

Section 27 introduces the minimal polynomial of a linear transformation.Semisimple and unipotent elements are discussed and it is shown that if F isperfect then each member g of GL(V) admits a Jordan decomposition; that isg can be written uniquely as the commuting product of a semisimple elementand a unipotent element.

25 Tensor productsIn this section G is a group, F is a field, and V a finite dimensional vectorspace over F.

Let (V1: 0 < i < m) be vector spaces over F, and denote by L(V1, ... , Vm;Vo) the set of all maps a: Vl x x V. -, Vo such that for each i, 1 <i < m, and each choice of V j E Vj, j i, the map B : Vi -+ Vo defined byv18 = (v1..... vm)a is an F-linear transformation. Such maps are called

Change of field of a linear representation

Let n : G -+ GL(V, F) be an FG-representation, E a subfield of F , and K an extension field of F . Then V is also a vector space over E with GE(V, F) 5 GL(V, E), so n also defines an EG-representation. Further, by a tensoring process discussed in section 25, n induces a KG-representation nK on a K- space VK. This chapter investigates the relationship among these representations. It will often be very useful to extend F to K by passing from n to n K . For example several results at the end of chapter 9 are established in this way.

n is said to be absolutely irreducible if n is irreducible for each extension K of F , and F is said to be a splitting jield for G if every irreducible FG- representation is absolutely irreducible. It develops in section 25 that n is absolutely irreducible precisely when F = EndFc(V) and in section 27 that if G is finite then a splitting field is obtained by adjoining a suitable root of unity to F. It's particularly nice to work over a splitting field. For example in section 27 it is shown that, over a splitting field, the irredycible representations of the direct product of groups are just the tensor products of irreducible representations of the factors.

Section 26 investigates representations over finite fields, where change of field goes very smoothly. Lemma 26.6 summarizes many of the relationships involved.

Section 27 introduces the minimal polynomial of a linear transformation. Semisimple and unipotent elements are discussed and it is shown that if F is perfect then each member g of GL(V) admits a Jordan decomposition; that is g can be written uniquely as the commuting product of a semisimple element and a unipotent element.

25 Tensor products In this section G is a group, F is a field, and V a finite dimensional vector space over F.

Let (Vi: 0 5 i ( m) be vector spaces over F, and denote by L(Vl, . . . , Vm; Vo) the set of all maps a : Vl x . . - x Vm -+ Vo such that for each i, 1 5 i I. m, and each choice of v j E Vj, j # i , the map 8: -+ Vo defined by vie = (vl, . . . , vm)a is an F-linear transformation. Such maps are called

118 Change of field of a linear representation

m-linear. L(VI, ... , Vm; Vo) is a vector space under F via

v(a+0)=va+v0v(act) = a(vu) a, P E L(Vi, ..., V.; Vo).

A tensorproduct of V1, ..., Vm is an F-space T together with 7r E L( Vl , ... ,Vm; T) with the following universal property: whenever U is an F-space anda E L(Vi, ... , Vm; U), there exists a unique 3 E HomF(T, U) with n$ = a.

-(25.1) Tensor products exist and are unique up to isomorphism.

Proof. See for example page 408 in Lang [La].

Because of 25.1 there is a unique tensor product of V1, ..., Vm which is de-noted by Vl ® . . . ® Vm or ®m 1 V. Write v1 ® ® vm for the image of(vi, ..., vm) under the map (denoted by r above) associated to the tensorproduct. The elements vl ® ... ® Vm, V1 E V1. are called fundamental tensors.It is easy to verify from the universal property that:

(25.2) Vl ® . . . ® Vm is generated as an F-space by the fundamental tensors.

Here are some elementary properties of the tensor product; they can be foundfor example in Lang, Chapter 16.

(25.3) Let (Vi: 1 < i < m) be F-spaces. Then(1) V1 ® V2 = V2 ® V1.

(2) (Vi®V2)®V3V1®(V20V3)(3) ((DUEI U) ® V (DUEI(U ® V) for any direct sum ®UEI U of F-

spaces.(4) Let Xi be a basis of Vi, i = 1, 2. Then

X 1 ®X2 = {X 1 ®x2: Xi E X1 }

is a basis for Vl ® V2.(5) Let al EEndF(V1). Then there exists a unique map al ® ® an E

EndF(Vl ®...(9 Vm)With (VI ®...®vm)(a1(9 ...®am) = vial ®...®vmam.(6) Forvl,u1 1, 2, and aE F:

(VI + ul) ® V2 = (Vi (9 V2) + (ul ® V2),

and

VI ® (V2 + u2) = (VI (9 V2) + (v] 0 u2),

a(vl (9 V2) = avl ® V2 = V1 ® avg.

118 Change ofJield of a linear representation

m-linear. L(Vl, . . . , V,; Vo) is a vector space under F via

A tensorproduct of Vl , . . . , V, is an F-space T together with n E L(V1, . . . , V,; T) with the following universal property: whenever U is an F-space and a E L(V1, . . . , V,; U), there exists a unique /3 E HomF(T, U) with n/3 = a.

(25.1) Tensor products exist and are unique up to isomorphism.

Prooj See for example page 408 in Lang [La].

Because of 25.1 there is a unique tensor product of Vl , . . . , V, which is denoted by Vl @ . - @ V, or &. Write vl @ n . - @ v, for the image of (vl, . . . , v,) under the map (denoted by n above) associated to the tensor product. The elements vl @ . . . @ v, , vi E Vi, are called fundamental tensors. It is easy to verify from the universal property that:

(25.2) Vl @ . - . @ V, is generated as an F-space by the fundamental tensors.

Here are some elementary properties of the tensor product; they can be found for example in Lang, Chapter 16.

(25.3) Let (&: 1 5 i 5 m) be F-spaces. Then (1) v l @ v 2 ~ v 2 @ V l . (2) (Vl @ V2) @ v3 Vl @ (V2 @ V3). (3) (e,,, U) @ V % @,,,(U @ V) for any direct sum e,,, U of F-

spaces. (4) Let Xi be a basis of Vi, i = 1,2. Then

XI @ X2 = {XI 8 x 2 : ~ ~ E Xi}

is a basis for Vl @ V2. (5) Let ai E EndF(Vi). Then there exists a unique map a1 @ . - . @ a, E

End~(V1 @... @V,)with(vl @..-@v,)(al@...@a,) = vial @...@v,a, . (6) Forvi,ui E Q , i = 1,2,anda E F:

and

V l 8 (212 + u2) = (211 €3 212) + (211 €3 u2),

Tensor products 119

If 7r1: G --). GL(V; ), 1 < i < m, are FG-representations, then by 25.3.5 thereis an FG-representation 7r1 ®. . . ® zr,,, of G on V1 ® . . ® V,, defined byg(nl ®. (9 7r,,,) = gJr1 ®... ®g7r,,,, for g E G. nl ® ®7rm is the tensorproduct of the representations 7r1, ... , 7r,,, .

A special case of these constructions is of particular interest. Let K be anextension field of F. Then K is a vector space over F, so the tensor productK ® U can be formed for any F-space U. Let X and B be bases for U and Kover F, respectively. By 25.3 each member of K ® U can be written uniquelyas F-(b,x)EBxx ab,x(b (9 x), with ab,x E F. As ab,x(b (9 x) = ((ab,x)b) ® x with(ab,x )b E K, it follows that each member of K ® U is of the form Ex Ex (c® (9 x),cx E K. Indeed it turns out K ® U can be made into a vector space K ® F U =UK over K by defining scalar multiplication via:

a 1:(cx ®x)) _ (acx ®x) a, cx E K, X E X.(XEX XEX

These remarks are summarized in the following lemma; see chapter 16,section 3 in Lang [La] for example.

(25.4) Let K be an extension field of F and X a basis for a vector space U overF. Then UK = K OF U is a vector space over K with 1 ® X = { 1 ® x: x E X}

a basis for UK.

I

It will be useful to have the following well known property of this construction,which can be found on page 419 in Lang [La].

(25.5) If L > K > F is a tower of fields and U an F-space then L ®F UL ®K (K OF U)

Notice that, for g E EndF(U), 1 ® g E EndK(UK), where 1 ® g is the mapdefined in 25.3.5 with respect to the identity map 1 on K. That is 1 ® g: a ®x H a ® xg. In this way EndF(U) is identified with a subalgebra of EndK(UK).Further if jr: G --- GL(U) is an FG-representation then we obtain a KG-representation .rrK: G --± GL(UK) defined by 7rK = 1 ®7r, where 1 is thetrivial representation of G on K. Equivalently (1 (9 x)(girK) = 10 xg7r foreach x E X, g E G. Observe that if U is finite dimensional then Mx(gir)

M(i®x)(gJrK)Recall the definition of enveloping algebra in section 12.

(25.6) Let 7r: G --- EndF(V) = E be an FG-representation, A the envelopingalgebra of 7r in E, and K an extension of F. Then the enveloping algebra of7rK in EndK(VK) is isomorphic to AK as a K-space.

If ni: G -+ GL(Vj), 1 5 i m, are FG-representations, then by 25.3.5 there is an FG-representation nl 8. . . @ nm of G on Vl @ . . . @ Vm defined by g(n1 8 . - - @ n m ) = g n l @. . .@gnm, fo rg t G.n l @...@n,isthetensor product of the representations nl , . . . , n,.

A special case of these constructions is of particular interest. Let K be an extension field of F. Then K is a vector space over F , so the tensor product K @ U can be formed for any F-space U. Let X and B be bases for U and K over F , respectively. By 25.3 each member of K @ U can be written uniquely

as C(b,x)EBxX ab.x(b @ withab,x E F. AS ab,x(b @ X) = ((ab,x)b) @ X with (ab,,)b E K, it follows that eachmember of K 8 U is of the form ~,, ,(c, @ x), c, t K. Indeed it turns out K @ U can be made into a vector space K @ U = lJK over K by defining scalar multiplication via:

These remarks are summarized in the following lemma; see chapter 16, section 3 in Lang [La] for example.

(25.4) Let K be an extension field of F and X a basis for a vector space U over F.ThenuK = K m F Ui~avectorspaceoverKwith1 @ X = (1 @ x : x E X) a basis for uK.

"i It will be useful to have the following well known property of this construction, which can be found on page 419 in Lang [La].

(25.5) If L > K > F is a tower of fields and U an F-space then L U Z

L @K (K @F U).

Notice that, for g t EndF(U), 1 @ g t E ~ ~ K ( u ~ ) , where 1 8 g is the map defined in 25.3.5 with respect to the identity map 1 on K. That is 1 @ g: a @

x ++ a @ xg. In this way EndF(U) is identified with a subalgebra of ~ n d ~ ( ~ ~ ) . Further if n: G -+ GL(U) is an FG-representation then we obtain a KG- representation n K : G -+ G L ( U ~ ) defined by n = 1 @ n , where 1 is the trivial representation of G on K. Equivalently (1 @ ~ ) ( ~ n ~ ) = 1 @ xgn for each x E X, g E G. Observe that if U is finite dimensional then Mx(gn) =

M(l@x,(gn K>

Recall the definition of enveloping algebra in section 12.

(25.6) Let n: G -+ EndF(V) = E be an FG-representation, A the enveloping algebra of n in E, and K an extension of F. Then the enveloping algebra of n in ~ n d K (vK) is isomorphic to as a K-space.

Proof. We may regard E and Endx(V') = EK as the rings Fnxn and Knxnrespectively, with E the set of matrices in Ex whose entries are in F. Now

A ag(g7r):ag E F} so AK = bg(g7r):bg c K .gEG gEG }

But, as matrices, g7r and g7r K are the same, so, as G is a group, Ax is thesubalgebra of Ex generated by G7rx. That is Ax is the enveloping algebra of7r x

Let G1 and G2 be groups and 7ri an FGi-representation. Denote by 7r1 ®7r2the tensor product it 1 ® n2 where ni is the representation of G 1 x G2 with : rirestricted to Gi equal to 7ri and ni restricted to G3_i trivial. This is a smallabuse of notation which will hopefully cause no problem. The convention isused in the proof of the next lemma.

Notice that if G 1 = G2 = G then G is diagonally embedded in G 1 x G2 viathe map g H (g, g), and if we identify G with this diagonal subgroup via theisomorphism, then the tensor product representation 7r, ® 7r2: G -> GL(ViV2) of G is the restriction of the tensor product representation of G1 x G2 tothis diagonal subgroup.

(25.7) Let K be a Galois extension of F, F the Galois group of K over F, andV an FG-module. Then

(1) There is a unique F(r x G)-representation on VK with (y, g): a ® v Hay 0 vg for each a K, v V,y F, g G.

(2) If W is a KG-submodule of VK with Wr = W, then W = UK for someFG-submodule U of V.

Proof. The representation in (1) is just the tensor product representation a ® fiof r x G where a is the action of r on K, and $ is the representation of Gon V.

Assume W is as in (2), and extend a basis Z = (zi: 1 m(aijxl + wi), for wi E W and aij E K. Let y e F. Then

MY = xi - 1:(aij)Yx' = (aij - aijy)x + wi.i>m j>m

But by hypothesis wiy e W, so, as Y is a basis for Vx, aij = aijy for all i, j.

120 Change ofjield of a linear representation

Prooj We may regard E and EndK(vK) = E~ as the rings FnXn and KnXn, respectively, with E the set of matrices in E K whose entries are in F. Now

But, as matrices, g n and g n K are the same, so, as G is a group, AK is the subalgebra of E generated by G n K . That is is the enveloping algebra of n K .

Let G1 and G2 be groups and ni an FGi-representation. Denote by nl 8 n2 the tensor product itl @ it2 where iti is the representation of G 1 x G2 with fti restricted to Gi equal to ni and iti restricted to G34 trivial. This is a small abuse of notation which will hopefully cause no problem. The convention is used in the proof of the next lemma.

Notice that if G1 E Gz E G then G is diagonally embedded in G1 x G2 via the map g H (g, g), and if we identify G with this diagonal subgroup via the isomorphism, then the tensor product representation nl 8 n2: G -+ GL(Vl 8 V2) of G is the restriction of the tensor product representation of GI x G2 to this diagonal subgroup.

(25.7) Let K be a Galois extension of F, r the Galois group of K over F , and V an FG-module. Then

(1) There is aunique F ( r x G)-representation on VK with (y, g): a @ v H

a y @ v g f o r e a c h a ~ K , v ~ V , y E ~ , ~ E G . (2) If W is a KG-submodule of VK with W r = W, then W = uK for some

FG-submodule U of V.

Prooj The representation in (1) is just the tensor product representation a 8 B of r x G where a is the action of r on K, and ,f3 is the representation of G on V.

Assume W is as in (2), and extend a basis Z = (zi: 1 i i _( m) for W inside of X' U Z to a basis Y = (zl, . . . , zm, x;,,, . . . , x:) of vK, where xf = 1 @ xi, and X = {xi: 1 5 i 5 n) is an F-basis for V. For i 5 m, xf = Cj,m(aijxl + wi), for wi E W and aij E K. Let y E r . Then

But by hypothesis wi y E W, so, as Y is a basis for vK, aij = aij y for all i, j.

Tensor products 121

Hence aij E Fix([') = F. As X' is a basis for VK, (wi: 1 m

Thus W = UK where U is the subspace of V generated by (vi: 1 < i < m).

An irreducible FG-module V is absolutely irreducible if VK remains irre-ducible for each extension K of F.

(25.8) Let V be an irreducible FG-module. Then V is absolutely irreducibleif and only if F = EndFG(V).

Proof. Assume F = EndFG(V). Then, by 12.16, E = EndF(V) is the envelop-ing algebra for G on V. So, if K is an extension of F and A' the envelopingalgebra of G in E' = EndK(VK), then, by 25.4 and 25.6, dimK(A') _dimF(A) = n2 = dimK(E'), so A' = E. In particular A', and hence also G,is irreducible on VK.

Conversely assume V is absolutely irreducible. Then VK is irreduciblewhere K is the algebraic closure of F. By 12.17, E' = EndK(VK) is theenveloping algebra for G in E', so, by 25.6, n2 = dimK(E') = dimF(A). LetD = EndFG(V). By 12.16, A = Dmxm where m = dimD(V). Then n =dimF(V) = mk, k = dimF(D) and dimF(A) = m2k. So m2k2 =n2 = m2kand hence k = 1. That is F = D.

If 7r : G -> GL(V) is an FG-representation, X is a basis for V, and a E Aut(F),then 7r°: G - GL(V) is the FG-representation with Mx(g7r°) = Mx(g7r)°.Here if A = (aij) E F""" then A° = (a ). Notice that if n° is the represen-tation defined with respect to a different basis of V, then ir° is equivalentto 7r° by a remark after 13.1. So 7r° is independent of X, up to equiva-lence. I'll sometimes write V° for V regarded as an FG-module with respectto 7r°.

Recall the character of an FG-representation 7r is the function X: G -± Fdefined by X (g) = Tr(g7r ).

Let V be an FG-module and k a field with F < k < EndFG(V). Then theaction of k on V makes V into a k-space. Further that k-space structure extendsthe F-space structure and is preserved by G, so we can regard V as a kG-module. Similarly if K is a subfield of F then V is certainly a K-space and Gpreserves that K-structure. So V is also a KG-module.

Hence ai, E Fix(r) = F . As X' is a basis for vK, (w,: 1 5 i _( m ) is a basis for W and we have shown w, = 1 @ vi, where

Thus W = uK where U is the subspace of V generated by (vi: 1 _( i ( m).

An irreducible FG-module V is absolutely irreducible if vK remains irreducible for each extension K of F .

(25.8) Let V be an irreducible FG-module. Then V is absolutely irreducible if and only if F = EndFG(V).

Proof. Assume F = EndFG (V). Then, by 12.16, E = EndF(V) is the enveloping algebra for G on V. So, if K is an extension of F and A' the enveloping algebra of G in E' = ~ n d ~ ( V ~ ) , then, by 25.4 and 25.6, dimK(Af) = dimF(A) = n2 = dimK(E1), so A' = E'. In particular A', and hence also G, is irreducible on vK.

Conversely assume V is absolutely irreducible. Then vK is irreducible where K is the algebraic closure of F . By 12.17, E' = ~ n d K ( v ~ ) is the enveloping algebra for G in E', so, by 25.6, n2 = dimK(Ef) = dimF(A). Let

'r D = EndFG(V). By 12.16, A 2 DmXm where m = dimD(V). Then n =

dimF(V) = mk, k = dimF(D) and dimF(A) = m2k. So m2k2 = n2 = m2k and hence k = 1. That is F = D.

If n : G + GL(V) is an FG-representation, X is abasis for V, and a E Aut(F), then n u : G + GL(V) is the FG-representation with Mx(gnU) = Mx(gn)". Here if A = (aij) E Fnxn then A" = (a;). Notice that if it" is the representation defined with respect to a different basis of V, then it" is equivalent to n u by a remark after 13.1. So n u is independent of X, up to equivalence. I'll sometimes write V" for V regarded as an FG-module with respect to n u .

Recall the character of an FG-representation n is the function X: G -+ F defined by ~ ( g ) = Tr(gn).

Let V be an FG-module and k a field with F 5 k 5 EndF&). Then the action of k on V makes V into a k-space. Further that k-space structure extends the F-space structure and is preserved by G, so we can regard V as a kG- module. Similarly if K is a subfield of F then V is certainly a K-space and G preserves that K-structure. So V is also a KG-module.

(25.9) Let a: G -+ GL(V) be an irreducible FG-representation such that K =EndFG(V) is a field, and let ,8 be the representation of G on V regarded as aKG-module, and X the character of P. Then

(1) K = F[X], where F[X] is the F-subalgebra of K generated by theelements X (g), g E G.

(2) Assume L is a normal extension of K and a E Gal(K/F)#. ThenL ®K V ° is not L G-isomorphic to L ®K V V.

Proof. Let X be a basis for V over K, m = I X1, and M the enveloping algebraof a in EndF(V). By definition ga = g18 for each g E G and M is the F-subalgebra generated by Ga. Thus M consists of the elements Eag(ga),ag E F. By 12.16, the map >ag(ga) i-+ >agMx(g,8) is an isomorphism ofM with the ring Kmxm of all m by m matrices over K. Hence for each x c Kthere is A E M with Tr(A) = x, and A = EgEG agMx(gf) for some ag c F.Now x = EgEG agX(g) E F[X]. So (1) is established.

Assume the hypothesis of (2) and let E be the fixed field of or. As Cr # 1,E 0 K. As K = EndFG(V) we may assume E = F. Let y be an extension ofor to L and U = L ®K V. Then L ®K V° = UY and the character of G on U isstill the character X of P. Thus if U is LG-isomorphic to UY then X = XY, soX (g) is contained in the fixed field k of y. But k n K = F, so, by (1), K = F,a contradiction.

(25.10) Let V be an irreducible FG-module, k a Galois extension of F, andK = EndFG(V). Then

(1) VK = ®aEA Wa for some irreducible kG-module W and some A CGal(k/F) = F with r = ANr(W), where Nr(W) = {y E F : WY = W).

(2) Let U be an irreducible kG-module. Then V is an FG-submodule of U(regarded as an FG-module) precisely when U is kG-isomorphic to W° forsome or E r.

(3) If k < K then A = I' and W° is kG-isomorphic to V for some or E F.(4) If K is a Galois extension of F then A is a set of coset representatives

for Nr(W) in 1.

Proof. Let F = Gal(k/F). By 25.7, F x G is represented on VK. Let W bean irreducible kG-submodule of Vk and M = (Wy: y E F). If M 0 Vk then,by 25.7.2, M = Uk, for some FG-submodule U of V. As 0 0 M # Vk,0 # U # V contradicting the irreducibility of V. Hence Vk = M and then(1) follows from 12.5. Also Vk is generated as an F-module by the copies(a V: a E V) of V, so Vk is a homogeneous FG-module and hence eachsummand W° is the sum of copies of V, as an FG-module. This gives halfof (2).

122 Change ofjeld of a linear representation

(25.9) Let a: G -+ GL(V) be an irreducible FG-representation such that K = EndFG(V) is a field, and let p be the representation of G on V regarded as a KG-module, and x the character of p. Then

(1) K = F[x], where F[x] is the F-subalgebra of K generated by the elements x (g), g E G.

(2) Assume L is a normal extension of K and a E Gal(K/F)#. Then L BK V" is not LG-isomorphic to L BK V.

Proof. Let X be a basis for V over K, m = 1x1, and M the enveloping algebra of a in EndF(V). By definition ga = gp for each g E G and M is the F- subalgebra generated by Ga. Thus M consists of the elements E U , ( ~ ~ ) , a, E F. By 12.16, the map E a, (ga) H E a, Mx(g/?) is an isomorphism of M with the ring KmXm of all m by m matrices over K. Hence for each x E K there is A E M with Tr(A) = x , and A = E,,, a,Mx(gj?) for some a, E F. Now x = CgEG a,x(g) E F [ X ] SO (I) is established.

Assume the hypothesis of (2) and let E be the fixed field of a. As a # 1, E # K. As K = EndEG(V) we may assume E = F. Let y be an extension of a to L and U = L gK V. Then L BK V" = UY and the character of G on U is still the character x of p. Thus if U is LG-isomorphic to UY then x = x Y , SO

x (g) is contained in the fixed field k of y. But k n K = F , so, by (I), K = F , a contradiction.

(25.10) Let V be an irreducible FG-module, k a Galois extension of F , and K = EndFG(V). Then

(1) VK = eaEA Wa for some irreducible kG-module W and some A Gal(k/F) = r with r = ANr(W), where Nr(W) = {y E r : WY G W).

(2) Let U be an irreducible kG-module. Then V is an FG-submodule of U (regarded as an FG-module) precisely when U is kG-isomorphic to W" for some a E r.

(3) If k 5 K then A = r and W" is kG-isomorphic to V for some a E r. (4) If K is a Galois extension of F then A is a set of coset representatives

for Nr (W) in I'.

Proof. Let r = Gal(k/F). By 25.7, r x G is represented on VK. Let W be an irreducible kG-submodule of vk and M = (Wy: y E r). If M # Vk then, by 25.7.2, M = uk, for some FG-submodule U of V. As 0 # M # Vk, 0 # U # V contradicting the irreducibility of V. Hence vk = M and then (1) follows from 12.5. Also vk is generated as an F-module by the copies (aV:a E k#) of V, so vk is a homogeneous FG-module and hence each summand Wa is the sum of copies of V, as an FG-module. This gives half of (2).

Representations over finite fields 123

Assume U is an irreducible kG-module and V an FG-submodule of U. LetX be an F-basis of V. As U is irreducible and V an FG-submodule of U, Xgenerates U as a kG-module. Also 1 (9 X is a basis for Vk over k, so we candefine:

a:Vk --> U

ax(1 x) H axx ax E k.

XEX XEX

Then a is a surjective kG-homomorphism, so Vk/ker(a) = U as a kG-module.Hence (1) and 12.5 imply U = W° for some o E A. So (2) holds.

In (3) we may regard V as an irreducible kG-module, so, by (2), V - W°for some Q E A. Then, by (1), 1Al = dimF(V)/dimk(V) _ Ik: Fl IFI soA = F and (3) is established.

To prove (4) let L be the extension generated by k and K and assume K isGalois over F. Then L is Galois over K, and, by (3), 25.5, and 25.3.3:

V L- L ®K V K= ® L ®K U°°eGal(K/F)

with U KG-isomorphic to V. Now L ®K U° is an irreducible LG-modulefor each a E Gal(K/F), since, by 25.8, V is absolutely irreducible as a KG-module. Further if o # r then L ®K U° L ®K U` by 25.9.2. On the otherhand if a and b are distinct members of A with W° = Wb then L ®k Wa LlL ®kWb so some irreducible occurs in VL''with multiplicity greater than 1,a contradiction. This completes the proof of (4).

A splitting field for a finite group G is a field F with the property that everyirreducible FG-representation is absolutely irreducible. Notice that by 25.8an irreducible FG-module V is absolutely irreducible precisely when F =EndFG(V). Hence, by 12.17:

(25.11) If F is algebraically closed then F is a splitting field for each finitegroup.

It will turn out in section 27 that if G is a finite group then a splitting field forG is obtained by adjoining a suitable root of unity to GF(p) for any prime p.

26 Representations over finite fieldsThe hypotheses of section 25 are continued in this section. In addition assumeF is of finite order.

The following observations make things go particularly smoothly when Fis finite.

Assume U is an irreducible kG-module and V an FG-submodule of U . Let X be an F-basis of V. As U is irreducible and V an FG-submodule of U , X generates U as a kG-module. Also 1 B X is a basis for vk over k, SO we can define:

Then a is a surjective kG-homomorphism, so Vk/ker(a) =" U as a kG-module. Hence (1) and 12.5 imply U =" W" for some a E A. So (2) holds.

In (3) we may regard V as an irreducible kG-module, so, by (2), V % W" for some a E A. Then, by (I), IAl = dim~(V)/ dimk(V) = Ik : FI = Il? 1 so A = l? and (3) is established.

To prove (4) let L be the extension generated by k and K and assume K is Galois over F. Then L is Galois over K, and, by (3), 25.5, and 25.3.3:

with U KG-isomorphic to V. Now L B K U" is an irreducible LG-module for each a E Gal(K/F), since, by 25.8, V is absolutely irreducible as a KG- module. Further if a # s then L BK U" 9 L BK U T by 25.9.2. On the other hand if a and b are distinct members of A with W a S wb then L Bk W a 2

L Bk wb SO some irreducible occurs in vL %ith multiplicity greater than 1, a contradiction. This completes the proof of (4).

A splitting jield for a finite group G is a field F with the property that every irreducible FG-representation is absolutely irreducible. Notice that by 25.8 an irreducible FG-module V is absolutely irreducible precisely when F = EndFc(V). Hence, by 12.17:

(25.11) If F is algebraically closed then F is a splitting field for each finite group.

It will turn out in section 27 that if G is a finite group then a splitting field for G is obtained by adjoining a suitable root of unity to GF(p) for any prime p .

26 Representations over finite fields The hypotheses of section 25 are continued in this section. In addition assume F is of finite order.

The following observations make things go particularly smoothly when F is finite.


(26.1) (1) Each finite dimensional division algebra over F is a finite field, andhence a Galois extension of F.

(2) If V is an irreducible FG-module then EndFG(V) is a finite Galois ex-tension of F.

The first remark follows from the well known facts that finite division ringsare fields, and that every finite field is Galois over each of its subfields. Thesecond remark is a consequence of the first and the hypothesis that V is offinite dimension over F.

(26.2) Let V be an irreducible FG-module, k a finite extension of F, andP = Gal(k/F). Then

(1) Vk = ®QEA W° for some irreducible kG-module W and any set A ofcoset representatives for Nr(W) in P.

(2) Let U be an irreducible kG-module. Then V is an FG-submodule of Uprecisely when U is kG-isomorphic to W° for some a E F.


Let jr: G -* GL(V) be an FG-representation and K a subfield of F. We say ncan be written over K if there exists an F-basis X of V such that each entry ofMX(gr)isinKforeach g E G.

(26.3) Let jr: G -- GL(V) be an irreducible FG-representation, K a subfieldof F, and (a) = Gal(F/K). Then the following are equivalent:

(1) rr can be written over K.(2) V = F ®KU for some irreducible KG-submodule U of V.(3) V is FG-isomorphic to V°.

Proof. The equivalence of (1) and (2) is trivial as is the implication (1) im-plies (3). Assume (3) and let U be an irreducible KG-submodule of V. ThenF = Gal(F/K) = NF(V) as f' = (a) and V = V°. Hence, by 26.2, OF = V.That is (3) implies (2).

(26.4) Let V be an irreducible FG-module, K a subfield of F, U an irreducibleKG-submodule of V, and E = NF(U). Then V = F ®E U.

Proof. Let U F = F ®E U and h = Gal(F/E). By 26.2, UF = ®QEA VQwhere A is a set of coset representatives for A = NF(V) in F. Let L bethe fixed field of A and W an irreducible LG-submodule of V; as V is

Change ofjield of a linear representation

(26.1) (1) Each finite dimensional division algebra over F is a finite field, and hence a Galois extension of F.

(2) If V is an irreducible FG-module then EndFG(V) is a finite Galois extension of F.

The first remark follows from the well known facts that finite division rings are fields, and that every finite field is Galois over each of its subfields. The second remark is a consequence of the first and the hypothesis that V is of finite dimension over F.

(26.2) Let V be an irreducible FG-module, k a finite extension of F , and r = Gal(k/F). Then

(1) vk = eaGA WO for some irreducible kG-module W and any set A of coset representatives for Nr(W) in r.

(2) Let U be an irreducible kG-module. Then V is an FG-submodule of U precisely when U is kG-isomorphic to Wu for some a E r.


Let n : G -+ GL(V) be an FG-representation and K a subfield of F . We say n can be written over K if there exists an F-basis X of V such that each entry of Mx(gn) is in K for each g E G.

(26.3) Let n: G -, GL(V) be an irreducible FG-representation, K a subfield of F , and (a) = Gal(F/K). Then the following are equivalent:

(1) n can be written over K. (2) V = F @ K U for some irreducible KG-submodule U of V. (3) V is FG-isomorphic to V".

Proof. The equivalence of (1) and (2) is trivial as is the implication (1) implies (3). Assume (3) and let U be an irreducible KG-submodule of V. Then r = Gal(F/K) = Nr(V) as r = ( m ) and V Z V". Hence, by 26.2, uF = V. That is (3) implies (2).

(26.4) Let V be an irreducible FG-module, K a subfield of F , U an irreducible KG-submodule of V, and E = NF(U). Then V = F B E U.

Proof. Let uF = F BEU and r = Gal(F/E). By 26.2, uF = eaEA Va, where A is a set of coset representatives for A = Nr(V) in r. Let L be the fixed field of A and W an irreducible LG-submodule of V; as V is

Representations over finite fields 125

a homogeneous EG-module we may assume U < W. If L F then, byinduction on IF : E1, L ®E U = W, while, by 26.3, V = F ®L W, SoV = F ®E U. Thus we may take L = F. But then A = I so

dimE(U) = dimF(UF) = I F : EI dimF(V) = dimE(V),

so U = V. Hence F = NF(U) = E, and the lemma holds.

(26.5) Let V be an irreducible FG-module. Then the following are equivalent:

(1) V can be written over no proper subfield of F.(2) V is an irreducible KG-module for each subfield K of F.(3) NAut(F)(V) = 1.

Proof. This follows from 26.3 and 26.4.

An FG-module V is condensed if V is absolutely irreducible and can be writtenover no proper subfield of F.

Theorem 26.6. Let p be a prime, Fp the field of order p, F p its algebraicclosure, A the set of finite subfields of Pp, and Il the set of pairs (F, V) whereF E A and V is an (isomorphism type of an) irreducible finite dimensionalFG-module. Define a relation,, on 1 by (F, V)T(K, U) if F < K and V is anFG-submodule of U. Let - be the equivalence relation on 1 generated by T.Then

(1) (F, V)T(K, U) if and only if F < K and U is a summand of K OF V-(2) Let A be an equivalence class of ^-. Then, for each F E A, AF =

{(F, V) E A} is nonempty and Aut(F) is transitive on AF. In particularAFp I = 1 and the map A H AFp is a bijection between the set of equiva-lence classes of and the isomorphism classes of irreducible finite dimen-sional FpG modules.

(3) If (F, V), (K, U) E 1 with F < K then (F, V) - (K, U) if and only if(F, V°)/(K, U), for some a e Aut(F).

(4) In each equivalence class A of ^- there exists a unique F E A such thatthe members of AF are condensed. Indeed for (Fp, V) E A, F = EndFpG(V)and (F, V) E A with V a condensed FG-module.

Proof. Part (1) follows from 26.2. Let (F, V) E Q. If E < F then there is anirreducible EG-module VE of V, and we saw during the proof of 25.10 that Vis a homogeneous EG-module, so VE is determined up to isomorphism. WriteVp for VFp.

Let (K, U) E 0. Claim (F, V) (K, U) if and only if Vp = Up. The suffi-ciency of Vp = Up is immediate from the definition of to prove necessity

epresentations overjnite jelds 125

a homogeneous EG-module we may assume U I W. If L # F then, by induction on IF : E 1, L gE U 2 W, while, by 26.3, V Z F gL W, So V Z F g E U.ThuswemaytakeL= F.ButthenA = 1so

so U = V. Hence F = NF(U) = E, and the lemma holds.

(26.5) Let V be an irreducible FG-module. Then the following are equivalent:

(1) V can be written over no proper subfield of F. (2) V is an irreducible KG-module for each subfield K of F .

(3) NAU~(F)(V) = 1.

ProoJ This follows from 26.3 and 26.4.

An FG-module V is condensed if V is absolutely irreducible and can be written over no proper subfield of F .

Theorem 26.6. Let p be a prime, Fp the field of order p, Fp its algebraic closure, A the set of finite subfields of E,, and a the set of pairs (F, V) where F E A and V is an (isomorphism type of an) irreducible finite dimensional FG-module. Define a relation r on 0 by (F, V)r(K, U) if F 5 K and V is an FG-submodule of U. Let -- be the equivaleqce relation on generated by r . Then

(1) (F, V)r(K, U) if and only if F (. K and U is a summand of K @,P V. (2) Let A be an equivalence class of --. Then, for each F E A, AF =

((F, V) E A) is nonempty and Aut(F) is transitive on AF In particular lAFp 1 = 1 and the map A I+ AFp is a bijection between the set of equivalence classes of -- and the isomorphism classes of irreducible finite dimensional Fp G modules.

(3) If (F, V), (K, U) E with F 5 K then (F, V) -- (K, U) if and only if (F, Vu)r(K, U), for some a E Aut(F).

(4) In each equivalence class A of -- there exists a unique F E A such that the members of AF are condensed. Indeed for (Fp , V) E A, F = EndFpG(V) and (F, V) E A with V a condensed FG-module.

Proof. Part (I) follows from 26.2. Let (F, V) E a. If E F then there is an irreducible EG-module VE of V, and we saw during the proof of 25.10 that V is a homogeneous EG-module, so VE is determined up to isomorphism. Write V, for VF,.

Let (K, U) E a. Claim (F, V) -- (K, U) if and only if Vp = Up. The suffi- ciency of Vp = Up is immediate from the definition of --; to prove necessity

it suffices to take (F, V)/(K, U) and to show Vp = Up. But this follows fromthe last paragraph.

Let A be an equivalence class of - and (Fp, V) E A. By 26.2, for each F E

A, VF = (DaEA Wa for some set A of coset representatives for NA t(F)(W) inAut(F). By (1) and the claim,

AF = {(F, W°): a E Aut(F)}.

That is (2) holds.To prove (3) observe that if (K, U) E 0 and F < K then (F, UF)/(K, U).

Then, by (2), (F, W) - (K, U) if and only if W° = UF for some a E Aut(F).So (3) holds.

Let F = EndFPG(V). Then (F, V) E A and, by 25.8, V is an absolutelyirreducible FG-module. By 25.10.3, 25.10.4, and 26.5, V can be written overno proper subfield of F, so V is condensed.

Finally suppose (K, U) is another condensed member of A. To completethe proof of (4) we must show K = F. Let k be the subfield of F. generatedby K and F. Use 26.5 and the fact that V is condensed as an FG-module toconclude V 5t V° for a E Aut(F)#; then by 26.2,

F ®GF(p) V = ® VQ.a EAut(F)

Then by 25.3 and 25.5,

k ®GF(p) V = k ®GF(p) (F ®GF(p) V) = k OF ( ® V y 1yEAu[(F) ///

® (k OFV').yEAut(F)

As V is absolutely irreducible as an FG-module, k OF V Y is irreducible foreach y. Hence k ®FP V has exactly IF : Fp I irreducible summands. But, bysymmetry between K and F, k®FP V also has (K : Fp I irreducible summands,soK=F.

Theorem 26.6 defines an equivalence relation on the finite dimensional repre-sentations of G over finite fields of characteristic p. This equivalence relationhas the property that each class contains a representative over each finite fieldof characteristic p. Hence we can think of such a representation as writtenover any finite field of characteristic p. However the lemma suggests that toeach class there is associated a field F over which the representation is bestwritten: namely the unique field over which the representation is condensed.I will refer to F as the field of definition of the representation. F can alsobe described as the smallest field over which the representation is absolutely

126 Change offield of a linear representation

it suffices to take (F, V)r(K, U) and to show Vp 2 Up. But this follows from the last paragraph.

Let A be an equivalence class of -- and (F,, V) E A. By 26.2, for each F E

A, VF = eaEA Wa, for some set A of coset representatives for NA"~(F)(W) in Aut(F). By (1) and the claim,

AF = {(F, Wa):a E Aut(F)}.

That is (2) holds. To prove (3) observe that if (K, U) E C2 and F 5 K then (F, UF)r(K, U).

Then, by (2), (F, W) -- (K, U) if and only if W" = UF for some a E Aut(F). So (3) holds.

Let F = EndFp&'). Then (F, V) E A and, by 25.8, V is an absolutely irreducible FG-module. By 25.10.3, 25.10.4, and 26.5, V can be written over no proper subfield of F , so V is condensed.

Finally suppose (K, U) is another condensed member of A. To complete the proof of (4) we must show K = F. Let k be the subfield of F , generated by K and F. Use 26.5 and the fact that V is condensed as an FG-module to conclude V 2 V" for a E Aut(F)#; then by 26.2,

Then by 25.3 and 25.5,

As V is absolutely irreducible as an FG-module, k @F VY is irreducible for each y. Hence k gFp V has exactly IF : Fp I irreducible summands. But, by symmetry between K and F , kBFp V also has IK : F,J irreducible summands, so K = F .

Theorem 26.6 defines an equivalence relation on the finite dimensional representations of G over finite fields of characteristic p. This equivalence relation has the property that each class contains a representative over each finite field of characteristic p. Hence we can think of such a representation as written over any finite field of characteristic p. However the lemma suggests that to each class there is associated a field F over which the representation is best written: namely the unique field over which the representation is condensed. I will refer to F as the field of definition of the representation. F can also be described as th

Minimal polynomials 127

irreducible. Equivalently F = EndF,G(V ), where V is the unique FpG-modulein the class.

27 Minimal polynomialsIn section 27 F is a field, V is a finite dimensional vector space over F, and Gis a group.

Suppose for the moment that A is a finite dimensional algebra over F andF[x] is a polynomial ring. For a E A and f (x) _ Ym o b;x` E F[x], definef (a) o bra` E A. Then the map aa: F[x] A such that aa: f H f (a),is an F-algebra homomorphism. The assumption that A is finite dimensionalforces ker(aa) 54 0 since F[x] is of infinite dimension. As F[x] is a prin-cipal ideal domain (PID), ker(aa) is a principal ideal, and indeed there isa unique monic polynomial fa(x) with ker(aa) = (fa). By definition fa isthe minimal polynomial of a. By construction fa divides each polynomialwhich annihilates a. Further fa is monic of degree at most dimF(A), sincedimF(A) > dimF(F[x]/(fa)) = deg(fa).

Applying these observations to the n2-dimensional F-algebra EndF(V), itfollows that each g E EndF(V) has a minimal polynomial min(g) = min(g, F,V). Observe that if n: A B is an F-algebra isomorphism then a and anhave the same minimal polynomial. In particular if X is a basis for V thenMX: EndF(V) -+ Fn"n is such an isomorphism, so min(g) = min(Mx(g)). IfK is an extension of F, then Ml®X(1 (9 g) = MX(g), so the minimal poly-nomial of 1 ® g over K divides the minimal polynomial of g over F. Indeedan easy application of rational canonical form shows the two minimal poly-nomials are equal. (Reduce to the case where V is a cyclic FG-module. Thenthere is a basis X = (x;: 1 < i < n) for V in which xjg = xj+1 for i < n andxng = - En o aj+lxj, where f (x) = x" + rz=o a,x` is the minimal polynomial for g over F. Then f (1 (9 g) = 0 and if h(x) E K[x] properly divides fthenh(1 (9 g) Oas

[1 ® x1:1 < i < deg(h) + 1}

is linearly independent and 1 ®xj+1 = (1 (9 xl)g`.) Thus we have shown:

(27.1) The minimal polynomial of a linear transformation is unchanged byextension of the base field. That is, if K is an extension of F and g E EndF ( V ),

then min(g, F, V) = min(1 (9 g, K, VK).

Let g E EndF(V) and a E F. We say a is a characteristic value for g if thereexists v E V# with vg = av. We call v a characteristic vector for a.

(27.2) Let g E EndF(V) and a E F. Then a is a characteristic value for g ifand only if a is a root of the minimal polynomial of g.


irreducible. Equivalently F = EndF,,c(V), where V is the unique F,G-module in the class.

27 Minimal polynomials In section 27 F is a field, V is a finite dimensional vector space over F , and G is a group.

Suppose for the moment that A is a finite dimensional algebra over F and F[x] is a polynomial ring. For a E A and f (x) = Cy=o bixi E F[x], define f (a) = CrZ0 biai E A. Then the map a,: F [x] + A such that a,: f H f (a), is an F-algebra homomorphism. The assumption that A is finite dimensional forces ker(a,) # 0 since F[x] is of infinite dimension. As F[x] is a principal ideal domain (PID), ker(a,) is a principal ideal, and indeed there is a unique monic polynomial f,(x) with ker(a,) = (f,). By definition fa is the minimal polynomial of a . By construction fa divides each polynomial which annihilates a . Further fa is monic of degree at most dimF(A), since

d im~(A) > dim~(F[xIl(f,)) = deg(f,). Applying these observations to the n2-dimensional F-algebra EndF(V), it

follows that each g E EndF(V) has a minimal polynomial min(g) = min(g, F , V). Observe that if n : A + B is an F-algebra isomorphism then a and a n have the same minimal polynomial. In particular if X is a basis for V then Mx: EndF(V) + FnXn is such an isomorphism, so min(g) = min(Mx(g)). If K is an extension of F , then MIBx(l @ g) = Mx(g), so the minimal polynomial of 1 @ g over K divides the minimal polynomial of g over F. Indeed an easy application of rational canonical form shows the two minimal poly-

I nomials are equal. (Reduce to the case where V is a cyclic FG-module. Then there is a basis X = (xi: 1 ( i ( n) for V in which xlg = xi+l for i < n and xng = - ~yz i ai+lxi, where f (x) = xn + ~ : i i alxi is the minimal polynomial for g over F . Then f (1 @ g) = 0 and if h(x) E K[x] properly divides f then h(l @ g) # 0 as

(1 @xi: 1 5 i (deg(h)+ 1)

is linearly independent and 1 @ xi+l = (1 @ x ~ ) ~ ' . ) Thus we have shown:

(27.1) The minimal polynomial of a linear transformation is unchanged by extension of the base field. That is, if K is an extension of F and g E EndF(V), thenmin(g, F, V) =min(l @ g , K, vK).

Let g E EndF(V) and a E F. We say a is a characteristic value for g if there exists v E V# with vg = av. We call v a characteristic vector for a.

(27.2) Let g E EndF(V) and a E F. Then a is a characteristic value for g if and only if a is a root of the minimal polynomial of g.

Proof. Let f (x) = min(g). If a is not a root of f then (f, x - a) = 1 sothere exist r, s E F[x] with f r + (x - a)s = 1. Now if v E V# with vg = avthen v = v1 = v(f (g)r(g) + (g - aI)s(g)) = 0, since f (g) =0 = v(g - aI).This contradiction shows characteristic values of g are roots of f.

Conversely if a is a root of f, then f = (x - a)h, for some h E F[x]. If a isnot a characteristic value of g then ker(g - a I) = 0 so (g -a I)-1 exists. Hence,as 0 = f (g) = (g - aI)h(g), we also have h(g) = 0. But then f divides h, acontradiction.

g E EndF(V) is semisimple if the minimal polynomial of g has no repeatedroots.

(27.3) Let g E EndF(V) and assume min(g) splits over F. Then g is semisimpleif and only if g is diagonalizable.

Proof. This is Exercise 9.2.

(27.4) Let S be a finite subset of commuting elements of EndF (V), and assumethe minimal polynomial of each number of S splits over F. Then there existsa basis X of V such that for each s E S the following hold:

(1) MX(s) is lower triangular.(2) The entries on the main diagonal of Mx(s) are the eigenvalues of min(s).(3) Ifs is semisimple then Mx(s) is diagonal.

Proof. Induct on n + IS 1. If n = 1 the result is trivial, so take n > 1. If someg E S is a scalar transformation then by induction on BSI, the result holds forS - {g}, and then also for S. So no member of S is a scalar transformation.

Let g E S. By 27.2, g possesses a characteristic value a1. Let V1 be theeigenspace of a1 for g. Then S C_ C(g) < N(V1). As g is not a scalar, V1 0 V,so, by induction on n there is a basis X 1 for V1 with MX, (SI v) as claimed inthe lemma. In particular there is a 1-dimensional subspace U of V1 fixed by S.

If g is semisimple, then, by 27.3, V = ®im-t1 Vi, where Vi is the eigenspaceof the characteristic value ai of g. Now, choosing a basis Xi for V, as in thelast paragraph, we see that X = U` 1 Xi is a basis with the desired properties.

So we can assume no member of S is semisimple. Thus (3) is established.Finally S acts on V/ U = V and, by induction on n, there is a basis X for Vwith MX(SIv) triangular. Now pick X = (xi: 1 < i < n) with U = (xi) and(ii: 1 < i < n) = X; then Mx(s) is triangular for each s E S.

g E EndF(V) is nilpotent if g' = 0 for some positive integer m, and g isunipotent if g = I + h for some nilpotent h E EndF(V).


Proof. Let f (x) = min(g). If a is not a root of f then (f, x - a) = 1 so there exist r, s E F[x] with f r + (x - a)s = 1. Now if v E V' with vg = av then v = v 1 = v( f (g)r(g) + (g - aI)s(g)) = 0, since f (g) = 0 = v(g - a I). This contradiction shows characteristic values of g are roots of f .

Conversely if a is a root o f f , then f = (x - a)h, for some h E F[x]. If a is not acharacteristic value of g then ker(g - a I ) = 0 so (g-a I)-' exists. Hence, as 0 = f (g) = (g - aI)h(g), we also have h(g) = 0. But then f divides h, a contradiction.

g E EndF(V) is semisimple if the minimal polynomial of g has no repeated roots.

(27.3) Let g E EndF(V) and assume min(g) splits over F. Then g is semisimple if and only if g is diagonalizable.

Proof. This is Exercise 9.2.

(27.4) Let S be a finite subset of commuting elements of EndF(V), and assume the minimal polynomial of each number of S splits over F . Then there exists a basis X of V such that for each s E S the following hold:

(1) Mx(s) is lower triangular. (2) The entries on the main diagonal of Mx(s) are the eigenvalues of min(s). (3) If s is semisimple then Mx(s) is diagonal.

Proof. Induct on n + I SI. If n = 1 the result is trivial, so take n > 1. If some g E S is a scalar transformation then by induction on IS/, the result holds for S - (g}, and then also for S. So no member of S is a scalar transformation.

Let g E S. By 27.2, g possesses a characteristic value al. Let Vl be the eigenspace of a l for g. Then S C(g) 5 N(Vl). As g is not a scalar, Vl # V, so, by induction on n there is a basis X1 for Vl with Mx, (SIv,) as claimed in the lemma. In particular there is a 1-dimensional subspace U of Vl fixed by S.

If g is semisimple, then, by 27.3, V = Vi, where Vi is the eigenspace of the characteristic value ai of g. Now, choosing a basis Xi for Vi as in the last paragraph, we see that X = UyZl Xi is a basis with the desired properties.

So we can assume no member of S is semisimple. Thus (3) is established. Finally S acts on V/U = V and, by induction on n, there is a basis X for V with Mz(SIv) triangular. Now pick X = (xi: 1 5 i 5 n) with U = ( X I ) and (fi: 1 5 i 5 n) = X; then Mx(s) is triangular for each s E S.

g E EndF(V) is nilpotent if g" = 0 for some positive integer m, and g is unipotent if g = I + h for some nilpotent h E EndF(V).

(27.5) Let g E EndF(V) and n = dimF(V). Then(1) The following are equivalent:

(i) g is nilpotent, unipotent, respectively.(ii) min(g) = x', (x - 1)', respectively, for some positive integer m.(iii) There exists a basis X for V such that MX(g) is lower triangular

and all entries on the main diagonal are 0, 1, respectively.(2) Unipotent elements are of determinant 1, and hence nonsingular.(3) If g is nilpotent and semisimple then g = 0.(4) If g is unipotent and semisimple then g = I.(5) Let char(F) = p > 0. Then gP" = 1 if g is unipotent. Conversely if

gP= 1 for some positive integer m, then g is unipotent.(6) If char(F) = 0 and g E GL(V) is of finite order then g is semisimple.(7) If char(F) = p > 0 and g E GL(V) is of finite order m then g is semisim-

ple if and only if (m, p) = 1.

Proof. All parts of the lemma are reasonably straightforward, but I'll makea couple of remarks anyway. If g E GL(V) is of finite order m the minimalpolynomial of g divides x'" -1, which has no multiple roots if (m, char(F)) = 1.Hence (6) and half of (7) hold. Parts (4) and (5) imply the remaining half of(7), since any power of a semisimple element is semisimple.

Recall a field F is perfect if char(F) = 0 or char(F) = p > 0 and F = FP,where FP is the image of F under the p-power map a i-+ aP. For examplefinite fields are perfect as are algebraically closed fields. We need the followingelementary fact, which appears for example as the Corollary on page 190 ofLang [La].

(27.6) If F is perfect then every polynomial in F[x] is separable.

(27.7) If F is perfect and a E EndF(V), then there exists f E F[x] and a posi-tive integer e with (f, f') = 1 (where f' is the derivative of f) and fe(a) = 0.If ,B E EndF(V) with f (,B) = 0 then ,B is semisimple.

Proof. Let min(a) = flm 1 fee' with fi irreducible. Letm

e = max{el: l <i < ml and f = ffe (a) = 0 and, by 27.6, fi has no repeated roots. Hence f has no repeated

roots, so (f, f') = 1. Also, if f (,B) = 0 then min(p) divides f, and hence hasno repeated roots, so ,B is semisimple.

Minimal polynomials

(27.5) Let g E EndF(V) and n = dim~(V). Then (1) The following are equivalent:

(i) g is nilpotent, unipotent, respectively. (ii) min(g) = xm, (x - l)m, respectively, for some positive integer m.

(iii) There exists a basis X for V such that Mx(g) is lower triangular and all ent'ries on the main diagonal are 0, 1, respectively.

(2) Unipotent elements are of determinant 1, and hence nonsingular. (3) If g is nilpotent and semisimple then g = 0. (4) If g is unipotent and semisimple then g = I . (5) Let char(F) = p > 0. Then gP" = 1 if g is unipotent. Conversely if

gP"' = 1 for some positive integer m, then g is unipotent. (6) If char(F) = 0 and g E GL(V) is of finite order then g is semisimple. (7) If char(F) = p > 0 andg E GL(V) is of finite order m then g is semisim-

ple if and only if (m, p) = 1.

Proof. All parts of the lemma are reasonably straightforward, but I'll make a couple of remarks anyway. If g E GL(V) is of finite order m the minimal polynomial of g divides xm - 1, which has no multiple roots if (m, char(F)) = 1. Hence (6) and half of (7) hold. Parts (4) and (5) imply the remaining half of (7), since any power of a semisimple element is semisimple.

Recall a field F is perfect if char(F) = 0 or char(F) = p > 0 and F = FP, where F P is the image of F under the p-power map a H up. For example finite fields are perfect as are algebraically closed fields. We need the following elementary fact, which appears for example as the Corollary on page 190 of Lang [La].

(27.6) If F is perfect then every polynomial in F [x] is separable.

(27.7) If F is perfect and a E EndF(V), then there exists f E F [x] and a positive integer e with (f, f ') = 1 (where f ' is the derivative o f f ) and f e(a) = 0. If B E EndF(V) with f (B) = 0 then B is semisimple.

Proof. Let min(a) = ny!Jei with f; irreducible. Let m

e=max(e; : l ii i m } and f =nfi. i=l

Then f e(a) = 0 and, by 27.6, fi has no repeated roots. Hence f has no repeated roots, so (f, f ') = 1. Also, if f (B) = 0 then min(B) divides f , and hence has no repeated roots, so B is semisimple.


The proof of the following lemma comes from page 71 of [Ch 2].

(27.8) Let F be perfect and a E EndF(V). Then there exist Y E EndF(V)with a = ,B + y, ,B semisimple, and y nilpotent. Further ,B = t(a) for somet(x) E F[x].

Proof. Choose f and e as in the last lemma. Then there exist h, h1 E F[x]with 1 = f'h + f h 1. Define an F-algebra homomorphism o: F[x] -+ F[x]by(go)(x) = g(x - f (x)h(x)). Observe go _ (g-g' f h) mod f 2, so inparticularf O = f - f'f h = f - f (1 - f h 1) = 0 mod f 2. Then proceeding by induc-tion on m and using the fact that 0 is a homomorphism, f 0' = 0 mod f2-.Choose m with 2' > e. Then:

(27.8.1) fo' =_ 0 mod fe, so (f¢')(a) = 0.

Next, for g(x) = Tj a,x' E F[x],

gok = (aixi) ok = >ai(xgk)` = g(xok),

as q5k is a homomorphism, so

(27.8.2) gok = g(xok) for each g E F[x].

Also xo = x - f h = x mod f , so proceeding by induction on k, and using thefact that is a homomorphism with f 0 = 0 mod f , we conclude:

(27.8.3) xqk = x mod f.

We can now complete the proof of the lemma. Let t = x0"', ,B = t(a), andy = a - P. Then .f (fl) = .f (t)(a) = .f (xom)(a) = (fom)(a) = 0, by 27.8.2and 27.8.1, respectively. Hence, by 27.7, ,B is semisimple. Also, by 27.8.3,(x - t) = 0 mod f , so ye = (a - ,B)e = (x - t)e(a) = 0. Thus y is nilpotent.

(27.9) Let F be perfect and a E GL(V). Then(1) There exist as, a E GL(V) with as, semisimple, a, unipotent, and

a = asa =a = µ = µ with , p c GL(V), semisimple, and µ unipotent,

then= as and /-t = au.(3) There exist polynomialst(x), v(x) E F[x] withal = t(a)andau = v(a).

Proof. By 27.8, a = + y, with P, y c EndF(V), B semisimple, y nilpotent,and = t(a), for some t(x) E F[x]. As ,B = t(a), $ E Z(C(a) fl EndF(V)).Let F be the algebraic closure of F. By 27.4 and 27.5 there is a basis X ofF with MX(,B) diagonal and Mx(y) strictly lower triangular. From this it isevident that det(,B) = det(er). Thus, as a is nonsingular, so is, P.

130 Change ofjeld of a linear representation

The proof of the following lemma comes from page 71 of [Ch 21.

(27.8) Let F be perfect and a E EndF(V). Then there exist j3, y E EndF(V) with a = j3 + y, j3 semisimple, and y nilpotent. Further j3 = t(a) for some t(x) E F[x].

Proof. Choose f and e as in the last lemma. Then there exist h, hl E F[x] with 1 = f'h + f hl . Define an F-algebra homomorphism4: F[x] + F[x] by (g@)(x) = g(x - f (x)h(x)). Observe g@ (g-g'f h) mod f 2, so in particular f 4 = f - f 'fh = f - f (1 - f h l ) = Omod f 2. Then proceeding by induction on m and using the fact that @ is a homomorphism, f @m = 0 mod f 2m.

Choose m with 2m > e . Then:

(27.8.1) f 4" = 0 mod f ', so (f @")(a) = 0.

Next, for g(x) = xi aixi E F[xl,

as @k is a homomorphism, so

(27.8.2) g@k = g ( ~ ~ k ) for each g E F[x].

Also x4 = x - f h = x mod f , so proceeding by induction on k, and using the fact that @ is a homomorphism with f @ = 0 mod f , we conclude:

(27.8.3) x @ ~ = x mod f.

We can now complete the proof of the lemma. Let t = x@", j3 = t(a), and y = a - j3. Then f (j3) = f (t)(a) = f (x@")(a) = (f @")(a) = 0, by 27.8.2 and 27.8.1, respectively. Hence, by 27.7, j3 is semisimple. Also, by 27.8.3, (X - t) = 0 mod f , so ye = (a - j3le = (x - t)e(a) = 0. Thus y is nilpotent.

(27.9) Let F be perfect and a! E GL(V). Then (1) There exist a,, a, E GL(V) with a,, semisimple, a, unipotent, and

a = asau = auas. (2) If a = cp = pc with c , p E GL(V), ( semisimple, and p unipotent,

then ( = a, and p = a,. (3) Thereexistpolynomialst(x), v(x) E F[x] witha, = t(a) anda, = v(a).

Proof. By 27.8, a = j3 + y, with j3, y E EndF(V), j3 semisimple, y nilpotent, and j3 = t(a), for some t(x) E F[x]. As j3 = t(a), j3 E Z(C(cf) f l EndF(V)). Let be the algebraic closure of F. By 27.4 and 27.5 there is a basis X of F with Mx(j3) diagonal and Mx(y) strictly lower triangular. From this it is evident that det(j3) = det(a). Thus, as a is nonsingular, so is j3.


Let as = ,B and a.= I -I- P-1 y. Asp-' and y commute and y is nilpotent,,B-1y is nilpotent, so a is unipotent. By construction a = asa = So

(1) and (3) hold.Suppose and µ are as in (2). As ,B E Z(C(a)), and µ commute with ,B

and a,,. By 27.4, ,B and can be simultaneously diagonalized over F, and hence,B-1 is diagonalizable over F, so,8-1 is semisimple by 27.3. Similarly

a is both semisimple and

unipotent, so, by 27.5.4, ,B = and a = µ.

as and a are called the semisimple part of a and the unipotent part of a,respectively, and the decomposition a = asa = is called the Jordan de-composition of a. As a consequence of 27.5 and 27.9 we have:

(27.10) Let a E GL(V) be of finite order. Then as and a are powers of a.If char(F) = 0 then a = as, while if char(F) = p > 0 then Ias _ aI p, andlaul = Ialp.

(27.11) Let F be perfect and a c EndF(V), b a characteristic value of a in F,U the eigenspace of b for a in V, and K an extension of F. Then UK is theeigenspace of b for 1 ® a in VK.

Proof. This is essentially an application of Jordan Form; I sketch a proof.Recall the map f H f(a) is an F-algebra representation of F[x] on V withkernel (M(x)), where M(x) = min(a). Let M(x) = Fl'= 1 pi(x)el be the primefactorization of M. From the theory of modules over a principal ideal do-main (cf. Theorems 3 and 6 on pages 390 and 397 of Lang [La]) we knowV = ®i-1 V(i), where V (i) = ker(pi (a)ej ). Indeed as the polynomials p; (x )e'are relatively prime the same holds in VK, so as V(i)K is contained in thekernel (VK)(i) of p{(x)ei on VK, we conclude V(i)K = (VK(i)). Thus withoutloss M = (x - b)e.

Again from the theory of modules over a PID, V = ®l=1 Vi, where V,v,F[x] is a cyclic module for F[x] with annihilator (x - b)ei, e = e1 > e2 >

> e5 > 1, and the invariants el are uniquely determined. As VK =VK is a module with invariants e1, we conclude the el are also the invariants ofF[x] on VK, and it remains to observe that U = ®;=1 U;, with UU = U fl V,of dimension 1.

(27.12) Let 7r: G -+ GL(V) be an FG-representation and K an extension ofF. Then dimK (CVK (g7rK)) = dimF (Cv (g7r )) for each g c G.


Let a, = j3 and a, = I + j3-' y. As j3-' and y commute and y is nilpotent, j3-' y is nilpotent, so a, is unipotent. By construction a = a,a, = auas. So (1) and (3) hold.

Suppose 6 and p are as in (2). As j3 E Z(C(a)), 6 and p commute with j3 and a,. By 27.4, j3 and 6 can be simultaneously diagonalized over F , and hence j3-'6 is diagonalizable over F , so j3-'6 is semisimple by 27.3. Similarly a,p-' is unipotent. Finally as Pa, = a = c p , j3-'6 = a&-' is both semisimple and unipotent, so, by 27.5.4, j3 = 6 and a, = p .

as and a, are called the semisimple part of a and the unipotent part of a , respectively, and the decomposition a = a,a, = auas is called the Jordan decomposition of a. As a consequence of 27.5 and 27.9 we have:

(27.10) Let a E GL(V) be of finite order. Then as and a, are powers of a . If char(F) = 0 then a = a,, while if char(F) = p > 0 then lasl = lalp~ and

I a u l = Ialp.

(27.11) Let F be perfect and a E EndF(V), b a characteristic value of a in F , U the eigenspace of b for a in V, and K an extension of F . Then uK is the eigenspace of b for 1 €3 a in vK.

/.

Proof. This is essentially an application of Jordan Form; I sketch a proof. Recall the map f H f (a) is an F-algebra representation of F[x] on V with kernel (M(x)), where M(x) = rnin(a). Let M(x) = niZl pi(x)" be the prime factorization of M. From the theory of modules over a principal ideal domain (cf. Theorems 3 and 6 on pages 390 and 397 of Lang [La]) we know V = @;=, V(i), where V(i) = ker(~i(a)~l) . Indeed as the polynomials pi(x)'' are relatively prime the same holds in VK, SO as ~ ( i ) ~ is contained in the kernel (vK)(i) of pi(x)" on vK, we conclude ~ ( i ) ~ = (vK (i)). Thus without loss M = (x - b)'.

Again from the theory of modules over a PID, V = @:=, Vi, where Vi = viF[x] is a cyclic module for F[x] with annihilator (x - b)" , e = el 2 ez L

- . L e, 3 1, and the invariants ei are uniquely determined. As vK = viK is a module with invariants ei, we conclude the ei are also the invariants of F[x] on v K , and it remains to observe that U = Ui, with Ui = U n Vi of dimension 1.

(27.12) Let n : G + GL(V) be an FG-representation and K an extension of F. Then dimK(Cv~(gnK)) = dimF(Cv(gn)) for each g E G.

Recall the definition of a splitting field in section 25.

(27.13) Let G be a finite group of exponent m and r: G -- GL(V) an FG-representation. Let k = m if char(F) = 0 and k = m p, if char(F) = p > 0. Letn = dimF(V). Then

(1) Let X be the character of r and g E G. Then X(g) is a sum of n kthroots of unity.

(2) F is a splitting field for G if F is finite and contains a primitive kth rootof unity.

Proof. Let F be the algebraic closure of F. By 27.4 there is a basis X for VFsuch that Mx(g) is triangular. By 27.10, g = gsg with gs semisimple and 1gsIdividing k, with g, unipotent, and with gs and g powers of g. Thus the entrieson the main diagonal of are 1 and so the entries of Mx(g) are the sameas those of MX(gs). In particular if a is such an entry then ak = 1 as (gs)k = 1.So (1) holds.

Assume F is finite, F contains co, a primitive kth root of 1, V is an irreducibleFG-module, and let K = EndFG(V). By 26.1.2, K is a finite field extending Fand hence containing co. Let r be the character of V regarded as a KG-module.The argument of (1) shows Vr(g) E F for each g e G. As i/r(g) E F for eachg e G, we conclude from 25.9.1 that K = F. Thus F is a splitting field for Gby 25.8, completing the proof.

Recall that given representations a and P of groups G and H, respectively,there is a representation a ®P of G x H. The definition of a ® f appearsjust before the statement of Lemma 25.7. This representation appears in thestatement of the next two lemmas.

(27.14) Let G < GL(V ), M = CGL(V)(G), and assume V is a homogeneousFG-module. Write I = Irr(G, V, F) for the set of irreducible FG-submodulesof V and choose Vi e 1, 1 < i < m, with V = ®ml Vi. Let K = EndFG (VI )be a field and A = HomFG(VI, V). Then

(1) There exists Y = (ai: 1 < i < m) c A with Viai = Vi and a1 = 1.(2) A is a KG-module and Y is a K-basis for A. Y induces a unique K-

space structure on V extending the F-structure such that ai E HomKG(VI, Vi)for each i. This structure is preserved by G.

(3) The map n: M - GL(A; K) defined by xir:,8 H 18x, x e M, P E A,is an isomorphism. M preserves the K-space structure on V.

Change ofjield of a linear representation

Proof. This is a direct consequence of 27.1 1.

Recall the definition of a splitting field in section 25.

(27.13) Let G be a finite group of exponent m and n: G + GL(V) an FG- representation. Let k = m if char(F) = 0 and k = m,~ if char(F) = p > 0. Let n = dimF(V). Then

(1) Let x be the character of n and g E G. Then ~ ( g ) is a sum of n kth roots of unity.

(2) F is a splitting field for G if F is finite and contains a primitive kth root of unity.

Proof. Let F be the algebraic closure of F . By 27.4 there is a basis X for V' such that Mx(g) is triangular. By 27.10, g = gsgu with g, semisimple and Igsl dividing k, with g, unipotent, and with g, and g, powers of g. Thus the entries on the main diagonal of Mx(g,) are 1 and so the entries of Mx(g) are the same as those of Mx(gs). In particular if a is such an entry then ak = 1 as (gs)k = 1. So (1) holds.

Assume F is finite, F contains o, a primitive kth root of 1, V is an irreducible FG-module, and let K = EndFG(V). By 26.1.2, K is a finite field extending F and hence containing w. Let + be the character of V regarded as a KG-module. The argument of (1) shows +(g) E F for each g E G. As +(g) E F for each g E G, we conclude from 25.9.1 that K = F. Thus F is a splitting field for G by 25.8, completing the proof.

Recall that given representations cr and B of groups G and H, respectively, there is a representation a @I B of G x H. The definition of cr @I B appears just before the statement of Lemma 25.7. This representation appears in the statement of the next two lemmas.

(27.14) Let G 5 GL(V), M = CcL(v)(G), and assume V is a homogeneous FG-module. Write I = Irr(G, V, F ) for the set of irreducible FG-submodules of v and choose % E I, 1 5 i 5 m, with V = 6. Let K = EndFG(V1) be a field and A = HomFG(V1, V). Then

(1) There exists Y = (ai: 1 5 i 5 m) E A with VIai = and a1 = 1. (2) A is a KG-module and Y is a K-basis for A. Y induces a unique K-

space structure on V extending the F-structure such that cri E Hom~G(v1, Vi) for each i . This structure is preserved by G.

(3) The map n: M + GL(A; K) defined by xn: B H Bx, x E M, B E A, is an isomorphism. M preserves the K-space structure on V.

(4) The map 7/r: A# ± I defined by 1:,8 H Vl,8 is a surjection and definesa bijection

0: S(A) SG(V)

BH(b:beB)between the set S(A) of all K-subspaces of A and the set SG(V) of all FG-submodules of V. 0 is a permutation equivalence of the actions of M on S(A)and SG(V).

(5) The map B: M x G -+ GL(V, K) defined by

(x, g)8: v H vxg

is a K(M x G)-representation whose image is GM and which is equivalent tothe tensor product of the representations of M on A and G on Vl over K.

Proof. As V is homogeneous there exists an isomorphism al: Vl -+ Vi ofFG-modules. Composing ai with the inclusion Vi c V we may regard ai as amember of A. Choose ar = 1. Then (1) holds. As K = EndFG(VI) and ai is anFG-isomorphism, K = EndFG (Vi) for each i and ai is also a KG-isomorphism.Thus we have a unique scalar multiplication of K on Vi extending that of Fsuch that ai e HomKG(V1, Vi), so there is a K-space structure on V extendingthat on F. It is defined by:

m m F

a (uai) = a(uai), u e Vi, a E K.i=1 i=1

Next A = HomFG (V1, ®m 1 Vi) = ®m j HomFG (Vi , Vi) is isomorphic toKm as an F-space. Similarly HomKG(V1, V) is isomorphic to Km as a K-space and is an F-subspace of A, so A = HomKG(V1, V) is also a K-space.Also, as Y is a K-linearly independent subset of A of order m, Y is a K-basisfor A, so (2) holds.

Evidently, for P E A and x c M, the composition ,8x is also in A, so themap r in (3) is a well-defined KM-representation. If aix = ai then, as G isirreducible on Vi, Vi < Cv(x). Hence r is faithful. Let (vj: 1 < j < d) be aK-basis for V1. Then

X = (vjai: 1 < j < d, l < i < m)

is a K-basis for V. An element of the general linear group GL(A, K) on A (re-garded as a K-space) may be regarded as an m by m matrix (aid) with respect tothe basis Y of A. Given such an element define x e M by vjai x = Ek vjaikak.Then x c M with x 7r = (aid), and x preserves the K-structure on V. Hence ris an isomorphism and (3) holds.


(4) The map +: A# + I defined by +: ,B H V1,B is a surjection and defines a bijection

between the set S(A) of all K-subspaces of A and the set SG(V) of all FG- submodules of V. 4 is a permutation equivalence of the actions of M on S(A)

The map 8 : M x G + GL(V, K) defined by

(x, g)8: u H uxg

is a K(M x G)-representation whose image is GM and which is equivalent to the tensor product of the representations of M on A and G on Vl over K.

Proof. As V is homogeneous there exists an isomorphism al: Vl + Vi of FG-modules. Composing ai with the inclusion Vi E V we may regard ai as a member of A. Choose al = 1. Then (1) holds. As K = EndFG(Vl) and ai is an FG-isomorphism, K = EndFG(X) for each i and ai is also a KG-isomorphism. Thus we have a unique scalar multiplication of K on extending that of F such that ai E HomKG(Vl, Vi), so there is a K-space structure on V extending that on F. It is defined by:

Next A = H0mFG(V1, vi) = HOmFG(Vi, Vi) is isomorphic to Km as an F-space. Similarly HomKG(Vl, V) is isomorphic to Km as a K- space and is an F-subspace of A, so A = HomKG(Vl, V) is also a K-space. Also, as Y is a K-linearly independent subset of A of order m, Y is a K-basis for A, so (2) holds.

Evidently, for ,B E A and x E M, the composition ,Bx is also in A, so the map n in (3) is a well-defined KM-representation. If six = ai then, as G is irreducible on Vi, Vi 5 Cv(x). Hence n is faithful. Let (uj: 1 ( j d) be a K -basis for Vl . Then

is a K-basis for V. An element of the general linear group GL(A, K) on A (regarded as a K-space) may be regarded as an m by m matrix (aij) with respect to the basis Y of A. Given such an element define x E M by U j f f i X = xk ujaikffk. Then x E M with x n = (aij), and x preserves the K-structure on V. Hence n is an isomorphism and (3) holds.

Evidently *maps A# into I and the induced map 0 takes S(A) into SG(V) andpreserves inclusion. It is also clear that 0 is an injection from the set S1(A) of1 -dimensional subspaces of A into I. Let W E I and let 7ri: W -k V, be the i thprojection. 7ri is trivial or an isomorphism by Schur's Lemma, and there existsan isomorphism f: Vl W. Then ai = ftiai 1 E K and a = >2aiai E Awith a* = Via = W, so 0: S1(A) -k I is a bijection. For each B, D E S(A),(B + D)O = BO + DO; from this remark and its predecessor it is not difficultto complete the proof of (4).

Finally, by (3), M preserves the K-space structure on V. Hence the map 0in (5) is indeed a well-defined K(M x G)-representation whose image is GM.The map vjai H vj ® ai induces an equivalence of 0 with the tensor productrepresentation

(x, g): ai ® vi ra aix ® vj g

of M x G on A ® V1, so (5) holds.

(27.15) Let Gi, i = 1, 2, be groups, F a splitting field for G1 and G2, and Aia collection of representatives for the equivalence classes of finite dimensionalirreducible F Gi -representations. Then the map (7r1, 7r2) H ni ®7r2 is a bijec-tion between A 1 x L2 and the set of equivalence classes of finite dimensionalirreducible F(G1 x G2)-modules.

Proof. Let7ri c Li with module Vi. By 27.14.4 and 27.14.5 there is a bijectionbetween the F(G1 x G2)-submodules of V1 ® V2 and the FG2-submodulesof V2, so, as 7r2 is irreducible, so is 7r1 ®7r2. Conversely let 7r: G1 x G2 -*GL(V) be an irreducible FG-representation. By Clifford's Theorem, 12.13, Vis a homogeneous FGi-module, so, by 27.14.5, 7r is equivalent to 7ri ®7r2 forsome 7ri E Ai. Indeed 7ri is determined up to equivalence by the equivalenceclass of irreducible FGi-submodules of V. So the lemma holds.

(27.16) Let G be a finite group and 7r : G -- GL(V) an irreducible FG-represen-tation. Then Z(G7r) = (z) is a cyclic group of order relatively prime to thecharacteristic of F and, if F contains a primitive IzIth root of unity w, then zacts on V by scalar multiplication via a power wk of w with (Iz1, k) = 1.

Proof. By 12.15, Z(G7r) is cyclic, say Z(G7r) = (z). By Exercise 4.3, n = JzIis relatively prime to char(F). So we can assume w is a primitive nth root of 1in F. Now z satisfies the polynomial f (x) = x" - 1 so its minimal polynomialdivides f and hence has roots powers of w. So by 27.2 wk is a characteristicvalue for z for some 0 < k < n, and then by Clifford's Theorem z acts by scalarmultiplication via wk on V. Thus n = 1z I _ I wk 1, so (k, n) = 1.

Id of a linear representation

preserves inclusion. It is also clear that q5 is an injection from the set Sl (A) of I-dimensional subspaces of A into I. Let W E I and let ni: W + Vi be the i th projection. ni is trivial or an isomorphism by Schur's Lemma, and there exists an isomorphism B: Vl + W. Then ai = Bnia;' E K and a = C a p i E A with a$ = Vla = W, so q5: Sl(A) + I is a bijection. For each B, D E S(A), (B + D)@ = Bq5 + Dq5; from this remark and its predecessor it is not difficult to complete the proof of (4).

Finally, by (3), M preserves the K-space structure on V. Hence the map 8 in (5) is indeed a well-defined K(M x G)-representation whose image is GM. The map ujai ++ uj €9 ai induces an equivalence of 8 with the tensor product representation

of M x G on A €9 Vl, so (5) holds.

(27.15) Let Gi, i = 1,2, be groups, F a splitting field for G 1 and G2, and Ai a collection of representatives for the equivalence classes of finite dimensional irreducible F Gi -representations. Then the map (771 ,772) t+ n1 €9 772 is a bijection between A x A2 and the set of equivalence classes of finite dimensional irreducible F(Gl x G2)-modules.

Proof. Let ni E Ai with module Vi. By 27.14.4 and 27.14.5 there is a bijection between the F(Gl x G2)-submodules of Vl €9 V2 and the FG2-submodules of V2, SO, as 772 is irreducible, so is 771 €3 772. Conversely let n : G1 x G2 + GL(V) be an irreducible FG-representation. By Clifford's Theorem, 12.13, V is a homogeneous FGi-module, so, by 27.14.5, IT is equivalent to ni €9 for some ni E Ai. Indeed ni is determined up to equivalence by the equivalence class of irreducible FGi-submodules of V. So the lemma holds.

(27.16) Let G be a finite group and n : G + GL(V) an irreducible FG-representation. Then Z(Gn) = (z) is a cyclic group of order relatively prime to the characteristic of F and, if F contains a primitive 1~1th root of unity w , then z acts on V by scalar multiplication via a power wk of w with (lzl, k) = 1.

Proof. By 12.15,Z(Gn) is cyclic, say Z(Gn) = (z). By Exercise 4.3, n = lzl is relatively prime to char(F). So we can assume o is a primitive nth root of 1 in F. Now z satisfies the polynomial f ( x ) = x n - 1 so its minimal polynomial divides f and hence has roots powers of w. So by 27.2 wk is a characteristic value for z for some 0 < k < n, and then by Clifford's Theorem z acts by scalar multiplication via wk on V. Thus n = lz 1 = lok 1, so (k, n) = 1.

(27.17) Let V be an irreducible FG-module. Assume G is finite and a semi-direct product of H a_ G by X of prime order p. Assume dim(V)p(dim(Cv(X))). Then V is a homogeneous FH-module and if F is finite oforder prime to p then V is an irreducible FH-module.

Proof. By Clifford's Theorem, 12.13, V is the direct sum of the homogeneouscomponents (V,: 1 < i < r) of H on V, and X permutes these componentstransitively. So as X = (x) is of prime order p either V is a homogeneousFH-module or p = r and X is regular on the components. But in the lat-ter case Cv(X) _ 1 vx`: v c V,}, so dim(V) = pdim(Cv(X)) contraryto hypothesis.

So assume F is finite of order q prime to p. By 27.13 there exists a finiteextension K of F which is a splitting field for H and contains a primitivepth root of 1. By 26.2, VK = ®aEA W° for some A C Gal(K/F). By 27.12,dim(CvK(X))=dim(Cv(X)). So dim(W); pdim(CW(X)). Further if H isirreducible on W then, by 26.2, { Wa: a E A) is the set of H-homogeneouscomponents of VK. But if 0 U is an FH-submodule of V then 0 UK is aGal(K/F)-invariant KH-submodule of VK, so Wa < UK for some a, and thenVK = UK. That is, H is irreducible on V. So, replacing (V, F) by (W, K), wemay assume F is a splitting field for H and contains a primitive pth root of1. The existence of the pth root forces q =_ 1 mod p. Then by 27.14.4 theset I of irreducible H-submodules of V Is of order (qm - 1)/(q - 1)m mod p, where dim(V) = m dim(U) for U E I. As V is an irreducibleFG-module, V = (Ux) is the sum of at most p conjugates of U, so m < p,with equality only if V = ®p 1 Ux' . This last case is out by an argument inthe last paragraph, so m < p. Further as V is an irreducible FG-module either{V } = I or X is fixed point free on I, and we may assume the latter. But then,by 5.14, m =III = 0 mod p, contradicting m < p.

(27.18) Let p and q be primes with q > p, G a group of order pq, X E Sylp (G),and V a faithful FG-module with (pq, char(F)) = 1 and Cv(X) = 0. Then Gis cyclic.

Proof. Extending F if necessary, we may assume with 27.12 that F containsa primitive qth root of 1. By Exercise 2.5, G has a normal Sylow q-groupH. As (pq, char(F)) = 1, V is the direct sum of irreducible FG-modulesby Maschke's Theorem, so H is faithful on one of these irreducibles, andhence we may assume V is an irreducible FG-module. So, by 27.17, V is ahomogeneous FH-module. Hence 27.16 says H acts by scalar multiplicationon V, so H < Z(G). Thus G = HX is cyclic.


(27.17) Let V be an irreducible FG-module. Assume G is finite and a semidirect product of H 2 G by X of prime order p. Assume dim(V) # p(dim(Cv(X))). Then V is a homogeneous FH-module and if F is finite of order prime to p then V is an irreducible FH-module.

Proof. By Clifford's Theorem, 12.13, V is the direct sum of the homogeneous components (Vi: 1 5 i I r ) of H on V, and X permutes these components transitively. So as X = (x) is of prime order p either V is a homogeneous FH-module or p = r and X is regular on the components. But in the latter case CV(X) = (EL, vxi: v E V,), SO dim(V) = p dim(Cv(X)) contrary to hypothesis.

So assume F is finite of order q prime to p. By 27.13 there exists a finite extension K of F which is a splitting field for H and contains a primitive pth root of 1. By 26.2, vK = eaEA Wa for some A G Gal(K/F). By 27.12, dim(Cv~ (X)) = dim(Cv(X)). So dim( W) # p dim(Cw(X)). Further if H is irreducible on W then, by 26.2, (Wa: a E A) is the set of H-homogeneous components of vK. But if 0 # U is an FH-submodule of V then 0 # uK is a Gal(K/F)-invariant KH-submodule of vK, SO Wa 5 uK for some a , and then vK = uK. That is, H is irreducible on V. So, replacing (V, F ) by (W, K), we may assume F is a splitting field for H and contains a primitive pth root of 1. The existence of the pth root forces q E 1 modp. Then by 27.14.4 the set I of irreducible H-submodules of V !is of order (qm - l)/(q - 1) = m mod p, where dim(V) = m dim(U) for U E I. As V is an irreducible FG-module, V = (uX) is the sum of at most p conjugates of U, so m ( p, with equality only if V = $:, uX'. This last case is out by an argument in the last paragraph, so m < p. Further as V is an irreducible FG-module either (V) = I or X is fixed point free on I, and we may assume the latter. But then, by 5.14, m = 1 11 = 0 mod p, contradicting m < p.

(27.18) Let p and q be primes with q > p, G a group of order pq, X E Syl,(G), and V a faithful FG-module with (pq, char(F)) = 1 and CV(X) = 0. Then G is cyclic.

Proof. Extending F if necessary, we may assume with 27.12 that F contains a primitive qth root of 1. By Exercise 2.5, G has a normal Sylow q-group H. As (pq, char(F)) = 1, V is the direct sum of irreducible FG-modules by Maschke's Theorem, so H is faithful on one of these irreducibles, and hence we may assume V is an irreducible FG-module. So, by 27.17, V is a homogeneous FH-module. Hence 27.16 says H acts by scalar multiplication on V, so H ( Z(G). Thus G = HX is cyclic.


Remarks. The classical theory of linear representations of finite groups con-siders representations over the complex numbers where things go relativelysmoothly. Unfortunately many questions about finite groups require consider-ation of representations over fields of prime characteristic, particularly finitefields. For example we've seen that the study of representations in the categoryof groups of a group G on an elementary abelian p-group E is equivalent to thestudy of GF(p)G-representations on E regarded as a GF(p)-space. Represen-tation theory over such less well behaved fields requires the kind of techniquesintroduced in this chapter. Lemma 27.18 provides one application of thesetechniques, and we will encounter others in section 36.

Little use is made in this book of the Jordan decomposition studied in section27. It is however fundamental to the study of groups of Lie type as linear groupsor algebraic groups.

Exercises for chapter 91. Let U, V, and W be FG-modules.

(1) Define 0: L(U, V; W) -> HomF((U, HomF(V, W))) by v(u(a4))) _(u, v)a for u E U, V E V, and a E L(U, V; W). Prove 0 is an isomor-phism of F-spaces.

(2) Prove 0 is an isomorphism of L(U, V; F) with HomF(U, V*), whereV* is the dual of V.

(3) G preserves f E L(U, V ; F) if f (ug, vg) = f (u, v) for each g E G.Prove G preserves f if and only if f0 E HomFG(U, V*).

(4) Let 0 be an automorphism of F of order at most 2 and LG (V, V B) theset of sesquilinear forms on V with respect to 0 which are preservedby G. Assume V is an irreducible FG-module. Prove LG(V, V9) 0

if and only if V is isomorphic to (Vs)* as an FG-module, in whichcase each member of LG(V, V9)# is nondegenerate. If V is absolutelyirreducible prove the members of LG(V, V B) are similar, if 0 = 1 eachmember is symmetric or each is skew symmetric, and if 101 = 2 somemember is hermitian symmetric.

2. Prove Lemma 27.3. (Hint: Use the theory of modules over a PID as in theproof of 27.11.)

3. Let n1 and n2 be FG-representations, Xi the character of nl, and X thecharacter of 71 0 72. Prove X = X1 X2, that is, for each g E G, X(g) _X1(g)X2(g)

4. Let G be a finite group and X the character of a complex G-representation.Prove X(g) = X*(g) = X(g-1) for each g E G, where X* is the characterof the dual representation and ft) is the complex conjugate of X(g).

5. (Spectral Theorem) Let V be a finite dimensional vector space over thecomplex numbers and f a positive definite unitary form on V. Then for


Remarks. The classical theory of linear representations of finite groups considers representations over the complex numbers where things go relatively smoothly. Unfortunately many questions about finite groups require consideration of representations over fields of prime characteristic, particularly finite fields. For example we've seen that the study of representations in the category of groups of a group G on an elementary abelian p-group E is equivalent to the study of GF(p)G-representations on E regarded as a GF(p)-space. Represen- tation theory over such less well behaved fields requires the kind of techniques introduced in this chapter. Lemma 27.18 provides one application of these techniques, and we will encounter others in section 36.

Little use is made in this book of the Jordan decomposition studied in section 27. It is however fundamental to the study of groups of Lie type as linear groups or algebraic groups.

Exercises for chapter 9 1. Let U, V, and W be FG-modules.

(1) Define 4: L(U, V; W) + HOmF((U, HOmF(V, W))) by V(u(a!@)) =

(u, v)a! for u E U, v E V, and a! E L(U, V; W). Prove 4 is an isomorphism of F-spaces.

(2) Prove 4 is an isomorphism of L(U, V; F ) with HOmF(U, V*), where V* is the dual of V.

(3) G preserves f E L(U, V; F ) if f (ug, vg) = f (u, v) for each g E G. Prove G preserves f if and only if f 4 E HomFG(U, V*).

(4) Let 6 be an automorphism of F of order at most 2 and LG(V, v') the set of sesquilinear forms on V with respect to 6 which are preserved by G. Assume V is an irreducible FG-module. Prove LG(V, v') # 0 if and only if V is isomorphic to (Ve)* as an FG-module, in which case each member of LG(V, ve)# is nondegenerate. If V is absolutely irreducible prove the members of LG(V, v') are similar, if 6 = 1 each member is symmetric or each is skew symmetric, and if 16 1 = 2 some member is hermitian symmetric.

2. Prove Lemma 27.3. (Hint: Use the theory of modules over a PID as in the proof of 27.11 .)

3. Let nl and n2 be FG-representations, xi the character of ni, and x the character of nl 8 n2. Prove x = ~ 1 x 2 ; that is, for each g E G, ~ ( g ) =

x1(g)x2(g>. 4. Let G be a finite group and x the character of a complex G-representation.

Prove X(g) = x*(g) = x(~ - ' ) for each g E G, where X * is the character of the dual representation and X(g) is the complex conjugate of ~ ( g ) .

5. (Spectral Theorem) Let V be a finite dimensional vector space over the complex numbers and f a positive definite unitary form on V. Then for

each g E O(V, f) there exists an orthonormal basis for (V, f) consistingof characteristic vectors for g. In particular every element of O(V, f) issemisimple.

6. Let (Vi, fi ), i = 1, 2, be 2-dimensional symplectic spaces over a field Fand let V = V1 ® V2. Let O(V1) be the group of similarities g of thatis g E GL(VV) with fi(xg, yg) = X(g) fi(x, y) for all x, y E Vi, and someX(g) E P. Prove(1) There exists a unique nondegenerate symmetric bilinear form f =

f, ® f2 on V such that

f(vl ® v2, U1 ® u2) = fi(vi, ui).f2(v2, u2), ui, vi E V.

(2) There is a unique quadratic form Q on V associated to f withQ(vl ®v2) = 0 for all vi E Vi.

(3) (V, Q) is a 4-dimensional hyperbolic orthogonal space.(4) Let Di = O(Vi, fi), Gi = O(Vi, fi), and it the tensor product rep-

resentation of A = O1 X A2 on V (cf. the convention before 25.7).ProveA7r < O(V, Q)with(gi, g2)ir E O(V, Q) if and only ifk(gl))1(g2)-1. ker(ir) = {(,LI,-1I):.l E F#}.

(5) Let a: (V1, fl) -> (V2, f2) be an isometry. Prove there is a uniquet E GL(V) with (u (& va)t = v ® ua. Prove t is a transvection or re-flection in O(V, Q), (O17r)` = O2ir, and (GI7r)` = G27r.

(6) O(V, Q) = (on)(t)(7) Q(V, Q) = (G1G2)ir = SL2(F) * SL2(F), unless IFS = 2.

7. If it is an irreducible FG-representation and Or E Aut(F), then 7r° is anirreducible FG-representation. If X is the character of it then X° is thecharacter of 7r°, where X°(g) = (X (g))' for g E G.

8. Let V be an n-dimensional vector space over a field F of prime charac-teristic p and x an element of order p in GL(V). Assume n > p. Provedim(Cv(x)) > 1.

9. Let V be a finite dimensional vector space over a field F, f a nontrivialsesquilinear form on V with respect to an automorphism 9 of finite orderm, and G = O(V, f). Assume G is irreducible on V. Prove that either(1) V is FG-isomorphic to Ve and G preserves a nondegenerate bilinear

form on V, or(2) m is even and V is FG-isomorphic to V02 but not to VB. Further

V = F ®K U and G preserves a nondegenerate hermitian symmetricform on U, where K is the fixed field of 92 and U is an irreducibleKG-submodule of V.(Hint: Use Exercise 9.1 and the fact that (V*)* is FG-isomorphic to V.)

10. Let it be an irreducible CG-representation and or a 1-dimensional CG-representation. Prove it ® or is an irreducible CG-representation.


each g E O(V, f ) there exists an orthonormal basis for (V, f ) consisting of characteristic vectors for g. In particular every element of O(V, f ) is semisimple.

6. Let (K, fi), i = 1,2, be 2-dimensional symplectic spaces over a field F and let V = Vl @ V2. Let A(Vi) be the group of similarities g of K; that is g E GL(K) with fi(ng, yg) = h(g)fi(n, y) for all n, y E K, and some h(g) E F'. Prove (1) There exists a unique nondegenerate symmetric bilinear form f =

f l @ f2 on V such that

(2) There is a unique quadratic form Q on V associated to f with Q(vl @ v2) = 0 for all vi E V,.

(3) (V, Q) is a 4-dimensional hyperbolic orthogonal space. (4) Let A, = A(Vi, fi), Gi = O(V,, f,), and n the tensor product rep-

resentation of A = Al x A2 on V (cf. the convention before 25.7). Prove An 5 A(V, Q) with (gl, g2)n E O(V, Q) if and only if h(g1) = ~ ( ~ 2 ) - ' . ker(n) = (@I, h-' I): h E F'}.

(5) Let a : (Vl, fl) + (V2, f2) be an isometry. Prove there is a unique t E GL(V) with (u @ va)t = v @ ua. Prove t is a transvection or reflection in O(V, Q), (Aln)' = A2n, and (Gin)' = G2n.

(6) A(V, Q) = (An)(t). (7) O(V, Q) = (G1G2)n Z SL2(F) * SL2(F), unless I Fl = 2.

7. If n is an irreducible FG-representation and a E Aut(F), then n u is an irreducible FG-representation. If x is the character of n then xu is the character of n u , where xU(g) = (~ (g ) ) " for g E G.

8. Let V be an n-dimensional vector space over a field F of prime characteristic p and n an element of order p in GL(V). Assume n > p. Prove dim(Cv (n)) > 1.

9. Let V be a finite dimensional vector space over a field F , f a nontrivial sesquilinear form on V with respect to an automorphism 6 of finite order m, and G = O(V, f ). Assume G is irreducible on V. Prove that either (1) V is FG-isomorphic to V' and G preserves a nondegenerate bilinear

form on V, or (2) rn is even and V is FG-isomorphic to V" but not to v'. Further

V = F mK U and G preserves a nondegenerate hermitian symmetric form on U , where K is the fixed field of o2 and U is an irreducible KG-submodule of V. (Hint: Use Exercise 9.1 and the fact that (V*)* is FG-isomorphic to V.)

10. Let n be an irreducible CG-representation and a a I-dimensional CG- representation. Prove n @ a is an irreducible CG-representation.

10

Presentations of groups

A group F is free with free generating set X if it possesses the followinguniversal property: each function a: X H of X into a group H extendsuniquely to a homomorphism of F into H. We find in section 28 that for eachcardinal C there exists (up to isomorphism) a unique free group F with freegenerating set of cardinality C. Less precisely: F is the largest group generatedby X.

If W is a set of words in the alphabet X U X-1, it develops that there is also alargest group G generated by X with w = 1 in G for each w E W. This is thegroup Grp(X : W) generated by X subject to the relations w = 1 for w E W.

In section 29 we investigate Grp(X: W) when X = {x1, ... , x, } is finite andW consists of the words (xi xj)'"ii = 1, for suitable integral matrices (mid). Sucha group is called a Coxeter group. For example finite symmetric groups areCoxeter groups. We find that Coxeter groups admit a representation ir: GO(V, Q) where (V, Q) is an orthogonal space over the reals and Xir consists ofreflections. If G is finite (V, Q) turns out to be Euclidean space. Finite Coxetergroups are investigated via this representation in section 30, which developsthe elementary theory of root systems.

The theory of Coxeter groups will be used extensively in chapter 14 to studythe classical groups from a geometric point of view.

28 Free groupsAn object G in an algebraic category A is said to be free with free generatingset X if X is a subset of G and, whenever H is an object in A and a: X His a function from X into H, there, exists a unique morphism P: G H of Ginto H extending a. This section discusses free groups.

But first recall that a monoid is a set G together with an associative binaryoperation on G possessing an identity 1. Here's an example of a monoid. LetX be a set. A word in X is a finite sequence xlx2 ... X, with xi in X; n is thelength of the word. The empty sequence is allowed and denoted by 1. Let Mbe the set of words in X and define the product of two words x1 ... x" andyi ym to be the word xl ... x"yl ... Ym of length n + m. Observe that M isa monoid with identity the empty sequence 1. Indeed

(28.1) M is a free monoid with free generating set X.


A group F is free with free generating set X if it possesses the following universal property: each function a : X + H of X into a group H extends uniquely to a homomorphism of F into H. We find in section 28 that for each cardinal C there exists (up to isomorphism) a unique free group F with free generating set of cardinality C. Less precisely: F is the largest group generated by X.

If W is a set of words in the alphabet XU X-', it develops that there is also a largest group G generated by X with w = 1 in G for each w E W. This is the group Grp(X : W) generated by X subject to the relations w = 1 for w E W.

In section 29 we investigate Grp(X: W) when X = {XI, . . . , x,) is finite and W consists of the words = 1, for suitable integral matrices (mij). Such a group is called a Coxeter group. For example finite symmetric groups are Coxeter groups. We find that Coxeter groups admit a representation n : G + O(V, Q) where (V, Q) is an orthogonal space over the reals and Xn consists of reflections. If G is finite (V, Q) turns out to be Euclidean space. Finite Coxeter groups are investigated via this representation in section 30, which develops the elementary theory of root systems.

The theory of Coxeter groups will be used extensively in chapter 14 to study the classical groups from a geometric point of view.

28 Free groups An object G in an algebraic category A is said to befree with free generating set X if X is a subset of G and, whenever H is an object in A and a: X + H is a function from X into H, there,exists a unique morphism B: G + H of G into H extending a. This section discusses free groups.

But first recall that a monoid is a set G together with an associative binary operation on G possessing an identity 1. Here's an example of a monoid. Let X be a set. A word in X is a finite sequence ~ 1 x 2 . . . X, with xi in X; n is the length of the word. The empty sequence is allowed and denoted by 1. Let M be the set of words in X and define the product of two words xl . . . x, and yl . . . y, to be the word xl . . . x,yl. . . y, of length n + m. Observe that M is a monoid with identity the empty sequence 1. Indeed

(28.1) M is a free monoid with free generating set X.

Free groups 139

For if H is a monoid and a: X H is a function then a can be extendedto a morphism /3: M H defined by (xl ... x7f)p = xla ... x7za. /3 is welldefined as each word has a unique representation as a product of members ofX. Evidently /3 is the unique extension of a. Indeed in general in any algebraiccategory if X is a generating set for an object G and a: X H is a functionthen there is at most one extension fi of a to a morphism of G into H. This isbecause if f' is another extension then K = {g E G: gj3 = g/3'} is a subobjectof G containing X.

Next assume X = Y U Y-1 with Y fl Y-1 = 0 and y i-+ y-1 is a bijection ofY with Y-1. Set (y-1)-1 = y for each y E Y; thus x H x-1 is a permutationof X of order 2. Define two words u and w to be adjacent if there exist wordsa, b E M and x E X such that u = axx-lb and w = ab, or vice versa.Thus adjacency is a reflexive and symmetric relation. Define an equivalencerelation on M by u w if there exists a sequence u = u1, ... , u =w of words such that u, and u;+l are adjacent for each i, 1 < i < n. Thatis - is the transitive extension of the adjacency relation. Write w for theequivalence class of a word w under - and let F be the set of equivalenceclasses.

(28.2) If u, v, w E M with u v then uw - vw and wu - wv.

Proof. There is a sequence u = u1, ... , u, = v of words with u, adjacent toui+1. Observe uiw = wi is adjacent to w;+1 and uw = w1, ... , w = VW, SOUW - VW.

Now define a product on F by uv = ii-D. By 28.2 this product is well defined.Further the product of the equivalence classes of the elements xn 1, ... , xi 1 isan inverse for XI ... Xn, so F is a group. Hence

(28.3) F is a group and w i-+ w is a surjective monoid homomorphism of Monto F.

(28.4) F is a free group with free generating set Y.

Proof. Observe first that Y generates F. This follows from 28.3 together withthe fact that X generates M and X= Y U Y-1.

Now let H be a group and a: Y - H a function. Define /B: X H by y/3 =ya and y-l Q = (ya)-1 for y E Y. As M is a free monoid on X, /3 extends to amorphism y: M H. Define S: F H by 08 = wy. I must show S is welldefined; that is if u - v then uy = vy. It suffices to assume u is adjacent to v,say u = axx-1b and v = ab. Then uy = (axx-1b)y = ayx/B(xp)-l by = ayby,

For if H is a monoid and a: X + H is a function then a can be extended to a morphism fi: M + H defined by (xl . . . x,)fi = xla . . . x,a. fi is well defined as each word has a unique representation as a product of members of X. Evidently fi is the unique extension of a. Indeed in general in any algebraic category if X is a generating set for an object G and a : X + H is a function then there is at most one extension fi of a to a morphism of G into H . This is because if fi' is another extension then K = {g E G: gfi = gfi'} is a subobject of G containing X.

Next assume X = Y U Y-' with Y f l Y-' = 0 and y H y-l is a bijection of Y with Y-'. Set (y-l)-l = y for each y E Y; thus x H X-l is a permutation of X of order 2. Define two words u and w to be adjacent if there exist words a , b E M and x E X such that u = axx-'b and w = ab, or vice versa. Thus adjacency is a reflexive and symmetric relation. Define an equivalence relation -- on M by u -- w if there exists a sequence u = ul, . . . , u, = w of words such that ui and ui+l are adjacent for each i , 1 5 i < n. That is -- is the transitive extension of the adjacency relation. Write for the equivalence class of a word w under -- and let F be the set of equivalence classes.

(28.2) If u , v, w E M with u -- v then uw -- vw and wu -- wv.

Proof. There is a sequence u = u 1, . . . , u, += v of words with ui adjacent to ui+l. Observe U ~ W = wi is adjacent to wi+l and uw = wl, . . . , w, = vw, so UW " V W .

Now define a product on F by ii3 = iZ. By 28.2 this product is well defined. Further the product of the equivalence classes of the elements x;', . . . , xcl is an inverse for X1 . . . X,, so F is a group. Hence

(28.3) F is a group and w H W is a surjective monoid homomorphism of M onto F.

(28.4) F is a free group with free generating set P.

Proof. Observe first that Y generates F. This follows from 28.3 together with the fact that X generates M and X = y U P-'.

Now let H be a group and a : E + H a function. Define fi: X + H by yfi = ya and y-lfi = (ya)-' for y E Y. As M is a free monoid on X, fi extends to a morphismy:M + H.Define6:F + HbyW6 =wy.Imustshow6iswell defined; that is if u -- v then u y = v y . It suffices to assume u is adjacent to v, sayu=axx-'bandv=ab.Thenuy =(axx-'b)y =ayxfi(xfi)-'by =ayby,

140 Presentations of groups

as desired. Evidently b is a homomorphism extending a. As f generates F, anearlier remark shows b is the unique extension of a.

Lemma 24.8 shows that for each set S there exists a free group with freegenerating set S. The universal property implies:

(28.5) Up to isomorphism there exists a unique free group with free generatingset of cardinality C for each cardinal C.

If W C M is a set of words in X, write Grp(Y : W) for the group F/N, whereN = (WF) is the normal subgroup of F generated by the subset W of F.Grp(Y : W) is the group generated by Y subject to the relations w = 1 forw c W. That is Grp(Y : W) is the largest group generated by the set Y in whichw = 1 for each w c W. To be more precise Grp(Y : W) = F/N = G with Yand W identified with

(yN: y c Y) and (w- N: w c W),

respectively. As Y generates F, Y generates G. As w E N, w = 1 in G foreach w E W. So G is generated by Y and each of the words in W is trivial. I'llalso say the relation w = 1 is satisfied in G to indicate that w = 1 in G. G isthe largest group with these properties in the following sense:

(28.6) Let a: Y - Ya be a function of Y onto a set Ya, H a group generatedby Ya, and W a set of words w = y8, ... ynn in Y UY-l with wa = (y, a)" ...(yna)8n = 1 in H for each w E W. (That is H is generated by Ya and satisfiesthe relations w = 1 for w E W.) Then a extends uniquely to a surjectivehomomorphism of Grp(Y : W) onto H.

Proof. Let F be the free group on Y. Then there exists a unique homomorphism8: F - H of F onto H extending a. Let N = (WF) and b: v H vN thenatural map of F onto G = F/N. Then N = ker(b). For w = xl ... xn E Wwith xt E X, 1 = wa = xla ... xna = xl/' ... xn _ (XI, ... xnV = 4,so w E ker(b). Thus N < ker(8), as N is the smallest normal subgroup of Fcontaining W. But as N < ker(8), induces a homomorphism y: G - H with

= by. As is surjective so is y. Also, for y c Y, ya = y = yby = yy.

A presentation for a group G is a set Y of generators of G together with aset W of words in Y U Y-l such that the relation w = 1 is satisfied in Gfor each w E W and the homomorphism of Grp(Y : W) onto G described inlemma 28.6 is an isomorphism. I'll summarize this setup with the statementG = Grp(Y : W). Every group has at least one presentation; namely:


as desired. Evidently 6 is a homomorphism extending a . As Y generates F , an earlier remark shows 6 is the unique extension of a.

Lemma 24.8 shows that for each set S there exists a free group with free generating set S. The universal property implies:

(28.5) Up to isomorphism there exists a unique free group with free generating set of cardinality C for each cardinal C.

If W 2 M is a set of words in X, write Grp(Y : W) for the group F IN , where N = (wF) is the normal subgroup of F generated by the subset w of F. Grp(Y : W) is the group generated by Y subject to the relations w = 1 for w E W. That is Grp(Y : W) is the largest group generated by the set Y in which w = 1 for each w E W. To be more precise Grp(Y : W) = F I N = G with Y and W identified with

(jiN: y E Y) and (wN:w E W),

respectively. As Y generates F , Y generates G. As w E N, w = 1 in G for each w E W. So G is generated by Y and each of the words in W is trivial. I'll also say the relation w = 1 is satisJied in G to indicate that w = 1 in G. G is the largest group with these properties in the following sense:

(28.6) Let a : Y 4 Y a be a function of Y onto a set Ya, H a group generated byYa,andWasetofwordsw = y:' . . . y F i n Y ~ Y - l withwa = (yla)" ... (y,a)" = 1 in H for each w E W. (That is H is generated by Y a and satisfies the relations w = 1 for w E W.) Then a extends uniquely to a surjective homomorphism of Grp(Y : W) onto H.

Proof. Let F be the free group on Y. Then there exists a unique homomorphism j?: F 4 H of F onto H extending a . Let N = (WF) and 6: v ++ vN the natural map of F onto G = FIN. Then N = ker(6). For w = xl . . . x, E W with xi E X, 1 = wa = xla .. .x,a = xlj? . . .x,j? = (XI, . . .xn)j? = wj?, so w E ker(j?). Thus N 5 ker(j?), as N is the smallest normal subgroup of F containing W. But as N ( ker(j?), j? induces a homomorphism y : G 4 H with j? =6y.Asj?issurjectivesois y.Also,fory E Y, ya = yj? = yay = yy.

A presentation for a group G is a set Y of generators of G together with a set W of words in Y U Y-' such that the relation w = 1 is satisfied in G for each w E W and the homomorphism of Grp(Y : W) onto G described in lemma 28.6 is an isomorphism. I'll summarize this setup with the statement G = Grp(Y : W). Every group has at least one presentation; namely:

Coxeter groups 141

(28.7) For each group G,

G = Grp(G: xy(xy)-1 = 1, x, y E G)

is a presentation for G.

Proof. Let g H k be a bijection of G with a set G, let F be the free groupon G, let W be the set of words xy(xy)-1, x, y E G, and let N = (WF).Evidently G satisfies the relations defined by W so the map g H g extends to ahomomorphism a of F onto G with N < ker(a). It remains to show N = ker(a).Assume otherwise and let v = 1 be a word in ker(a) - N of minimallength n. As 1 E N, n > 0. If n = 1 then x1 = 1 and v = a = 1.1x1 1 E W,

contrary to the choice of v. Hence n > 2 so v = x1x2u for some word u of lengthn - 2. Now w E W C N, sow-1v = .1x2 u E ker(a) - N.As (xix2)u is of length at most n - 1, the choice of v of minimal length iscontradicted.

Here's a slightly more nontrivial example. The dihedral group of order 2n isdefined to be the semidirect product of a cyclic group X = (x) of order n bya group Y = (y) of order 2, with respect to the automorphism xy = x-1. Thecase where n = oc and x is the infinite cyclic group is also allowed. Denotethe dihedral group of order 2n by D2n. Dihedral 2-groups have already beendiscussed in the chapter on p-groups.

(28.8) D2n = Grp(x, y : xn = y2 = 1 = xyx).

If n = oo the relation xn = 1 is to be ignored. The proof of 28.8 is easy. LetD=XY=D2nand

By 28.6 there is a homomorphism a of G onto D with xa = x and ya = y.Then n = Ix I divides I I, so, as xn = 1, 1 I = n and a: X - X is anisomorphism, where X = (x). Similarly, setting Y = (y), a: Y - Y is anisomorphism. 1 = XXY so V = Thus X <G = (x, y), so G = XY.Hence, as a restricted to X and Y is an isomorphism, a itself is an isomorphism.

29 Coxeter groupsDefine a Coxeter matrix of size n to be an n by n symmetric matrix with 1 s onthe main diagonal and integers of size at least 2 off the main diagonal. To eachCoxeter matrix M = (mid) of rank n there is associated a Coxeter diagram:this diagram consists of n nodes, indexed by integers 1 < i < n, together withan edge of weight mid - 2 joining distinct nodes i and j, 1 < i < j < n. We

(28.7) For each group G,

G = G ~ ~ ( G : X ~ ( X ~ ) - ' = 1, X, y E G)

is a presentation for G.

Proof. Let g ++ g be a bijection of G with a set G, let F be the free group on G , let W be the set of words , ?YE)- ' ,x , y E G, and let N = (WF). Evidently G satisfies the relations defined by W so the map g ++ g extends to a homomorphisma of F onto G with N _( ker(a). It remains to show N = ker(a). Assume otherwise and let v = Zl . . . Zn be a word in ker(a) - N of minimal length n. As 1 E N, n > 0. If n = 1 then xl = 1 and v = jll = flx12;' E W, contrary to the choice of v. Hence n > 2 so v = ZIZzu for some word u of length n - 2. Now w = x l x 2 ( ~ ) - ' E W C N , so w-'v = m u E ker(a) - N . As ( m ) u is of length at most n - 1, the choice of v of minimal length is contradicted.

Here's a slightly more nontrivial example. The dihedral group of order 2n is defined to be the semidirect product of a cyclic group X = (x) of order n by a group Y = (y) of order 2, with respect to the automorphism xY = x-'. The case where n = co and x is the infinite cyclic group is also allowed. Denote the dihedral group of order 2n by Dzn. Dihedral 2-groups have already been discussed in the chapter on p-groups. 4

If n = co the relation xn = 1 is to be ignored. The proof of 28.8 is easy. Let D = X Y = Dzn and

By 28.6 there is a homomorphism a of G onto D with xa = x and Ya = y. Then n = 1x1 divides 1x1, so, as 3" = 1, 1x1 = n and a : ji 4 X is an isomorphism, where X = (2). Similarly, setting Y = (y), a: 4 Y is an isomorphism. 1 = 3 3 Y so ZY = (3)-'. Thus X I! G = (x, ), so G = XY. Hence, as a restricted to X and is an isomorphism, a itself is an isomorphism.

29 Coxeter groups Define a Coxeter matrix of size n to be an n by n symmetric matrix with 1 s on the main diagonal and integers of size at least 2 off the main diagonal. To each Coxeter matrix M = (mij) of rank n there is associated a Coxeter diagram: this diagram consists of n nodes, indexed by integers 1 5 i _( n, together with an edge of weight mij - 2 joining distinct nodes i and j, 1 _( i < j 5 n. We

will be most concerned with Coxeter matrices with the following diagrams:

A1 2o-o n-1 n

n

1 2 n-2 n-1 nCn o-o o-®

n-11 2 n-2

Dn o-o 0n

Thus a diagram of type An defines a Coxeter matrix of size n with mid = 3 ifJi - j I = 1 and mid = 2 if Ii - j I > 1. Similarly a diagram of type Cn definesa matrix with mid = 3 if Ji - j I = 1 and i, j < n, mn-l,n = mn,n-, = 4, andmid=2ifIi-j!>1.

A Coxeter system with Coxeter matrix M = (mid) of size n is a pair (G, S)where G is a group, S = (si: 1 < i < n) a family of elements of G, and

G = Grp(S: (sisj)m,i = 1, 1 < i < n, 1 < j < n).

G is a Coxeter group if there exists a family S such that (G, S) is a Coxetersystem.

In the remainder of this section let (G, S) be a Coxeter system with matrixM = (m1, .) of size n. Let S = (si: 1 < i < n). Notice

s2 = (SiSi)mii = 1.

(29.1) Let T be the set of conjugates of members of S under G and for eachword r = r1 ... rn in the alphabet S and each t E T define

N(r, t) = (i: t = ri < i < m}.

Then, if r1 ... rn = rl . rk in G, we have (N(r, t)j - IN(r', t)j mod 2 foreach t E T.

Proof. Let A be the set product (±1} x T and for s E S define sn c Sym(A)by (E, t)sn = (EE(s, t), t''), where 8(s, t) _ -1 ifs = t and 8(s, t) = +1 ifs # t. Observe that sn is an involution. I'll show (sinsjn)mii = 1 for all i, j;hence, by 28.6, n extends to a permutation representation n of G on A. Inparticular n is a homomorphism so, if g = r1 ... rm c G and r1 E S, then

(E, t)gn = (e, t)rln ... rnn = E r i ,tri...r;_1) t8)

i=1


will be most concerned with Coxeter matrices with the following diagrams:

Thus a diagram of type A, defines a Coxeter matrix of size n with mij = 3 if li - j I = 1 and mij = 2 if li - j 1 > 1. Similarly a diagram of type C, defines amatrixwithmij = 3 i f l i - jl = l and i , j <n,m,-I,, = m ,,,- 1 =4,and m i j = 2 i f l i - j ] > 1.

A Coxeter system with Coxeter matrix M = (mij) of size n is a pair (G, S ) where G is a group, S = (si: 1 _( i ( n ) a family of elements of G, and

G = G r p ( S : ( ~ ~ s ~ ) ~ ' j = 1, 1 _ ( i I n , 1 5 j ( n ) .

G is a Coxeter group if there exists a family S such that (G, S ) is a Coxeter system.

In the remainder of this section let (G, S) be a Coxeter system with matrix M = (mij) of size n. Let S = (si: 1 5 i 5 n). Notice

(29.1) Let T be the set of conjugates of members of S under G and for each word r = rl . . . rm in the alphabet S and each t E T define

Then, if r l . ..rm = ri . . . r i in G , we have IN(r,t)l = IN(rl, t)l mod 2 for eacht E T.

Proof. Let A be the set product (f 1) x T and for s E S define s n E Sym(A) by ( E , t ) sn = ( ~ 6 ( s , t ) , tS ) , where 6(s, t ) = -1 if s = t and 6(s, t ) = +1 if s # t . Observe that s n is an involution. I'll show (sinsjn)"lj = 1 for all i, j ; hence, by 28.6, n extends to a permutation representation n of G on A. In particular n is a homomorphism so, if g = rl . . . rm E G and ri E S, then

Coxeter groups 143

Further 3(ri, t''...ri-') = -1 exactly when i E N(r, t). So (e, t)g7r =(E(-l)IN(r,t)I, t8) and hence IN(r, t) I mod 2 depends only on g and not on r.

It remains to show (si7rsj7r)m"j = 1; equivalently a = (si rsj7r)m'j fixes each(e, t) E A. Now (e, t)a = (Es,

t(s'si)»")= (EE, t), as (sisj)m'j = 1, where of

course

8 = 8(si, t)E(sj, ts')....

So we must show 8 = 1. But 8(si, t(S,sj )k) _ -1 for some 0 < k < mij, precise-ly when t = (si s j )2ksi . Also 8(sj, t(S'Sj )ks') -1 precisely when t = (si sj )2k+1 si .

Further, by Exercise 10.1, (si, sj) is either dihedral of order 2m, where m dividesmij, or of order at most 2. Hence if t 0 ((sisj))si then all terms in 8 are +1,while if t E ((sisj))si exactly (2mij)/IsisjI terms are -1, as t = (sisj)dIs"'jIsifor 0 < d < (2mij )/ Isis j 1.

(29.2) The members of S are involutions.

Proof. In the proof of 29.1 a homomorphism 7r of G into Sym(A) was con-structed for which s7r was an involution for each s E S. So 2 = Isn I dividesis (. But of course, as (G, S) is a Coxeter system, s? = 1.

If H is a group with generating set R then the length of h E H with respect toR is the minimal length of a word w in the alphabet R U R-1 such that w = hin H. Denote this length by 1(h) = lR(h).

(29.3) Let g c G and r = r1 ... rm a word in the alphabet S with g = r in G.Define

r7(r) = I It E T: IN(r, t) I = 1 mod 2}1

in the notation of 29.1. Then 1](r) = l(g).

Proof. By 29.1, if r' = g then 1](r) =.1](r'), while by definition of l(g) thereis r' = ri ... rk with k = l(g) and r' = g in G. So, without loss, m = l(g).So evidently 1](r) < m = l(g). If 1](r) < m there are i, j, 1 < i < j <

r,_I ...riri = rj , so riri+t ... rj-1 = Yi+1 ... rj gr1 ... r1-1ri ... rj_1rj ... rm = r1 ... ri_1ri+1 ... rj_1rj+1 ... rm is of length atmost m - 2, contrary to the choice of r.

(29.4) Let H be a group generated by a set R of involutions. Then (H, R) is aCoxeter system precisely when the following Exchange Condition is satisfied:

Exchange Condition: If ri E R, 0 < i < n, and h = r1 ... r E H with1(h) = n and l(roh) < n, then there exists 1 < k < n such that ror1 .. rk-1 =r1 ... rk in H.

Further 6(ri, tr'...'i-1) = - 1 exactly when i E N(r, t ) . So ( E , t )gn =

(~( - l ) I~( ' , ' ) l , tg) and hence IN(r, t)l mod 2 depends only on g and not on r. It remains to show (s, nsjn)"'l = 1; equivalently a = (s, ns, n)"ll fixes each

( E , t ) E A. Now ( E , t )a = ( ~ 6 , t ( S~S~)"" l ) = ( ~ 6 , t ) , as ( s , ~ ~ ) " ~ ] = 1, where of course

So we must show 6 = 1. But 6(si, t('1'1)~) = - 1 for some 0 5 k < m,j, precisely when t = (sisj)2ksi. Also 6(sj , ~ ( S ~ S J ) ~ ' ~ ) = - 1 precisely when t = (s is j )2k+1~i. Further, by Exercise 10.1, (s, , s j ) is either dihedral of order 2m, where m divides mij, or of order at most 2. Hence if t $2 ((sisj))si then all terms in 6 are + 1, while if t E ((sisj))si exactly (2mij)/ lsisj I terms are - 1 , as t = ( S ~ S ~ ) ~ ~ ' ~ ' J lsi for 0 5 d < (2mIJ)/lsisjl.

(29.2) The members of S are involutions.

Proof. In the proof of 29.1 a homomorphism n of G into Sym(A) was con- structed for which s n was an involution for each s E S. So 2 = Isn I divides Is!. But of course, as (G, S ) is a Coxeter system, s? = 1.

If H is a group with generating set R then the length of h E H with respect to R is the minimal length of a word w in the abhabet R U R-' such that w = h in H . Denote this length by l (h) = l ~ ( h ) .

(29.3) Let g E G and r = rl . . . rm a word in the alphabet S with g = r in G. Define

~ ( r ) = I{t E T : IN(r, t)l = 1 mod 211

in the notation of 29.1. Then ~ ( r ) = l(g).

Proof. B y 29.1, if r' = g then ~ ( r ) = .q(rl), while by definition of l (g) there is r' = ri . . . ri with k = l (g) and r' = g in G. So, without loss, m = l(g). So evidently ~ ( r ) 5 m = l(g). If ~ ( r ) < m there are i , j, 1 5 i 5 j 5 m, with r"-~...r~

rj-1 ... r~ = r . , so riri+l.. .rj-1 = ri+l.. .rj and hence g =

I J

rl . . . ri-lri . . . rj-lrj . . . rm = rl . . . ri-lri+l. . . rj-lrj+l. . . rm is of length at most m - 2, contrary to the choice of r.

(29.4) Let H be a group generated by a set R of involutions. Then ( H , R ) is a Coxeter system precisely when the following Exchange Condition is satisfied:

Exchange Condition: If ri E R, 0 5 i 5 n , and h = rl . . .r, E H with l (h) = n and l(roh) 5 n, then there exists 1 5 k 5 n such that rorl . . . rk-1 =

r l . . .rk in H.

Proof. Suppose first that (H, S) is a Coxeter system and let rt, h, satisfy thehypothesis of the Exchange Condition. Then, setting r = rori ... rn, ri(r) <l(roh) < n + 1 by 29.3, so there are i, j, 0 < i < j < n with r;

'-'.__Y° =and hence r, ... rj_1 = rt+i ... rj. By 29.3, ri(ri ... r,) = 1(h) = n, so i = 0.Thus the Exchange Condition is satisfied.

Conversely suppose (H, R) satisfies the Exchange Condition. Let a: R -+ Xbe a function into a group X such that, for each r, s E R, (rasa)I's' = 1. It willsuffice to show a extends to a homomorphism of H into X.

Let h E H, n = l(h), and r1 ... rn = h = s1 ... sn with ri, si c R. Claim

r1a ... rna = sla ... sna and {rt: 1 < i < n) = {s,:1 < i < n).

Assume not and pick a counterexample with n minimal. Then l (si h) = n -1 <l(h), so, by the Exchange Condition, s1r1 ... rk-1 = r1 ... rk for some k. Hences1r1 ... rk-lrk+l ... rn = h = s1 ... sn so r1 ... rk_1rk+1 ... rn = S2 ... Sn.Thus, by minimality of n, r1a ... rk_1ark+la ... rna = sea ... sna and{S2, ... , Sn) = {ri: i 0 k). Also if k < n then, by minimality of n, siaria ...rk_la = r1a ... rka and {Si, r1, ..., rk_11 = jr, ..., rk), which combined withthe last set of equalities establishes the claim. So k = n, sir, ... rn-i = h, and{ri , ... , rn_i) = {s2, ..., sn ) . Similarly rise ... sn_1 = h and {s1, ... , Sn_1) ={r2 ... , rn ). In particular jr, .... rn) _ {s1 .... sn ), establishing half the claim.Replacing r1 .... rn, and si, ..., sn, by si, r1, ... , rn-i and r1, si, ...,sn_i, and continuing in this manner, we obtain (sire)n/2=(rise)n/2 or

ri(Siri)(n-1)/2 = Si(risi)(n-1)/2, with equality of images under a failing inthe respective case. It follows that (sire)n = 1, so the order m of s1r1 in Hdivides n. But by hypothesis the order of slarla divides m, so equality ofimages under a does hold, a contradiction.

So the claim is established. Since Coxeter systems satisfy the ExchangeCondition we can record:

(29.5) Let g c G with 1(g) = m and r1, tt c S with r1 ... rn = t1 ... to = g.Then {rt:1 . X by ha = r1a ... rna,for h = r1 ... rn with n = 1(h) and r, E R. The claim shows a to be welldefined. Let's see next that (rh)a = raha for r c R. If l(rh) = 1(h) + 1this is clear, so assume not. Then, by the Exchange Condition, rr1 ... rk-1 =r1 ... rk for some k < n. By the claim, raria ... rk-1a = r1a ... rka. Alsorh = r1 . . . rk-irk+1 . . rn is of length at most n - 1. As 1(h) = n, we concludel(rh) = n - 1. So (rh)a = r1a ... (rk-i)a(rk+1)a ... rna = raria ... rna =raha, establishing the second claim.

It remains to show gaha = (gh)a for g, h c H. Assume not and choose acounter example with 1(g) minimal. By the last paragraph, 1(g) > 1, so g = rk,


Proof. Suppose first that ( H , S ) is a Coxeter system and let ri, h , satisfy the hypothesis of the Exchange Condition. Then, setting r = rorl . . . r,, ~ ( r ) 5 l(roh) < n + 1 by 29.3, so there are i, j, 0 5 i < j 5 n with r,?-"..ro = rrj-'.'-ro J and hence ri . . . rj-1 = ri+l . . . rj. By 29.3, v(rl . . . r,) = l (h) = n, so i = 0. Thus the Exchange Condition is satisfied.

Conversely suppose ( H , R ) satisfies the Exchange Condition. Let a!: R + X be a function into a group X such that, for each r, s E R, (rasa)lrsl = 1 . It will suffice to show a! extends to a homomorphism of H into X.

Let h E H, n = l(h), and rl . . . r, = h = sl . . . s, with ri, si E R. Claim

r l a . . .r,a = sla!. . .s,a and (ri: 1 5 i 5 n ] = (si: 1 5 i 5 n] .

Assume not and pick a counterexample with n minimal. Then l(sl h ) = n - 1 < l(h), so, by the Exchange Condition, slrl . . . rk-1 = rl . . . rk for some k. Hence slrl . . .rk-lrk+l.. .r, = h = sl . . .s, so r l . . .rk-lrk+l .. .rn = s2.. .s,. Thus, by minimality of n , rla . . . rk-lark+la.. .rna = s p . . . s,a and ($2, . . . , s,] = {ri: i # k ] . Also if k < n then, by minimality of n, slurla!. . . rk-la = r1a . . . rka and Isl, rl , . . . , rk-1) = (rl . . . , rk], which combined with the last set of equalities establishes the claim. So k = n, slrl . . . r,-1 = h, and { r l , . . . , rnv l ) = (s2, . . . , s,). Similarly rlsl . . . s,-1 = hand {sl , . . . , s,-1) = {r2 . . . , r,). In particular {rl . . . , r,) = (sl . . . , s,), establishing half the claim. Replacing rl . . . , r,, and sl, . . . , s,, by sl, r l , . . . , r,-1 and rl , sl , . . . , s,-1, and continuing in this manner, we obtain (slr1)"I2 =(rls1)"I2 or r l ( ~ ~ r ~ ) ( " - ' ) / ~ = ~ l ( r l s ~ ) ( ~ - ~ ) / ~ , with equality of images under a failing in the respective case. It follows that (slrl)" = 1, so the order m of slrl in H divides n. But by hypothesis the order of slarla divides m, so equality of images under a does hold, a contradiction.

So the claim is established. Since Coxeter systems satisfy the Exchange Condition we can record:

(29.5) Let g E G with L(g) = m and ri, ti E S with rl . . . rm = tl . . . tm = g. Then (ri: 1 5 i 5 m ) = {ti: 1 5 i 5 m) .

Now back to the proof of 29.4. Define a : H -+ X by ha = rla! . . . ma!,

for h = rl . . . r, with n = l (h ) and ri E R. The claim shows a to be well defined. Let's see next that (rh)a = ruha for r E R. If l (rh) = l (h) + 1 this is clear, so assume not. Then, by the Exchange Condition, rrl . . . rkPl = rl . . . rk for some k 5 n. By the claim, rurla . . . rk-la = rla . . . rka. Also rh = rl . . . rk-lrk+l . . . r,, is of length at most n - 1. As l (h) = n, we conclude l (rh) = n - 1. So (rh)a! = rla!. . . ( ~ ~ - ~ ) a ( r ~ + ~ ) a . . ..,,a = rurla!. . .rna = raha!, establishing the second claim.

It remains to show gaha = (gh)a for g, h E H . Assume not and choose a counter example with 1 ( g ) minimal. By the last paragraph, 1 ( g ) > 1, so g = rk ,

Coxeter groups 145

r E R, k E H with l(k) = l(g) - 1. Then (gh)a = ra(kh)a = rakaha =gaha by minimality of l(g), completing the proof.

Let V be an n-dimensional vector space over the reals R with basis X =(xi:1 < i < n) and let Q be the quadratic form on V with (xi,xj) =-cos(7r/mid), where ( , ) is the bilinear form determined by Q as in sec-tion 19. Thus, for u, v E V, Q(v) _ (v, v)/2 and

(u, v) = Q(u) + Q(v) - Q(u + v).

(29.6) (1) Q(xi) = 1/2.(2) (xi, xj) < 0 for i 0 j, with (xi, xj) = 0 if and only if mii = 2.

Proof. mii = 1, so Q(xi) = (xi, xi)/2 = - cos(h)/2 = 1/2. Similarly ifi # j then mil > 2, so (xi, xj) cos(7r/mid) < 0 with equality if and onlyif mil = 2.

By 29.6 and 22.6.2 there is a unique reflection ri on V with center (xi) ; moreovervri = v - 2(v, xi)xi for v c V.

(29.7) For i 0 j, rirj is of order mil, (ri, rf) = D2,,,,i, and if mil > 2 then(ri, rj) is irreducible on (xi, xj). '

Proof. Let U = (xi, xj), D = (ri, rj), m = mid, and 9 = 7r/m. Observe that,for a, b c III,

2Q(axl + bx2) = a2 - 2ab cos 9 + b2 = (a - b cos 0)2 + b2(sin 9)2 > 0,

with equality precisely when a = b = 0. Thus Q is a positive definite quadraticform on U, so in particular U is a nondegenerate subspace of V and henceV = U ® U1. But U1 < xk < Cv(rk) for k = i and j, so U1 < Cv(D).Hence D is faithful on U.

As Q is positive definite on U, (xk, xk) = 1,and(xi, xj) _ -cos9, (U, Q) isisometric to 2-dimensional Euclidean space R2 with the standard inner productand with xi = (1, 0) and xj = (cos(7r -0), sin(7r -0)) in the standard coordinatesystem. Thus ri and rj are the reflections on R2 through the vertical axis andthe axis determined by 7r/2-0, respectively. Hence ri rj is the rotation throughthe angle -27r/m, and therefore is of order m as desired.

Thus the first claim of 29.7 is established and the second is a consequenceof the first and Exercise 10.1. (Xk) and xk fl U are the only nontrivial propersubspaces of U fixed by rk and if m > 2 then, by 29.6.2, (xi) 0 x fl U, so Dis irreducible on U.

145

r E R, k E H with l(k) = l(g) - 1. Then (gh)a = ra(kh)a = rakaha =

guha by minimality of l(g), completing the proof.

Let V be an n-dimensional vector space over the reals R with basis X =

(x,: 1 5 i 5 n) and let Q be the quadratic form on V with (x,, x,) = - cos(n/m,,), where ( , ) is the bilinear form determined by Q as in section 19. Thus, for u, v E V, Q(v) = (v, v)/2 and

(29.6) (1) Q(x~) = 112. (2) (xi, xj) 5 0 for i # j, with (xi, xj) = 0 if and only if mij = 2.

Proof. mii = 1, so Q(xi) = (xi, xi)/2 = - cos(n)/2 = 112. Similarly if i # j then mij > 2, so (xi, xj) = - cos(n/mij) 5 0 with equality if and only if mij = 2.

By 29.6 and 22.6.2 thereis auniquereflection ri on V withcenter (xi); moreover vri = v - 2(v, xi)xi for v E V.

(29.7) For i # j, rirj is of order mij , (ri , r j ) E D2m,j, and if mij > 2 then (ri , r j ) is irreducible on (xi, xi). I

Proof. Let U = (xi, xj), D = (ri, r j ) , m = mij, and 8 = n lm. Observe that, for a , b E R,

2Q(axl + bx2) = a2 - 2ab cos 8 + b2 = (a - b cos 812 + b2(sin8)2 2 0,

with equality precisely when a = b = 0. Thus Q is a positive definite quadratic form on U , so in particular U is a nondegenerate subspace of V and hence V = U @ u'. But U' 5 xk 5 CV(rk) for k = i and j, so U' 5 Cv(D). Hence D is faithful on U .

As Q is positivedefiniteon U , (xk, xk) = 1, and(xi, xj) = - cos 8, ( U , Q) is isometric to 2-dimensional Euclidean space R2 with the standard inner product and withxi = (1,O) andx, = (cos(n -@), sin(n -8)) in the standardcoordinate system. Thus r, and r, are the reflections on R2 through the vertical axis and the axis determined by n /2 - 8, respectively. Hence rir, is the rotation through the angle -2n/m, and therefore is of order m as desired.

Thus the first claim of 29.7 is established and the second is a consequence of the first and Exercise 10.1. (xk) and xk f l U are the only nontrivial proper subspaces of U fixed by rk and if m > 2 then, by 29.6.2, (x,) # xf f? U , so D is irreducible on U .

(29.8) Let W = (ri: 1 < i < n) be the subgroup of O(V, Q) generated bythe reflections (ri: 1 < i < n). Then there exists a surjective homomorphisma: G -+ W with sia = ri for each i. In particular a is an RG-representationwhich identifies S with a set of reflections in O(V, Q).

Proof. This is immediate from 29.7 and 28.6.

(29.9) (1) S is of order n.(2) For each i j, Isis j = mi j and (si, s j) = D2m;j

Proof. The map a: (si, s j) -+ (ri, r j) induced by the map of 29.8 is a sur-jective homomorphism. By Exercise 10.1 and 29.7, (ri, r j) =Grp(ri, r: r? =r = (ri r j )mii) so, as (si, s j) satisfies these relations, 28.6 says there is a ho-momorphism Y of (ri, rj) onto (si, sj) with rkp = Sk. Then /3 = a-1 so a isan isomorphism and 29.7 implies (2). As r, rj for i j, (1) holds.

Let 0 = {1, ... , n} be the set of nodes of the Coxeter diagram of (G, S). Thegraph of the diagram is the graph on 0 obtained by joining i to j if the edgebetween i and j in the Coxeter diagram is of weight at least 1, or equivalentlyifmij>3.

(29.10) Let (Ok: 1 < k < r) be the connected components of the graph of theCoxeter diagram 0 of (G, S) and let Gk = (Si: i E Ok). Then

(1) G = G1 x x G, is the direct product of the subgroups Gk, 1 < k < r.(2) V is the orthogonal direct sum of the subspaces Vk = [Gka, V], 1 <-

k < r.

Proof. If i and j are in distinct components of 0 then Isis j I = mij = 2, so[si, sj] = 1. Thus G is the central product of the subgroups Gk, 1 < k < r,and hence there is a surjective homomorphism P of G1 x x Gr = D ontoG with sip = si for each i. Conversely S satisfies the Coxeter relations inD, so by 28.6 there is a homomorphism y of G onto D with si y = si. Theny = P-1, so /3 is an isomorphism and (1) holds. Similarly, by 29.6.2, xj E x; ,so, if j E Da and i E Ab, then

Va=([V,rk]:kEA.)=(xk:kEDa)<Vb,and hence (2) holds.

Because of 29.10 it does little harm to assume the graph of the Coxeter diagramof (G, S) is connected. In that event (G, S) is said to be an irreducible Coxetersystem.


(29.8) Let W = (ri: 1 ( i ( n ) be the subgroup of O(V, Q ) generated by the reflections (ri: 1 5 i ( n). Then there exists a surjective homomorphism a : G + W with sia = ri for each i . In particular a is an RG-representation which identifies S with a set of reflections in O(V, Q).

Proof. This is immediate from 29.7 and 28.6.

(29.9) (1) S is of order n . (2) For each i # j, Jsisj I = mij and (s i , s j ) DZmij.

Proof. The map a : (s i , s j ) + (r i , r j ) induced by the map of 29.8 is a surjective homomorphism. By Exercise 10.1 and 29.7, (ri , r j ) =Grp(ri , rj: r? = r: = ( r i r j )m~l ) SO, as (si , s j ) satisfies these relations, 28.6 says there is a homomorphism p of (r i , r j ) onto (si , s j ) with rkp = sk. Then p = a-' so a is an isomorphism and 29.7 implies (2). As ri # rj for i # j, (1) holds.

Let A = { I , . . . , n ) be the set of nodes of the Coxeter diagram of (G, S). The graph of the diagram is the graph on A obtained by joining i to j if the edge between i and j in the Coxeter diagram is of weight at least 1, or equivalently if mij 2 3.

(29.10) Let ( A k : 1 5 k 5 r ) be the connected components of the graph of the Coxeter diagram A of (G, S ) and let G k = (si: i E A k ) . Then

(1) G = G1 x . . x G, is the direct product of the subgroups G k , 1 5 k 5 r . (2) V is the orthogonal direct sum of the subspaces Vk = [Gka , V ] , 1 (

k 5 r .

Proof. If i and j are in distinct components of A then / s i s j / = mij = 2, so [si, s j ] = 1. Thus G is the central product of the subgroups G k , 1 ( k 5 r , and hence there is a surjective homomorphism p of GI x . . - x G, = D onto G with sip = si for each i . Conversely S satisfies the Coxeter relations in D, so by 28.6 there is a homomorphism y of G onto D with si y = si. Then y = p-', so ,9 is an isomorphism and (1) holds. Similarly, by 29.6.2, xj E x l , so, if j E A, and i E Ab, then

v, = ( [ V , rk] : k E A a ) = ( ~ k : k E A,) ( v:, and hence (2) holds.

Because of 29.10 it does little harm to assume the graph of the Coxeter diagram of (G, S ) is connected. In that event (G, S ) is said to be an irreducible Coxeter system.

Coxeter groups 147

(29.11) Assume (G, S) is an irreducible Coxeter system. Then(1) G acts absolutely irreducibly on V/ V -L.(2) If W is finite then (V, Q) is nondegenerate.

Proof. Let U be a proper RG-submodule of V. For Si E S, [V, si] = (xi) isof dimension 1 so either xi E U or U < Cv (si) = xi . Thus either there existsi E 0 with xi E U or U < niEAXi = V1, and I assume the former. Claimxj E U for each j E 0, so that V = (xj: j E 0) < U, contradicting U proper.As the graph of 0 is connected it suffices to prove xj E U for mid > 2. But,for such j, (si, sj) is irreducible on (xi, xj) by 29.7, so as xi E U and U isG-invariant, xj E U.

I've shown G is irreducible on V = V/V 1. F = EndRG (V) acts on [ V , si ](xi), so, for a E F, axi = bxi for some b E R', and hence, as G is irreducibleon V and centralizes a, a acts as a scalar transformation via b on V. That isF = R. So, by 25.8, G is absolutely irreducible on V.

Suppose W is finite. Then, by Maschke's Theorem, V = V -L Z for someEf G-submodule Z of V. V is Ef G-isomorphic to Z and [V, si] 1, so [Z, si]

1. Hence (xi) _ [Z, si] < Z, so V = (xi: 1 < i < n) < Z. Thus V1 = 0and (2) holds.

(29.12) Assume (G, S) is an irreducible Coxeter system and W is finite. Then(V, Q) is isometric to n-dimensional Euclidean space under the usual innerproduct.

Proof. Let h be the bilinear form on V which makes X into an orthonormalbasis and define g: V x V -+ R by g(u, v) = F_wEW h(uw, vw). It is straight-forward to check that g is a symmetric bilinear form on V preserved by G. Asthe quadratic form of h is positive definite, so is the form P of g, so (V, g)is nondegenerate. But, by 29.11, V is an absolutely irreducible RG-module,so, by Exercise 9.1, P = aQ for some a E fl8#. As P is positive definiteand Q(xi) = 1/2 > 0, a > 0, so Q is positive definite. By 19.9 there isa basis Y = (yi: 1 0 so, adjusting by a suitable scalar, we can takeQ(yi) = 1, since every positive member of E is a square in R. Thus Y is an or-thonormal basis for (V, Q), so (V, Q) is Euclidean space under the usual innerproduct.

Let0={1,...,n}andforJCOletSj=(SJ:j EJ)andGj=(Sj).Thesubgroups GJ and their conjugates under G are the parabolic subgroups of theCoxeter system (G, S).

Coxeter groups

(29.11) Assume (G, S ) is an irreducible Coxeter system. Then (1) G acts absolutely irreducibly on V / v'. (2) If W is finite then ( V , Q ) is nondegenerate.

Proof. Let U be a proper RG-submodule of V . For si E S , [ V , si] = ( x i ) is of dimension 1 so either xi E U or U ( Cv(si) = x l . Thus either there exists i E A with xi E U or U 5 niEAx; = v', and I assume the former. Claim xj E U for each j E A , so that V = ( x j : j E A ) _( U , contradicting U proper. As the graph of A is connected it suffices to prove xj E U for mij > 2. But, for such j, (s i , s j ) is irreducible on (x i , x j ) by 29.7, so as xi E U and U is G-invariant, xj E U .

I've shown G is irreducible on v = v/v'. F = EndRc@) acts on [ v , si] = ( x i ) , so, for a E F, axi = bxi for some b E R#, and hence, as G is irreducible on v and centralizes a , a acts as a scalar transformation via b on v. That is F = R. So, by 25.8, G is absolutely irreducible on v.

Suppose W is finite. Then, by Maschke's Theorem, V = V' CB Z for some RG-submodule Z of V . v is RG-isomorphic to Z and [ v , si] # 1, so [Z, si] # 1. Hence (x i ) = [Z, si] 5 Z, so V = (x i : 1 ( i ( n) 5 Z. Thus V' = 0 and (2) holds.

(29.12) Assume (G, S ) is an irreducible Coxeter system and W is finite. Then ( V , Q ) is isometric to n-dimensional Euclidean space under the usual inner product.

Proof. Let h be the bilinear form on V which makes X into an orthonormal basis and define g: V x V + R by g(u, v ) = CWEw ~ ( u w , V W ) . It is straightforward to check that g is a symmetric bilinear form on V preserved by G. As the quadratic form of h is positive definite, so is the form P of g, so ( V , g) is nondegenerate. But, by 29.11, V is an absolutely irreducible RG-module, so, by Exercise 9.1, P = aQ for some a E R#. As P is positive definite and Q(xi) = 112 > 0, a > 0, so Q is positive definite. By 19.9 there is a basis Y = (y i : 1 ( i 5 n) for V with (y i , y j ) = 0 for i # j. As Q is positive definite, Q(yi) > 0 so, adjusting by a suitable scalar, we can take Q(yi) = 1, since every positive member of R is a square in R. Thus Y is an orthonormal basis for ( V , Q) , so ( V , Q ) is Euclidean space under the usual inner product.

Let A = { I , . . . , n} and for J c A let S j = ( S J : j E J } and Gj = ( S j ) . The subgroups G j and their conjugates under G are the parabolic subgroups of the Coxeter system (G, S).

(29.13) Let J, K C A and g E Gj. Then(1) If l (g) = m and g = si, ... Si,,, with Si, E S then ik E J for each 1 < k < in.(2) (Gj, Sj) is a Coxeter system with Coxeter matrix Mj = (Mid), i, j E J.(3) (Gd, GK) = GJUK and Gi fl GK = GJnK(4) If Gj = GK then J = K.

Proof. Let g = sat ... sak with ak E J and k minimal subject to this con-straint. Claim k = m. By induction on k, l(sa,g) = k - 1, so by the ExchangeCondition either k = l(g) or sa, ... sa,_, = saZ ... sar for some t, in whichcase g = sae ... sa,_,sa,+,sak, contrary to minimality of k. Hence (1) followsfrom 29.5.

Next (1) says the Exchange Condition is satisfied by (Gj, Sj), so (Gj, Sj)is a Coxeter system by 29.4. 29.9 says Mj is the Coxeter matrix of (Gj, Sj).

The first remark in (3) and the inclusion GJnK < Gj fl GK are trivial. Part(1) gives the inclusion Gj fl GK < GJnK.

If Gj = GK then Gj = GJnK by (3), so we may assume K C J. By (2)we may assume J = A. Hence G = Gj = GK, so, by (1), S = SK. HenceK = A = J by 29.9.1.

30 Root systemsIn this section V is a finite dimensional Euclidean space over a field F equal tothe reals or the rationals. That is V is an n-dimensional space over F togetherwith a quadratic form Q such that (V, Q) possesses an orthonormal basis.Hence Q is positive definite. Let (, ) be the bilinear form defined by Q. Forv E V# there exists a unique reflection with center (v) by 22.6.2; denote thisreflection by r,,.

A root system is a finite subset E of V# invariant under W (E) = (r : v E E)and such that I (v) fl E I < 2 for each v E E. Observe that if v E E then -v =vr E E, so, as ((v) fl EI < 2, (v) fl E = {v, -v). We call W(E) the Weylgroup of E.

Here's one way to obtain root systems:

(30.1) Let G be a finite subgroup of O(V, Q) generated by a G-invariant setR of reflections. Let E consist of those v E V# with Q(v) = 1/2 and (v) thecenter of some member of R. Then E is a root system and G = W(E).

Proof. For v E E, (v) is the center of some r E R, so r = r,,. Thus G = (R) =W(E), so it remains to show E is a root system. If u = av E E then, by defi-nition of E, 1/2 = Q(u) = a2Q(v) = a2/2, so a = ±1. Thus I E fl (v) I < 2.


(29.13) Let J, K 2 A and g E Gj. Then (1) Ifl(g)=mandg=si, ... s i , , , w i t h s i , ~ S t h e n i k ~ J f o r e a c h l ~ k ~ m . (2) (Gj, SJ) is a Coxeter system with Coxeter matrix M j = (mij), i , j E J.

(3) (GJ, GK) = GJUK and GJ n '% = G J ~ K . (4) I fGJ=&then J = K.

Proof. Let g = s,, . . . s,, with a k E J and k minimal subject to this constraint. Claim k = m. By induction on k, l(s,,g) = k - 1, so by the Exchange Condition either k = l(g) or s,, . . . s ,,-, = s,, . . . s,, for some t, in which case g = s, . . . s ,,-, s,,+,s,,, contrary to minimality of k. Hence (1) follows from 29.5.

Next (1) says the Exchange Condition is satisfied by (Gj, SJ), so (Gj, SJ) is a Coxeter system by 29.4. 29.9 says Mj is the Coxeter matrix of (Gj, SJ).

The first remark in (3) and the inclusion GJnK 5 G j f l GK are trivial. Part (1) gives the inclusion GJ n GK I GJnK.

If G j = GK then G j = G j n ~ by (3), so we may assume K 2 J. By (2) we may assume J = A. Hence G = Gj = GK, SO, by (I), S = SK. Hence K = A = J by 29.9.1.

30 Root systems In this section V is a finite dimensional Euclidean space over a field F equal to the reals or the rationals. That is V is an n-dimensional space over F together with a quadratic form Q such that (V, Q) possesses an orthonormal basis. Hence Q is positive definite. Let ( , ) be the bilinear form defined by Q. For v E V# there exists a unique reflection with center (v) by 22.6.2; denote this reflection by r,.

A root system is a finite subset C of V# invariant under W(C) = (r, : v E C) andsuchthatI(v) n C I 5 2foreachv E C.Observethatifv E C then-v= vr, E C, so, as I(v) n Cl I 2, (v) n C = {v, -v). We call W(C) the Weyl group of C.

Here's one way to obtain root systems:

(30.1) Let G be a finite subgroup of O(V, Q) generated by a G-invariant set R of reflections. Let C consist of those v E V# with Q(v) = 112 and (v) the center of some member of R. Then C is a root system and G = W(C).

Proof. For v E C, (v) is the center of some r E R, so r = r,. Thus G = (R) = W(C), so it remains to show C is a root system. If u = av E C then, by definition of C, 112 = Q(u) = a 2 ~ ( v ) = a2/2, so a = f 1. Thus / C n (v)I 5 2.

Root systems 149

Also if g E G then, as R is G-invariant, rg E R and (vg) is the center of rgwith Q(vg) = Q(v) = 1/2, so vg E E. Hence G = W(E) acts on E. FinallyI E J = 21R1 < 2G I< oo. So E is a root system.

By 29.12 and 30.1, every finite Coxeter group can be represented as the Weylgroup of some root system. We'll see later in this section that on the one handthis representation is faithful and on the other that the Weyl group of each rootsystem is a finite Coxeter group. Thus the finite Coxeter groups are preciselythe Weyl groups of root systems.

For the remainder of this section let E be a root system and W = W(E) itsWeyl group. The elements of E will be called roots.

(30.2) (1) (E) is a nondegenerate subspace of V and W centralizes El, so Wis faithful on (E).

(2) The permutation representation of W on E is faithful.(3) W is finite.

Proof. Let U = (E). Then H = CW(U) = CW(E) and W/H is faithful onE, so W/H is finite. Thus to prove (2) and (3) it suffices to show H = 1.As Q is positive definite, U is nondegenerate. Thus V = U ® U'. But, forv E E, r centralizes vl Q U1, so W = (r,,: v E E) centralizes U1. HenceH centralizes V, so H = 1, completing the proof of the lemma.

An ordering of V is a total ordering of V preserved by addition and multipli-cation by positive scalars; that is if u, v, w E V with u > v, and 0 < a E F,then u + w > v + w and au > av. I leave the following lemma as an exercise.

(30.3) (1) If < is an ordering on V and V+ = {v E V: v > 01 then

(i) V+ is closed under addition and multiplication by positive scalars, and(ii) for each v E V#, I{v, -v) n V+) = 1.

(2) If S CV# satisfies (i) and (ii) of (1) then there exists a unique orderingofVwith S=V+.

(3) If T C V# satisfies (1.i) and I {v, -v) n T I < 1 for each v E V#, thenT C V+ for some ordering of V.

A subset P of E is a positive system if P = E+ = E n V+ for some orderingof V. A subset Tr of E is a simple system if Tr is linearly independent and eachv E E can be written v = ExE7r axx such that either 0 < ax E F for all x E 7ror 0 > ax E F for all x E r.

149

Also if g E G then, as R is G-invariant, rg E R and (vg) is the center of rg with Q(vg) = Q(v) = 112, so vg E C. Hence G = W(C) acts on C. Finally 1x1 = 21RI 5 21GI < GO. So C is aroot system.

By 29.12 and 30.1, every finite Coxeter group can be represented as the Weyl group of some root system. We'll see later in this section that on the one hand this representation is faithful and on the other that the Weyl group of each root system is a finite Coxeter group. Thus the finite Coxeter groups are precisely the Weyl groups of root systems.

For the remainder of this section let C be a root system and W = W(C) its Weyl group. The elements of C will be called roots.

(30.2) (1) (C) is a nondegenerate subspace of V and W centralizes CL, so W is faithful on (C) .

(2) The permutation representation of W on C is faithful. (3) W is finite.

Proof. Let U = (C). Then H = Cw(U) = Cw(C) and W/H is faithful on C, so W/H is finite. Thus to prove (2) and (3) it suffices to show H = 1. As Q is positive definite, U is nondegenerate. Thus V = U @ u'. But, for v E C, r, centralizes vL 2 u I , so W = (r,: v E C) centralizes u I . Hence H centralizes V, so H = 1, completing the groof of the lemma.

An ordering of V is a total ordering of V preserved by addition and multiplication by positive scalars; that is if u , v, w E V with u > v, and 0 < a E F, then u + w > v + w and a u 2 av. I leave the following lemma as an exercise.

(30.3) (1) If 5 is an ordering on V and V+ = {v E V: v > 0) then

(i) V+ is closed under addition and multiplication by positive scalars, and (ii) for each v E v#, I{v, -v) n V+I = 1.

(2) If S 2 V# satisfies (i) and (ii) of (1) then there exists a unique ordering of V with S = V+.

(3) If T 2 V# satisfies (1.i) and I{v, -v) n TI 5 1 for each v E v#, then T 5 V+ for some ordering of V.

A subset P of C is apositive system if P = C+ = C n V+ for some ordering of V. A subset n of C is a simple system if n is linearly independent and each v E C can be written v = Ex,, a,x such that either 0 5 a, E F for all x E n

orO>a, E F forallx e n .

(30.4) Each simple system is contained in a unique positive system and eachpositive system contains a unique simple system.

Proof. If n is a simple system let T consist of those elements FxER axx E V#with ax > 0. By 30.3.3, T C V+ for some ordering of V. By definition ofsimple systems and 30.3.1, P = T n E is the unique positive system containingn. Let S be the set of those v E P such that, if v = EyEY Cyy for some Y C Pand 0 < cy E F, then Y = {v}. Evidently S c n. I'll show n c S which willprove n = S is the unique simple system in P.

So assume v E n and v = EyEYCyy for some Y C P and 0 < Cy E F.Now, for Y E Y, Y = ExEK byxx for some 0 < byx E F, so v = Ey X Cybyxx

and hence, by the linear independence of n, 0 = Ey Cybyx for X E n - (v).Therefore byx = 0 for x # v, soy = by v for each y E Y. Then, as P fl (v) _(v), Y = (v), completing the proof of the claim.

It remains to show that if P is a positive system then P contains a simplesystem. In the process I'll show:

(30.5) If n is a simple system and x and y are distinct members of n, then(x,y)<0.

Indeed, returning to the proof of 30.4, let n be a subset of P minimal subjectto P being contained in the nonnegative linear span of n. I'll show n satisfies30.5 and then use this fact to show n is linearly independent and hence a simplesystem. This will complete the proof of 30.4.

Suppose (x, y) > 0. By 22.6.2, z = yrx = y-2(x, y)x/(x, x), so z = y-cxwith c > 0. Now z = ten at with at > 0 if z c P and at < 0 if z c -P. Ifz c P, consider the equation:

(1 - ay)y = att + (ax + c)x.t54x,y

The right hand side is in V+, so, as y c V+, 1 - ay > 0. But then y is in thenonnegative span of n - (y), contradicting the minimality of n. If z c -Pconsider:

(ax + Ox = (-at)t + (1 - ay)yt54x,y

and apply the same argument for a contradiction.So n satisfies 30.5 and it remains to show n is linearly independent. Suppose

0 = EXEn axx and let a = (x c n: ax > 0) and,8 = {y c n: ay < 0}. ThenEXEa axx = z = EyEf(-ay)y so 0 < (z, z) = (F_xE« axx, EyE,s(-ay)y) <

0 by 30.5. Hence as Q is positive definite z = 0. But if ax # 0 for some x


(30.4) Each simple system is contained in a unique positive system and each positive system contains a unique simple system.

Proof. If n is a simple system let T consist of those elements Ex,, a,x E V# with a, > 0. By 30.3.3, T E V+ for some ordering of V. By definition of simple systems and 30.3.1, P = T n X is the unique positive system containing n . Let S be the set of those v E P such that, if v = EYE cY y for some Y E P and 0 < cy E F , then Y = {v}. Evidently S E n . I'll show n E S which will prove n = S is the unique simple system in P.

So assume v E n and v = C,, cY y for some Y P and 0 < cy E F. Now,fory E Y,y = ~,,,b,,xforsomeO 5 by, E F , s o v = ~, , , cyb , ,x and hence, by the linear independence of n, 0 = Cy c,b,, for x E n - (v}. Therefore by, = 0 for x # v, so y = b,,v for each y E Y. Then, as P f l (v) = (v}, Y = (v}, completing the proof of the claim.

It remains to show that if P is a positive system then P contains a simple system. In the process I'll show:

(30.5) If n is a simple system and x and y are distinct members of n , then (x, Y) 5 0.

Indeed, returning to the proof of 30.4, let n be a subset of P minimal subject to P being contained in the nonnegative linear span of n . I'll show n satisfies 30.5 and then use this fact to show n is linearly independent and hence a simple system. This will complete the proof of 30.4.

Suppose (x, y) > 0. By 22.6.2, z = yr, = y -2(x, y)x/(x, x), so z = y -cx with c > 0. Now z = C,,, at t with a, > 0 if z E P and at 5 0 if z E -P. If z E P , consider the equation:

(1 - aY)y = C at t + (a, + c)x. t#x,y

The right hand side is in V+, so, as y E V+, 1 - a, > 0. But then y is in the nonnegative span of n - (y}, contradicting the minimality of n . If z E -P consider:

(a, + C)X = C (-at) + (1 - aY)y t#x,y

and apply the same argument for a contradiction. So n satisfies 30.5 and it remains to show n is linearly independent. Suppose

O=CXEITa,xandle ta = ( x En :a , >0}and/3 = ( y En :a , <O}.Then

Exes axx = z = EyED(-a , )~ so 0 5 (z,z) = (Ex,, axx, CYEB(-ay)y) 5 0 by 30.5. Hence as Q is positive definite z = 0. But if a, # 0 for some x

Root systems 151

then z E V+, a contradiction. So n is linearly independent and the proof iscomplete.

For the remainder of this section let n be a simple system and P the positivesystem containing n.

(30.6) For each v E P there is x E n with (v, x) > 0.

Proof. Write v = EX E, axx with ax > 0 and observe 0 < (v, v) = >x ax (v, x).

(30.7) For each x E n, xrx = -x and rx acts on P - (x).

Proof. Let r = rx. By definition of r, xr = -x. Let V E P - {x}, so thatv = EYE, ayy with ay > 0. As (x)= (x) fl P, aZ > 0 for some z E n - (x).Now

yr = > ayyr ay(y - 2(y, x)x) ayy + bxY Y Y54x

for some b E F. As aZ > 0 it follows that yr E P from the definition of simplesystem.

(30.8) W is transitive on simple systems and on positive systems.

Proof. By 30.3, W permutes positive systems, while it is evident from thedefinitions that W permutes simple systems. By 30.4 it suffices to show W istransitive on positive systems. Assume not and let P and R be positive systemsin different orbits of W, and subject to this constraint with ( P fl (-R)J = nminimal. As R # P, n > 0. Let n be the simple system in P. By 30.4 thereis x E n fl (-R). By 30.7, I Prx fl (-R)j = n - 1, so, by minimality ofn, R E (Prx)W = PW, a contradiction.

For V E E, V = EZE7[ axx for unique ax E F; define the height of v to beh(v) = ExEn ax. Evidently the height function h depends on n. Notice forv E P that h(v) > 0 and h(-v) = -h(v) < 0. Also h(x) = 1 for x E n.

(30.9) (1) h(v) > l for each v E P - n.(2) W = (rx: x E 7r).(3) Each member of E is conjugate to an element of n under W.

Proof. Let G = (rx: x E n) and v E P. Pick U EP fl vG such that h(u) isminimal. By 30.6 there is y E n with (u, y) > 0. Then ury = u - cy with

Root systems

then z E V + , a contradiction. So n is linearly independent and the proof is complete.

For the remainder of this section let n be a simple system and P the positive system containing n .

(30.6) For each v E P there is x E n with ( v , x ) > 0.

Proof. Write v = Ex,, a,x with a, 2 0 and observe 0 < ( v , v ) = Ex a, ( v , x).

(30.7) For each x E n, xr, = -x and r, acts on P - ( x } .

Proof. Let r = r,. By definition of r, xr = -x. Let v E P - {x } , so that v = EYE, ayy with ay 2 0. As ( x } = ( x ) f l P, a, > 0 for some z E n - ( x } . Now

for some b E F. As a, > 0 it follows that vr E P from the definition of simple system.

(30.8) W is transitive on simple systems and on positive systems.

Proof. By 30.3, W permutes positive systems, while it is evident from the , definitions that W permutes simple systems. By 30.4 it suffices to show W is

transitive on positive systems. Assume not and let P and R be positive systems in different orbits of W, and subject to this constraint with IP f l (-R)I = n minimal. As R # P, n > 0. Let n be the simple system in P. By 30.4 there is x E n f l (-R). By 30.7, IPr, n (-R)I = n - 1, so, by minimality of n, R E (Pr,) W = P W , a contradiction.

For v E X, v = E,,, a,x for unique a, E F; define the height of v to be h(v) = Ex,, a,. Evidently the height function h depends on n. Notice for v E P that h(v) > 0 and h(-v) = -h(v) < 0. Also h(x) = 1 for x E n.

(30.9) (1) h (v ) > 1 for each v E P - n. (2) W = (r,:x E n). (3) Each member of X is conjugate to an element of n under W.

Proof. Let G = (r,: x E n) and v E P. Pick u E P fl vG such that h(u) is minimal. By 30.6 there is y E n with (u , y) > 0. Then ur, = u - cy with

c = 2(u, y) > Oandifu 0 y then, by 30.7,ury E PnvG andh(ury) = h(u)-c < h(u), a contradiction. So u = y E jr. In particular if v 0 then h(v) >h(y) = 1, so (1) holds. Further it follows that each member of E is conjugateto an element of Tr under G, so that (3) holds. Finally v = yg for some g E G,so rU = ryg = (ry)g E G and hence W = (r,,: v E P) = G.

For W E W let N(w) = I Pw n (-P)I. Then N(I)=O and, by 30.7, N(rx) = 1for x E Tr. Let R = Jr,,: x E Tr} and l = lR the length function with respect toR; that is, for w E W, l(w) is the minimal length of a word u in the membersof R such that u = w. It will develop shortly that the functions l and N agreeon W.

(30.10) Let W E W and X E r. Then(1) N(rxw) = N(w) + 1 if xw > 0, and(2) N(rw) = N(w) - 1 if xw < 0.

Proof. Prxw = ({-x} U (P - {x}))w = {-xw} U (Pw - {xw}) by 30.7, so

Prxw n (-P)(Pw n (-P)) U {-xw} if xw > 0,

=(Pw n (-P)) - {xw) if xw < 0.

(30.11) Let ri E R, 0 < i < n and w = r1 ... r,, E W with N(w) = n andN(row) < n. Then there exists k, 1 < k < n, such that rorl ... rk_1 = r1 ... rk.

Proof. By 30.10, xw < 0, where x E Tr with ro = rx. Let k be minimalsubject to xr1 ... rk < 0. Then, by 30.7, y = xr1 ... rk_1 E Tr and rk = ry. Thus

...rkr oYxrl ... rA_ I = - Yy = rk, which can be rewritten as r0r1 rk_p 1 = Y1 ... Yk.

(30.12) 1(w) = N(w) for each w E W.

Proof. I've already observed that 1 (1) = 0 = N (1). Then 30.10 and inductionon 1(w) implies 1(w) > N(w). Suppose that the lemma is false and pick u =rorl ... rn E W with 1(u) = n + 1 > N(u), and subject to this constraint withn minimal. Let w = r1 ... rn . As 1(u) = n + 1, 1(w) = n, and, by minimalityof 1(u), N(w) = 1(w). As N(row) = N(u) < n + 1, there exists k withror1 ... rk_1 = r1 ... rk by 30.11. Thus u = ro ... rk_1rk ... rn = r1 ... r rk+l

rn = r1 . . . rk-lrk+1 . . . rn is of length at most n - 1, contrary to the choiceof U.

(30.13) (W, R) is a Coxeter system.


c = 2(u, y) > 0 andif u # y then, by 30.7, ur, E P n v G and h(ur,) = h(u)- c < h(u), a contradiction. So u = y E n. In particular if v 6 n then h(v) > h (y ) = 1 , so ( 1 ) holds. Further it follows that each member of Z is conjugate to an element of n under G, so that (3 ) holds. Finally v = yg for some g E G, so r, = r,, = (r,), E G and hence W = (r,: v E P ) = G.

For w E W let N(w) = IPw fl (-P)I. Then N ( l ) = 0 and, by 30.7, N(r,) = 1 for x E n. Let R = {r, : x E n] and 1 = 1~ the length function with respect to R; that is, for w E W , l ( w ) is the minimal length of a word u in the members of R such that u = w. It will develop shortly that the functions 1 and N agree on W .

(30.10) Let w E W and x E n. Then (1) N(r,w) = N ( w ) + 1 if xw > 0, and (2 ) N(r,w) = N ( w ) - 1 if xw < 0.

Proof. Pr,w = ( ( - X I U ( P - (x}))w = ( -xw} U (Pw - { x w } ) by 30.7, so

(Pw n ( - P) ) U ( -xw] if xw > 0 , Pr,w n ( - P ) =

(Pw fl ( -P ) ) - ( x w ] if xw < 0.

(30.11) Let ri E R,O 5 i 5 n and w = rl . . . r , E W with N ( w ) = n and N(row) 5 n. Then there exists k, 1 5 k 5 n, such that rorl . . . rk-1 = rl . . . rk.

Proof. By 30.10, xw < 0, where x E n with ro = r,. Let k be minimal subject to xrl . . . rk < 0. Then, by 30.7, y = xrl . . . rk-1 E n and rk = r,. Thus

r;l".rk-l - - r,, ,... ,,-, = r, = rk, which can be rewritten as rorl . . . rk-1 = rl . . . rk.

(30.12) l (w ) = N(w) for each w E W .

Proof. I've already observed that l(1) = 0 = N(1). Then 30.10 and induction on l (w) implies l (w ) 2 N(w) . Suppose that the lemma is false and pick u = rorl . . . r, E W with l (u) = n + 1 > N(u) , and subject to this constraint with n minimal. Let w = rl . . . r,. As l (u) = n + 1 , l (w) = n, and, by minimality of l(u), N (w) = l(w). As N(row) = N(u ) < n + 1, there exists k with rorl . . .rk-1 = rl . . .rk by 30.11. Thus u = ro.. .rk-irk.. .rn = rl . . .rzrk+l . . . r, = rl . . . rk-lrk+l . . . rn is of length at most n - 1 , contrary to the choice of u.

(30.13) (W , R ) is a Coxeter system.

i


c = 2(u, y) > Oandifu : y then, by 30.7,ury E PnvG andh(ury) = h(u)-c < h(u), a contradiction. So u = y E r. In particular if v 0 r then h(v) >h(y) = 1, so (1) holds. Further it follows that each member of E is conjugateto an element of 7r under G, so that (3) holds. Finally v = yg for some g E G,so rU = ryg = (ry)g E G and hence W = (ru: v E P) = G.

For W E W let N(w) = I Pw n (-P)I. Then N(l)=0 and, by 30.7, N(rx) =1for x E r. Let R = Jr,,: x E r) and l = lR the length function with respect toR; that is, for w E W, l(w) is the minimal length of a word u in the membersof R such that u = w. It will develop shortly that the functions l and N agreeon W.

(30.10) Let W E W and X E r. Then

Root systems

Proof. This is immediate from 29.4, 30.11, and 30.12.

(30.14) W is regular on the positive systems and on the simple systems.

153

Proof. By 30.8 and 30.4 it suffices to show NW(P) = 1. But NW(P) = {w EW: N(w) = 0) so 30.12 completes the proof.

Let D consist of those v E V such that (v, x) > 0 for all x E Jr.

(30.15) (1) vW n D is nonempty for each v E V.(2) (d, u) > O for each d E D and u E P.

(1) N(r,,w) = N(w) + 1 if xw > 0, and Proof. Pick Z E v W maximal with respect to the ordering defining P. Then,

(2) N(row) = N(w) -1 if xw < 0. for x E Jr, z > zrx = z - 2(z, x)x, so (z, x) > 0 as x > 0. That is z E

Proof. Prxw = ({-x) U (P - (x}))w = {-xw) U (Pw - {xw)) by 30.7, so

Prxw n (-P) =(Pw n (-P)) u {-xw} if xw > 0,

(Pw n (-P)) - {xw) if xw < 0.

vW n D. Let d E D and u E P. Then u = XEn axx with a, > 0 so(d, u) = >x ax(d, x) > 0.

(30.16) Let d E D. Then Cw(d) is the Weyl group of the root systems E n dland has simple system 7r n dl.

(30.11) Let ri E R, 0 < i < n and w = rl ... r E W with N(w) = n andN(row) < n. Then there exists k, l < k < n, such that rori ... rk-t = ri . . . rk.

Proof. By 30.10, xw < 0, where x E Tr with ro = rx. Let k be minimalsubject to xr1 ... rk < 0. Then, by 30.7, y = xr1 ... rk-1 E 7r and rk = ry. Thusr'' rk = r r y , which can be rewritten as r r rk_p xr ... i = y = k 0 l rk

(30.12) 1(w) = N(w) for each w E W.

Proof. I've already observed that 1(1) = 0 = N(1). Then 30.10 and inductionon 1(w) implies 1(w) > N(w). Suppose that the lemma is false and pick u =rori ... rn E W with 1(u) = n + 1 > N(u), and subject to this constraint withn minimal. Let w = rl ... rn. As l(u) = n + 1, 1(w) = n, and, by minimalityof 1(u), N(w) = 1(w). As N(row) = N(u) < n + 1, there exists k withrorl ... rk_1 = rl ... rk by 30.11. Thus u = ro ... rk_lrk ... rn = ri ... rkrk+l... rn = ri ... rk_irk+l ... rn is of length at most n - 1, contrary to the choiceof U.

(30.13) (W, R) is a Coxeter system.

Proof. U = (ry: x E Tr n d') < Cw(d), so assume w E Cw(d) - U and sub-ject to this constraint with n = 1(w) minimal. w : 1 so n > 0. Then N(w) =n > 0 so there is x E zr with xw < 0. Now, by 30.15.2, 0 > (xw, d) =(x, dw-1) = (x, d), so x E dl by another application of 30.15.2. Thus rx E Uand, by 30.10, l(rxw) < n, so, by minimality of n, rxw E U. But then w E U.

So U = Cw(d). Evidently E n dl is a root system and an easy calculationusing 30.15.2 shows Ir n dl is a simple system.

(30.17) Let S C V. Then Cw(S) is the Weyl group of the root system E n S'.

Proof. Let U = (rn: v EE n Sl). It suffices to show U = Cw(S). Cw(S) =Cw((S)) = Cw(So) where So is a basis for (S), so without loss S is finite.Replacing S by a suitable conjugate under W and appealing to 30.15, we maytake d E Dn S. Let G = Cw(d) and E0 = E n dl. Then, by 30.16, G = W(Eo)and Cw(S) = CG(S), while, by induction on the order of E, CG(S) = U-

(30.18) Let (G, S) be a Coxeter system with G finite. Then the representationa of 29.8 is faithful and G is the Weyl group of a root system.

Proof. By 29.12 and 30.1, Ga is the Weyl group of a root system. By 29.7and 30.13, Ga is a Coxeter group of type M, so I Gal _ I G 1. Hence a is anisomorphism.

(30.19) Let H be the symmetric group on SZ = {1, ... , n), (e) = 12, andG = (e)wroH the wreath product of (e) by H. Choose notation so that CH(e)is the stabilizer in H of n E Q. Let G1 = H, S1 = {(1, 2), (2, 3), ... , (n -1, n)), G2 = G, S2 = S1 U (e), S3 = S1 U {(n - 1, n)e), and G3 = (S3). Then(Gi, Si) is a Coxeter system of type An-1, Cn, Dn, for i = 1, 2, 3, respectively.Further G3 is of index 2 in G2 and is the semidirect product of E2'-, by S.

Proof. Let V be n-dimensional Euclidean space with orthonormal basic X =(xi: 1 < i < n), ei the reflection with center (xi), and E = (ei: 1 < i < n).Represent H on V via xih = Xih for h E H. Then G = (H, E) < O(V) and G isthe wreath product (e) w rn H, where e = en. Hence G is the semidirect productof E - E2,1, by H = Sn, and G3 is the semidirect product of D = E fl G3 byH where D = (ei ej: 2 < i < j < n) = E2-, . The transposition (i, j) is thereflection with center (xih), where xi j = xi - xj. Also Gi = (Si) and Si is a setof reflections, so, by 30.1, Gi = W(Ei) where Ei is the root system consistingof the Gi-conjugates of x12 if i = 1 or 3, and {x12, xn) if i = 2. Thus

E1={±xxij:1<i<j<n},

E3 = E1 U (±(xi +x3):1 < i < j < n),

and

E2=E3U{±xi:2<i <n}.

Next ni is a simple system for Ei, where ni = {xi,i+1: 1 < i < n), 7r2 =71 U(xn), and 73 = nl U {xn_1 + xn}. Hence, as Si = {rx: x E ni), (Gi, Si) isa Coxeter system by 30.13. Finally it is evident that the Coxeter diagram of(Gi, Si) is An- 1, C,, Dn, for i = 1, 2, 3, respectively.

For J C r let Vj _ (J) and Wj = (rj: j E J). Recall the subgroups WJ andtheir conjugates under W are called parabolic subgroups of W.

(30.20) Let 0 0 J C r. Then(1) Ej = E fl vi is a root system with simple system J and Weyl group

Wr .

(2) Wi = Cw(VJ ).

Proof. Wi < Cw(VjL) = U and, by 30.17, U is the Weyl group of the rootsystem Ej.Tr is linearly independent and J spans Vj, so J is a basis of Vj.


Proof. By 29.12 and 30.1, G a is the Weyl group of a root system. By 29.7 and 30.13, G a is a Coxeter group of type M, so /Gal = IGI. Hence a is an isomorphism.

(30.19) Let H be the symmetric group on 52 = (1, . . . , n), (e) Z Z2, and G = (e)wrnH the wreath product of (e) by H . Choose notation so that CH(e) ' is the stabilizer in H of n E 52. Let G1 = H, S1 = ((1,2), (2,3), . . . , (n - 1, n)), G2 = G, S2 = S1 U (e), S3 = S1 U {(n - 1, n)e), and G3 = (S3). Then (Gi, Si) is a Coxeter system of type A,-1, C,, D,, for i = 1,2,3, respectively. Further G3 is of index 2 in G2 and is the semidirect product of E2"-1 by S,.

Proof. Let V be n-dimensional Euclidean space with orthonormal basic X = (xi: 1 ( i 5 n), ei the reflection with center (xi), and E = (ei: 1 5 i ( n). Represent H on V via xi h = xih for h E H. Then G = (H, E ) 5 O(V) and G is the wreath product (e) w rn H , where e = en. Hence G is the semidirect product of E Z E2", by H Z S,, and G3 is the semidirect product of D = E n G3 by H where D = (eiej: 2 ( i _( j ( n) Z E2"-1. The transposition (i, j ) is the reflection with center (xij), where xij = xi -xi. Also Gi = (Si) and Si is a set of reflections, so, by 30.1, Gi = W(Ci) where Ci is the root system consisting of the Gi-conjugates of xl2 if i = 1 or 3, and 1x12, x,) if i = 2. Thus

and .

Next ni is a simple system for Xi, where ni = (xi,i+l: 1 5 i < n), n2 =nl U

( ~ n ) r and n3 = nl U (x,- 1 + x,]. Hence, as Si = {r, : x E xi) , (Gi, Si) is a Coxeter system by 30.13. Finally it is evident that the Coxeter diagram of (Gi, Si) is A,-l, C,,, D,, for i = 1, 2, 3, respectively.

For J J n let Vj = (J) and Wj = (rj: j E J ) . Recall the subgroups Wj and their conjugates under W are called parabolic subgroups of W.

(30.20) Let 0 # J J n . Then (1) CJ = C f l VJ is a root system with simple system J and Weyl group

WJ. (2) WJ = cW(vJ').

Proof. Wj ( cw(vJ') = U and, by 30.17, U is the Weyl group of the root system C J. n is linearly independent and J spans Vj, so J is a basis of Vj.

Root systems 155

Hence each member of Ej is a linear combination of the members of J, so, as 7ris a simple system, J is a simple system for Ej. Thus U = (rj: j c- J) = Wj.

Remarks. The discussion of Coxeter systems in section 29 follows that ofBourbaki [Bo] and Suzuki [Su]. The presentation of root systems given heredraws heavily on the appendix of Steinberg [St].

Coxeter groups and root systems play an important role in branches of math-ematics other than finite group theory, most particularly in the study of Liealgebras, Lie groups, and algebraic groups. We will find in chapters 14 and 16that they are crucial to the study of the finite groups of Lie type.

Exercises for chapter 101. Prove D2n =Grp (x, y: x2 = y2 = (xy)n = 1). Prove every group generated

by a pair of distinct involutions is a dihedral group.2. Prove lemma 30.3.3. Let E be a root system, 7r a simple system for E, P the positive system of

7r, R = (ra: a E 7r), and W = (R) the Weyl group of E. Prove(1) There exists a unique wo E W with Pwo = -P.(2) wo is the unique element of W of maximal length in the alphabet R.

Further 1(wo) = I P I and wo is an involution.(3) rwo = -7t and Rw0 = R.

4. Let E be an irreducible root system, 7r a simple system for E, P the positivesystem for ,r, J C n, E j, the subset of E, spanned by J, and lU = P - Ej.Prove (E) _ (i).

5. Let E be a root system, P a positive system for E, and w E W. Prove foreach a E P that Pw contains exactly one of a and -a.

6. Assume the hypothesis of Exercise 10.3 and let wo be the element of W ofmaximal length in the alphabet R. Define a relation < on W by u < w ifw = xu with 1(w) = 1(x) + 1(u). Prove(1) < is a partial order on W.(2) wo is the unique maximal element of W. That is w < wo for all w E W.(3) For r = ra E R and W E W the following are equivalent:

(a) rw < w.(b) l(rw) < 1(w).(c) aw < 0.(d) w = rl ... rn with r = r1 and l(w) = n.

(4) If u < w and r E R with l(ur) < 1(u) and l(wr) < 1(w) then ur < wr.(Hint: To prove (2) let wo 0 w E W and use Exercise 10.3.2, 30.10, and30.12 to show there exists r E R with 1(w) < l(rw).)

Root systems 155

Hence each member of C J is a linear combination of the members of J, so, as n is a simple system, J is a simple system for El . Thus U = ( r j : j E J ) = W l .

Remarks. The discussion of Coxeter systems in section 29 follows that of Bourbaki [Bo] and Suzuki [Su]. The presentation of root systems given here draws heavily on the appendix of Steinberg [St].

Coxeter groups and root systems play an important role in branches of mathematics other than finite group theory, most particularly in the study of Lie algebras, Lie groups, and algebraic groups. We will find in chapters 14 and 16 that they are crucial to the study of the finite groups of Lie type.

Exercises for chapter 10 1. Prove D2, =Grp ( x , y: x2 = Y 2 = ( ~ y ) ~ = 1). Prove every group generated

by a pair of distinct involutions is a dihedral group. 2. Prove lemma 30.3. 3. Let C be a root system, n a simple system for C , P the positive system of

n, R = (r,: a E n), and W = ( R ) the Weyl group of C . Prove (1) There exists a unique wo E W with Pwo = -P. (2) wo is the unique element of W of maximal length in the alphabet R.

Further 1 (wo) = I P I and wo is an involution. (3) nwo = -n and RwO = R.

4. Let C be an irreducible root system, n a simple system for C , P the positive system for n, J c n , C j , the subset of C, spanned by J, and $ = P - C j . Prove ( C ) = ($).

5 . Let C be a root system, P a positive system for C , and w E W. Prove for each a E P that Pw contains exactly one of a and -a.

6. Assume the hypothesis of Exercise 10.3 and let wo be the element of W of maximal length in the alphabet R. Define a relation 5 on W by u 5 w if w = xu with l (w ) = l ( x ) + l(u). Prove (1) 5 is a partial order on W. (2) wo is the unique maximal element of W. That is w 5 wo for all w E W. (3) For r = r, E R and w E W the following are equivalent:

(a) rw 5 w .

(b) Qrw) 5 Qw). (c) a w < 0. (d) w = rl . . .r, with r = rl and l (w) = n.

(4) If u 5 w and r E R with l (ur ) 5 l (u) and l (wr) 5 l (w) then ur 5 wr. (Hint: To prove (2) let wo # w E W and use Exercise 10.3.2, 30.10, and 30.12 to show there exists r E R with l (w) < l(rw).)

11

The generalized Fitting subgroup

We've seen that the composition factors of a finite group control the structureof the group in part, but that control is far from complete. Section 31 introducesa tool for studying finite groups via composition factors `near the bottom' ofthe group. The generalized Fitting subgroup F*(G) of a finite group G is acharacteristic subgroup of G generated by the small normal subgroups of Gand with the property that CG(F*(G)) < F*(G). This last property supplies arepresentation of G as a subgroup of Aut(F*(G)) with kernel Z(F*(G)). G canbe effectively investigated via this representation because F*(G) is a relativelyuncomplicated group whose embedding in G is particularly well behaved.

It turns out that F*(G) is a central product of the groups OP(G), p r= 7r(G),with a subgroup E(G) of G. To define E(G) requires some terminology. Acentral extension of a group X is a group Y together with a surjective homo-morphism of Y onto X whose kernel is in the center of Y. The group Y willalso be said to be a central extension of X. A group L is quasisimple if L isperfect and the central extension of a simple group. The components of G areits subnormal quasisimple subgroups, and E(G) is the subgroup of G genera-ted by the components of G. It develops that E(G) is a central product of thecomponents of G.

Recall that if p is a prime then a p-local subgroup of G is the normalizerin G of a nontrivial p-subgroup of G. The local theory of groups investigatesfinite groups from the point of view of p-locals. A question of great interest inthis theory is the relationship between the generalized Fitting subgroup of Gand that of its local subgroups. Section 31 contains various results about suchrelationships. In the final chapter we'll get some idea of how such results areused to classify the finite simple groups.

If F*(G) is a p-group, it can be particularly difficult to analyze the structureof G. One tool for dealing with such groups is the Thompson factorization.Lemma 32.5 shows that, if G is solvable and the Thompson factorization fails,then the structure of G is rather restricted. This result will be used in laterchapters to prove the Thompson Normal p-Complement Theorem and theSolvable Signalizer Functor Theorem. The Normal p-Complement Theoremwill be used in turn to establish the nilpotence of Frobenius kernels.

Finally the importance of components focuses attention on quasisimplegroups. In section 33 we find there is a largest perfect central extension


We've seen that the composition factors of a finite group control the structure of the group in part, but that control is far from complete. Section 3 1 introduces a tool for studying finite groups via composition factors 'near the bottom' of the group. The generalized Fitting subgroup F*(G) of a finite group G is a characteristic subgroup of G generated by the small normal subgroups of G and with the property that CG(F*(G)) 5 F*(G). This last property supplies a representation of G as a subgroup of Aut(F*(G)) with kernel Z(F*(G)). G can be effectively investigated via this representation because F*(G) is a relatively uncomplicated group whose embedding in G is particularly well behaved.

It turns out that F*(G) is a central product of the groups O,(G), p E n ( G ) , with a subgroup E(G) of G. To define E(G) requires some terminology. A central extension of a group X is a group Y together with a surjective homomorphism of Y onto X whose kernel is in the center of Y. The group Y will also be said to be a central extension of X. A group L is quasisimple if L is perfect and the central extension of a simple group. The components of G are its subnormal quasisimple subgroups, and E(G) is the subgroup of G generated by the components of G. It develops that E(G) is a central product of the components of G.

Recall that if p is a prime then a p-local subgroup of G is the normalizer in G of a nontrivial p-subgroup of G. The local theory of groups investigates finite groups from the point of view of p-locals. A question of great interest in this theory is the relationship between the generalized Fitting subgroup of G and that of its local subgroups. Section 31 contains various results about such relationships. In the final chapter we'll get some idea of how such results are used to classify the finite simple groups.

If F*(G) is a p-group, it can be particularly difficult to analyze the structure of G. One tool for dealing with such groups is the Thompson factorization. Lemma 32.5 shows that, if G is solvable and the Thompson factorization fails, then the structure of G is rather restricted. This result will be used in later chapters to prove the Thompson Normal p-Complement Theorem and the Solvable Signalizer Functor Theorem. The Normal p-Complement Theorem will be used in turn to establish the nilpotence of Frobenius kernels.

Finally the importance of components focuses attention on quasisimple groups. In section 33 we find there is a largest perfect central extension

The generalized Fitting subgroup 157

7r: G G of each perfect group G. G is the universal covering group ofG and ker(7r) is the Schur multiplier of G. In particular if G is simple then G isthe largest quasisimple group with G as a homomorphic image and the Schurmultiplier of G is the center of G. As an illustration of this theory, the coveringgroups and Schur multipliers of the finite alternating groups are determined.

The Schur multiplier can be defined for nonperfect groups using an alternatedefinition requiring homological algebra. The presentation given here is grouptheoretic and restricted to perfect groups; it follows Steinberg [St].

31 The generalized Fitting subgroupIn this section G is a finite group. A group X is quasisimple if X = V andX/Z(X) is simple.

(31.1) Let X be a group such that X/Z(X) is a nonabelian simple group. ThenX = X(') Z(X) and XM is quasisimple.

Proof. Let Y = X M and X* = X/Z(X). Now Y* < X * and X * is simple soY* =1 or X*. In the latter case X = YZ(X) and in the former X* is abelian,contrary to hypothesis.

So X = YZ(X). Thus X/ Y(1) is abelian so Y = YO). Further Y/Z(Y) = X* issimple, so Y is quasisimple.

(31.2) Let X be a quasisimple group and H < < X. Then either H = X orH < Z(X).

Proof. If H Z(X), X = HZ(X ), so that X/H is abelian and hence X =XM < H.

The components of a group X are its subnormal quasisimple subgroups. WriteComp(X) for the set of components of X. Set E(X) = (Comp(X)).

(31.3) If H < < X then Comp(H) = Comp(X) n H.

(31.4) Let L E Comp(G) and H < < G. Then either L E Comp(H) or[L, H] =1.

Proof. Let G be a minimal counterexample. If L = G the lemma holds by31.2, so by 7.2 we may take X = (Lc) # G. Similarly if H = G the lemmais trivial, so take Y = (Ho) G. X n Y < X, so, by minimality of G, eitherL E Comp(X n Y) or [L, x n Y] = 1. In the first case L E Comp(Y) by 31.3

n: (? + G of each perfect group G. (? is the universal covering group of G and ker(n) is the Schur multiplier of G. In particular if G is simple then (? is the largest quasisimple group with G as a homomorphic image and the Schur multiplier of G is the center of (?. As an illustration of this theory, the covering groups and Schur multipliers of the finite alternating groups are determined.

The Schur multiplier can be defined for nonperfect groups using an alternate definition requiring homological algebra. The presentation given here is group theoretic and restricted to perfect groups; it follows Steinberg [St].

31 The generalized Fitting subgroup In this section G is a finite group. A group X is quasisimple if X = x(') and X/Z(X) is simple.

(31.1) Let X be a group such that X/Z(X) is a nonabelian simple group. Then X = x(')z(x) and x(') is quasisimple.

Proof. Let Y = x(') and X* =X/Z(X). Now Y* 9 X* and X* is simple so Y* = 1 or X*. In the latter case X = YZ(X) and in the former X* is abelian, contrary to hypothesis.

So X = YZ(X). Thus X/Y(') is abelian so Y = ~ ( ' 1 . Further Y/Z(Y) E X* is simple, so Y is quasisimple.

(31.2) Let X be a quasisimple group and H 99 X. Then either H = X or H 5 Z(X).

Proof. If H $ Z(X), X = HZ(X), so that X/H is abelian and hence X = x(') 5 H .

The components of a group X are its subnormal quasisimple subgroups. Write Comp(X) for the set of components of X. Set E(X) = (Comp(X)).

(31.3) If H 9 9 X then Comp(H) = Comp(X) n H.

(31.4) Let L E Comp(G) and H 9 9 G. Then either L E Comp(H) or [L, HI = 1.

Proof. Let G be a minimal counterexample. If L = G the lemma holds by 31.2, so by 7.2 we may take X = ( L ~ ) # G. Similarly if H = G the lemma is trivial, so take Y = ( H ~ ) f G. X n Y 9 X, so, by minimality of G, either L E Comp(X n Y) or [L, X n Y] = 1. In the first case L E Comp(Y) by 31.3

158 The generalized Fitting subgroup

and then, as H < Y < G, the lemma holds by minimality of G. In the second[Y, L, L] < [YnX,L]=1,so,by8.9, [Y,L]=1.

(31.5) Distinct components of G commute.


(31.6) Let L E Comp(G) and H an L-invariant subgroup of G. Then(1) Either L E Comp(H) or [L, H] =1.(2) If H is solvable then [L, H] = 1.(3) If R < G then either L E Comp([R, L]) or [R, L] = 1.

Proof. Part (1) follows from 31.4 applied to LH in the role of G. Then (1)implies (2). If R < G then [L, R] is L-invariant by 8.5.6, so by (1) eitherL E Comp([L, R]) or [R, L, L] = 1. In the latter case [R, L] = 1 by 8.9.

(31.7) Let E = E(G), Z = Z(E), and E* = E/Z. Then(1) Z = (Z(L): L E Comp(G)).(2) E* is the direct product of the groups (L*: L E Comp(G)).(3) E is a central product of its components.

The Fitting subgroup of G, denoted by F(G), is the largest nilpotent normalsubgroup of G. O,,.(G) denotes the largest solvable normal subgroup of G.

(31.8) Let G be a finite group. Then F(G) is the direct product of the groups(OP(G): P E n(G)).

Proof. See 9.11.

(31.9) O.(CG(F(G))) = Z(F(G)).

Proof. Let Z = Z(F(G)). G* = G/Z, and H = O,,(CG(F(G))). AssumeH*= 1 and let X* be a minimal normal subgroup of H*. Then X* is a p-group for some prime p, so X = PZ, where P E SyIP(X). X centralizes Z,so P <X. Thus P <OP(G) < F(G), So P < CF(G)(F(G)) = Z, contradictingP*=X*01.

(31.10) If G is solvable then CG(F(G)) < F(G).

Define the socle of G to be the subgroup generated by all minimal normalsubgroups of G, and write Soc(G) for the socle of G.


and then, as H 5 Y < G, the lemma holds by minimality of G. In the second [Y, L, LI I [Y n x , LI = 1, SO, by 8.9, [Y, LI = 1.

(31.5) Distinct components of G commute.

Proof. This is a direct consequence of 31.2 and 3 1.4.

(31.6) Let L E Comp(G) and H an L-invariant subgroup of G. Then (1) Either L E Comp(H) or [L, HI = 1. (2) If H is solvable then [L, HI = 1. (3) If R _( G then either L E Comp([R, L]) or [R, L] = 1.

Proof. Part (1) follows from 31.4 applied to L H in the role of G. Then (1) implies (2). If R 5 G then [L, R] is L-invariant by 8.5.6, so by (1) either L E Comp([L, R]) or [R, L, L] = 1. In the latter case [R, L] = 1 by 8.9.

(31.7) Let E = E(G), Z = Z(E), and E* = E/Z. Then (1) Z = (Z(L): L E Comp(G)). (2) E* is the direct product of the groups (L*: L E Comp(G)). (3) E is a central product of its components.

The Fitting subgroup of G, denoted by F(G), is the largest nilpotent normal subgroup of G. O,(G) denotes the largest solvable normal subgroup of G.

(31.8) Let G be a finite group. Then F(G) is the direct product of the groups

(Op(G): P E dG) ) .

Proof. See 9.11.

Proof. Let Z = Z(F(G)). G* = G/Z, and H = O,(CG(F(G))). Assume H*# 1 and let X* be a minimal normal subgroup of H*. Then X* is a p- group for some prime p, so X = PZ, where P E Sylp(X). X centralizes 2, so P 9 X. Thus P 5 O,(G) 5 F(G), So P 5 CF(G)(F(G)) = Z, contradicting P* = X* # 1.

(31.10) If G is solvable then CG(F(G)) 5 F(G).

Define the socle of G to be the subgroup generated by all minimal normal subgroups of G, and write Soc(G) for the socle of G.

(31.11) Let Z = Z(F(G)), G* = G/Z, and S* = Soc(CG(F(G))*). ThenE(G) = SM and S = E(G)Z.

Proof. Let H = CG(F(G)). By 31.9, O,0(H*) =1. So, by 8.2 and 8.3, eachminimal normal subgroup of H* is the direct product of nonabelian simplegroups and by 31.3 these factors are components of H*. Thus S* < E(H*). LetK* be a component of H*. By 31.1, K = K(' )Z with KM quasisimple. By 31.3,K(' E Comp(G), so S < E(G)Z. By 31.6.2, E(G) < H. Let L E Comp(G), andM= (LH). Then M* is a minimal normal subgroup of H* by 31.4, so M < S.Thus S = E(G)Z. By 31.1, E(G) = Ski).

Define the generalized Fitting subgroup of G to be F*(G) = F(G)E(G).

(31.12) F*(G) is a central product of F(G) with E(G).

Proof. See 31.6.2.

(31.13) CG(F*(G)) < F*(G).

Proof. Let H = CG(F*(G)), K = CG(F(G)), Z = Z(F(G)), and G* = G/Z.Then H* < K*, so if H* 01 then 10 H* n Soc(K*), and then, by 31.11,H n E(G) 0 Z, a contradiction.

Recall Op',E(G) is defined by Op',E(G)/OP,(G) = E(G/OP,(G)).

(31.14) Let p be a p-subgroup of G. Then

(1) Op',E(NG(P)) < CG(OP(G)), and(2) if P < Op(G) then OP(F*(NG(p))) = OP(F*(G)).

Proof. Let X = OP,,E(NG(p)). To prove (1), it suffices to show X centralizesR = COp(G)(P) by the A x B Lemma. But [R, OP,(X)] < R n OP-(X) = 1and [R, X] < OP,(X) by 31.6.2, so [R, X] = 1 by coprime action, 18.7.

So take P<OP(G). Then Yo=OP(F*(G))<Y=OP(F*(NG(P))). If Lis a component of NG(P) then [L, OP(G)] = 1 by (1), and, as Yo < Y, ei-ther [L, Yo] = 1 or L E Comp(Yo) by 31.4. In the first case L <_ CG(F*(G)) =Z(F(G)), a contradiction. So E(NG(P)) = E(G). Let q 0 p and Q = Oq(Y).We must show Q < O9 (G). Passing to G/O9 (G) and appealing to coprimeaction, 18.7, we may take Oq(G) = 1, and it remains to show Q =1. But, by(1) and as Yo < Y, Q < CG(O9(F*(G)))=CG(F*(G))=Z(F(G)), so indeedQ=1.


(31.11) Let Z = Z(F(G)), G* = G/Z, and S* = Soc(CG(F(G)>*). Then E(G) = S('! and S = E(G)Z.

Proof. Let H =CG(F(G)). By 31.9, O,(H*) = 1. So, by 8.2 and 8.3, each minimal normal subgroup of H* is the direct product of nonabelian simple groups and by 31.3 these factors are components of H*. Thus S* 5 E(H*). Let K * b e a c o m p o n e n t o f ~ * . ~ ~ 3 1 . 1 , K = ~ ( ' ) ~ w i t h ~ ( ' ) ~ u a s i s i m p l e . ~ y 3 1 . 3 , K(') E Comp(G), so S I E(G)Z. By 31.6.2, E(G) 5 H. Let L E Comp(G), and M = ( L H ) . Then M* is a minimal normal subgroup of H* by 31.4, so M ( S. Thus S = E(G)Z. By 3 1.1, E(G) = s(').

Define the generalized Fitting subgroup of G to be F*(G) = F(G)E(G).

(31.12) F*(G) is a central product of F(G) with E(G).

Proof. See 31.6.2.

Proof. Let H = CG(F*(G)), K = CG(F(G)), Z = Z(F(G)), and G* = G/Z. Then H* 9 K*, so if H* # 1 then 1 # H* n Soc(K*), and then, by 31.11, H n E(G) # Z, a contradiction.

I Recall O,J,~(G) is defined by Op,,E(G)/Op,(G) = E(G/O,,(G)).

(31.14) Let p be a p-subgroup of G. Then

(1) O~~ ,E(NG(P>> 5 CG(O~(G)), and (2) if P 5 O,(G) then OP(F*(NG(p))) = Op(F*(G)).

Proof. Let X = 0 ,1 ,~ (N~(p ) ) . To prove (I), it suffices to show X centralizes R = Co,(G)(P) by the A x B Lemma. But [R , Op,(X)] ( R n Op!(X) = 1 and [R , XI 5 Op,(X) by 31.6.2, so [R , XI = 1 by coprime action, 18.7.

So take P 5 Op(G). Then Yo = Op(F*(G)) 5 Y = OP(F*(NG(P))). If L is a component of NG(P) then [ L , Op(G)] = 1 by (I), and, as Yo ( Y, either [L , Yo] = 1 or L E Comp(Yo) by 3 1.4. In the first case L ( CG (F* (G)) = Z(F(G)), a contradiction. So E(NG(P)) = E(G). Let q # p and Q = 0, (Y). We must show Q 5 0, (G). Passing to G/Oq (G) and appealing to coprime action, 18.7, we may take Oq(G) = 1, and it remains to show Q = 1. But, by (1) and as Yo 5 Y, Q 5 CG(Oq(F*(G))) = CG(F*(G)) = Z(F(G)), so indeed Q = 1.

(31.15) If G is solvable and P is a p-subgroup of G then Op, (NG (P)) < Op, (G).

Proof. Passing to G/Op,(G) and appealing to coprime action, 18.7, we maytake Op,(G) = 1, and it remains to show Op'(NG(P)) = 1. This follows from31.10 and 31.14.1.

(31.16) Let F*(G) = Op(G) for some prime p. Then F*(NG(P)) = Op(NG(P))for each p-subgroup P of G.

Proof. This follows from 31.13 and 31.14.1.

The Schreier Conjecture says Out(L) is solvable for each finite simple group L.

(31.17) Let Op,(G) = 1 and P a p-subgroup of G. Then(1) Op',E(NG(P)) fixes each component of G.(2) If each component of G satisfies the Schreier conjecture, then

Op',E(NG(P))°° < E(G).

Proof. Let H = NG(P), X = Op,(H), H* = H/X, and Y = Op',E(H)°°. LetK < X or K < H with K* E Comp(H*), and subject to these constraints pickK minimal subject to moving a component of G. Let P < Po E Sylp(CG(K)).As K satisfies the same hypothesis with respect to Po, we may take P = Po. Inparticular, by 31.14, Op(G) < P. Let R E Sylp(H fl E(G)).

Suppose first K < X. Then K is a q-group for some prime q, and by co-prime action, 18.7, there exists an R-invariant Sylow q-group Q of Op,(H).Replacing K by a suitable conjugate, we may assume K < Q. By 24.4, Q =[R, Q]CQ(R). Now [R, Q] < [E(G), Q] < E(G), so [R, Q] fixes each com-ponent of G. Hence we may take [K, R] =1. Thus, by choice of P, R < P.So P fl E(G) E Sylp(E(G)) by Exercise 3.2. As Op,(G) = 1, p c 7r (L) foreach L E Comp(G), so 1 # P fl L E Sylp(L) by 6.4. Then P fl L Z(L), soL = [E(G), P fl L] is K-invariant.

So K X. By 31.4, either K* < [K*, R*] or [K*, R*] = 1. In the lattercase by coprime action, 18.7, K =Op-(K)CK(R) so, by minimality of K,[K, R] = 1. But then an argument in the last paragraph supplies a contradiction.In the former, K < X [K, R]:< XE(G), so K fixes each component of G.

Let L be a component of G. I have shown Y < N(L). If L satisfies theSchreier conjecture, then Auty(L) < Inn(L) as Y = YO°, so Y < LC(L). Hence,under the hypothesis of (2), Y < E(G)C(E(G)) and then, by 31.14, Y <E(G)C(F*(G)) < F*(G), so Y < F*(G)O° = E(G).


(31.15) If G is solvable and P is ap-subgroup of G then O,I(NG(P)) 5 O,!(G).

Proof. Passing to G/O,l(G) and appealing to coprime action, 18.7, we may take O,f(G) = 1, and it remains to show O,~(NG(P)) = 1. This follows from 31.10 and 31.14.1.

(31.16) Let F*(G) = O,(G)forsomeprime p.Then F*(NG(P)) = O,(NG(P)) for each p-subgroup P of G.

Proof. This follows from 3 1.13 and 3 1.14.1.

The Schreier Conjecture says Out(L) is solvable for each finite simple group L.

(31.17) Let O,t(G) = 1 and P a p-subgroup of G. Then (1) O,J,~(NG(P)) fixes each component of G. (2) If each component of G satisfies the Schreier conjecture, then

O p f , ~ ( N ~ ( P > ) " L E(G).

Proof. Let H = NG(P), X = O,!(H), H* = H/X, and Y = O,I,~(H)*. Let K ( X or K ( H with K* E Comp(H*), and subject to these constraints pick K minimal subject to moving a component of G. Let P ( Po E Sylp(CG(K)). As K satisfies the same hypothesis with respect to Po, we may take P = Po. In particular, by 31.14, Op(G) 5 P. Let R E Syl,(H n E(G)).

Suppose first K 5 X. Then K is a q-group for some prime q, and by coprime action, 18.7, there exists an R-invariant Sylow q-group Q of O,f(H). Replacing K by a suitable conjugate, we may assume K ( Q. By 24.4, Q = [R, Q]CQ(R). Now [R, Q] 5 [E(G), Q] ( E(G), so [R, Q] fixes each component of G. Hence we may take [K, R] = 1. Thus, by choice of P , R 5 P. So P n E(G) E Syl,(E(G)) by Exercise 3.2. As O,/(G) = 1, p E n(L) for each L E Comp(G), so 1 # P fl L E Syl,(L) by 6.4. Then P n L $ Z(L), so L = [E(G), P n L] is K-invariant.

So K $ X. By 31.4, either K* 5 [K*, R*] or [K*, R*] = 1. In the latter case by coprime action, 18.7, K =O,f(K)CK(R) so, by minimality of K, [K, R] = 1. But then an argument in the last paragraph supplies a contradiction. In the former, K ( X[K, R] 5 XE(G), so K fixes each component of G.

Let L be a component of G. I have shown Y 5 N(L). If L satisfies the Schreier conjecture, then Autr(L) 5 Inn(L) as Y = Y*, so Y ( LC(L). Hence, under the hypothesis of (2), Y 5 E(G)C(E(G)) and then, by 31.14, Y ( E(G)C(F*(G)) 5 F*(G), so Y ( F*(G)* = E(G).

G is balanced for the prime p if O p, (CG M) < Op- (G) for each X of order pin G.

(31.18) Let Op- (G) = 1, x of order pin G, L E Comp(G), and Y = Op'(CG (x)).Then

(1) If L # [L, x] then [L, Y] = 1 and either L E Comp(CG(x)) or L # Lxand C[L,x](x)(' = K E Comp(CG(x)) with K a homomorphic image of L.

(2) If L = [L, x] andAuty(x)L(L) is balancedfortheprime p, then [Y, L] = 1.

Proof. Assume L # [L, x]. Then either [L, x] = 1 or L # Lx. In the firstcase L E Comp(C(x)), so [L, Y] = 1. In the second let M = [L, x] and M* =M/Z(M). By 8.9, [M, L] # 1, so, by 31.4, L E Comp(M), and we concludeM = (0)) is the central product of the groups (Lx': 0 < i < p) from 31.5.Hence, by Exercise 3.5, K = CM(x)(l) is a homomorphic image of L. As L isquasisimple, so is its homomorphic image K. Also M < < G, so K < < CG (x),and hence K E Comp(C(x)). So [K, Y] = 1. Y acts on L* by 31.17, so,as [K*, Y] = 1, [L*, Y] = 1 by Exercise 3.5. Then [L, Y] = 1 by 31.6,completing the proof of (1).

So assume L = [L, x], let U = Y(x)L, and let rr: U --> AutG(L) be theconjugation map. Let P = Op(U)(x) so that P E Sylp(ker(7r)(x)). By 31.14,[P, Y] = 1, so, as Cu(P) < Cv(x), Y < Op,(Cv(P)) = Op,(Nv(P)). Then,as P E Sylp(ker(7r)(x)), NUn((x7r)) = Nu(P)7r by a Frattini Argument, soYrr < Op' (Cu. (x 7r)). Hence, if Autu (L) is balanced for the prime p, then[Y, L] = 1, so that (2) holds.

(31.19) Let Op-(G) = 1 and assume AutH(L) is balanced for the prime p foreach L E Comp(G) and each H < G with L < H. Then G is balanced for theprime p.

Proof. Let X be a subgroup of G of order p and Y = Op'(CG(X)). I must showY = 1. By 31.18 and the hypothesis on the components of G, [Y, E(G)] = 1.By 31.14, [Y, Op(G)] =1. So Y < CG(F*(G)) = Z(F(G)), so, as Op,(G) = 1,Y=1.

Here's technical lemma to be used in chapter 15.

(31.20) Let A be an elementary abelian r-group acting on a solvable r'-groupG and let a E A. Let p E r c 7r (G), p" = rr' U {p}, and P an A-invariant p-subgroup of G.


G is balanced for the prime p if O,l(CG(X)) < Opj(G) for each X of order p in G.

(31.18) Let Op~(G) = 1, x of order p in G, L E Comp(G), and Y = Op~(CG(~)) . Then

(1) If L # [L, x] then [L, Y] = 1 and either L E Comp(CG(x)) or L # Lx and CcL,xl(~)( l ) = K E Comp(CG(x)) with K a homomorphic image of L.

(2) If L = [L, x] a n d A ~ t ~ ( , ) ~ ( L ) is balancedfortheprime p, then [Y, L] = 1.

Proof. Assume L # [L, x]. Then either [L , x] = 1 or L # Lx . In the first case L E Comp(C(x)), so [L, Y] = 1. In the second let M = [L, x] and M* =

M/Z(M). By 8.9, [M, L] # 1, so, by 31.4, L E Comp(M), and we conclude M = (L(")) is the central product of the groups (Lx': 0 5 i < p) from 31.5. Hence, by Exercise 3.5, K = cM(x)(') is a homomorphic image of L. As L is quasisimple, so is its homomorphic image K. Also M 1 1 G, so K 1 1 CG (x), and hence K E Comp(C(x)). So [K, Y] = 1. Y acts on L* by 31.17, so, as [K*,Y] = l ,[L*,Y] = 1 by Exercise 3.5. Then [L,Y] = 1 by 31.6, completing the proof of (1).

So assume L = [L, x], let U = Y (x) L, and let n : U + AutG(L) be the conjugation map. Let P = O,(U)(x) so that P E Syl,(ker(n)(x)). By 31.14, [P, Y] = 1, so, as CU(P) 5 CU(x), Y 5 OPf(CU(P)) = Op~(NU(P)). Then, as P E Syl,(ker(n)(x)), Nuk((xn)) = Nu(P)n by a Frattini Argument, so Y n < Op~(Cur(xn)). Hence, if Autu(L) is balanced for the prime p, then [Y, L] = 1, so that (2) holds.

(31.19) Let O,<(G) = 1 and assume AutH(L) is balanced for the prime p for each L E Comp(G) and each H ( G with L 1 H. Then G is balanced for the prime p.

Proof. Let X be a subgroup of G of order p and Y = Op,(CG(X)). I must show Y = 1. By 3 1.18 and the hypothesis on the components of G, [Y, E(G)] = 1. By 31.14, [Y, O,(G)] = 1. So Y 5 CG(F*(G)) = Z(F(G)), so, as O,,(G) = 1, Y = l .

Here's technical lemma to be used in chapter 15.

(31.20) Let A be an elementary abelian r-group acting on a solvable r'-group G and let a E A. Let p E n C n(G), pk = n' U {p], and P an A-invariant p- subgroup of G.

(1) Suppose P < Op(K) for some a-invariant 7r-subgroup K of G such thatCK(a) is a Hall 7r-subgroup of CG(a). Then P for each a-invariantsubgroup N of G with N = (P, CN(a)).

(2) Assume A is noncyclic and let A be the set of hyperplanes ofA. Assume f o r each B E A that Cp(B) <On((CG(a), Cp(B))). ThenP <On((CG(a), P)).

(3) If B is a noncyclic subgroup of CA(P) and P < O,r((CG ((a, b)), P))for each b E B#, then P < On((CG(a), P)).

Proof. Let G be a minimal counterexample to (1), (2), or (3). Without loss,A = (a) in (1) and A = (B, a) in (3). In each case it suffices to assume G =(P, CG(a)) and prove P < Q, (G), where a = p" in (1) and a = 7r in (2) and (3).

Let H be a minimal A-invariant normal subgroup of G and G* = G/H. ThenP* < O,,(G*) = S* by minimality of G and coprime action 18.7.4, and H isa q-group for some prime q. Hence if P < O,,(HP) we are done, so this is notthe case, and in particular q V a.

Let I = CG(a). Then CH(a) < I. We show [P, CH(a)] = 1. Then G =(I, P) < N(CH(a)), so, by minimality of H, CH(a) (aor H. In the lattercase [P, H] = 1, contradicting P O,(PH). Thus CH(a) = 1.

Now to verify that [P, CH(a)] =1. In (1), as q E 7r and CH(a) < I, CH(a)is contained in each Hall 7r-group of CG(a) by Hall's Theorem 18.5, andhence in K. So [CH(a), P] <Op(K) flOq(G)=1. In (2), [Cp(B), CH (a)]O,r(GB) fl Oq(GB) =1, where GB = (CG(a), Cp(B)) and B E A. Hence,by Exercise 8.1, P = (Cp(B): B E A) < C(CH(a)). Finally, in (3),[P, CH((a, b))] <O,r(Gb) fl Oq(Gb) = 1, where Gb = (P, CG((a, b))). So,again by Exercise 8.1, CH(a) _ (CH((a, b)): b E B#) C(P).

We've shown CH(a) = 1. In (2) and (3) let R be an A-invariant Hall7r-subgroup of S containing P. By a Frattini Argument, G = HNG(R). AsCH(a) =1, CG(a) = CG*(a) = NG(R) fl C(a), so CG(a) <, NG(R). But thenG = (CG(a), P) < N(R), so [P, H] < R fl H = 1, a contradiction.

This leaves (1). Here we may take P =Op(K) < K, so, if U is a Hall7r'-group of I, then R = (PI) = (P(inK)U) _ (PU) and R < (I, P) = G. Thuswe may assume H < R. Let X be an a-invariant Hall q'-subgroup of US.As CH(a) = 1, CUS(a) is a q'-group, so X is the unique a-invariant Hall q'-subgroup of US by Exercise 6.2. Hence (P, U) < X. But then H < R < X, acontradiction.

32 Thompson factorizationIn this section p is a prime and G is a finite group. Denote by M(G) the set ofelementary abelian p-subgroups of G of p-rank m p(G). Set J(G) _ (. ((G)).


(1) Suppose P 5 O,(K) for some a-invariant n-subgroup K of G such that CK(a) is a Hall n-subgroup of CG(a). Then P 1 OPr(N) for each a-invariant subgroup N of G with N = (P , CN(a)).

(2) Assume A is noncyclic and let A be the set of hyperplanes of A. Assume for each B E A that Cp(B) 5 O,((CG(a), Cp(B))). Then

P 5 O"((CG(~>, P)). (3) If B is a noncyclic subgroup of CA(P) and P 5 O,((CG((a, b)), P ) )

for each b E B', then P 5 0, ((CG(a), P)).

ProoJ Let G be a minimal counterexample to (I), (2), or (3). Without loss, A = (a) in (1) and A = (B, a ) in (3). In each case it suffices to assume G = (P , CG(a)) and prove P 5 O,(G), where a = p" in (1) and a = n in (2) and (3).

Let H be a minimal A-invariant normal subgroup of G and G* = GIH. Then P* 5 O,(G*) = S* by minimality of G and coprime action 18.7.4, and H is a q-group for some prime q. Hence if P 5 O,(HP) we are done, so this is not the case, and in particular q $ a.

Let I=CG(a) . Then CH(a)<]I. We show [P, CH(a)]=l. Then G = (I, P ) 5 N(CH(a)), so, by minimality of H, CH(a) = 1 or H. In the latter case [P, HI = 1, contradicting P $ O,(PH). Thus CH(a) = 1.

Now to verify that [P, CH(a)] = 1. In (I), as q E n and CH(a) <] I, CH(a) is contained in each Hall n-group of CG(a) by Hall's Theorem 18.5, and hence in K. So [CH(a), PI 5 O,(K) n Oq(G) = 1. In (21, [Cp(B), C H ( ~ ) ] < 0, (GB) no, (GB) = 1, where GB = ( C G ( ~ ) , CP(B)) and B E A. Hence, by Exercise 8.1, P = (Cp(B): B E A) 5 C(CH(a)). Finally, in (3), [P, C H ( ( ~ , b ) ) l 5 O,(Gb) n Oq(Gb) = 1, where Gb = (P , C G ( ( ~ , b))). So, again by Exercise 8.1, CH(a) = (CH((a, b)): b E B') 5 C(P).

We've shown CH(a) = 1. In (2) and (3) let R be an A-invariant Hall n-subgroup of S containing P . By a Frattini Argument, G = HNG(R). As CH(a) = 1, CG(a) 2 CG*(a) 2 NG(R) i l C(a), so CG(a) < NG(R). But then G = (CG(a), P) 5 N(R), so [P, HI 5 R i l H = 1, a contradiction.

This leaves (1). Here we may take P =Op(K) <] K, so, if U is a Hall nl-group of I , then R = (PI) = (p(InK)') = (P') and R <] (I, P) = G. Thus we may assume H 5 R. Let X be an a-invariant Hall ql-subgroup of US. As CH(a) = 1, Cus(a) is a ql-group, so X is the unique a-invariant Hall q1- subgroup of US by Exercise 6.2. Hence (P, U) 5 X. But then H 5 R 5 X, a contradiction.

32 Thompson factorization In this section p is a prime and G is a finite group. Denote by d ( G ) the set of elementary abelian p-subgroups of G of p-rank mp(G). Set J(G) = ( d ( G ) ) .

Thompson factorization 163

We call J(G) the Thompson subgroup of G. Of course J(G) depends on thechoice of p.

(32.1) (1) J(G) char(G).(2) If A is in a (G) and A < H < G then ,((H) c ,((G).(3) Let P E Sylp(G). Then J(P) = J(Q) for each p-subgroup Q of G con-

taining J(P).

If V is a GF(p)G-module define GJ'(G, V) to consist of the nontrivial elemen-tary abelian p-subgroups A of G such that

m(A) + m(Cv(A)) > m(B) + m(Cv(B))

for each B < A. Notice that if B is a nontrivial subgroup of A E °J'(G, V) forwhich this inequality is an equality, then B is in GJ'(G, V). A subset _1:1P ofP(G, V) is stable if G permutes °J' via conjugation and, whenever A is in GJ'

and B is a nontrivial. subgroup of A with

m(A) + m(Cv(A)) = m(B) + m(Cv(B)),

then B is in °J' As a final remark note that if A is in °J'(G, V) then m(A) >m(V /Cv (A)); to see this just take B =1 in the inequality defining membershipin °J'(G, V).

(32.2) Let V be a normal elementary abelian p-subgroup of G, G* = G/ CG(V ),and GJ' = {A*: A E ,1(G) and A* 11. Then GJ' is a stable subset of 7)(G*, V).

Proof. Let A be in a(G) and CA(V) < B < A. A0 = ACv(A) is an elemen-tary abelian p-group, so, as A E , (G), m(Ao) < m(A). Hence A = A0 sinceA < A0. Thus Cv(A) = A fl V. Similarly m(BCv(B)) < m(A). As CB(V)CA(V), we have

m(BCv(B)) = m(B*) + m(CA(V)) + m(Cv(B)) - m(A fl v)

while

m(BCv(B)) < m(A) = m(A*) + m(CA(V))

so as Cv (A) = A fl V it follows that

m(A*) + m(Cv(A)) > m(B*) + m(Cv(B))

with equality only if m(BCv(B)) = m(A). Thus if A* 1 then A* E °J'(G, V)while if m(B*) + m(Cv(B)) = m(A*) + m(Cv(A)) then m(A) = m(BCv(B)),so B0 = BCv(B) is in ,1(G) and hence B* = Bo is in GJ'(G*, V).

Thompson factorization

We call J ( G ) the Thompson subgroup of G . Of course J ( G ) depends on the choice of p.

(32.1) (1) J ( G ) char(G). (2) If A is in d ( G ) and A 5 H 5 G then d ( H ) G d ( G ) . (3) Let P E Syl,(G). Then J ( P ) = J ( Q ) for each p-subgroup Q of G con-

taining J ( P ) .

If V is a GF(p)G-module define P ( G , V) to consist of the nontrivial elementary abelian p-subgroups A of G such that

m(A) + m(Cv(A)) 1 m(B) + m(Cv(B))

for each B 5 A. Notice that if B is a nontrivial subgroup of A E P ( G , V) for which this inequality is an equality, then B is in P ( G , V). A subset P of P ( G , V) is stable if G permutes P via conjugation and, whenever A is in P and B is a nontrivial. subgroup of A with

then B is in 9 As a final remark note that if A is in P ( G , V ) then m(A) 2 m(V/Cv(A)); to see this just take B = 1 in the inequality defining membership in P ( G , V).

(32.2) Let V be a normal elementary abelian p-subgroup of G , G* = G I CG(V), and P = {A*: A E d ( G ) and A* # 1). Then P is a stable subset of P ( G * , V).

Prooj Let A be in d ( G ) and CA(V) 5 B 5 A. A. = ACV(A) is an elementary abelian p-group, so, as A E d ( G ) , m(Ao) 5 m(A). Hence A = A. since A 5 Ao. Thus CV(A) = A il V. Similarly m(BCv(B)) 5 m(A). As CB(V) = CA(V), we have

while

so as Cv(A) = A i l V it follows that

with equality only if m(BCV(B)) = m(A). Thus if A* # 1 then A* E P ( G , V) while if m(B*) + m(Cv(B)) = m(A*) + m(Cv(A)) then m(A) = m(BCv(B)), so Bo = BCv(B) is in d ( G ) and hence B* = B,* is in P ( G * , V).

(32.3) Let G be a solvable group, V a faithful GF(p)G-module, and P aSylow p-group of G. Assume Op(G) = 1, GJ' is a stable subset of -OP(G, V)and H=Then p<3,H=HI x . x H,,, V=[V,H] ®CV(H)with [V, H] = [V, Hl] (D . . . ®[V, H,,], G permutes (H,:1 m(B)+m(Cv(B)) withB E °J' or B = 1 in case of equality. So by minimality of A either Al I= p orthe inequality is strict, and in that event, as m(A) = m(B) + 1, it follows thatU = CV(B).

Suppose Al I> p and let K = [F(G), A]. As Op(G) = 1, F(G) is a p'-groupand A is faithful on F(G) by 31.10, so K 1. By Maschke's Theorem, V =[V, K] ®CV(K), and, as K 0 1, 0 [V, K]. A acts on K and hence on [V, K],so C[V,K](A) 0. Thus U Cv(K). On the other hand, by Exercise 8.1,K = (CK(B): IA:BI = p) and by the last paragraph, U=CV(B),soCK(B) <N(U). Thus K < N(U), so, by Exercise 3.6, K = [K, A] < CG(U), contra-dicting U Cv(K).

So JAI = p for each AEI'. As A E P(G, V), m(V/CV(A)) <m(A)=1,so A is generated by a transvection. Let S2 consist of the subgroups L =(A1, A2) with Ai E F and 10 OP(L) < F(G). For each Ao E F there is Lo E 0with Ao < Lo as [Ao, F(G)] 1. Let L = (A, B) E S2 and W = [V, L]. W =[V, A] + [V, B] and Cv(L)=CV(A) fl CV(B), so, as 1=m([V, A])=m(V/CV(A)), m(W) <2> m(V/CV(L)). If 0 o CW(L) then A and B cen-tralize CW(L), W/CW(L), and V/W, SO L centralizes CW(L), W/CW(L), andV/ W, which is impossible by Exercise 3.1 since L is not a p-group. There-fore W = [V, A] ® [V, B] is of rank 2 and V = W ® Cv(L). In particular L isfaithful on W so L < GL(W) = GL2(p). Now A fixes only [V, A] amongst theset 0 of p + 1 points of W and hence is transitive on the remaining p points.So as B moves [V, A], L is transitive on 0. Thus as L contains the group Aof transvections with center [V, A], L contains all transvections in GL(W), so,by 13.7, L = SL(W) = SL2(p). In particular, as G is solvable, p < 3 by 13.8.

Next let Y = O"(L) and M = NG(Y). Notice M acts on W and CV(L)as W = [V, Y] and Cv(L) = Cv (Y). Suppose L Lo E S2 with Yo =OP(L0) < M. Then either W = [V, Yo] or [V, Yo] < CV(Y), since Yo is irre-ducible on [V, Yo]. If W = [V, Yo] then Cv(Y) = CV(Yo), so L = SL(W) = Lo,contrary to the choice of Lo. Thus [V, Vol < CV(Y), so W < Cv(L) and, byExercise 3.6, [L, Lo] < CG(W) fl CG(CV(L)) = CG(V) = 1

Let A be a maximal set of commuting members of S2 and D = (A). By the lastparagraph, D is the direct product of (I: I E A) and V = [V, D] ® Cv (D) with[V, D] _ ®IEO[V, I]. So it remains to show S2 = A and J = (-opfl P) < D.


(32.3) Let G be a solvable group, V a faithful GF(p)G-module, and P a Sylow p-group of G. Assume Op(G) = 1, P i s a stable subset of P ( G , V ) and H = ( P ) # 1 . T h e n p ( 3 , H = H 1 x . - . x H,, V = [ V , H ] @ C V ( H ) with [V , HI = [V , H I ] @ . . . @ [V , H,], G permutes (Hi: 1 5 i _( n) , Hi S SL2(p), m([V, Hi]) = 2, and ( P f l P ) is a Sylow p-group of H.

Proo$ Let r consist of the minimal members of P, pick A E r, and set U = Cv(A). If B is a hyperplane of A then m(A) +m(U) L m(B)+m(Cy-(B)) with B E P or B = 1 in case of equality. So by minimality of A either / A / = p or the inequality is strict, and in that event, as m(A) = m(B) + 1, it follows that U = Cv(B).

Suppose ] A / > p and let K = [F(G), A]. As Op(G) = 1 , F(G) is a pf-group and A is faithful on F(G) by 3 1.10, so K # 1. By Maschke's Theorem, V = [V , K] @ Cv(K), and, as K # 1,0 # [V , K] . A acts on K andhence on [V , K ] , so CIV,Kl(A) # 0. Thus U $ Cv(K) . On the other hand, by Exercise 8.1, K = (CK(B): / A : BI = p ) andby thelastparagraph, U = Cv(B), so C K ( B ) 5 N(U). Thus K 5 N(U) , so, by Exercise 3.6, K = [ K , A] 5 CG(U), contradicting U $ Cv(K) .

So / A / = p for each A E r. As A E P(G, V ) , m(V /Cv(A) ) ( m ( A ) = 1, so A is generated by a transvection. Let 52 consist of the subgroups L = ( A l , Az) with Ai E r and 1 # OP(L) ( F(G). For each A. E r there is Lo E C2 with A. ( Lo as [Ao, F(G)] # 1. Let L = ( A , B ) E 52 and W = [V, L]. W = [V, A] + [V, B] and Cv(L) = Cv(A) fl Cy-(B), so, as 1 = m([V, A]) = m(V/Cy-(A)), m(W) ( 2 m(V/CV(L)) . If 0 # Cw(L) then A and B centralize Cw(L), W/Cw(L) , and V / W , so L centralizes Cw(L) , W/Cw(L) , and V / W , which is impossible by Exercise 3.1 since L is not a p-group. There- fore W = [V, A] @ [V, B] is of rank 2 and V = W @ CV(L). In particular L is faithful on W so L 5 GL(W) 2 GL2(p). Now A fixes only [V , A] amongst the set 8 of p + 1 points of W and hence is transitive on the remaining p points. So as B moves [V , A] , L is transitive on 8. Thus as L contains the group A of transvections with center [V, A], L contains all transvections in GL(W), so, by 13.7, L = SL(W) E SL2(p). In particular, as G is solvable, p ( 3 by 13.8.

Next let Y = Op(L) and M = NG(Y). Notice M acts on W and C V ( L ) as W = [V, Y ] and Cv(L) = Cv(Y). Suppose L # Lo€ 52 with Yo= OP(Lo) 5 M . Then either W = [V, Yo] or [V, Yo] 5 Cv(Y), since Yo is irreducible on [V, Yo]. If W = [V, Yo] then CV(Y) = Cv(Yo), so L = SL(W) = Lo, contrary to the choice of Lo. Thus [V , Yo] 5 Cv(Y) , so W 5 Cy-(L) and, by Exercise 3.6, [ L , Lo] _( CG(W) i'l CG(Cv(L)) = CG(V) = 1

Let A be a maximal set of commuting members of 52 and D = ( A ) . By the last paragraph, D is the direct product of ( I : I E A ) and V = [V , Dl @ Cv(D) with [V , Dl = @,,,[V, I ] . So it remains to show 52 = A and J = ( P n P) 5 D.

Thompson factorization 165

If S2 0 then, by Exercise 11.4, there is L E c - 0 with Y =OP(L) <NF(G)(Op(D)). Thus Y permutes E = ([V, I]: I E 0) and, as m([V, Y]) = 2,it follows that f acts on each member of E, and hence also on each member of0. But now by the next-to-last paragraph and maximality of A, L E 0, contraryto the choice of L.

So S2 = A. Finally assume J D and let E E °J' with E minimal subjectto END. Let Z=(WE). As E is abelian, CE(W)=CE(Z) and E/NE(L)is regular on WE. Let S be a set of coset representatives for NE(L) in E.Then Z = ®SES Ws so Cz(E) = (USES ws: w E CW(NE(L))] is of rank s =m(CW/NE(L)). Therefore m(Z/CZ(E)) = m(Z) - s = 2pa - s, where a =m(E/NE(L)). Also m(NE(W)/CE(W)) = 3 < 1 with s = 1 in case of equality.Further s < m(W) = 2. Thus if a > 0 then 2pa - s > a + S. Finally, ifE NE(L), we conclude

m(Cv(CE(Z))) - m(Cv(E)) > m(Z) - m(Cz(E))

= 2pa - s > a + S = m(E) - m(CE(Z)),

contradicting E E ?7 (G, V).Therefore E acts on each member of A. Hence, as L = OP'(GL(W)),

E <DCG([D, V])=X. From the action of D on [D, V] it follows thatm(E/CE([D, V]) < m([D, V]/C[D,v](E)), so as °J' is stable eitherCE([D, V]) = 1 or CE([D, V]) is in GJ'. In either case, by minimality ofE, CE([D, V]) < CD([D, V]) = 1. Now m(E) <m([D, VIIC[D,v] (E)), so asE E °J'(G, V), E < CX (Cv (D)) = D.

This completes the proof of the lemma. Notice that, as D = H < G, S2 is theset of all subgroups of G generated by a pair of noncommuting members of F.

(32.4) Let F*(G) = Op(G), P E Sylp(G), Z = c21(Z(P)), V = (ZG) and G* _G/CG(V ). Then V is elementary abelian and Op(G*) = 1.

Proof. F*(G) = Op(G) < P < C(Z), so, by 31.13, Z < S21(Z(Op(G))) = Vo.So, as Vo < G, V = (ZG) < Vo and hence, as Vo is elementary abelian, so isV. Let K* =Op(G*). By 6.4, Q = P fl K E Sylp(K), so, as K* is a p-group,K = CG(V)Q < CG(Z). Hence, as K < G, V = (ZG) <Z(K),soK<CG(V).That is K* = 1.

Lemma 32.4 supplies a tool for analyzing groups G with F*(G) = Op(G).Namely, as V is elementary abelian, we can regard V as a vector space overGF(p) and the representation of G on V by conjugation makes V into afaithful GF(p)G*-module. These observations are used in conjunction with

Thompson factorization

If 52 # A then, by Exercise 11.4, there is L E 52 - A with Y = OJ'(L) ( NF(G)(OP(D)). Thus Y permutes Z = ( [ V , I ] : I E A ) and, as m([V, Y ] ) = 2, it follows that Y acts on each member of Z, and hence also on each member of A. But now by the next-to-last paragraph and maximality of A , I, E A, contrary to the choice of L.

So 52 = A. Finally assume J $ D and let E E P with E minimal subject to E $ D. Let Z = (wE) . As E is abelian, C E ( W ) = C E ( Z ) and E / N E ( L ) is regular on W E . Let S be a set of coset representatives for NE(L) in E. Then Z = esEs Ws so C Z ( E ) = (C,,, ws: w E C w ( N ~ ( L ) ) ) is of rank E = m(Cw/NE (L)) . Therefore m(Z /Cz (E) ) = m ( Z ) - E = 2pa - E , where a = m(E/NE(L)) . Also m(NE(W) /CE(W) ) = 6 ( 1 with E = 1 in case of equality. Further E ( m(W) = 2. Thus if a > 0 then 2pa - E > a + 6. Finally, if E # NE(L) , we conclude

contradicting E E P ( G , V ) . Therefore E acts on each member of A. Hence, as L =o~'(GL(w)) ,

E 5 DCG([D, V ] ) = X . From the action of D on [D , V ] it follows that m(E/CE([D , V ] ) ( m([D, V]/CID,vl(E)) , so as P is stable either CE([D, V ] ) = 1 or CE([D, V ] ) is in P. In either case, by rninimality of E , CE([D , V I ) 5 ~ D ( [ D , v l>= 1. Now m(E) 5 m([D, V]/C[D,v](E)) , so as E E P(G, V ) , E 5 C,y(Cv(D)) = D.

This completes the proof of the lemma. Notice that, as D = H G, 52 is the set of all subgroups of G generated by a pair of noncommuting members of r .

(32.4) Let F*(G) = Op(G), P E Sylp(G), Z = Q1(Z(P)) , V = (zG) and G* = G/CG(V) . Then V is elementary abelian and Op(G*) = 1.

Proof. F*(G)=Op(G) ( P I C(Z) , so, by 31.13, Z 5 521(Z(Op(G)))= VO. So, as Vo L] G , V = (zG) ( Vo and hence, as Vo is elementary abelian, so is V . Let K* = Op(G*). By 6.4, Q = P f l K E Sylp(K), so, as K* is a p-group, K = CG(V)Q 5 CG(Z). Hence, as K L] G , V = (zG) 5 Z ( K ) , so K ( CG(V). That is K* = 1.

Lemma 32.4 supplies a tool for analyzing groups G with F*(G) = Op(G). Namely, as V is elementary abelian, we can regard V as a vector space over GF(p) and the representation of G on V by conjugation makes V into a faithful GF(p)G*-module. These observations are used in conjunction with

lemmas 32.2 and 32.3 in the proof of the next two lemmas. Both are versionsof Thompson Factorization.

(32.5) (Thompson Factorization) Let G be a solvable group with F(G) =Op(G), let P E Sylp(G), Z = 01(Z(P)), V = (ZG), and G* = GI CG(V ). Theneither

(1) G = NG(J(P))CG(Z), or(2) p < 3, J(G)* is the direct product of copies of SL2(p) permuted by G,

and J(P)* E Sylp(J(G)*).

Proof. If J(P)* 1 then, by 32.2 and 32.4, G*, V, sl(G)* satisfies the hy-pothesis of 32.3, and hence (2) holds by 32.3. So assume J(P) < D = CG(V)-By 32.1 and a Frattini Argument, G = NG(J(P))D < NG(J(P))CG(Z).

(32.6) (Thompson Factorization) Let G be a solvable group with F(G) _Op(G). Let p be odd and if p = 3 assume G has abelian Sylow 2-subgroups.Then G = NG(J(P))CG(Q1(Z(P))) for P E Sylp(G).

Proof. This follows from 32.5 and the observation that SL2(3) has nonabelianSylow 2-groups.

33 Central extensionsA central extension of a group G is a pair (H, ir) where H is a group andir: H G is a surjective homomorphism with ker(ir) < Z(H). H is also saidto be a central extension of G. Notice that the quasisimple groups are preciselythe perfect central extensions of the simple groups.

A morphism a: (G1, Ti)r-+ (G2, ir2) of central extensions of G is a grouphomomorphism a: G1 G2 with nt =a7r2. A central extension (G, lr) of Gis universal if for each central extension (H, a) of G there exists a uniquemorphism a: (G, Jr) -* (H, a) of central extensions.

(33.1) Up to isomorphism there is at most one universal central extension of agroup G.

Proof. If (Gi, iri), i = 1, 2, are universal central extensions of G then thereexist morphisms of central extensions ai: (G;, ,r,) (G3_1, 7r3-0. As aia3-iand 1 are morphisms of (Gi, 7ri) to (Gi, ni ), the uniqueness of such a morphismsays ala2 = 1 = a2a1. Thus ai is an isomorphism.


lemmas 32.2 and 32.3 in the proof of the next two lemmas. Both are versions of Thompson Factorization.

(32.5) (Thompson Factorization) Let G be a solvable group with F(G) = Op(G), let P E Sylp(G), Z = nl(Z(P)), V = ( zG) , and G* = G/CG(V). Then either

(1) G = NG(J(P))CG(Z), or (2) p 5 3, J(G)* is the direct product of copies of SL2(p) permuted by G,

and J(P)* E Sylp(J(G)*).

Proof. If J(P)* # 1 then, by 32.2 and 32.4, G*, V, d (G)* satisfies the hy- '

pothesis of 32.3, and hence (2) holds by 32.3. So assume J ( P ) 5 D = CG(V). By 32.1 and a Frattini Argument, G = NG(J(P))D 5 NG(J(P))CG(Z).

(32.6) (Thompson Factorization) Let G be a solvable group with F(G) = Op(G). Let p be odd and if p = 3 assume G has abelian Sylow 2-subgroups. Then G = NG(J(P))CG(S21(Z(P))) for P E Syl,(G).

Proof. This follows from 32.5 and the observation that SL2(3) has nonabelian Sylow 2-groups.

33 Central extensions A central extension of a group G is a pair (H, n ) where H is a group and n : H + G is a surjective homomorphism with ker(n) 5 Z(H). H is also said to be a central extension of G. Notice that the quasisimple groups are precisely the perfect central extensions of the simple groups.

A morphism a: (GI, nl) + (G2, n2) of central extensions of G is a group homomorphism a: G1 + G2 with nl = an2. A central extension (e, n ) of G is universal if for each central extension (H, a ) of G there exists a unique morphism a : (e, n ) += (H, a ) of central extensions.

(33.1) Up to isomorphism there is at most one universal central extension of a group G.

Proof. If (Gi, xi), i = 1,2, are universal central extensions of G then there exist morphisms of central extensions ai: (Gi, ni) + (G34, n3-i). As aia3-i

and 1 are morphisms of (Gi , ni) to (Gi, ni), the uniqueness of such a morphism says ala2 = 1 = ~~2~x1 . Thus ai is an isomorphism.

Central extensions 167

(33.2) If (G, ir) is a universal central extension of G then both G and G areperfect.

Proof. Let H = G x (G/&)) and define a: H G by (x, y)a = xir. Then(H, a) is a central extension of G and a1: (G, ir) (H, a), i = 1, 2, are mor-phisms, where xa1= (x, 1) and xa2 = (x, x&)). So, by the uniqueness ofsuch a morphism, aI = a2 and hence G = 0(1). Thus G is perfect, so, by 8.8.2,G = Gir is perfect.

(33.3) Let G be perfect and (H, .7r) a central extension of G. Then H =ker(ir)HO) with HO perfect.

Proof. By 8.8, H(1)7r = (Hir)(1) = G(1), so, as G is perfect, H(1)7r = G. HenceH = ker(.7r)H(1). As (H, ir) is a central extension, ker(,r) < Z(H), so H/H(2) =Z(H/H(2))(H(')/H(2)) is abelian, and hence HO = H(2) by 8.8.4. Thus HOis perfect.

(33.4) G possesses a universal central extension if and only if G is perfect.

Proof. By 33.2, if G possesses a universal central extension then G is perfect.Conversely assume G is perfect. Let g H g be a bijection of G with a set a

and let F be the free group on G. Let r be the set of words xy(xy)-1, x, y E G,and let M be the normal subgroup of F generated by F. Next let A be the set ofwords [w, i], w E P, Z E G, and let N be the normal subgroup of F generated byA. As M 4 F, N = [M, fl :4 F and M/N < Z(F/N) by 8.5.2. Then by 28.6and 28.7 there is a unique homomorphism ir: F/N G, with (xN)7r =x forall x E G, and indeed M/N = ker(ir). Therefore (F/N, ir) is a central extensionof G.

Let (H, a) be a central extension of G. For X E G, let h(x) E H withh(x)or = x. Then, for x, y, z E G, w = h(x)h(y)h(xy)-1 E ker(a) < CH(h(z)),so [w, h(z)] = 1. Hence by 28.6 there exists a unique homomorphisma: F/N H with (xN)a = h(x) for each x E G. Notice a: (F/N, ir) - (H, a)is a morphism.

Now let G = (FIN)('). By 33.3, FIN= ker(7r)O and G is perfect. Hence(G, ir) is also a central extension of G, and a: (G, ir) (H, a) a morphism.Suppose ,B: (G, ir) (H, a) is a second morphism, and define y: G H byuy = ua(uf)-1, for u E G. Then aQ =7r =for, so Gy c Z(H). Thus (uv)y =

(uv)a((uv),B)-1 = uava(v,8)-1(u$)-1 = uavy(u,8)-1 = ua(u,8)-lvy =uyvy, so y is a homomorphism. Moreover Gy is abelian, so, as G is per-fect, y is trivial by 8.8.4. Thus a = P.


(33.2) If (6, n ) is a universal central extension of G then both 6 and G are perfect.

Proof. Let H = 6 x (6/6(')) and define a : H + G by ( x , y)a = x n . Then ( H , a ) is a central extension of G and ai: (6, n ) + ( H , a ) , i = 1,2, are morphisms, where xal = ( x , 1 ) and xa2 = ( x , ~ 6 " ) ) . So, by the uniqueness of such a morphism, a , = a2 and hence 6 = 6"). Thus 6 is perfect, so, by 8.8.2, G = 6n is perfect.

(33.3) Let G be perfect and ( H , n ) a central extension of G . Then H = ker(n)HC1) with H(') perfect.

Proof. By 8.8, H( ' )n = ( H n ) ( ' ) = G('), so, as G is perfect, H( ' )n = G. Hence H = ker(n)HC'). As ( H , n ) is acentralextension, ker(n) 5 Z ( H ) , so H / H ( ~ ) = z ( H / H ( ~ ) ) ( H ( ' ) / H ( ~ ) ) is abelian, and hence H(') = HC2) by 8.8.4. Thus H(') is perfect.

(33.4) G possesses a universal central extension if and only if G is perfect.

Proof. By 33.2, if G possesses a universal central extension then G is perfect. Conversely assume G is perfect. Let g F+ g be a bijection of G with a set (?

and let F be the free group on (?. Let r be the set of words X jilxy)-', x , y E G , and let M be the normal subgroup of F generated by r. Next let A be the set of words [w , Z ] , w E r, z E G , and let N be the normal subgroup of F generated by A. As M a F, N = [ M , F ] F and M I N 5 Z ( F / N ) by 8.5.2. Then by 28.6 and 28.7 there is a unique homomorphism n: F I N + G , with (XN)n = x for all x E G , and indeed M I N = ker(n). Therefore ( F I N , n ) is a central extension of G .

Let ( H , a ) be a central extension of G. For x E G , let h(x ) E H with h(x )a = x . Then, for x , y, z E G , w = h ( ~ ) h ( ~ ) h ( x ~ ) - ' E ker(a) 5 C ~ ( h ( z ) ) , so [w, h(z)] = 1. Hence by 28.6 there exists a unique homomorphism a : FIN+ H with (XN)a = h ( x ) foreachx E G . Noticea: ( F I N , n ) + ( H , a ) is a morphism.

Now let 6 = ( F I N ) ( ' ) . By 33.3, FIN = ker (n)6 and 6 is perfect. Hence ( 6 , n ) is also a central extension of G , and a : (6, n ) + ( H , a ) a morphism. Suppose p: ( 6 , n ) + ( H , a ) is a second morphism, and define y: + H by u y = ua(up)-', for u E 6. Then a a = n = p a , so GY Z ( H ) . Thus (uv )y = ( u v ) a ( ( u v ) ~ ) - ' = uava(v,!?)-' (up)-' = u a v y (up)-' = ua(up)-'v y = u y v y , so y is a homomorphism. Moreover 6 is abelian, so, as 6 is perfect, y is trivial by 8.8.4. Thus a = p.

(G, 7r) has been shown to be a universal central extension, so the proof iscomplete.

If G is a perfect group and (G, 7r) its universal central extension, then G iscalled the universal covering group of G and ker(7r) the Schur multiplier of G.Notice that, by 33.2, G is perfect.

A perfect central extension or covering of a perfect group G is a centralextension (H, a) of G with H perfect.

(33.5) Let (H, a) be a central extension of a group G, and (K, ,8) a perfectcentral extension of H. Then (K, Pa) is a perfect central extension of G.

Proof. ,8a: K G is the composition of surjective homomorphisms and hencea surjective homomorphism. Let X E ker(,da) and y E K. x,d E ker(a) <Z(H), so [x, y],8 = [x,d, yd] = 1. Thus [x, y] E ker(b) < Z(K). Thus [ker(da),K, K] = 1, so, by 8.9, ker(8a) < Z(K).

(33.6) Let (H, a) and (K, ,8) be central extensions of a group G with K perfect,and y: (H, a) - (K, ,8) a morphism of central extensions. Then (H, y) is acentral extension of K.

Proof. y: H K is a homomorphism with a = y,8. The latter fact impliesker(y) < Z(H), so itremains to show y is a surjection. As a = y,8 is a surjection,K = (Hy)ker(,8), so, as ker(b) < Z(K), Hy 4 K and K/Hy is abelian. HenceK = Hy as K is perfect.

(33.7) Let G be the covering group of a perfect group G and let (H, a) be aperfect central extension of G. Then a is an isomorphism.

Proof. Let ir: G G be the universal covering. By 33.5, (H, an) is a per-fect central extension of G, so by the universal property there is a morphism8: (G, 7r) (H, an). Then ,8a r =7r so by the uniqueness property of univer-sal extensions, ,8a = 1. Hence ,8: G H is an injection, while ,8 is a surjectionby 33.6. Thus fi is an isomorphism and as ,8a = 1, a =,8-1 is too.

(33.8) Let G be perfect, (G, 7r) the universal central extension of G, and (H, a)a perfect central extension of G. Then

(1) There exists a covering a: G H with r = aa.(2) (G, a) is the universal central extension of H.(3) The Schur multiplier of H is a subgroup of the Schur multiplier of G.


( e , n ) has been shown to be a universal central extension, so the proof is complete.

If G is a perfect group and (e, n ) its universal central extension, then is called the universal covering group of G and ker(n) the Schur multiplier of G. Notice that, by 33.2, is perfect.

A perfect central extension or covering of a perfect group G is a central extension (H, a ) of G with H perfect.

(33.5) Let (H, a ) be a central extension of a group G , and (K, p) a perfect central extension of H. Then (K, pa ) is a perfect central extension of G.

Proof. pa: K + G is the composition of surjective homomorphisms and hence a surjective homomorphism. Let x E ker(pa) and y E K. xp E ker(a) 5 Z(H), so [x, y]p = [xp, yp] = 1. Thus [ x , y] E ker(p) 5 Z(K). Thus [ker(pa), K, K l = 1, so, by 8.9, ker(pa) 5 Z(K).

(33.6) Let (H, a ) and (K, p) be central extensions of a group G with K perfect, and y: (H, a ) + (K, p) a morphism of central extensions. Then (H, y) is a central extension of K.

Proof. y: H 4 K is a homomorphism with a = yp. The latter fact implies ker(y) <_ Z(H), so it remains to show y is a surjection. As a = yp is a surjection, K = (H y)ker(p), so, as ker(p) 5 Z(K), H y 9 K and K/H y is abelian. Hence K = H y as K is perfect.

(33.7) Let e be the covering group of a perfect group G and let (H, a ) be a perfect central extension of e. Then a is an isomorphism.

Proof. Let n : e + G be the universal covering. By 33.5, (H, a n ) is a perfect central extension of G, so by the universal property there is a morphism p: (G, n ) + (H, an) . Then ,!?an = n so by the uniqueness property of universal extensions, pa = 1. Hence p: e + H is an injection, while ,3 is a surjection by 33.6. Thus p is an isomorphism and as pa = 1, a = p-' is too.

(33.8) Let G be perfect, ( e , n ) the universal central extension of G , and (H, a ) a perfect central extension of G. Then

(1) There exists a covering a : (? + H with n = aa. (2) (6, a ) is the universal central extension of H. (3) The Schur multiplier of H is a subgroup of the Schur multiplier of G.

(4) If Z(G) =1 then Z(G) is the Schur multiplier of G, and Z(H) = ker(a)ker(7r)/ker(a) is the quotient of the Schur multiplier of G by the Schur multi-plier of H.

Proof. By the universal property there exists a morphism a: (G, 7r) (H, a).Then 7r = as and, by 33.6, a is a covering. Let (H, ,B) be the universal cover-ing of H. By the universal property there is a morphism y: (H, ,B) (G, a).By 33.6 and 33.7, y is an isomorphism so (2) holds. Now (3) and (4) arestraightforward.

(33.9) Let G be a group with G/Z(G) finite. Then GM is finite.

Proof. Let n = IG/Z(G)I. For Z E Z(G) and g, h E G, [g, hz] = [g, h] =[gz, h], so the set 0 of commutators is of order at most n2. I claim each g E G(1)

can be expressed as a word g = xl ... x in the members of 0 of length in < n3.

The claim together with the finiteness of 0 show G(1) is finite.It remains to establish the claim. Pick an expression for g of minimal length

m. If m > n3 then, as I 0 < n2, there is some d E 0 with I' = f i : x1 = d} oforderk > n. Asxix1+l =xa+1xx `+' withxx`+' E 0 we can assume F = [1, ... , k}.Hence it remains to show that do+1 can be written as a product of n commu-tators, since then the minimality of m will be contradicted. Let d = [x, y]. AsIG/Z(G)I = n, do E Z(G), so d"+1 = (dn)Xd = (dn-')Xdxd = (dx)n-'[x2,

y]by 8.5.4. In particular do+1 is a product of n commutators.

(33.10) Let G be a perfect finite group. Then the universal covering group ofG and the Schur multiplier of G are finite.


(33.11) Let (H, a) be a perfect central extension of a finite group G, p a prime,and P E Sylp(H). Then P n ker(a) < ((P).

Proof. Passing to H/((D(P) nker(a)) we may assume 1(P) nker(a) = 1 andit remains to show x = P n ker(a) = 1. But as P/(D (P) = P* is elementaryabelian there is a complement Y* to X* in P*. Then P = X x Y so P splitsover X. Hence by Gaschiitz' Theorem, 10.4, H splits over X. Hence, as H isperfect and X < Z(H), X = 1.

(33.12) Let G be a perfect finite group and M the Schur multiplier of G. Then.7r(M) c 7r(G).

Central extensions

(4) If Z(G) = 1 then z((?) is the~churmulti~lier of G, and Z(H) = ker(a) S

ker(n)/ker(a) is the quotient of the Schur multiplier of G by the Schur multiplier of H.

Proof. By the universal property there exists a morphism a : (e, n ) + (H, a ) . Then n = aa and, by 33.6, a is a covering. Let (H, B) be the universal covering of H. By the universal property there is a morphism y: (fi, ,!?) + ((?, a). By 33.6 and 33.7, y is an isomorphism so (2) holds. Now (3) and (4) are straightforward.

(33.9) Let G be a group with G/Z(G) finite. Then G(') is finite.

Proof. Letn = IG/Z(G)I.Forz E Z(G)andg,h E G,[g,hz] = [g,h] = [gz, h], so the set A of commutators is of order at most n2. I claim each g E G(') can be expressed as a word g = XI . . . x, in the members of A of length m 5 n3. The claim together with the finiteness of A show G(') is finite.

It remains to establish the claim. Pick an expression for g of minimal length m. If m > n3 then, as 1 A1 5 n2, there is some d E A with r = {i: x, = d] of orderk > n . A s ~ , x , + ~ = x ~ + ~ x ~ + ' withx?+' E A wecanassumer = { I , . . . , k]. Hence it remains to show that dn+' can be written as a product of n commutators, since then the minimality of m will be contradicted. Let d = [x, y]. As IG/Z(G)I = n, dn E Z(G), so dn+' = (dn)xd = (dn-')xdxd = (dx)n-' [x2, y] by 8.5.4. In particular dn+' is a product of n commutators.

(33.10) Let G be a perfect finite group. Then the universal covering group of , G and the Schur multiplier of G are finite.


(33.11) Let (H, a ) be a perfect central extension of a finite group G, p a prime, and P E Syl,(H). Then P n ker(a) 5 @(P).

Proof. Passing to H/(@(P) n ker(a)) we may assume @(P) n ker(a) = 1 and it remains to show X = P n ker(a) = 1. But as P/@(P) = P* is elementary abelian there is a complement Y* to X* in P*. Then P = X x Y so P splits over X. Hence by Gaschiitz' Theorem, 10.4, H splits over X . Hence, as H is perfect and X 5 Z(H), X = 1.

(33.12) Let G be a perfect finite group and M the Schur multiplier of G. Then

n(M) S: n(G).

(33.13) Let (H, or) be a perfect central extension of a finite group G with ker(a )a p-group, let Go be a perfect subgroup of G containing a Sylow p-group of G,let Ho = Q-1(Go), and ao = or, H0: Ho Go. Then (Ho, ao) is a perfect centralextension of Go with ker(ao) = ker(o ). Hence a Sylow p-group of the Schurmultiplier of G is a homomorphic image of a Sylow p-group of the Schurmultiplier of Go.

Proof. Evidently (Ho, ao) is a central extension of Go with ker(a) = ker(ao),so, by 33.3, Ho = ker(a)Ho" with Hot) perfect. As Go contains a Sylow- p-group of G, Ho contains a Sylow p-group P of H and, as ker(a) is a p-group,ker(or) < P. Then, by 33.11, ker(a) < (D(P), so Ho = Thus P --0

P fl Ho = t(P)(P fl Ho1)) by the modular property, 1.14, so P = P fl Hot) <H' ) by 23.1. Thus Ho PHo1) = Hot>, so Ho is perfect, completing the proof.

(33.14) If G is a perfect finite group with cyclic Sylow p-groups then the Schurmultiplier of G is a p'-group.

Proof. Let (H, a) be a perfect central extension of G with ker(a) a p-group;I must show ker(or) = 1. Let P E Sylp(H), so that Z =ker(a) < P and P/Zis cyclic. By 33.11, Z < 4)(P), so PI 4)(P) is cyclic and hence, by 23.1, Pis cyclic. At this point I appeal to 39.1, which says some h E H induces anontrivial p'-automorphism on P. But then, by 23.3, [S21(P), h] 0 1, so, asZ < Z(H), z n S21(P) = 1. As P is cyclic and Z < P this says Z = 1.

The section on the generalized Fitting group focused attention on quasisimplegroups. Observe that the finite quasisimple groups are precisely the perfectcentral extensions of the finite simple groups. Hence, for each finite simplegroup G, the universal covering group of G is the largest quasisimple groupwith G as its simple factor, and the center of any such quasisimple group isa homomorphic image of the Schur multiplier of G. Thus it is of particularinterest to determine the covering groups and Schur multipliers of the finitesimple groups. This section closes with a description of the covering groupsand Schur multipliers of the alternating groups.

(33.15) Let G = A, n > 5, G the universal covering group of G, and 2 =Z(O) the Schur multiplier of G. Represent G on X = (1, ... , n). Then

(1) 2 = 716 if n = 6 or 7, while 2 =712 otherwise.


Proof. This is a consequence of 33.1 1.

(33.13) Let (H, a ) be a perfect central extension of a finite group G with ker(a) a p-group, let Go be a perfect subgroup of G containing a Sylow p-group of G, let Ho = a - ' ( ~ o ) , and a0 = a l ~ , : Ho + Go. Then (Ho, 00) is a perfect central extension of Go with ker(ao) = ker(a). Hence a Sylow p-group of the Schur multiplier of G is a homomorphic image of a Sylow p-group of the Schur multiplier of Go.

Proof. Evidently (Ho, 00) is a central extension of Go with ker(a) = ker(ao), so, by 33.3, Ho = ke r ( a )~ i ' ) with H:') perfect. As Go contains a Sylow p- group of G, Ho contains a Sylow p-group P of H and, as ker(a) is a p-group, ker(a) i P. Then, by 33.1 1, ker(a) ( @(P), so Ho = @(P)H:'). Thus P = P i l Ho = @(P)(P n H,'") by the modular property, 1.14, so P = P n H:') 5 H~'') by 23.1. Thus Ho = P H:') = H:'), so Ho is perfect, completing the proof.

(33.14) If G is aperfect finite group with cyclic Sylow p-groups then the Schur multiplier of G is a pl-group.

Proof. Let (H, a ) be a perfect central extension of G with ker(a) a p-group; I must show ker(a) = 1. Let P E Syl,(H), so that Z = ker(a) ( P and P / Z is cyclic. By 33.1 1, Z _( @(P), so P/@(P) is cyclic and hence, by 23.1, P is cyclic. At this point I appeal to 39.1, which says some h E H induces a nontrivial pl-automorphism on P. But then, by 23.3, [Q1(P), h] # 1, so, as Z(Z(H), Z i l Q1(P)=l . As P iscyclicand Z ( P this says Z = 1.

The section on the generalized Fitting group focused attention on quasisimple groups. Observe that the finite quasisimple groups are precisely the perfect central extensions of the finite simple groups. Hence, for each finite simple group G, the universal covering group of G is the largest quasisimple group with G as its simple factor, and the center of any such quasisimple group is a homomorphic image of the Schur multiplier of G. Thus it is of particular interest to determine the covering groups and Schur multipliers of the finite simple groups. This section closes with a description of the covering groups and Schur multipliers of the alternating groups.

(33.15) Let = A,, n > 5, e the universal covering group of e, and 2 = ~ ( e ) the Schur multiplier of e. Represent on X = (1, . . . , n). Then

(1) 2 G z6 if n = 6 or 7, while 2 E Z2 otherwise.

(2) Let i be a 2-element in G such that the image i of i in G is an involution.Then t has 2k cycles of length 2, i is an involution if k is even, and i is of order4 if k is odd.

(3) If n = 6 or 7 then 31+2 is a Sylow 3-group of G.

This result is usually proved using homological algebra, but I'll take a grouptheoretical approach here.

There are two parts to the proof. First show the order of the Schur multiplierof An is at least 2 (or 6 if n = 6 or 7). Second show the multiplier has order atmost 2 (or 6) and establish 33.15.2 and 33.15.3. Exercise 11.5 handles part one(unless n = 6 or 7 where the proof that 3 divides the order of the multiplier isomitted). The second part is more difficult and appears below.

Assume for the remainder of the section that (G, n) is a perfect centralextension of G = An and write S for the image Sn of S C G. Let Z = Z(G)and assume Z is a nontrivial p-group for some prime p. Let P E Sylp(G). Itwill suffice to show p < 3, IZI = p, and 33.15.2 and 33.15.3 hold. Assumeotherwise and choose G to be a counter example to one of these statementswith n minimal. The idea of the proof is simple: exhibit a perfect subgroup Hof G containing P, use the induction assumption to show the multiplier of His a p'-group, and hence obtain a contradiction from 33.13. When p = 2, andsometimes when p = 3, the situation is more complicated but the same generalidea works.

I begin a series of reductions.

(33.16) n > 2p.

Proof. By 33.14, P is not cyclic.

(33.17) n > 5.

Proof. Assume n = 5. By 33.16, p = 2. Assume I Z I = 2.P < H < G withH = A4, so P - E4 and H is transitive on P#. By 33.11, (D(P) = Z, so by1.13 there is an element of order 4 in P. Hence, as H is transitive on P#, everyelement in P - Z is of order 4. So, by Exercise 8.4, P is quaternion of order 8.Notice this gives 33.15.2 in this case, so it remains to show I Z I < 2. Now P =(g, h) so PM = (z) where z = [g, h]. If U is of index 2 in Z we've seen thatP/ U - Q8, so z V U. Thus Z/d)(Z) is cyclic so Z is cyclic. So, as Z/ U is theunique subgroup of P/ U of order 2, P also has a unique involution. Hence, byExercise 8.4, P is quaternion. So, as IP: Z(P) I = 4, P - Q8, That is I ZI = 2.


(2) Let i be a 2-element in 6 such that the image t of in 6 is an involution. Then f has 2k cycles of length 2, f is an involution if k is even, and 1 is of order 4 if k is odd.

(3) If n = 6 or 7 then 31+2 is a Sylow 3-group of 6.

This result is usually proved using homological algebra, but I'll take a group theoretical approach here.

There are two parts to the proof. First show the order of the Schur multiplier of A, is at least 2 (or 6 if n = 6 or 7). Second show the multiplier has order at most 2 (or 6) and establish 33.15.2 and 33.15.3. Exercise 11.5 handles part one (unless n = 6 or 7 where the proof that 3 divides the order of the multiplier is omitted). The second part is more difficult and appears below.

Assume for the remainder of the section that (G, n ) is a perfect central extension of e =A, and write for the image Sn of S E G. Let Z = Z(G) and assume Z is a nontrivial p-group for some prime p. Let P E Syl,(G). It will suffice to show p 5 3, IZI = p , and 33.15.2 and 33.15.3 hold. Assume otherwise and choose G to be a counter example to one of these statements with n minimal. The idea of the proof is simple: exhibit a perfect subgroup H of containing p, use the induction assumption to show the multiplier of H is a p'-group, and hence obtain a contradiction from 33.13. When p = 2, and sometimes when p = 3, the situation is more complicated but the same general idea works.

I begin a series of reductions.

Proof. By 33.14, p is not cyclic.

Proof. Assume n = 5. By 33.16, p = 2. Assume IZI = 2.P ( H ( 6 with H E A4, so p E E4 and H is transitive on P#. By 33.1 1, @(P) = Z, so by 1.13 there is an element of order 4 in P . Hence, as H is transitive on p#, every element in P - Z is of order 4. So, by Exercise 8.4, P is quaternion of order 8. Notice this gives 33.15.2 in this case, so it remains to show IZI ( 2. Now p = (8, A ) so P(') = (z) where z = [g, h]. If U is of index 2 in Z we've seen that P / U E Q g , so z 4 U. Thus Z/@(Z) is cyclic so Z is cyclic. So, as Z/U is the unique subgroup of P/ U of order 2, P also has a unique involution. Hence, by Exercise 8.4, P is quaternion. So, as I P : Z(P)I = 4, P E Q g , That is IZI = 2.

For S C G or G write Fix(S) and M(S) for the set of points in X fixed by Sand moved by S, respectively.

(33.18) Let 713 = Y < G with IM(Y)I = 3. Then Y = (y) x Z for some yof order 3, and if 8<n=,49 or 10 and p = 3 then 03' (CG (Y)) =-Y x L withL = An-3

Proof. Y < H - A5 with IM(H)I = 5. Without loss p = 3, so, by 33.17, H =Z x HO). Thus Y = (y) x Z where y E H(l). Also, if 8 < n 9 or 10, thenO3'(CG(Y)) _ (y) x k, k - An_3, and, by minimality of n, K = Z x K(l).

(33.19) If p = 3 then n > 6.

Proof. Assume p = 3 and n = 6. Then E9 - P = (y, g) with y and g moving3 points. By 33.18 we may take y and g of order 3. Let R = (y, g). Then Ris of class at most 2 so, by 23.11, R is of exponent 3. By 33.11, R = P andZ < t(P). PM = ([g, y]) so, as P is of exponent 3, Z = P(l) = Z(P). ThusP is extraspecial of order 27 and exponent 3, so, by 23.13, P - 31+2 . Henceall parts of 33.15 hold in this case.

I now consider two cases: Case I: n is not a power of p; Case II: n is a powerof p. In Case I P is not transitive on X so there is a partition 171, = [X 1, X2)of X with P acting on X 1 and X2. Let H be the subgroup acting on X 1. ThenH = (a)Ho with Ho = H1 X H2, where Hi = Ox,_, - An; is the subgroupof G fixing each point of X3_;, ni = IXi1, and (a) Hi - Sn; (unless nI or n2is 1). In case II, P is transitive on X and there is a partition °J' of X into psubset Xi, 1 < i < p, of order n/p, with P transitive on °J'. The subgroup ofG preserving ?P contains a subgroup H with P < H and H - (An/P)wr Apif p 2, while H is of index 2 in Sn12 wr Z2 if p = 2. Notice that, by 33.16,n/p > p.

Observe next that:

(33.20) One of the following holds:

(1) K = Op'(H) is perfect.(2) p = 2 and, in Case I, n1 > 1 < n2-(3) p = 3 and n1 or n2 is 3 or 4 in Case I.

Moreover, by minimality of n, Exercises 11.2 and 11.3, and 33.13, if k isperfect then p < 3, and if p = 3 and Case I holds, then n I or n2 is 6 or 7. Inparticular:


For S c G or G write Fix(S) and M(S) for the set of points in X fixed by S and moved by S, respectively.

(33.18) Let Z3 Z 5 G with I M ( ~ ) I = 3. Then Y = (y) x Z for some y of order 3, and if 8 5 n # 9 or 10 and p = 3 then 03'(cc(E)) =.Y x L with L E An-3.

Proof. E 5 H Z A5 with I M ( H ) I = 5. Without loss p = 3, so, by 33.17, H = Z x ~ ( ' 1 . Thus Y = (y) x Z where y E H('). Also, if 8 5 n # 9 or 10, then 03'(cG(y)) = (7) x K, K Z An-3, and, by rninimality of n, K = Z x K(').

(33.19) If p = 3 then n > 6.

Proof. Assume p = 3 and n = 6. Then E9 Z P = (9, g ) with y and g moving 3 points. By 33.18 we may take y and g of order 3. Let R = (y, g). Then R is of class at most 2 so, by 23.1 1, R is of exponent 3. By 33.1 1, R = P and Z 5 @(P). P(') = ([g, y]) so, as P is of exponent 3, Z = P(') = Z(P). Thus P is extraspecial of order 27 and exponent 3, so, by 23.13, P Z 31+2 . Hence all parts of 33.15 hold in this case.

I now consider two cases: Case I: n is not a power of p; Case 11: n is a power of p. In Case I P is not transitive on X so there is a partition P = (XI, X2) of X with P acting on XI and X2. Let H be the subgroup acting on XI. Then H = (ii)Ao with Ho = H I x Hz, where Hi = Gx,-< Z A , is the subgroup of G fixing each point of X3+, ni = /Xi 1, and (ii) Hi Z S,, (unless n 1 or n2 is 1). In case 11, P is transitive on X and there is a partition P of X into p subset Xi, 1 5 i 5 p, of order n lp , with P transitive on P. The subgroup of G preserving P contains a subgroup H with P 5 H and H Z (AnIp)wr A, if p # 2, while H is of index 2 in Sn12 wr Z2 if p = 2. Notice that, by 33.16,

nip 2 P. Observe next that:

(33.20) One of the following holds:

(1) K = OP'(H) is perfect. (2) p = 2 and, in Case I, nl > 1 < n2. (3) p = 3 and nl or n2 is 3 or 4 in Case I.

Moreover, by minimality of n, Exercises 11.2 and 11.3, and 33.13, if K is perfect then p 5 3, and if p = 3 and Case I holds, then nl or n2 is 6 or 7. In particular:

Central extensions

(33.21) p < 3 and if p = 3 and Case I holds then n1 or n2 is 3, 4, 6, or 7.

(33.22) p = 2.

173

Proof. If not p = 3. Suppose Case I holds. Then n1 or n2 is 3, 4, 6, or 7; sayn1. Then P acts on a subset Xi of X1 of order 3, so replacing X1 by Xi wemay take n1 = 3. But now 33.18 implies K = (y) x Z x L if 8 < n 0 9 or 10,contradicting 33.11. Hence n =7, 9, or 10. As n is not a power of 3, n 0 9,while if n = 10 we could have chosen n1 = 1, n2 = 9. Son = 7, where wechoose n1 = 1, so that A6 and, by 33.13 and minimality of n, IZ = 3 andP=31+2

So Case II holds. Here k = (a)Ko where Ko is the direct product of 3copies of An13, and a is of order 3 with IM(a)I =n. Then a = bg for someb E KO, and, by minimality of n and Exercise 11.2, K0 = Z x K0(1) if n > 9.If n = 9 then, by 33.18, K0 = (Z, gi: 1 < i < 3) with I M(gi) = 3. Furthergi E Z(O3'(CG(CG(gi)))) so K0 is abelian. Finally there is a 2-groupT = [T, a] < G with Ko = [K0, T], so, as K0 is abelian, 24.6 implies K0T =Z X O3(K0T) and a of course acts on O3(K0T). Let L = Kol) if n > 9 andL = O3(K0T) if n = 9. We can choose b E L, so that a is of order 3. But nowP = Z x ((P fl L)(a)) contradicting 33.11.

This leaves p = 2. The proof here is similar to the case p = 3. If n is odd thenCase I holds with n j = 1, so H = An-1 is perfect and 33.13 and minimality of ncomplete the proof. Thus n is even and H = Ho U with Ho = HO = H 1 X H2,where, in Case I, Hi = Ani, ni is even, and U = (u) = Z2, while, in Case II,n is a power of 2. Hi An/2 and U = (u, v) = E4. We can take Ug < Hofor some g E G. Let ni = n/2 in Case II. By minimality of n, Hi = Z * HI(1)

with Zi = HH(1) fl Z of order at most 2, unless ni < 4. If ni = 4 thenHi <KA5with IM(K)l=5,andK=Z*KU),soHi=Z*O2(Hi)withZi = 02(Hi) fl Z of order at most 2. Let Li = 02 (Hi) in either case. If ni > 4set Li = Hi(1)

If n j > 4 < n2 then there is yi ELI with I M(yi) =4 and yi = Y2 for somea E G. Notice Zi = (y?) so, as Z < Z(G), Z2 = Zl = Z1. Thus Z0 = Z fl L0 =Z1 where L0 = L 1 L2. This of course also holds if ni < 4 for some i, since thenni = 2 and Hi = 1. L0 char Ho so U acts on L0, and we may take Ug < L0, soZ fl U9 < Z1. Thus H = Z * ULo with Z fl UL0 = Z1, so, by 33.11, Z = Z1is of order 2. Thus 33.15.1 is established and it remains to establish 33.15.2.

Let t be a 2-element in G with I an involution in G. As G = An consists ofeven permutations on X, t has 2k cycles of length 2 on X for some positiveinteger k. Then there is a partition X = Yo U Yl U . . . U Yk of X with Y0 =Fix(t), I Yi I = 4 for 1 0 there

Central extensions

(33.21) p ( 3 and if p = 3 and Case I holds then nl or n2 is 3,4,6, or 7.

ProoJ If not p = 3. Suppose Case I holds. Then nl or n2 is 3 ,4,6, or 7; say nl. Then P acts on a subset X/, of X1 of order 3, so replacing X1 by X/, we may take nl = 3. But now 33.18 implies K = (y) x Z x L if 8 _( n # 9 or 10, contradicting 33.1 1. Hence n = 7, 9, or 10. As n is not a power of 3, n # 9, while if n = 10 we could have chosen nl = 1, n2 = 9. So n = 7, where we choosenl = 1, so that K 2 A6 and, by 33.13 and minimality of n, /Z l= 3 and P 2 3l+2.

So Case I1 holds. Here K = (ii) K O where K O is the direct product of 3 copies of An/3r and ii is of order 3 with I M(ii)l= n. Then ii = Jg for some b E KO, and, by minimality of n and Exercise 11.2, KO = Z x K;') if n > 9. If n = 9 then, by 33.18, KO = (Z, g,: 1 ( i (3) with IM(gi)l = 3. Further g, E z(03'(cG(cG(gi)))) SO KO is abelian. Finally there is a 2-group 7 = [T, ii] _( with K O = [KO, 71, so, as KO is abelian, 24.6 implies KoT = Z x O~(KOT) and a of course acts on 03(KoT). Let L = K;') if n > 9 and L = O ~ ( K ~ T ) if n = 9. We can choose b E L, so that a is of order 3. But now P = Z x ((P n L)(a)) contradicting 33.1 1.

This leaves p = 2. The proof here is similar to the case p = 3. If n is odd then -

Case I holds with nl = 1, so H 2 A,-1 is perfect and 33.13 and minimality of n complete the proof. Thus n is even and H = with H0 = H0 = H1 x Hz, where, in Case I, Hi 2 A , , ni is even, and = (ii) 2 Z2, while, in Case 11, n is a power of 2. Hi 2 An12 and u = (ii, rir) 2 E4. We can take Ug ( Ho for some g E G. Let ni = n/2 in Case 11. By minimality of n, Hi = Z * H,(') with Zi = H:') n Z of order at most 2, unless ni ( 4. If ni = 4 then Hi I_( K 2 A5 with IM(K)I = 5, and K = Z * ~ ( ' 1 , so Hi = Z * O'(H,) with Zi = 0 2 ( ~ i ) n Z of order at most 2. Let Li = 0 2 ( ~ i ) in either case. If ni > 4 set L~ = H:').

I fn l > 4 ( n z thenthereis yi € L i with IM(yi)l=4and yf = y2for some a E G. Notice Zi = (y:) SO, as Z ( Z(G), Z2 = Zf = Z1. Thus Zo = Z n Lo = Z1 where Lo = L1 L2. This of course also holds if ni < 4 for some i, since then ni = 2 and Hi = 1. Lo char Ho so U acts on Lo, and we may take Ug ( Lo, so Z n Ug ( Z1. Thus H = Z * U b with Z n ULo=Z1, so, by 33.11, Z = Z1 is of order 2. Thus 33.15.1 is established and it remains to establish 33.15.2.

Let t be a 2-element in G with t an involution in G. As G = A, consists of even permutations on X, t has 2k cycles of length 2 on X for some positive integer k. Then there is a partition X = Yo U Yl U . . . U Yk of X with Yo = Fix(t), I Yi I = 4 for 1 ( i ( k, and t acting on Yi . We've seen that for i > 0 there

is Ki fixing X - Y; pointwise with Ki - SL2(3) and Ki centralizing Gy. Thus(Ki: 1 < i < k) = Kl * * Kk is a central product with identified centers andt = tl . tk with ti E Ki. As Ki SL2(3), ti is of order 4 with (t?) = Z, sot isof order 4 if and only if k is odd. Hence the proof of 33.15 is at last complete.

Remarks. I'm not sure who shares credit for the notion of `components'. Cer-tainly Bender [Be 1], Gorenstein and Walter [GW], and Wielandt [Wi 1] shouldbe included. I use Bender's notation of E(G) for the subgroup generated bythe components of a group G. The Gorenstein and Walter notation of L(G) isalso used in the literature to denote this subgroup. I believe Bender [Be 1] wasthe first to formally define the generalized Fitting subgroup.

Thompson factorization was introduced by Thompson in [Th 1 ] and [Th 2],although 32.5 was presumably first proved by Glauberman [G1 2]. Thereare analogues of Thompson factorization for arbitrary finite groups G withF*(G) = Op(G). Such results require deep knowledge of the GF(p)-represen-tations of nearly simple groups. In particular one needs to know the pairs (G, V)with G a nearly simple finite group and V an irreducible GF(p)G-module suchthat GJ'(G, V) is nonempty. Generalized Thompson factorization plays an im-portant role in the classification; see in particular the concluding remarks insection 48. Thompson factorization for solvable groups is used in this bookto establish the Thompson Normal p-Complement Theorem, 39.5, and thenilpotence of Frobenius kernels (cf. 35.24 and 40.8).

Exercises for chapter 111. Let r be a prime, G a solvable r'-group, and A an r-group acting on G. Prove

(1) If r C 7r (G) and H is aHall 7r -subgroup of G, then Op(H) < Oph(G)for each p E r, where p" = r' U { p).

(2) If [A, F(G)] = 1 then [A, G] = 1.(3) If A centralizes a Sylow p-group of G then [A, G] < Op,(G).

2. Let (Gi : i E I) be perfect groups and Gi the universal covering group of Gi.Prove the universal covering group of the direct product D of the groupsGi, i E I, is the direct product of the covering groups 6j, i E I, and hencethe Schur multiplier of G is the direct product of the Schur multipliers ofthe groups Gi, i E I.

3. Let H and G be perfect groups with universal covering groups h and G,let H act transitively on a set I, and let W = G wr1 H. Prove the universalcovering group of W is the semidirect product of a central product of I Icopies of G with H, and the Schur multiplier of W is the direct productof the multipliers of G and H.

4. Let G be a finite group, p a prime, Q a G-invariant collection of p-subgroups of G, P a p-subgroup of G, and A C P n Q with A C Np(O).


is Ki fixing X - Yi pointwise with Ki S SL2(3) and Ki centralizing Gyi . Thus (Ki: 1 5 i 5 k) = K1 * . - a Kk is a central product with identified centers and t = tl . . . tk with ti E Ki. As Ki S SL2(3), ti is of order 4 with (ti2) = 2, SO t is of order 4 if and only if k is odd. Hence the proof of 33.15 is at last complete.

Remarks. I'm not sure who shares credit for the notion of 'components'. Cer- tainly Bender [Be 11, Gorenstein and Walter [GW], and Wielandt [Wi 11 should be included. I use Bender's notation of E(G) for the subgroup generated by the components of a group G. The Gorenstein and Walter notation of L(G) is also used in the literature to denote this subgroup. I believe Bender [Be 11 was the first to formally define the generalized Fitting subgroup.

Thompson factorization was introduced by Thompson in [Th 11 and [Th 21, although 32.5 was presumably first proved by Glauberman [Gl 21. There are analogues of Thompson factorization for arbitrary finite groups G with F*(G) = Op(G). Such results require deep knowledge of the GF(p)-representations of nearly simple groups. In particular one needs to know the pairs (G, V) with G a nearly simple finite group and V an irreducible GF(p)G-module such that P (G , V) is nonempty. Generalized Thompson factorization plays an important role in the classification; see in particular the concluding remarks in section 48. Thompson factorization for solvable groups is used in this book to establish the Thompson Normal p-Complement Theorem, 39.5, and the nilpotence of Frobenius kernels (cf. 35.24 and 40.8).

Exercises for chapter 11 1. Let r be aprime, G a solvable rl-group, and A an r-group acting on G. Prove

(1) If n E n(G) and H is a Hall n-subgroup of G, then Op(H) 5 Oph(G) for each p E n , where pn = n1 U {p}.

(2) If [A, F(G)] = 1 then [A, GI = 1. (3) If A centralizes a Sylow p-group of G then [A, GI I Opt(G).

2. Let (Gi : i E I ) be perfect groups and Gi the universal covering group of Gi . Prove the universal covering group of the direct product D of the groups Gi, i E I , is the direct product of the covering groups G ~ , i E I , and hence the Schur multiplier of G is the direct product of the Schur multipliers of the groups Gi , i E I.

3. Let H and G be perfect groups with universal covering groups H and e, let H act transitively on a set I , and let W = G wrl H . Prove the universal covering group of W is the semidirect product of a central product of 1 11 copies of with I?, and the Schur multiplier of W is the direct product of the multipliers of G and H.

4. Let G be a finite group, p a prime, !2 a G-invariant collection of p- subgroups of G, P a p-subgroup of G, and A E P n !2 with A E Np(A).

Prove that either A = P fl Q or there exists X E (P fl Q) - A with X <Np(O).

5. Let U be a 2-dimensional vector space over C with basis X = (x, y) anddefine a, $, and y to 1be the elements of GL(V) such that

Mx(a) = t 0 f , Mx($) = (0 ) , Mx(Y) _ (0 -O(

l) ,

where i is in C `with i2 = -1. Let V = Ul ® .. ®U be the tensor productof n copies UJ of U and let sk in Endc(V) be defined (cf. 25.3.5) as follows:

s2J =(1®...®1®((1®$)+(a(9 Y))®y(9 ...(, Y)l',

1<j<n,

S2J-1=(1®...®1®(a+$)®y(9 ...®Y)/I, 1 <j Vin,

where (1 (9 $) + (a (9 y) is in the (j, j + 1)-position of s2j and a + $ isin the j-position of s211. Prove(1) sk = I and (sksk+l)3 = -I for 1 < k < 2n.(2) (sksj)2 = -I, for k, j with lk - j l > 1.(3) Let G be the subgroup of GL(V) generated by (sk: 1 < k < 2n). Prove

G/(-I) = Stn with -I E GM, and GM is quasisimple if n > 2.6. Let G be a nonabelian simple group and G the universal covering group

of G. Prove Aut(G) - Aut(G).7. Let G be a solvable group with F(G) = Op(G) = T, let R be a p-subgroup

of G containing T, let (NG(C): 1 C char R) < H < G, let X =J(ROp,F(G)), V = (Q1(Z(R))x), andXR* =XR/CXR(V). Prove either(1) Op(H) = T, or(2) p < 3, X* = Xi x . x X* with X = SL2(p), R fl X E Sylp(X),

and NG(R) permutes {X1: 1 < i < n).8. Let G be a finite group, r a prime, r a set of primes with r V r, and

E,.3-A<G.For Er.2=B<Gdefine

a(B) = aG(B) _ ({On(CG(b)), B]: b E B#) n On(CG(b))bEB#

Assume for each component L of G that A < NG(L) and for each H < Gwith L a H that QAUt (L)(D) = 1 for each E,.2 = D < AutH(L). Provea(B) =1 for each E,.2 - B < A if On(G) = 1.

9. Let G be a finite group, p a prime, and H < < G. Prove Op(G) <NG(OP(H)).

10. Let F be a geometry of rank 2 and G a flag transitive group of automor-phism of r (cf. section 3). For X E F let ZX be the pointwise stabilizerin G of UyErx Fy. Assume for each x E F and Y E FX that GX is finite,ZX # 1, GX # Gy, and NG,(H) < GX for each nontrivial normal subgroup


Prove that either A = P n Q or there exists X E ( P n Q) - A with X (

NP(A). 5. Let U be a 2-dimensional vector space over C with basis X = (x, y) and

define a , B, and y to be the elements of GL(V) such that

whereiisinCwithi2 = [email protected]@U,bethetensorproduct of n copies Ui of U and let sk in Endc(V) be defined (cf. 25.3.5) as follows:

where (1 @ B) + (a @ y) is in the ( j , j + 1)-position of szj and a + B is in the j-position of s2j-1. Prove (1) S: = I and ( s ~ s ~ + ~ ) ~ = -I for 1 ( k < 2n. (2) (sksj)2 = -I , for k , j with ( k - j 1 > 1. (3) Let G be the subgroup of GL(V) generated by (sk: 1 5 k < 2n). Prove

G/ (- I) E S2, with -I E G('), and G(') is quasisimple if n > 2. 6. Let G be a nonabelian simple group and the universal covering group

of G. Prove Aut(G) S ~ u t ( c ) . 7. Let G be a solvable group with F(G) = Op(G) = T, let R be a p-subgroup

of G containing T, let (NG(C): 1 # C char R) ( H 5 G , let X = J(ROp,F(G)), V = (Q~(Z(R))'), and XR* = XR/CxR(V). Prove either (1) Op(H) = T, or (2) p (3, X* = X; x . .. x X,* with Xi* E SL2(p), R n X E Sylp(X),

and NG(R) permutes {Xi: 1 ( i ( n]. 8. Let G be a finite group, r a prime, n a set of primes with r 4 n, and

E,J A ( G. For Er2 E B ( G define

a(B) = ~ G ( B ) = ([On(CG(b)). Bl: b E B') n on(c~(b ) ) . b€B#

Assume for each component L of G that A ( NG(L) and for each H ( G with L 9 H that CY~,,,,(~)(D) = 1 for each Er2 E D ( AutH(L). Prove a(B) = 1 for each Er2 E B ( A if O,(G) = 1.

9. Let G be a finite group, p a prime, and H I! I! G. Prove O,(G) 5

NG(OP(H)). 10. Let I' be a geometry of rank 2 and G a flag transitive group of automor-

phism of r (cf. section 3). For x E T let 2, be the pointwise stabilizer in G of UYErx I',. Assume for each x E r and y E r, that G, is finite, Z, # 1, G, # G,, and NG, (H) 5 G, for each nontrivial normal subgroup

H of Gx contained in Gy. Prove there exists a prime p such that for eachx c F and y c Fx, F*(Gx) and F*(Gxy) are p-groups and either Zx or Zyis a p-group.(Hint: Let x c F, y C Fx, z C Fy, and q a prime such that Zx is not a q-group. For H < G set B(H) = Oq (H)E(H) and let Qx = Gx,rz . ProveZx <4 Gy and GyZ. Then use 31.4 and Exercise 11.9 to show O(Gy),9(GyZ) < Gx. Conclude:

(*) B(Gy) < 9(Gxy) < 9(Qy) < 9(Gy)

Then (interchanging the roles of x and y if necessary) conclude Zx is ap-group for some prime p. Using (*) show

9(Zy) < 0(Qx) < O(Gx,y) = 9(Qy)

and hence conclude F*(Zy) is a p-group. Finally use (*) to show O(Gx) _B(Gxy) = O(Gy) for each prime q 54 p. This proof is due to P. Fan.)

11. (Thompson [Th 3]) Prove there exists a function f from the positive inte-gers into the positive integers such that, for each finite set X, each primitivepermutation group G on X, each x c X, and each nontrivial orbit Y of Gxon X, either

(a) IGxI < f(IYI), or(b) F*(Gx) is a p-group for some prime p.Remark: Let y E Y, (Gx, Gy), and apply Exercise 11.10 to the actionof G on the geometry F(G, ). The Sims Conjecture says that a functionf exists with IGxI< f(IYI) even when F*(Gx) is a p-group. The SimsConjecture is established in [CPSS] using the classification.


H of G, contained in G,. Prove there exists a prime p such that for each x E r and y E r,, F*(G,) and F*(G,,) are p-groups and either Z, or Z, is a p-group. (Hint: Let x E r , y E r,, z E r,, and q a prime such that Z, is not a q- group. For H 5 G set 0(H) = O,(H)E(H) and let Q, = G,J~. Prove Z, gl G, and G,,. Then use 31.4 and Exercise 11.9 to show O(Gy), O(Gy,) 5 G,. Conclude:

Then (interchanging the roles of x and y if necessary) conclude Z, is a p-group for some prime p. Using (*) show

and hence conclude F*(Z,) is a p-group. Finally use (*) to show 0(G,) = 0(G,,) = O(Gy) for each prime q # p. This proof is due to P. Fan.)

11. (Thompson [Th 31) Prove there exists a function f from the positive integers into the positive integers such that, for each finite set X, each primitive permutation group G on X, each x E X, and each nontrivial orbit Y of G, on X, either

(a) lGxl 5 f(lYI19 or (b) F*(G,) is a p-group for some prime p. Remark: Let y E Y, 9 = (G, , G,), and apply Exercise 1 1.10 to the action of G on the geometry r(G, 9 ) . The Sims Conjecture says that a function f exists with lGxl 5 f (IYI) even when F*(G,) is a p-group. The Sims Conjecture is established in [CPSS] using the classification.

12

Linear representations of finite groups

Chapter 12 considers FG-representations where G is a finite group, F is asplitting field for G, and the characteristic of F does not divide the orderof G. Under these hypotheses, FG-representation theory goes particularlysmoothly. For example Maschke's Theorem says each FG-representation isthe sum of irreducibles, while, as F is a splitting field for G, each irreducibleFG-representation is absolutely irreducible.

Section 34 begins the analysis of the characters of such representations. Wefind that, if m is the number of conjugacy classes of G, then G has exactly mirreducible characters (Xi: 1 < i < m), and that these characters form a basisfor the space of class functions from G into F.

A result of Brauer (which is beyond the scope of this book) shows that,under the hypothesis of the first paragraph, the representation theory of G overF is equivalent to the theory of G over C, so section 35 specializes to thecase F = C. The character table of G over C is defined; this is the m by mcomplex matrix (Xi(gj)), where (gj: 1 < j < m) is a set of representativesfor the conjugacy classes of G. Various numerical relations on the charactertable are established; among these the orthogonality relations of lemma 35.5are most fundamental. The concepts of induced representations and inducedcharacters are also discussed. These concepts relate the representations andcharacters of subgroups of G to those of G. Induced characters and relationsamong characters on the one hand facilitate the calculation of the charactertable, and on the other make possible the proof of deep group theoreticalresults. Specifically chapter 35 contains a proof of Burnside's pagb-Theorem,which says a group whose order is divisible by just two primes is solvable,and of Frobenius' Theorem on the existence of Frobenius kernels in Frobeniusgroups.

Section 36 applies some of the results in section 35 and previous chap-ters to analyze certain minimal groups and their representations. This sectionis in the spirit of the fundamental paper of Hall and Higman [HH], whichshowed how many group theoretic questions could be reduced to questionsabout the FG-representations of certain minimal groups, particularly exten-sions of elementary abelian or extraspecial p-groups by groups of primeorder.

Linear representations of finite groups

Chapter 12 considers FG-representations where G is a finite group, F is a splitting field for G, and the characteristic of F does not divide the order of G. Under these hypotheses, FG-representation theory goes particularly smoothly. For example Maschke's Theorem says each FG-representation is the sum of irreducibles, while, as F is a splitting field for G, each irreducible FG-representation is absolutely irreducible.

Section 34 begins the analysis of the characters of such representations. We find that, if m is the number of conjugacy classes of G, then G has exactly m irreducible characters (x, : 1 5 i _( m), and that these characters form a basis for the space of class functions from G into F .

A result of Brauer (which is beyond the scope of this book) shows that, under the hypothesis of the first paragraph, the representation theory of G over F is equivalent to the theory of G over C, so section 35 specializes to the case F = C. The character table of G over Cis defined; this is the m by m complex matrix (x,(g,)), where (gj: 1 5 j 5 m) is a set of representatives for the conjugacy classes of G. Various numerical relations on the character table are established; among these the orthogonality relations of lemma 35.5 are most fundamental. The concepts of induced representations and induced characters are also discussed. These concepts relate the representations and characters of subgroups of G to those of G. Induced characters and relations among characters on the one hand facilitate the calculation of the character table, and on the other make possible the proof of deep group theoretical results. Specifically chapter 35 contains a proof of Burnside's paqb-Theorem, which says a group whose order is divisible by just two primes is solvable, and of Frobenius' Theorem on the existence of Frobenius kernels in Frobenius groups.

Section 36 applies some of the results in section 35 and previous chapters to analyze certain minimal groups and their representations. This section is in the spirit of the fundamental paper of Hall and Higman [HH], which showed how many group theoretic questions could be reduced to questions about the FG-representations of certain minimal groups, particularly extensions of elementary abelian or extraspecial p-groups by groups of prime order.

178 Linear representations of finite groups

34 Characters in coprime characteristicIn this section G is a finite group and F is a splitting field for G with character-istic not dividing the order of G. Let R = F[G] be the group ring of G over F.

(34.1) R is a semisimple ring and(1) R is the direct product of ideals (Ri: 1 < i < m) which are simple as

rings. In particular Ri Rj = 0 for i # j and 1 = Em 1 e; , where ei =1 Ri .(2) Ri is isomorphic to the ring Fn' "n' of all ni by ni matrices over F.(3) (Ri: 1 < i < m) is the set of homogeneous components of R, regarded

as a right module over itself.(4) Let Si consist of the matrices in Ri with 0 everywhere except in the first

row. Then (Si: 1 < i < m) is a set of representatives for the equivalence classesof simple right R-modules.

(5) F=EndR(Si) and dimF(S()=ni.

Proof. As char(F) does not divide the order of G, R is semisimple by Maschke'sTheorem. Hence we can appeal to the standard theorems on semisimple rings(e.g. Lang [La], chapter 17) which yield 34.1, except that Ri is the ring D!' xni

of matrices over the division ring Di = EndR(Ti) where Ti is a simple submod-ule of the ith homogeneous component Ri. But by hypothesis F is a splittingfield for G, so F = EndR(T) for each simple R-module T by 25.8, completingthe proof.

Throughout this section m, Ri, Si, ni, and ei will be as in 34.1. Notice thatIGI = dimF(R) dimF(Ri) n?, which I record as:

(34.2) IGI=Ymin?.

I have already implicitly used the equivalence between FG-modules and R-modules discussed in section 12, and will continue to do so without furthercomment.

Let (Ci: 1 < i < m') be the conjugacy classes of G, gi E Ci, and definezi E R by zi = F-g,c.g. It develops in the next lemma that m = m'.

(34.3) (1) The number m of equivalence classes of irreducible FG-representa-tions is equal to the number of conjugacy classes of G.

(2) (zi: 1 < i < m) and (ei: 1 < i < m) are bases for Z(R) over F.

Proof. Recall ei =1 R,. As R is the direct product of the ideals (Ri : 1 < i < m),Z(R) = ®m 1 Z(Ri) = ®m i Fei, with the last equality following from theisomorphism Ri = Fn' Xni and 13.4.1. Thus if I show (zi: 1 < i < m') = B is

178 Linear representations of Jinite groups

34 Characters in coprime characteristic In this section G is a finite group and F is a splitting field for G with characteristic not dividing the order of G. Let R = F [GI be the group ring of G over F.

(34.1) R is a semisimple ring and (1) R is the direct product of ideals (Ri: 1 5 i 5 m) which are simple as

rings. In particular Ri Rj = 0 for i # j and 1 = CyZl ei, where ei = 1 R ~ . (2) Ri is isomorphic to the ring Fni Xnl of all ni by ni matrices over F. (3) (Ri: 1 5 i 5 m) is the set of homogeneous components of R, regarded

as a right module over itself. (4) Let Si consist of the matrices in Ri with 0 everywhere except in the first '

row. Then (Si: 1 5 i 5 m) is a set of representatives for the equivalence classes of simple right R-modules.

(5) F = EndR(Si) and dimF(Si) = ni.

Proof. As char(F) does not divide the order of G, Ris semisimple by Maschke's Theorem. Hence we can appeal to the standard theorems on semisimple rings (e.g. Lang [La], chapter 17) which yield 34.1, except that Ri is the ring DF of matrices over the division ring Di = EndR(z) where Ti is a simple submodule of the ith homogeneous component Ri. But by hypothesis F is a splitting field for G, so F = EndR(T) for each simple R-module T by 25.8, completing the proof.

Throughout this section m, Ri, Si, ni, and ei will be as in 34.1. Notice that IG/ = dimF(R) = EYE1 dimF(Ri) = ELl n?, which I record as:

I have already implicitly used the equivalence between FG-modules and R- modules discussed in section 12, and will continue to do so without further comment.

Let (Ci: 1 5 i 5 m') be the conjugacy classes of G, g; E Ci, and define zi E R by zi = EgEc,g. It develops in the next lemma that m = m'.

(34.3) (1) The number m of equivalence classes of irreducible FG-representations is equal to the number of conjugacy classes of G.

(2) (zi: 1 5 i 5 m) and (ei: 1 5 i 5 m) are bases for Z(R) over F .

Proof. Recall ei = 1 R ~ . As R is the direct product of the ideals (Ri: 1 5 i 5 m), Z(R) = @Y="=,(Ri) = @y="=,ei, with the last equality following from the isomorphism Ri Z Fni xni and 13.4.1. Thus if I show (zi : 1 5 i 5 m') = B is

Characters in coprime characteristic 179

also a basis for Z(R), the proof will be complete. As G is a basis for R, B islinearly independent, so it remains to show B spans Z(R) over F.

Let z = L-gEG agg E R, ag E F. Then Z E Z(R) precisely when zh = z foreach h E G, which holds in turn precisely when ag = a(gh) for all g, h E G. Thusz E Z(R) if and only if ag = ai is independent of the choice of g E Ci, orequivalently, when z = F_ aizi. So indeed B spans Z(R) over F.

A class function on G (over F) is a function from G into F which is constanton conjugacy classes. Denote by cl(G) = cl(G, F) the set of all class functionson G and make cl(G) into an F-algebra by defining:

g(a -1- 0) = ga + g,8 a, ,B E cl(G),

g(aa) = a(ga) a E F, g E G,

g(ao) = ga go.

Evidently dimF(cl(G)) = m. By 14.8, if r is an FG-representation and X itscharacter, then X is a class function. Let ni be the representation of G on Siby right multiplication and Xi its character. Then (Tri: 1 < i < m) is a set ofrepresentatives for the equivalence classes of irreducible FG-representations.By 14.8, equivalent representations have the same character, so f Xi : 1 < i < m)is the set of irreducible characters of G over F; that is the set of all charactersof irreducible FG-representations.

The degree of a representation 7r is just the dimension of its representationmodule. As this is also the trace of the d by d identity matrix which is in turnX (1), it follows that:

(34.4) Xi (1) = ni is the degree of yri.

Each group G possesses a so-called principal representation of degree 1 inwhich each element of G acts as the identity on the representation moduleV. As dim(V) = 1, the principal representation is certainly irreducible. Byconvention 7rl is taken to be the principal representation. Hence:

(34.5) Subject to the convention that 7r1 is the principal representation, Xi (g) _n1=1forallgEG.

Observe that there is a faithful representation of the F-algebra cl(G) on R de-fined by (F agg)a = F ag(ga), for a E cl(G) and F agg E R. Recall also thatni is just the restriction to G of the representation of R on Si and Xi (r) =Tr(r7ri )for r E R. Finally notice that Xi (r) is also the value of Xi at r obtained from therepresentation of cl(G) on R.

Characters in coprime characteristic 179

also a basis for Z(R), the proof will be complete. As G is a basis for R, B is linearly independent, so it remains to show B spans Z(R) over F.

Let z = CgEG agg E R, a, E F. Then z E Z(R) precisely when zh = z for each h E G, which holds in turn precisely when a, = a(,h, for all g, h E G. Thus z E Z(R) if and only if a, = ai is independent of the choice of g E Ci, or equivalently, when z = C aizi. So indeed B spans Z(R) over F .

A class function on G (over F ) is a function from G into F which is constant on conjugacy classes. Denote by cl(G) = cl(G, F ) the set of all class functions on G and make cl(G) into an F-algebra by defining:

Evidently dimF(cl(G)) = m. By 14.8, if n is an FG-representation and x its character, then x is a class function. Let n, be the representation of G on Si by right multiplication and xi its character. Then (ni: 1 5 i 5 m) is a set of representatives for the equivalence classes of irreducible FG-representations. By 14.8, equivalent representations have the same character, so {xi: 1 5 i 5 m) is the set of irreducible characters of G over F ; that is the set of all characters of irreducible FG-representations.

The degree of a representation n is just the dimension of its representation module. As this is also the trace of the d by d identity matrix which is in turn ~ ( l ) , it follows that:

(34.4) xi (1) = ni is the degree of ni .

Each group G possesses a so-called principal representation of degree 1 in which each element of G acts as the identity on the representation module V. As dim(V) = 1, the principal representation is certainly irreducible. By convention nl is taken to be the principal representation. Hence:

(34.5) Subject to the convention that nl is the principal representation, xl (g) = n l = 1 for a l l g ~ G .

Observe that there is a faithful representation of the F-algebra cl(G) on R defined by (C a,g)a = C a,(ga), for a E cl(G) and C a,g E R. Recall also that ni is just the restriction to G of the representation of R on Si and xi (r) = Tr(rni) for r E R. Finally notice that xi (r) is also the value of xi at r obtained from the representation of cl(G) on R.

(34.6) (1) RJ7i=0=Xi(Rj)for i 0 j.

(2) Xi (Ri) = F for each i.(3) Xi (ei r) = Xi (r) for each i and each r E R. In particular Xi (ei) = Xi (1) = ni.

Proof. By 34.1 RiRj =0 for i 0 j, so (1) holds. As ei = 1R;, (1) implies (3).Part (2) is an easy exercise given the description of Si in 34.1.4.

(34.7) The irreducible characters form a basis for cl(G) over F.

Proof. As dimF(cl(G)) =m = number of irreducible characters, it suffices toshow (Xi: 1 < i < m) is a linearly independent subset of cl(G). But this isimmediate from 34.6.1 and 34.6.3.

The sum of FG-representations was defined in section 12. An immediate con-sequence of that definition is that Xa + Xp = Xa+p for FG-representations a and$ and their sum a + P, where Xy denotes the character of the representationy. Denote by char(G) the Z-submodule of cl(G) spanned by the irreduciblecharacters of G. char(G) is the set of generalized characters of G. As eachFG-representation is the sum of the irreducible representations ni, 1 < i < m,(cf. 12.10), it follows from the preceding remarks that each character of G is anonnegative Z-linear combination of the irreducible characters. Thus charac-ters are generalized characters, although by 34.7 not all generalized charactersare characters.

Further if (mi : 1 mi Xi

is the character of the representation >i mi ni. Finally, by Exercise 9.3, Xa®p =X«Xp, so the product of characters is a character. Hence char(G) is a Z-subalgebra of cl(G). These remarks are summarized in:

(34.8) The Z-submodule char(G) of cl(G) spanned by the irreducible charac-ters is a Z-subalgebra of G. The members of char(G) are called generalizedcharacters. The characters are precisely the nonnegative Z-linear span of theirreducible characters, and hence a subset of the generalized characters.

Representations and characters of degree 1 are said to be linear.

(34.9) Let G be an extraspecial p-group of order pl+2n and Z = Z(G) _ (z).Then

(1) G has per linear representations.(2) G has p - 1 faithful irreducible representations (P1, ... , (pp-1. Notation

can be chosen so that zOi acts via the scalar wi on the representation module

180 Linear representations ofjinite groups

(34.6) (1) Rini=0=xi(Rj)fori # j. (2) x i (Ri )=F foreachi. (3) xi(eir) = xi(r) for each i and eachr E R. In particular xi(ei) = xi(1) = ni.

Prooj By 34.1 Ri Ri = 0 for i # j, so (1) holds. As ei = lRi , (1) implies (3). Part (2) is an easy exercise given the description of Si in 34.1.4.

(34.7) The irreducible characters form a basis for cl(G) over F

Prooj As dimF(cl(G)) = m = number of irreducible characters, it suffices to show (xi: 1 5 i 5 m) is a linearly independent subset of cl(G). But this is .

immediate from 34.6.1 and 34.6.3.

The sum of FG-representations was defined in section 12. An immediate consequence of that definition is that X, + xg = xa+g for FG-representations a and B and their sum a t- B, where X, denotes the character of the representation y. Denote by char(G) the Z-submodule of cl(G) spanned by the irreducible characters of G. char(G) is the set of generalized characters of G. As each FG-representation is the sum of the irreducible representations ni , 1 5 i 5 m, (cf. 12.10), it follows from the preceding remarks that each character of G is a nonnegative Z-linear combination of the irreducible characters. Thus characters are generalized characters, although by 34.7 not all generalized characters are characters.

Further if (mi : 1 5 i 5 m) are nonnegative integers not all zero then Ci mi xi

is the character of the representation Ci mini. Finally, by Exercise 9.3, xa@g = xaxg, SO the product of characters is a character. Hence char(G) is a Z- subalgebra of cl(G). These remarks are summarized in:

(34.8) The Z-submodule char(G) of cl(G) spanned by the irreducible characters is a Z-subalgebra of G. The members of char(G) are called generalized characters. The characters are precisely the nonnegative Z-linear span of the irreducible characters, and hence a subset of the generalized characters.

Representations and characters of degree 1 are said to be lineal:

(34.9) Let G be an extraspecial p-group of order and Z = Z(G) = ( z ) . Then

(1) G has p2n linear representations. (2) G has p - 1 faithful irreducible representations 41, . . . , Notation

can be chosen so that zq5i acts via the scalar w k n the representation module

Characters in characteristic 0 181

Vi of O , where w is some fixed primitive pth root of 1 in F. The Ot arequasiequivalent for 1 < i < p - 1.

(3) O; is of degree p".(4) G has exactly pen + p - 1 irreducible representations: those described

in (1) and (2).(5) Let E be the enveloping algebra of 0, and Y a set of coset representatives

for Z in G. Then E = FP""p" and Y is a basis for E over F.

Proof. By Exercise 12.1, G has exactly IG/G(1)I linear representations andeach is irreducible. As G is extraspecial of order pl+2n G(' = Z and IG/ZI _p , so (1) holds.en

For x E G - Z, x G = xZ, so G has m = p2n - 1 + p conjugacy classes.As G has p2n irreducible linear representations, this leaves exactly p - 1nonlinear irreducible representations, by 34.3. Let 0 be such a representation.By Exercise 12.1, Z = G(1) ker((P). But Z is the unique minimal normalsubgroup of G, so ker((P) = 1. That is 0 is faithful. By 27.16, zO = a((P)I forsome primitive pth root of unity a((P), say a((P) = w. By Exercise 8.5, there is anautomorphism a of G of order p -1 regular on Z#. Let Oj = ai-1 0 , 1 < i < p.Then

-1) -1)zo; = z'

PO= w

(i-1) I

where za =z' and f j('-'): 1 < i <p}=[i:1 < i < p} mod p. So, renumber-ing, we may take z0i = w` I, 1 < i < p. Hence, for i # j, Ot is not equivalentto Off, so we have found our remaining p - 1 irreducibles, and established (2)and (4).

Let d be the degree of (P. As each qt is quasiequivalent to 0, it too has degreed. We now appeal to 34.2 to conclude p1+2n = IGI = p2n + (p - 1)d2, keepingin mind that our first pen irreducibles are linear. Of course (3) follows fromthis equality.

Finally let E be the enveloping algebra of (P. As pn = deg((P), E = Fp' XP'by 12.16. In particular dimF(E) = pen. Let Y be a set of coset representativesfor Z in G. Then (Y I = IGIZI = pen = dimF (E), so it suffices to show Y spansE over F. But for e E E, e = L+gEG ag(go), for some ag E F. Further eachg E G is of the form z` y for some y E Y and (z` y)O = w` yO with w` E F, soe = Eg ag(g(k) = >y by y, for some by E F, as desired.

35 Characters in characteristic 0The hypothesis and notation of the last section are continued in this section. Inaddition assume F = C.


of q5i, where w is some fixed primitive pth root of 1 in F . The q5i are quasiequivalent for 1 5 i 5 p - 1.

(3) q5i is of degree pn. (4) G has exactly p2n t- p - 1 irreducible representations: those described

in ( 1 ) and (2). (5) Let E be the enveloping algebra of q5i and Y a set of coset representatives

for Z in G. Then E E F P " ~ P " and Y is a basis for E over F.

Proof. By Exercise 12.1, G has exactly 1 GIG(') I linear representations and each is irreducible. As G is extraspecial of order G(') = Z and I G/Z I = p2", SO ( 1 ) holds.

For x E G - Z, xG = xZ, SO G has m = p2n - 1 + p conjugacy classes. As G has p2n irreducible linear representations, this leaves exactly p - 1 nonlinear irreducible representations, by 34.3. Let q5 be such a representation. By Exercise 12.1, Z = G(') $ ker(q5). But Z is the unique minimal normal subgroup of G, so ker(q5) = 1. That is q5 is faithful. By 27.16, zq5 = a(q5)I for some primitive pth root of unity a(@), say a(@) = w. By Exercise 8.5, there is an automorphism a of G of order p - 1 regular on z'. Let cPi = a'-'@, 1 5 i < p. Then

.(,-I)

zq5i = z J q5 = mj(8- ' ) I

where za = z j and {j('-'): 1 5 i < p} = {i: 1 5 i < p } mod p. So, renumbering, we may take zq5i = wi I , 1 5 i < p. Hence, for i # j, q5i is not equivalent to q5j, SO we have found our remaining p - 1 irreducibles, and established (2)

I

and (4). Let d be the degree of q5. As each q5i is quasiequivalent to q5, it too has degree

d. We now appeal to 34.2 to conclude = IGI = P2n + ( p - l )d2 , keeping in mind that our first p2n irreducibles are linear. Of course (3) follows from this equality.

Finally let E be the enveloping algebra of q5. As pn = deg(q5), E E FP" xp"

by 12.16. In particular dimF(E) = p2". Let Y be a set of coset representatives for Z in G. Then I Y I = I G/Z/ = p2n = dimF (E ) , so it suffices to show Y spans E over F . But for e E E, e = C g E G a g ( g 4 ) , for some ag E F. Further each g E G is of the form z i y for some y E Y and (ziy)q5 = wiyq5 with wi E F , so e = C, ag(gq5) = Cy by y, for some by E F , as desired.

35 Characters in characteristic 0 The hypothesis and notation of the last section are continued in this section. In addition assume F = @.


(35.1) FG-representations 7r and 0 are equivalent if and only if they have thesame character.

Proof. By 14.8, equivalent representations have the same character. Converselyassume 7r and q have the same character X. Now 7r = > mi7ri and 0 _ > ki7rifor some nonnegative integers mi, ki. It suffices to show m, = ki for each i. But

mi Xi = X = ki Xi, so, by 34.7, m i = ki for each i.

The regular representation of G is the representation of G by right multiplica-tion on R.

(35.2) Let ei = F(ai,g)g, ai,g E F. Then

ai,g =X(eig-')/IGI =niXi(g-')/IGI,where X is the character of the regular representation of G.

Proof. Observe X(eig-1) = X(F-h(ai,h)hg-1) = F-h ai,hX(hg-1). But, by

Exercise 12.2, X (x) = 0 if x 1 and X (1) = IGI , so X (ei g-1) = IGI (ai,g), yield-ing the first equality in the lemma. Next, by Exercise 12.2, X = F-m i ni Xi, so

m

X(eig-')= Y. njxj(eig 1) = nixi(eig-1) = niXi(g-')j=1

with the last two equalities holding by 34.6.1 and 34.6.3, respectively.

We now define a hermitian symmetric sesquilinear form (, ) on cl(G) withrespect to the complex conjugation map on C. (Recall c denotes the complexconjugate of c in C.) Namely for X, 9 E cl(G) define

(X'0)= (x()())g/IGI.It is straightforward to check that (, ) is hermitian symmetric and sesquilinear.Indeed the next lemma shows the form is nondegenerate.

(35.3) The irreducible characters form an orthonormal basis for the unitaryspace (cl(G), ( , )). That is (Xi, Xj) = Sip

Proof. By Exercise 9.4, X (g) = X (g-1) for each character X and each g E G.By 35.2:

Xi(ej)lnj = (xi(g)xi(g1))I , Xj)

gEG

Hence (Xi, X j) = Si j by 34.6.1 and 34.6.3.

182 Linear representations ofJinite groups

(35.1) FG-representations x and C#I are equivalent if and only if they have the same character.

Proof. By 14.8, equivalent representations have the same character. Conversely assume x and C#I have the same character X. Now n = C mixi and C#I = C kixi for some nonnegative integers mi, ki. It suffices to show mi = ki for each i. But C mi xi = x = C ki xi, SO, by 34.7, mi = ki for each i.

The regular representation of G is the representation of G by right multiplication on R.

(35.2) Let ei = C(ai,,)g, ai,, E F . Then

ai,, = ~ ( e i g - ' ) l l ~ l = n i ~ i ( g - ' ) l l ~ l ,

where x is the character of the regular representation of G.

Proof. Observe X(eig-') = ~(E~(a i , h )hg - ' ) = Ehai,hx(hg-'). But, by Exercise 12.2, ~ ( x ) = 0 if x + 1 and ~ ( 1 ) = I G I , SO ~(e ig- ' ) = IGI(ai,,), yield- ing the first equality in the lemma. Next, by Exercise 12.2, x = Ey=l niXi, SO

m - 1 x(eig-')= C n j x j ( e i g ) = nixi(eigF1) = niXi(gP1)

j=1

with the last two equalities holding by 34.6.1 and 34.6.3, respectively.

We now define a hermitian symmetric sesquilinear form ( , ) on cl(G) with respect to the complex conjugation map on @. (Recall (5. denotes the complex conjugate of c in @.) Namely for X , 8 E cl(G) define

It is straightforward to check that ( , ) is hermitian symmetric and sesquilinear. Indeed the next lemma shows the form is nondegenerate.

(35.3) The irreducible characters form an orthonormal basis for the unitary space (cl(G), ( , )). That is (xi, xi) = 6ij.

Proof. By Exercise 9.4, ji(g) = ~ ( g - ' ) for each character x and each g E G. By 35.2:

Hence (xi, xi) = Jij by 34.6.1 and 34.6.3.

Recall (Ci: 1 < i < m) are the conjugacy classes of G, gi E Ci, and by conven-tion g1 = 1. The character table of G (over C) is the m by m matrix (Xi (g j )).Thus the rows of the character table are indexed by the irreducible charac-ters of G and the columns by the conjugacy classes of G. In particular thecharacter table is defined only up to a permutation of the rows and columns,except that by convention Xl is always the principal character and g1 = 1.Subject to this convention, the character table has each entry in the first rowequal to 1, while the entries in the first column are the degrees of the irreducibleCG-representations.

(35.4) Let A be the character table of G and B the matrix (I Ci I g j (gi) / I G 1).Then B = A-1.

Proof. (AB)ij = Ek Xi(gk)ICk1Xj(gk)/IGI = XI(g)Xj(g))/IGI = (Xi,Xj) = 8ij. Therefore AB = I is the identity matrix. So A is nonsingular andB = A-1.

Observe that, by 5.12, I Ci I = IG: CG(gi)I, so ICi I/IGI = ICG(gi)I in lemma35.4.

(35.5) (Orthogonality Relations) Let hk = ICkI be the order of the kth conju-gacy class of G. Then the character table of G satisfies the following orthogo-nality relations:

n 0 if is j,(1) hkXi(gk)XJ(gk) _

k=1 IGI if i = j,

(2)

(3)

n 0 ifXi(gk)Xi(gi) =

i=1 ICG(gk)I =IGI/hk if k=l,n

IGI = n2,i=1

(4) niXi(gk)=0 if k > 1,i=1

n

(5) >hkXi(gk)=0 if i > 1,k=1

n

(6) > hk I Xi (gk) 12 = IGIk=1

Characters in characteristic 0

Recall (Ci : 1 5 i 5 m) are the conjugacy classes of G, gi E Ci, and by convention gl = 1. The character table of G (over C) is the m by m matrix (xi(gj)). Thus the rows of the character table are indexed by the irreducible characters of G and the columns by the conjugacy classes of G. In particular the character table is defined only up to a permutation of the rows and columns, except that by convention ~1 is always the principal character and gl = 1. Subject to this convention, the character table has each entry in the first row equal to 1, while the entries in the first column are the degrees of the irreducible CG-representations.

(35.4) Let A be the character table of G and B the matrix (ICi 1 X j (gi)/l G I). Then B = A-' .

Proof. (AB)ij = xk xi(gk)lCklXj(gk)/lGI = (CgEG xi(g)Xj(g))/lGI = (xi, xi) = Jij. Therefore AB = I is the identity matrix. So A is nonsingular and B = A-l.

Observe that, by 5.12, ICil = IG: CG(gi)l, SO ICil/lGI = l~G(gi)l-' in lemma 35.4.

(35.5) (Orthogonality Relations) Let hk = lCk 1 be the order of the kth conjugacy class of G. Then the character table of G satisfies the following orthogo-

; nality relations:

Proof. Part (1) is a restatement of 35.3, while part (2) is a restatement of35.4. Parts (3) and (4) are the special cases of (2) with k = l =1 and 1 = 1,respectively (using Xi (1) = ni). Similarly (5) and (6) are just (1) with j = 1 andj = i, respectively.

The orthogonality relations may be interpreted as follows. Part (1) says theinner product of the rows of the character table (weighted by the factor hk)is 0 or I G J, while part (2) says the inner product of the columns (twisted bycomplex conjugation) is 0 or IGI/hk.

(35.6) Let h, g E G. Then h c gG if and only if Xi (g) = Xi (h) for each i, 1 <i < n.

Proof. If h E gG then Xi (g) = Xi (h) since characters are class functions. Con-versely assume Xi (g) = Xi (h) for each i, but h 0 gG. By Exercise 9.7, Xi is alsoan irreducible character, so Xi (g) = Xi (h). Hence, by 35.5.2:

n n

0= EXi(g)X1(h)= EXj(g)Xj(g) =I CG(9)1i_i i_i

which is of course a contradiction.

At this point we'll need a few facts about algebraic integers. Recall an algebraicinteger is an element of C which is a root of a monic polynomial in 71[x]. Thefollowing facts are well known and can be found for example in chapter 9 ofLang [La].

(35.7) (1) The algebraic integers form a subring of C.(2) 71 is the intersection of the algebraic integers with' Q.(3) JNorm(z)J > 1 for each algebraic integer z 0 0.(4) An element c c C is an algebraic integer if and only if there exists a

faithful 7/[c]-module which is finitely generated as a 71-module.

(35.8) ni divides n for each i.

Proof. Let a =n/ni. By 35.2, ei = Eg1G(ai,g)g with ai,g = Xi(g)la. Henceaei = F-gcG Xi(g)g. As e?=ei, also aei = Xi(g)gei. By 27.13.1, X(g)is the sum of JGJ-th roots of unity for each character X and each g E G. Let Mbe the 71-submodule of R generated by the elements

g E G, is a IGI-th root of 1).


Proof. Part ( 1 ) is a restatement of 35.3, while part (2 ) is a restatement of 35.4. Parts (3) and (4 ) are the special cases of ( 2 ) with k = l = 1 and 1 = 1 , respectively (using ~ ~ ( 1 ) = ni ) . Similarly (5) and (6) are just ( 1 ) with j = 1 and j = i , respectively.

The orthogonality relations may be interpreted as follows. Part ( 1 ) says the inner product of the rows of the character table (weighted by the factor hk) is 0 or I G 1, while part (2) says the inner product of the columns (twisted by complex conjugation) is 0 or I G l/hk.

(35.6) Let h, g E G. Then h E gG if and only if x i (g ) = xi(h) for each i , 1 5 i F n .

Proof. If h E gG then x i (g ) = xi(h) since characters are class functions. Con- versely assume x i (g ) = xi(h) for each i, but h $ g C . By Exercise 9.7, ji is also an irreducible character, so j i ( g ) = j i (h ) . Hence, by 35.5.2:

which is of course a contradiction.

At this point we'll need a few facts about algebraic integers. Recall an algebraic integer is an element of C which is a root of a monic polynomial in Z[x]. The following facts are well known and can be found for example in chapter 9 of Lang [La].

(35.7) ( 1 ) The algebraic integers form a subring of C. (2) Z is the intersection of the algebraic integers with Q. (3) ]Norm(z)l 2 1 for each algebraic integer z # 0. (4) An element c E (C is an algebraic integer if and only if there exists a

faithful Z[c]-module which is finitely generated as a Z-module.

(35.8) ni divides n for each i .

Proof. Let a = n / n i . By 35.2, ei = CgEG(a i , , )g with ai,, = ;Xi(g)/a Hence aei = CgEG ji(g)g . AS e: = ei, also aei = CgGG ;Xi(g)gei. By 27.13.1, ~ ( g ) is the sum of IGI-th roots of unity for each character x and each g E G. Let M be the Z-submodule of R generated by the elements

( { g e , : g E G, { is a /GI-th root of 1 ) .

Then M is a finitely generated 71-module and Xi(h)ghei c M, as XZ(h) is the sum of IGI-th roots of unity. Hence M is a 71[a]-submodule of R, and certainly M is faithful as 7/[a] < C and R is a vectorspace over C. Therefore a is an algebraic integer by 35.7.4, and then a is aninteger by 35.7.2.

Define the kernel of X to be

ker(X) = {g E G: X(g) = X(1)}.

(35.9) Let X be the character of a CG-representation a and let g E G. Then(1) IX(g)/X(1)I < 1 > INorm(X(g)/X(1))I with equality if and only if

ga = coI for some root of unity to.(2) ker(X) = ker(a) < G.

Proof. By 27.13.1,X(g)= Y_'=, wi,wheren = deg(a) = X(1)andcoiis arootof unity. Thus IX(g)I < IwiI < X(1) with equality if and only if wi =cois independent of i. Similarly, if Norm(X) is the norm of X then

Norm(X) _ fl X°EE

where E is the set of embeddings of Q(X) into Q. But X° = Yi w°, so IX°IX (1) and hence

INorm(X)I/n < [1 IX°I/n < 1°

with equality if and only if coi = co for all i. Recall in this last case from theproof of 27.13 that as C is algebraically closed, ga =col. So (1) holds. Also ifX (g) = n then co =1 so ga = I. Hence (2) holds.

(35.10) Let zi = Y-gECj g, aii = hi Xj (gi)l nj, and bilk = 11(g, h): g E Ci, h ECj,gh=gk)I.Then

(1) zi = Ein=1 aijej,

(2) ail = Xj (zi)l nj,(3) aitap = Ek bifkak1,(4) ail is an algebraic integer.

Proof. As R =ED' 1 Rj, zi = Y'i zij for suitable zip E R. By 34.3, zi EZ(R), so zij E Z(Rj). Now Rj = Cnj xnj so Z(Rj) = Cep is 1-dimensional andconsists of the scalar matrices in Cnj xnj . Thus zit = ci1 ej for some cij in C,and Xj(Zi) = Xj(Y-k cikek) = Ek cikXj(ek) = cijnj by 34.6. Thus to prove (1)


Then M is a finitely generated Z-module and urgei = {gaei = ChEG jii(h) {ghei EM, as jii(h) is the sum of /GI-th roots of unity. Hence M is a Z[a]- submodule of R, and certainly M is faithful as Z[a] 5 C and R is a vector space over C. Therefore a is an algebraic integer by 35.7.4, and then a is an integer by 35.7.2.

Define the kernel of x to be

(35.9) Let x be the character of a CG-representation a and let g E G. Then (1) Ix(g)/x(l)I 5 1 3 INorm(x(g)/x(1))1 with equality if and only if

ga = wI for some root of unity w. (2) kerO() = ker(a) 2 G.

Proof. By 27.13.1, ~ ( g ) = Cy=, wi, wheren = deg(a) = ~ ( 1 ) a n d q is aroot of unity. Thus Ix(g)l I CyZl lwil 5 x(1) with equality if and only if wi = w is independent of i. Similarly, if Norm(x) is the norm of x then

where C is the set of embeddings of Q(x) into Q. But xu = Ci w:, so ]xu I 5 x (1) and hence

with equality if and only if wi = w for all i. Recall in this last case from the proof of 27.13 that as C is algebraically closed, ga = w I . So (1) holds. Also if x (g) = n then w = 1 so ga = I . Hence (2) holds.

(35.10) Let zi =: CgEc, g, aij = hixj(gi)/nj, and bijk = I{(g, h): g E Ci, h E

Cj,gh=gk)l.Then (1) Zi = CyZl aijej, (2) aij =xj(zi)/nj, (3) ailajl = Ck bijkaklr (4) aij is an algebraic integer.

Proof. As R = Rj, zi = zij for suitable zij E Rj. By 34.3, zi E

Z(R), so zi j E Z(Rj). Now Rj = Cnj '"j so Z(Rj) = Cej is I-dimensional and consists of the scalar matrices in C n l X n ~ . Thus zij = cijej for some cij in C, and xj(zi) = xj (Ck cikek) = Ck cikxj(ek) = Cijn j by 34.6. Thus to prove (1)

and (2) it remains to show ai j = c; j . But ci j n j = X j (zi) = X j (Y-gEc, g) _Y-gEC, Xj(g) = hi xj(gi), so the proof of (1) and (2) is complete.

Next zizj e Z(R), so, as (zi: 1 < i < m) is basis of Z(R) by 34.3.2, ziz j =Y-k CijkZk. Also

zi

z j

= \gE g/ \gE h gE ghEcihECj

and, in the sum on the right, the coefficient of gk is bi jk. As zizj is a linearcombination of the Zk, and as G is a basis for R, the coefficient of x E Ck inthe sum on the right is equal to that of gk and that coefficient is Cijk. That isbi jk = Ci jk. Next

Zizj = t Zik ZilZjl\ k i k,I i

as RkRI = 0 for k 0 1. Also zilzjl = ailelajlel = ailajiei so zizj =Y-i ailajlei. On the other hand zizj = Y-k bijkZk = Y-k,l bijkaklel by the lastparagraph and (1), so Y-kbijkakl=ailaji, as (ei:1 < i < m) is a basis forZ(R). Hence (3) holds.

Fix (i, 1) and set a = ai,i. Then

(*) as j,l = E bi jkak,i 1 < j < n.k

Consider the following system of n equations in n unknowns x = (xi, ... , x,):

("") 0 = E bijkxk + (bi jj - a)xj 1 < j < n.k#j

Observe (**) has the solution a = (al,l, , a.,l) by (*). xi(gi) = ni 0 0, soal,i = hiXi(gl)/ni 0 0. So, as (0, ... , 0) is also a solution to (**), the matrixM of (**) is of determinant 0. Consider the matrix N with entries in l[x]obtained by replacing a by x in M. (Observe that by definition the bijks arenonnegative integers.) Let f (x) = det(N) E l[x]. Then f (a) = 0 and f is amonic polynomial, so a is an algebraic integer. Thus (4) holds.

(35.11) Let (ni, hj) = 1. Then either

(1) Xi(gj) = 0, or(2) Ixi(gj)I =ni, so gjker(xi) E Z(G/ker(Xi))

Proof. By 27.13.1 and 35.7.1, a = Xi (g j) is an algebraic integer, as is b = ai j =h ja/ni by 35.10.4. Assume a 0 0 and let f be the minimal polynomial of aover Q. Say f (x) _ yd_k=0 akxk, ak E Q. Now the field extensions 0(a) and0(b) are equal as r = hj/ni is in Q. So the minimal polynomial f (x) of bover 0 is also of degree d. As b is a root of g(x) = k=i akrd-kxk it follows


and (2) it remains to show aij = Cij. But Cijnj = xj(zi) = xj(CgeCi g) = xgECi xj(g) = hixj(gi), SO the proof of (1) and (2) is complete.

Next zizj E Z(R), so, as (zi: 1 5 i 5 m) is basis of Z(R) by 34.3.2, zizj = xk CijkZk- Also

and, in the sum on the right, the coefficient of gk is bijk. As zizj is a linear combination of the zk, and as G is a basis for R, the coefficient of x E Ck in the sum on the right is equal to that of gk and that coefficient is cijk. That is bijk = cijk. Next

as RkRl = 0 for k # 1. A~SO Zi lZj l =: ailelajlel = ailajlel So Z i Z j = xi ailajlel. On the other hand zizj = xk bijkzk = xk,l bijkaklei by the last paragraph and (I), so xk bijkakl = ailajl, as (ei: 1 ( i 5 m ) is a basis for Z(R). Hence (3) holds.

Fix (i, 1) and set a = ai,l. Then

Consider the following system of n equations in n unknowns x = (XI, . . . , x,):

Observe (**) has the solution a = ( a ~ , ~ , . . . , a,,l) by (*). xl(g1) =nl # 0, so a1.l = hlxl(gl)/nl # 0. So, as (0, . . . , 0) is also a solution to (**), the matrix M of (**) is of determinant 0. Consider the matrix N with entries in Z[x] obtained by replacing a by x in M. (Observe that by definition the bijks are nonnegative integers.) Let f (x) = det(N) E Z[x]. Then f (a) = 0 and f is a monic polynomial, so a is an algebraic integer. Thus (4) holds.

(35.11) Let (ni, h j) = 1. Then either

Proof. By 27.13.1 and 35.7.1, a = xi(gj) is an algebraic integer, asis b = aij = hja/ni by 35.10.4. Assume a # 0 and let f be the minimal polynomial of a over Q. Say f (x) = x f = o akxk, ak E Q. Now the field extensions Q(a) and Q(b) are equal as r = hj/ni is in Q. So the minimal polynomial f (x) of b over Q is also of degree d. As b is a root of g(x) = ~ f = ~ akrdPkxk, it follows

that g = f. As b is an algebraic integer, g e l[x], so akrd-k is an integer. Asr = hi /ni with (hj, ni) = 1, it follows that nd-k divides ak for each k. Hence

d

h(x) (akl rid k)xk E 7L[x].k=1

But a/ni is a root of h(x) and h(x) is monic, so a/ni is an algebraic integer. Onthe other hand (Norm (a/ni)l < 1 by 35.9.1. Thus (Norm (a)I =ni by 35.7.3,so, by 35.9.1, gj 7ri = wi. Therefore (2) holds, completing the proof.

(35.12) Assume hj = pe is a prime power for some j > 1 and G is simple.Then G is of prime order.

Proof. We may assume G is not of prime order. Then, as G is simple, Z(G) =1and ker(Xi) =1 for each i > 1. So, by 35.11, for each i, j > 1 either Xi (gj) = 0or p divides ni. Next, by the orthogonality relations:

n

0niXi(gj)=1+pci=1

for some algebraic integer c. Hence c = -p-1, impossible as -p-1 is not analgebraic integer, by 35.7.2.

(35.13) (Burnside's pagb-Theorem). Let I GI = pagb, with p, q prime. ThenG is solvable.

Proof. Let G be a minimal counterexample. If 1 0 H a G then G/H and Hare of order less then G and both are {p, q}-groups, so, by minimality of G,each is solvable. But now 9.3.2 contradicts the choice of G. So G is simple.

Let Q E Sylq (G). If Q =1 then G is a p-group and hence G is solv-able. So Q 0 1, so in particular there is gj E Z(Q)#. Then Q <CG(gj), sohj _ I G : CG (gj) I divides J G : Q J = pa . So, by 35.12, G is of prime order, andhence G is solvable.

(35.14) Let i/r E cl(G). Then

(1) 1r = Xi")`Xi, and

(2) (Y', Vf) = yin 1(Y', Xi)2.

Proof. As the irreducible characters are a basis for cl(G), * = Em t ai Xifor some complex numbers ai. By 35.3, (i, Xj) aiXi, Xi) =

Eiai(Xi,Xj)=ai,so(1)holds. Similarly riaiaj(Xi,Xj)= Y_ia?,so (2) holds.

We next consider induced representations. Let H < G, F a field, and a an FG-representation. It will be convenient to regard the image ha of h e H under


that g = f . As b is an algebraic integer, g E Z[x], so akrd-k is an integer. As r = hj/ni with (hj, ni) = 1, it follows that ntVk divides ak for each k. Hence

But a/ni is a root of h(x) and h(x) is monic, so a/ni is an algebraic integer. On the other hand ]Norm (a/ni)l 5 1 by 35.9.1. Thus INorm (a)[ =ni by 35.7.3, so, by 35.9.1, gjni = w l . Therefore (2) holds, completing the proof.

(35.12) Assume hi = pe is a prime power for some j > 1 and G is simple. Then G is of prime order.

Proof. We may assume G is not of prime order. Then, as G is simple, Z(G) = 1 and ker(xi) = 1 for each i > 1. So, by 35.1 1, for each i, j > 1 either xi(gj) = 0 or p divides ni. Next, by the orthogonality relations:

for some algebraic integer c. Hence c = -pP1, impossible as -p-' is not an algebraic integer, by 35.7.2.

(35.13) (Burnside's paqb-Theorem). Let [GI = paqb, with p , q prime. Then G is solvable.

Proof. Let G be a minimal counterexample. If 1 # H 9 G then G/H and H are of order less then G and both are {p, q}-groups, so, by minimality of G, each is solvable. But now 9.3.2 contradicts the choice of G. So G is simple.

Let Q E Sylq(G). If Q = 1 then G is a p - ~ o u p and hence G is solvable. So Q # 1, so in particular there is g j E z(Q)#. Then Q 5 CG(gj), so hi = IG: CG(gj)l divides IG: Ql =pa . So, by 35.12, G is of prime order, and hence G is solvable.

(35.14) Let @ E cl(G). Then (1) @ = ELI (@, ~ i ) ~ i , and (2) (@, @) = Ey=l(@y xi)'.

Proof. As the irreducible characters are a basis for cl(G), @ = xy=l aiXi for some complex numbers ai. BY 35.3, (@, xj) = ( x i aiXi, xj) =:

xi ai(xi, xj) =ai, SO (1) holds. Similarly (@, @) = xi aiaj()(i, xi) = Xi a?, so (2) holds.

We next consider induced representations. Let H 5 G, F a field, and a an FG-

a as a matrix rather than a linear transformation. Extend a to G by definingga = 0 for g E G - H. Let X = (x1: 1 < i < n) be a set of coset representativesfor H in G; hence n = IG : H1. For g E G define gaG = ((xigx 1)a) to be then by n matrix whose (i, j)-th entry is the deg(a) by deg(a) matrix (xjgxj-1)a.We can also regard gaG as a square matrix of size deg(a)n over F. If we takethis point of view then:

(35.15) aG is an FG-representation of degree deg (a) I G : H1.

Proof. Let u, v E G, A = uaG, B = vac, and C = (uv)aG. We must showAB = C. Regarding A and B as n by n matrices with entries from Fdxd (where

d = deg(a)) we have:

(AB)ij = AikBkj = (xiuxk 1)a(xkvx,j 1)a.k k

Most of the terms in the sum on the right are 0, since (xi uxk 1)a = 0 un-less xiuxk 1 E H. But xiuxkl E H precisely when Hxiu = Hxk, so (AB)ij = 0unless there exists some k with Hxi u = Hxk and Hxkv = Hxi. This holdsprecisely when Hxi u v = Hxj, and in that event there is a unique k withthe property; namely k is defined by Hxk = Hxiu. Moreover in this case(xi uxk 1)a (xk vxJ 1)a = (xi u vxj 1)a = C13 . Thus (AB )ij = C11 if Hxi u v =

Hx j and (AB)1 j = 0 otherwise. Finally if Hxi u v Hx1 then xi u vxj 1 V H, so

Ci j = (xi uvxi 1)a = 0 = (AB)i1. Hence AB = C, so aG is a homomorphism,and the proof is complete.

aG is called the induced representation of a to G.

(35.16) (1) Up to equivalence, aG is independent of the choice of coset rep-resentatives for H in G.

(2) If a and ,8 are equivalent FH-representations then aG and pG are equiv-alent FG-representations.

Proof. Let Y = (yi: 1 < i < n) be a second set of coset representatives forH in G. Then yi = hixi for some hi E H. Now, if 0 is the induced represen-tation of a to G defined with respect to Y, then (gO)ij = (hixigxj 'hi 1)a =(hia)(xigxj_1)a(hja)-1. Let B be then by n diagonal matrix over Fdxd withBii = hi a. Then (gO) = (gaG)B, so, by a remark at the beginning of section 13,0 is equivalent to aG as an FG-representation.

Similarly if a is equivalent to ,B there is D E Fd xd with (ha)D = hO for eachh E H. Let E be the n by n diagonal matrix over Fdxd with Eii = D for eachi. Then (gaG)E = g1G for each g E G, so (2) holds.


a as a matrix rather than a linear transformation. Extend a to G by defining g a = 0 for g E G - H. Let X = (xi : 1 5 i 5 n) be a set of coset representatives for H in G; hence n = I G : HI. For g E G define gaG = ((xigxil)a) to be the

n by n matrix whose (i, j)-th entry is the deg(a) by deg(a) matrix (xigxjl)a. We can also regard gaG as a square matrix of size deg(a)n over F. If we take this point of view then:

(35.15) aG is an FG-representation of degree deg (a)lG : HI.

Proof. Let u, v E G, A = uaG, B = vaG, and C = (uv)aG. We must show AB = C. Regarding A and B as n by n matrices with entries from F~~~ (where d = deg(a)) we have:

Most of the terms in the sum on the right are 0, since (xiuxil)a = 0 unless xiuxkl E H. But xiuxkl E H precisely when Hxiu = Hxk, SO (AB)ij = 0 unless there exists some k with Hxiu = Hxk and Hxkv = Hxj. This holds precisely when Hxiuv = Hxj, and in that event there is a unique k with the property; namely k is defined by Hxk = Hxiu. Moreover in this case (xi uxF1)a . (xkvxil)a = (xi uvxrl)a = cij. ~ h u s ( A B ) ~ ~ = cij if H X ~ uv =

Hxj and (AB)ij = 0 otherwise. Finally if Hxi uv # Hxj then xi uvx j1 $ H, so

Cij = (xiuvxil)a = 0 = (AB)ij. Hence AB = C, so aG is a homomorphism, and the proof is complete.

aG is called the induced representation of a to G.

(35.16) (1) Up to equivalence, aG is independent of the choice of coset representatives for H in G.

(2) If a and B are equivalent FH-representations then aG and BG are equivalent FG-representations.

Proof. Let Y = (yi: 1 5 i I: n) be a second set of coset representatives for H in G. Then yi = hixi for some hi E H. Now, if 8 is the induced representation of a to G defined with respect to Y, then = (h ix igx~ 'h j l )a =

(hia)(xigxi1)a(hja)-'. Let B be the n by n diagonal matrix over F~~~ with Bii = hia. Then (go) = so, by a remark at the beginning of section 13, 8 is equivalent to aG as an FG-representation.

Similarly if a is equivalent to B there is D E Fd Xd with (ha)D = h8 for each h E H. Let E be the n by n diagonal matrix over Fdxd with Eii = D for each i. Then (gaG)E = gBG for each g E G, so (2) holds.

(35.17) Let x be the character of a and extend x to G by defining x(g) = 0for g E G - H. Then

(1) The character X G of the induced representation aG is defined by

xG(S) = Y, x(Sv)) IHI.(VEG

(2) XG(g) =0 if g Hv for some v E G.

Proof XG(g) = Tr(aG) =1

Tr((gX'')a) = Ei=1 X(Sx' i). Moreoverx (gx; ') = 0 unless g E Hxi , so (2) holds. Finally G = U" 1 Hxi and, as x is aclass function, x(gxi ") = x(gx+ ') for h E H, so EyEHX; X(Sy t) = I H I X(Sx'

t )

Hence (1) holds.

X G is called the induced character of x to G.

(35.18) Let H < K < G and x a character of H. Then XG = (xK)G

Proof. Let 0 = XK. Then, for g E G,

eG(S) = (9gv) /IK- /IKIIHI.vEG gEG

UEK

Further the map (v, u) H vu is a surjection of G x K onto G whose fibres areof order IKI, so OG(g)=(E,,,EG X(Sw))/IHI = XG(8)

For * E cl(H) define t/,G: G -- F by 1/,G(g) = (EvEG *(gv))/IHI, where asusual *(x) = 0 for x E G - H. If ,/r is a character of H then, by 35.17 , G

is a character of G, and evidently the map * H ,/rG preserves addition andscalar multiplication, so as the irreducible characters form a basis for cl(H) weconclude:

(35.19) The map i/r H ,/1G is a linear transformation of cl(G) into cl(H)which maps characters to characters and generalized characters to generalizedcharacters.

The map i/r H 1116 is called the induction map of cl(G) into cl(H). Notice that35.17 and the definition of ,/1G show:

(35.20) Let * E cl(H). Then(1) *G(g)=0forgEG-(UEGH"),and(2) *G(1)=IG:HI,/r(1).


(35.17) Let x be the character of a! and extend x to G by defining ~ ( g ) = 0 for g E G - H. Then

(1) The character X G of the induced representation aG is defined by

Proof. xG(g) = ~ r ( a ~ ) = C:='=, ~r((g"i-l)a) = C;='=, x(~";'). Moreover X(gxi-l) = 0 unless g E H", so (2) holds. Finally G = U;='=, Hxi and, as x is a class function, x(gf I") = X(g-l) for h E H, so Z,,,, x(gy-I = I H I ~(g";'). Hence (1) holds.

X G is called the induced character of x to G.

(35.18) Let H 5 K 5 G and x a character of H. Then xG = (XK)G.

Proof. ~ e t 8 = x K . hen, for g E G,

Further the map (v, u) H vu is a surjection of G x K onto G whose fibres are

of order IKI, so OG(g) = (C,,, x(gW))/IHI = xG(g).

For $r E cl(H) define $rG: G -+ F by $rG(g) = (CVEG $r(gV))/I HI, where as usual $r(x) = 0 for x E G - H. If + is a character of H then, by 35.17 , $rG is a character of G, and evidently the map $r H $rG preserves addition and scalar multiplication, so as the irreducible characters form a basis for cl(H) we conclude:

(35.19) The map $r H $rG is a linear transformation of cl(G) into cl(H) which maps characters to characters and generalized characters to generalized characters.

The map $r H $rG is called the induction map of cl(G) into cl(H). Notice that 35.17 and the definition of $rG show:

(35.20) Let $r E cl(H). Then

(1) $rG(g) = O for g E G - (UVEG HV), and (2) $rG(l)= IG: Hl$r(l).


Recall that there are hermitian symmetric forms (, )H and (, )G defined oncl(H) and cl(G), respectively.

(35.21) (Frobenius Reciprocity Theorem) Let 1 E cl(H) and X E cl(G). Then

(V',XIH)H =(V'G,X)G

Proof.('G, X)c = Y *G(g)X(g) IGI

gEG

= CY- *(g"WO IGIIHIU

_ j(gv)X(gv) /IGHIv,gEG

since X is a class function. As the map g i-, g" is a permutation of G for eachv E G, it follows that (1G, X)G = (F-gEG *(g)X(g))/I HI. As fi(g) = 0 forg E G - H, this sum reduces to (EhEH 1(h)2(h))/I HI = (v', X I H)H

A subset T of G is said to be a TI-set in G if T n T9 c{ 1) for each g E G-NG (T ).

(35.22) Let T be a TI-set in G, H = NG(T), and r, 9 E cl(H) with 1 and 0equal to 0 on H - T. Then

(1) *G(t) _ fi(t) for each t E T#.(2) If 1r(1) = 0 then (1, O)H = (V" BG)G-

Proof. Let g E G#. Then lG(g) = EVEG i(g" ))/IHI with *(g")=0 unlessg" E T. Thus fG(g) = 0 unless g E T" for some u E G, in which case Exercise12.3 says T" is the unique conjugate of T containing g while g" E T if and only

if v E u`H. In particular if g E T then *G(g) _ (F-hEH *(gh))/I HI = f(g)Thus (1) holds. Also, as /,.G is a class function, fG(g) _ *(g") if g" E T forsome v E G, and G(g) = 0 otherwise.

Assume *(1) = 0. Then f G (l) = 0 by 35.20.2, so

(V'G,OG)G = ((g(g)) IGIgEG

(,(gv)(gv)) /IGIgEA


Recall that there are hermitian symmetric forms ( , )H and ( , )G defined on cl(H) and cl(G), respectively.

(35.21) (Frobenius Reciprocity Theorem) Let + E cl(H) and x E cl(G). Then

(+, XIHIH = (+G, X)G.

since j is a class function. As the map g F+ g" is a permutation of G for each v E G, it follows that ($G, X)G = (CgeG +(g)j(g))/l H I . As +(g) = 0 for g E G - H, this sum reduces to (CheH +(h)j(h))llHI = (+, XIHIH.

A subset T of G is said to be a TZ-set in G if T n Tg E 11) for each g E G -

NG(T>.

(35.22) Let T be a TZ-set in G, H = NG(T), and +, 8 E cl(H) with + and 8 equal to 0 on H - T. Then

(1) IlrG(t) = +(t) for each t E T'. (2) If +(l) = 0 then (+, 19)~ = (+G, o G ) ~ .

ProoJ Let g E G'. Then +G(g) = (CUeG +(gU))/I HI with +(gU) = 0 unless gU E T. Thus IlrG(g) = 0 unless g E Tu for some u E G, in which case Exercise 12.3 says Tu is the unique conjugate of T containing g while gU E T if and only if v E u-' H. In particular if g E T then +G(g) = (ChEH +(gh))/l H I = +(g). Thus (1) holds. Also, as +G is a class function, +G(g) = +(gu) if gU E T for some v E G, and +G(g) = 0 otherwise.

Assume +(l) = 0. Then +G(l) = 0 by 35.20.2, so


where A consists of those g E G# with 9'(g) E T for some v(g) E G. Then, byExercise 12.3,

(*G, OG)G = (*(t)e(t)I HI e)HET# 1

as*is0onH-T#.

A Frobenius group is a transitive permutation group G on a finite set X suchthat no member of G# fixes more than one point of X and some member of G#fixes at least one point of X. The following lemma is left as Exercise 12.4.

(35.23) (1) If G is a Frobenius group on a set X, X E X, and H = G, then His a proper nontrivial subgroup of T, H is a TI-set in G, and H = NG(H). LetK be the subset of G consisting of 1 together with the elements of G fixing no

points of X. Then I K I= I X I.(2) Assume H is a proper nontrivial subgroup of a finite group G such that

H is a TI-set in G and H = NG(H). Then G is faithfully represented as aFrobenius group by right multiplication on the coset space G/H.

Notice that, under the hypothesis of Lemma 35.23.1, the representation of Gon X is equivalent to the representation of G by right multiplication on thecoset space G/H by 5.8. Hence the permutation group theoretic hypothesesof 35.23.1 are equivalent to the group theoretic hypothesis of 35.23.2 by thatlemma. I'll refer to a group G satisfying either hypothesis as a Frobenius groupand call the subgroup H the Frobenius complement of G. Exercise 12.4 says His determined up to conjugacy in G. The subset K of 35.23.1 can be describedgroup theoretically by

K = G - (UH#).

gEG

K will be called the Frobenius kernel of G. The following important theoremof Frobenius shows K is a normal subgroup of G.

(35.24) (Frobenius' Theorem) The Frobenius kernel of a Frobenius group Gis a normal subgroup of G.

Proof. Let H and K be the Frobenius complement and kernel of G, respec-tively. The idea is to produce a character X of G with K = ker(X ); then K a Gby 35.9.2. This will be achieved by applying 35.22 to the TI-set H of G.


where A consists of those g E G' with g u ( g ) E T for some v ( g ) E G. Then, by Exercise 12.3,

A Frobenius group is a transitive permutation group G on a finite set X such that no member of G' fixes more than one point of X and some member of G' fixes at least one point of X . The following lemma is left as Exercise 12.4.

(35.23) (1) If G is a Frobenius group on a set X , x E X , and H = G,, then H is a proper nontrivial subgroup of T, H is a TI-set in G, and H = NG(H). Let K be the subset of G consisting of 1 together with the elements of G fixing no points of X. Then IKI = 1x1.

(2) Assume H is a proper nontrivial subgroup of a finite group G such that H is a TI-set in G and H = NG(H). Then G is faithfully represented as a Frobenius group by right multiplication on the coset space G/H.

Notice that, under the hypothesis of Lemma 35.23.1, the representation of G on X is equivalent to the representation of G by right multiplication on the coset space G/H by 5.8. Hence the permutation group theoretic hypotheses of 35.23.1 are equivalent to the group theoretic hypothesis of 35.23.2 by that lemma. I'll refer to a group G satisfying either hypothesis as a Frobenius group and call the subgroup H the Frobenius complement of G. Exercise 12.4 says H is determined up to conjugacy in G. The subset K of 35.23.1 can be described group theoretically by

K will be called the Frobenius kernel of G. The following important theorem of Frobenius shows K is a normal subgroup of G.

(35.24) (Frobenius' Theorem) The Frobenius kernel of a Frobenius group G is a normal subgroup of G.

Proof. Let H and K be the Frobenius complement and kernel of G, respectively. The idea is to produce a character x of G with K = kerO(); then K 9 G by 35.9.2. This will be achieved by applying 35.22 to the TI-set H of G.

Let Bi , 1 1.Vfi(1)=di0l(1)-Bi(1)=di-di = 0. Hence, by 35.22.2 and the orthogonality relations:

(1/fiG>V`G)G = (Y'i, Vj)H = (diet - ei, dje1 - ej) =didj + 8ij.

Next, by 35.19, if = E j 1 CJ X j for suitable integers ci j . By 35.14 andFrobenius reciprocity, ci1 = (iiG, X1)G = (Vi, X1IH)H = (fi, Oi) = di. Soc 1 = di. Also d? + 1 = ( iG, 1G) = _j1(Cij)2 = d? + Ej'_2(cij)2, so1lf ; = di X1 + Ei Xt(i) for some irreducible character Xt(i) with t(i) > 1, and someEi =±1. Next didj + 8ij =(riG VfG)G =didj + EiEj(Xt(i), Xt(j)), so the mapi H t(i) is an injection, and without loss we may take t(i) = i. Finally, by35.20.2, ii/ (1) = Yr1(1)IG: HI = 0, so, as iiiG(1) = di +EiXi(l) and Xi(1) _deg (rri) is a positive integer, it follows that Ei = -1 and Xi (1) = di.

Define X = rk_1 di Xi Then X (1) _ rk=1 d? = IHI by 35.5.3. By 35.20.1,Vf q (g) = 0 for g E K#, so Xi (g) = di -1/''iG (g) = di. Thus X (g) = Ei1 d? _X(1) for all g E K, so K c ker(X). On the other hand each member of G - Kis conjugate to an element of H#, so to show K = ker(X) (and complete theproof) it remains to show X(h) = 0 for h E H#. By 35.22, V/iG(h) = Y'i (h) =di - 9 (h), so Xi(h) = di - Vlf (h) = 9 (h). Therefore X(h) = >i di9i(h) = 0by 35.5.4.

(35.25) Let G be a Frobenius group with Frobenius complement H and kernelK. Then

(1) K < G and H is a complement to K in G. Thus G is a semidirect productofKbyH.

(2) H acts semiregularly on K; that is CH(x) = 1 for each x E K#, or equiv-alently CK (y) = 1 for each y E W.

(3) K is a regular normal subgroup of G in its representation as a Frobeniusgroup.

Proof. By Frobenius' Theorem, K a G. By definition of H and K, H fl K = 1.By 35.23.1, IKI=IG:HI, so, by 1.7, IKHI =IKIIHI=IGI.Thus G=HK.Hence (1) holds. Notice (1) and 15.10 imply (3), while (3) and 15.11 imply (2).

36 Some special actionsIn this section representation theory developed in previous sections is used toderive various results on the representations of certain minimal groups, andthese results are used in turn to prove a number of group theoretic lemmas.Some of these lemmas will be used in chapter 15 and one is used in 40.7 toprove the nilpotence of Frobenius kernels.

192 Linear representations of$nite groups

Let Oi, 1 5 i 5 k , be the irreducible characters of H , di = Oi(l), and, for 1 1 . $ i ( l ) =diOl( l ) -Oi( l )=di- di = 0. Hence, by 35.22.2 and the orthogonality relations:

Next, by 35.19, 11,: = Cy=l cijxj for suitable integers cij. By 35.14 and Frobenius reciprocity, cil = ($?, ~ 1 ) ~ = ($i , ~1 I H ) H = ($i , 81) = di. SO

2 ~ i l = di. Also d; + 1 = ($:, $:) = C y = l ( ~ i j ) 2 = di2 + C y = 2 ( ~ i j ) SO

$? = di X I + E i X t ( i ) for some irreducible character xt(i) with t ( i ) > 1, and some ~i = 5 1 . Next didj + S i j = ($:, $,?), =didj + E ~ E , ( x ~ ( ~ ) , x t( j)) , SO the map i H t ( i ) is an injection, and without loss we may take t ( i ) = i . Finally, by 35.20.2, $:(I) = $i(l) lG : HI = 0 , so, as $:(l) = di + ~ i x i ( 1 ) and xi(1) = deg (xi) is a positive integer, it follows that ~i = - 1 and xi ( 1 ) = di .

Define x = zfz1 dix i . Then x ( 1 ) = ~ f = , d; = 1 HI by 35.5.3. By 35.20.1, $7 ( g ) = 0 for g E K', so xi ( g ) = di - ( g ) = di . Thus x ( g ) = EL1 di2 = ~ ( 1 ) for all g E K , so K 5 kero(). On the other hand each member of G - K is conjugate to an element of H', so to show K = kero() (and complete the proof) it remains to show ~ ( h ) = 0 for h E H'. By 35.22, $?(h) = $i(h) = di - Oi(h), so x i (h ) = di - $:(h) = Oi(h). Therefore ~ ( h ) = xi diOi(h) = 0 by 35.5.4.

(35.25) Let G be a Frobenius group with Frobenius complement H and kernel K . Then

( 1 ) K 9 G and H is a complement to K in G . Thus G is a semidirect product of K by H .

(2) H acts semiregularly on K; that is C H ( x ) = 1 for each x E K', or equivalently CK(y) = 1 for each y E H'.

(3) K is a regular normal subgroup of G in its representation as a Frobenius group.

Proof. By Frobenius' Theorem, K 9 G. By definition of H and K, H n K = 1 . By 35.23.1, l K / = l G : HI, so, by 1.7, IKHI = lKllHl= IGI. Thus G = HK. Hence ( I ) holds. Notice ( 1 ) and 15.10 imply (3), while (3 ) and 15.1 1 imply (2).

36 Some special actions In this section representation theory developed in previous sections is used to derive various results on the representations of certain minimal groups, and these results are used in turn to prove a number of group theoretic lemmas. Some of these lemmas will be used in chapter 15 and one is used in 40.7 to prove the nilpotence of Frobenius kernels.

Some special actions 193

(36.1) Let p, q, and r be distinct primes, X a group of order r acting on an extra-special q-group Q with CQ (X) = Z(Q ), and V a faithful GF(p)XQ-module suchthat Cv(X)=0. Then r=2"+1 is a Fermat prime, q =2, and Q is of width n.

Proof. Replacing V by an irreducible XQ-submodule of [V, Z(Q)], we mayassume V is an irreducible XQ-module. Let F be a splitting field for XQ overGF(p); by 27.13 we may take F to be finite. Pass to VF = V ® GF(p) V. By 26.2,F is the direct sum of c Galois conjugates of an irreducible FXQ-module W.By 27.12, CW(X) = 0. In particular dimF(W) # r dimF(CW(X)), so by 27.17,W is an irreducible FQ-module. Therefore by 34.9, dimF(W) = qe, where eis the width of Q, and for any set Y of coset representatives for Z = Z(Q) inQ, Y is a basis for E =EndF(W) over F. As CQ(X) = Z, we may pick Y tobe invariant under X via conjugation and we may pick 1 E Y.

Next, XQ is a subgroup of E and hence acts on E via conjugation, and asY is a basis for E over F, E is the permutation module for the permutationrepresentation of X on Y. As CQ(X) = Z, X is semiregular on Y - {1}, so, byExercise 4.6.1, CE(X) is of dimension

d=1+ JYJ -1 =1+g2e-1r r

On the other hand we may assume F contains a primitive rth root of unity, so agenerator x of X can be diagonalized on W. Let ai, 1 < i < r, be the rth rootsof 1, and mi the multiplicity of ai as an eigenvalue of x. As x is diagonalizable,CE(x) is isomorphic as an F-algebra to the direct product of algebras Fm' xm;1 j

Therefore

T(mi-mj)2=(r-1)m?)- 2mimji>j i i>

=rd -q2e = r - 1.= r m2) - (mi)2i i

Pick l in the range 1 < l < r, and let s = s(l) = I{i : mi = mi}I. Ifs = r thenqe = dim(W) = rmi, impossible as r and q are distinct primes. So there areat least two multiplicities. Now since there are s(r - s) differences of the formml -mj with mi#mj,(*) r - 1 = T(mi - M j)2 > s(r - s)

i>j


(36.1) Let p , q, and r be distinct primes, X a group of order r acting on an extraspecial q-group Q with CQ(X)= Z(Q), and V a faithful GF(p)XQ-module such that Cv(X)=O. Then r =2"+ 1 is a Fermat prime, q =2, and Q is of width n.

Proof. Replacing V by an irreducible XQ-submodule of [V, Z(Q)], we may assume V is an irreducible XQ-module. Let F be a splitting field for XQ over GF(p); by 27.13 we may take F to be finite. Pass to vF = V @GF(p)V. By 26.2, F is the direct sum of c Galois conjugates of an irreducible FXQ-module W. By 27.12, Cw(X) = 0. In particular dimF(W) # r dimF(Cw(X)), so by 27.17, W is an irreducible FQ-module. Therefore by 34.9, dimF(W) = qe, where e is the width of Q, and for any set Y of coset representatives for Z = Z(Q) in Q, Y is a basis for E =EndF(W) over F. As CQ(X) = Z, we may pick Y to be invariant under X via conjugation and we may pick 1 E Y.

Next, XQ is a subgroup of E and hence acts on E via conjugation, and as Y is a basis for E over F , E is the permutation module for the permutation representation of X on Y. As CQ(X) = Z, X is semiregular on Y - { I } , so, by Exercise 4.6.1, CE(X) is of dimension

IYI - 1 q2e - 1

r d = l + = l + ~

On the other hand we may assume F contains a primitive rth root of unity, so a generator x of X can be diagonalized on W. Let ai, 1 5 i 5 r , be the rth roots of 1, and mi the multiplicity of ai as an eigenvalue of x. As x is diagonalizable, CE(x) is isomorphic as an F-algebra to the direct product of algebras Fml ""I, 1 - < i 5 r , so d = xi=, m?. Also qe = dim~(W) = xi mi. Thus

Therefore

Pick 1 in the range 1 5 1 5 r , and let s = s(1) = I{i : mi = ml}l. If s = r then qe = dim(W) = rml, impossible as r and q are distinct primes. So there are at least two multiplicities. Now since there are s(r - s) differences of the form mi -mi withmi #mi ,

with equality only if there are exactly two multiplicities m1 and Mk and Im1 -Mk l = 1. Then

(**) r(s - 1) < s2 - 1 =(S - 1)(s + 1),

soeithers=1 orr <s+1.Thenass <r,itfollowsthats=l orr- 1 and(**) is an equality. Thus (*) is also an equality, so by the remark after (*), thereare exactly two multiplicities and (mt - mkI = 1. Hence we may take s(l) =1,m = ml, s(k) = r - 1, and Mk = m + E, where c = f 1. Therefore

dim(W) m1=ml+(r- 1)mk=m +(r - 1)(m+E).

Let a1 =1. As Cv (X) = 0, m 1 = 0. As Z = CQ (X ), X does not act by scalarmultiplication on W, so s(1) r - 1. Thus s(1) =1 and Mk =1 fork > 1.Therefore qe = Ft m, = r - 1, completing the proof of the lemma.

(36.2) Let p, q, and r be distinct primes, X a group of order r acting faithfullyon a q-group Q, and V a faithful GF(p)XQ-module. If q = 2 and r is a Fermatprime assume Q is abelian. Then Cv (X) 0.

Proof. Assume otherwise and choose a counterexample with m(V) and I XQIminimal. Then XQ is irreducible on V by minimality of m(V), and, by mini-mality of IXQJ, X centralizes every proper subgroup of Q. The latter remarkand Exercise 8.10 imply Q is elementary abelian or Q is extraspecial andZ(Q)=CQ(X). Moreover in either case X is irreducible on Q/(D(Q). Nowby 36.1, Q is elementary abelian, and then Exercise 4.4 and 27.18 supply acontradiction.

(36.3) Let a be an involution acting on a solvable group G of odd order, letp E 7r c ,r(G), and p" = {p} U,r'. Assume K is an a-invariant subgroup ofG such that CK(a) contains a Hall ar-subgroup of CG(a) and X = [X, a] is ap-subgroup of OFn (K). Then X < O pn (G).

Proof. Take G to be a minimal counterexample. Then On r (G) = 1 and itremains to show X = 1. Let V be a minimal normal subgroup of G(a)contained in G and G* = G/ V. By minimality of G, X* < O pn (G*), soCX(V) < Opn(G) = 1. Also V is a q-group for some prime q, and asOp-(G) = 1, q E 7r - (P).

By coprime action, Exercise 6.2, X is contained in an a-invariant Hall 7r-subgroup H of K, and as CK (a) contains a Hall 7r-subgroup of CG(a), CH (a)is a Hall pr-subgroup of CG(a). As X < Op. (K), X < Op (H). Thus setting


with equality only if there are exactly two multiplicities ml and mk and Iml - mkl = 1. Then

so either s = 1 or r ( s + 1. Then a s s < r , it follows that s = 1 or r - 1 and (**) is an equality. Thus (*) is also an equality, so by the remark after (*), there are exactly two multiplicities and Iml - mkl= 1. Hence we may take s(1) = 1, m=ml,s(k)=r - l , andmk=m +€,where€= f 1.Therefore

dim(W) = mi =mi + (r - 1)mk =m + (r - l)(m + c). I

Let a1 = 1. As CV(X) = 0, ml = 0. As Z = CQ(X), X does not act by scalar .

multiplication on W, so s(1) # r - 1. Thus s(1) = 1 and mk = 1 for k > 1. Therefore qe = xi mi = r - 1, completing the proof of the lemma.

(36.2) Let p, q, and r be distinct primes, X a group of order r acting faithfully on a q-group Q, and V a faithful GF(p)XQ-module. If q = 2 and r is a Fermat prime assume Q is abelian. Then CV(X) # 0.

Proof. Assume otherwise and choose a counterexample with m(V) and /XQl minimal. Then XQ is irreducible on V by minimality of m(V), and, by minimality of I XQ 1, X centralizes every proper subgroup of Q. The latter remark and Exercise 8.10 imply Q is elementary abelian or Q is extraspecial and Z(Q) = CQ(X). Moreover in either case X is irreducible on Q/@(Q). Now by 36.1, Q is elementary abelian, and then Exercise 4.4 and 27.18 supply a contradiction.

(36.3) Let a be an involution acting on a solvable group G of odd order, let p E n g n(G), and p" = {p} U n'. Assume K is an a-invariant subgroup of G such that CK(a) contains a Hall n-subgroup of CG(a) and X = [X, a ] is a p-subgroup of OPn (K). Then X ( Opn (G).

Proof. Take G to be a minimal counterexample. Then OPx(G) = 1 and it remains to show X = 1. Let V be a minimal normal subgroup of G(a) contained in G and G* = G/V. By minimality of G, X* ( OPn(G*), so Cx(V) ( Opn(G) = 1. Also V is a q-group for some prime q, and as Op*(G) = 1, q E n - {p}.

By coprime action, Exercise 6.2, X is contained in an a-invariant Hall n- subgroup H of K, and as CK (a) contains a Hall n-subgroup of %(a), CH(a) is a Hall n-subgroup of &(a). As X _( Opr(K), X ( Op(H). Thus setting

Ko = (Cxv(a), X), X < Opn(Ko) by 31.20.1. Therefore (Ko, XV) satisfiesthe hypotheses of (K, G), so G = XV by minimality of (G1. In particular Xis irreducible on V and G is a pr-group. As G is a pr-group, CG(a) < K andX < Op(K), so [Cv(a), X] < V fl Op(K) = 1. Therefore, as X is faithfuland irreducible on V, Cv(a) = 1. But now 36.2 supplies a contradiction,completing the proof of the lemma.

Remarks. The representation theory in sections 34 and 35 is basic and belongsin any introductory course on finite groups. The results in section 36 are moretechnical. They are in the spirit of the fundamental paper of Hall and Higman[HH]. In particular lemma 36.1 is due to Shult [Sh] although the proof givenhere is from Collins [Co] and uses techniques of Hall and Higman [HH]. Lemma36.3 will be used in the proof of the Solvable 2-Signalizer Functor Theoremin chapter 15. In the first edition of this text, section 36 contained strongerresults used in the proof of the Solvable Signalizer Functor Theorem given inthe first edition. These results have been omitted, since they are unnecessaryfor 2-signalizers.

Exercises for chapter 121. Let G be a finite group and F a splitting field for G whose characteristic

does not divide the order of G. Prove(1) A character X of G is linear if and only if X is irreducible and G(1) <

ker(X ).

(2) If G = (g) is cyclic of order n then F contains a primitive nth root ofunity w and the irreducible characters of G are Xi, 1 < i < n, withXi (gi) _ a)ij

(3) G has exactly IG/GM) linear characters.2. Let 7r be the regular representation of a finite group G over a splitting field

F whose characteristic does not divide the order of G and let X be thecharacter of 7r. Prove(1) 7r is the representation induced by the regular permutation representa-

tion.(2) 7r = Fml ni7ri, where (nri: 1 < i < m) are the irreducible FG-represen-

tations and ni = deg(7ri).(3) X(g)=0forgEG#andX(1)=IGI.

3. Let G be a finite group, T C G, A= {tg: t E T#, g E G}, and H= NG(T).Prove(1) T is a TI-set in G if and only if, for each t E T#, tG fl T = tH and

CG(t) < H.

Some special actions

KO = (&(a), X), X 5 Opn (KO) by 3 1.20.1. Therefore (KO, XV) satisfies the hypotheses of (K, G), so G = XV by minimality of IGI. In particular X is irreducible on V and G is a n-group. As G is a n-group, CG(a) 5 K and X 5 Op(K), so [Cv(a), XI 5 V n Op(K) = 1. Therefore, as X is faithful and irreducible on V, Cv(a) = 1. But now 36.2 supplies a contradiction, completing the proof of the lemma.

Remarks. The representation theory in sections 34 and 35 is basic and belongs in any introductory course on finite groups. The results in section 36 are more technical. They are in the spirit of the fundamental paper of Hall and Higman [HH]. In particular lemma 36.1 is due to Shult [Sh] although the proof given here is from Collins [Co] and uses techniques of Hall and Higman [HH]. Lemma 36.3 will be used in the proof of the Solvable 2-Signalizer Functor Theorem in chapter 15. In the first edition of this text, section 36 contained stronger results used in the proof of the Solvable Signalizer Functor Theorem given in the first edition. These results have been omitted, since they are unnecessary for 2-signalizers.

Exercises for chapter 12 1. Let G be a finite group and F a splitting field for G whose characteristic

does not divide the order of G. Prove (1) A character x of G is linear if and only if x is irreducible and G(') 5

ker(x). (2) If G = (g) is cyclic of order n then F contains a primitive nth root of

unity w and the irreducible characters of G are xi, 1 5 i 5 n, with xi (gj) = wij .

(3) G has exactly IG/G(')I linear characters. 2. Let n be the regular representation of a finite group G over a splitting field

F whose characteristic does not divide the order of G and let x be the character of n . Prove (1) n is the representation induced by the regular permutation representa-

tion. (2) n = Cy=l nixi, where (xi: 1 5 i 5 m) are the irreducible FG-represen-

tations and ni = deg(ni). (3) x(g)=Oforg ~ G # a n d x ( l ) = I G I .

3. Let G be a finite group, T g G, A = {tg: t E T', g E G), and H = NG(T). Prove (1) T is a TI-set in G if and only if, for each t E T', tG n T = tH and

C G ( ~ ) 5 H.

(2) If T is a TI-set in G then the set of conjugates of T # under G partitions0, and, for each t E T #, I tG I= IG : H I I tH I

4. (1) Prove lemma 35.23.(2) Prove a finite group has at most one faithful permutation representation

as a Frobenius group, and conclude a Frobenius group has at most oneclass of Frobenius complements. (You may use the fact (proved in 40.8)that Frobenius kernels are solvable.)

5. Let G = S5 be the symmetric group on X = it, ... , 5}.(1) Use 15.3 to determine the conjugacy classes of G.(2) As in Exercise 5.1.2, show G has 2-transitive permutation represen-

tations of degree 5 and 6, and find their permutation characters (cf.Exercise 4.5).

(3) Find all linear characters of G.(4) Determine the character table of G.(Hint: Use (2) and Exercise 12.6 to determine two irreducible charactersof G. Then use the nonprincipal linear character from (3) and Exercises9.3 and 9.10 to produce two more irreducible characters. Finally, giventhese characters and the linear characters, use the orthogonality relations tocomplete the table.)

6. Let 7r be a permutation representation of the finite group G on a set X, athe CG-representation induced by r, and X the character of a (cf. Exercise4.5). We say X is the permutation character of Jr. Prove(1) (X, Xi) is the number of orbits of G on X.(2) If 7r is transitive then (X, X) is the permutation rank of G on X.(3) G is doubly transitive on X if and only if G = Xi +Xi for some i > 1. G

is of permutation rank 3 if and only if G = Xi + Xt + Xj for somei > j > 1. (Hint: See Exercise 4.5.)

7. Let a be an element of prime order r acting on an r'-group G. Let p bea prime, with p = 2 if r is a Fermat prime, let P = Op(G), and assumeCp(a) =1. Prove [a, O"(G)] < CG(P).

8. Let a be an element of prime order r acting on an r'-group G. Let 1 }X = [X, a] be a q-subgroup of G with X abelian if q = 2, and let p be aprime distinct from q. Prove [Op (G), X] = (C[op(G),x1(a)X?


(2) If T is a TI-set in G then the set of conjugates of T# under G partitions A, and, for each t E T', ltGl = IG : H I J ~ ~ I .

4. (1) Prove lemma 35.23. (2) Prove a finite group has at most one faithful permutation representation

as a Frobenius group, and conclude a Frobenius group has at most one class of Frobenius complements. (You may use the fact (proved in 40.8) that Frobenius kernels are solvable.)

5. Let G E S5 be the symmetric group on X = (1,. . . , 5 ) . (1) Use 15.3 to determine the conjugacy classes of G. (2) As in Exercise 5.1.2, show G has 2-transitive permutation represen-

tations of degree 5 and 6, and find their permutation characters (cf. Exercise 4.5).

(3) Find all linear characters of G. (4) Determine the character table of G. (Hint: Use (2) and Exercise 12.6 to determine two irreducible characters of G. Then use the nonprincipal linear character from (3) and Exercises 9.3 and 9.10 to produce two more irreducible characters. Finally, given these characters and the linear characters, use the orthogonality relations to complete the table.)

6. Let n be a permutation representation of the finite group G on a set X, a the CG-representation induced by n, and x the character of a (cf. Exercise 4.5). We say x is the permutation character of n. Prove (1) (x, xl) is the number of orbits of G on X. (2) If n is transitive then (x, X) is the permutation rank of G on X. (3) G is doubly transitive on X if and only if G = x1 +xi for some i > 1. G

is of permutation rank 3 if and only if G = x1 + X, + xj for some i > j > 1. (Hint: See Exercise 4.5.)

7. Let a be an element of prime order r acting on an r'-group G. Let p be a prime, with p = 2 if r is a Fermat prime, let P = Op(G), and assume Cp(a) = 1. Prove [a, OP(G)] 5 Cc(P).

8. Let a be an element of prime order r acting on an r'-group G. Let 1 # X = [X, a] be a q-subgroup of G with X abelian if q = 2, and let p be a prime distinct from q. Prove [O,(G), XI = (Cp,cc,,xl(a)x).

13

Transfer and fusion

If G is a finite group, H < G, and a: H A is a homomorphism of H into anabelian group A, then it is possible to construct a homomorphism V: G ->. Afrom a in a canonical way. V is called the transfer of G into A via a. If we canshow there exists g E G - ker(V), then, as G/ker(V) is abelian, g 0 Gf1 thecommutator group of G. In particular G is not nonabelian simple.

It is however in general difficult to calculate g V explicitly and decide whetherg E ker(V ). To do so we need information about the fusion of g in H; that isinformation about gG fl H. Hence chapter 13 investigates both the transfer mapand techniques for determining the fusion of elements in subgroups of G.

Section 38 contains a proof of Alperin's Fusion Theorem, which says thatp-local subgroups control the fusion of p-elements. To be somewhat moreprecise, if P is a Sylow p-subgroup of G then we can determine when subsetsof P are fused in G (i.e. conjugate in G) by inspecting the p-locals H of GwithPflHSylowinH.

Section 39 investigates normal p-complements. A normal p-complementfor a finite group G is a normal Hall p'-subgroup of G. Various criteria for theexistence of such objects are generated, The most powerful is the ThompsonNormal p-Complement Theorem, which is used in the next section to establishthe nilpotence of Frobenius kernels. Section 39 also contains a proof of theBaer-Suzuki Theorem which says a p-subgroup X of a finite group G iscontained in Op(G) if and only if (X, Xg) is a p-group for each g in G.

A group A is said to act semiregularly on a group G if A is a group of auto-morphisms of G with CG(a) =1 for each a c A. Such actions are investigatedin section 40.

37 TransferLet G be a finite group, H < G, and a: H A a homomorphism of H intoan abelian group A. Given a set X of coset representatives for H in G, defineV:G->. Aby

gV =fl((xg)(xg)_1)a

xEX

where xg denotes the unique member of X in the coset Hxg. We say V is thetransfer of G into A via a.

Transfer and fusion

If G is a finite group, H ( G, and a: H + A is a homomorphism of H into an abelian group A, then it is possible to construct a homomorphism V: G -+ A from a in a canonical way. V is called the transfer of G into A via a. If we can show there exists g E G - ker(V), then, as G/ker(V) is abelian, g 4 G(') the commutator group of G. In particular G is not nonabelian simple.

It is however in general difficult to calculate g V explicitly and decide whether g E ker(V). To do so we need information about thefusion of g in H; that is information about gG n H. Hence chapter 13 investigates both the transfer map and techniques for determining the fusion of elements in subgroups of G.

Section 38 contains a proof of Alperin's Fusion Theorem, which says that p-local subgroups control the fusion of p-elements. To be somewhat more precise, if P is a Sylow p-subgroup of G then we can determine when subsets of P arefised in G (i.e. conjugate in G) by inspecting the p-locals H of G with P n H Sylow in H.

Section 39 investigates normal p-complements. A normal p-complement for a finite group G is a normal Hall p'-subgroup of G. Various criteria for the existence of such objects are generated, The most powerful is the Thompson Normal p-Complement Theorem, which is used in the next section to establish the nilpotence of Frobenius kernels. Section 39 also contains a proof of the Baer-Suzuki Theorem which says a p-subgroup X of a finite group G is contained in O,(G) if and only if (X, X g ) is a p-group for each g in G.

A group A is said to act semiregularly on a group G if A is a group of automorphisms of G with &(a) = 1 for each a E A'. Such actions are investigated in section 40.

37 Transfer Let G be a finite group, H 5 G, and a: H + A a homomorphism of H into an abelian group A. Given a set X of coset representatives for H in G, define V :G+A by

where xg denotes the unique member of X in the coset Hxg. We say V is the transfer of G into A via a.

198 Transfer and fusion

(37.1) The transfer map V is independent of the choice of the set X of cosetrepresentatives.

Proof. Let Y be a second set of coset representatives for H in G. Then there isa bijection x H y(x) of X with Y such that y(x) = h(x)x for some h(x) E H.For y E Y and g E G, write yg for the member of Y in Hyg. Observe that{y(x)g} = Hy(x)gfY = HxgfY = Hxgf1Y = {y(xg)}; that is y(x)g = y(xg).Also

y(x)g = h(x)xg = h(x)xg(xg)-1(xg)

= h(x)xg(xg)-1h(xg)-1y(xg)

so that

Therefore

y(x)g(y(x)g)-1 = h(x)xg(xg)-1h(xg)-1

fl(yg(yg)-1)a = fl(y(x)g(y(x)9)-1)ayEY xEX

= fl h(x)a(xg(xg)-1)ah(xg)-1a =fl(xg(xg)_1)a

xEX xEX

with the last equality holding as A is abelian and the map x i-± xg is a permu-tation of X. So the lemma holds.

(37.2) The transfer V is a group homomorphism of G into A.

Proof. Lets, t E G. Observe x(st) = (xs)1. Hence

(st)V = fl((xst)(xst)-1)a = fl(xs(xs)-1(xs)t(xst)-1)a

xEX XEX

_ fl(xs(xs)-1)a((xs)t(xst)-1)axEX

(fl(xs(x)_1)a) (fl(xt(xi)_1)a) =sVtV,xEX

using the fact that A is abelian and the map x H xg is a permutation on X.

(37.3) Let (Hx1 gj: 0 < j < nl ), 1 < i < r, be the cycles of g E G on the cosetspace G/H. Pick X = {x1gi: 1 < i < r, 0 < j < n1). Then

(1) (g"1)x+`EHfor 1<i <r.(2)

Et-1nl=1G:HI.

(3) gV = ];=1((gn')x' )a.

198 Transfer andfision

(37.1) The transfer map V is independent of the choice of the set X of coset representatives.

Proof. Let Y be a second set of coset representatives for H in G. Then there is a bijection x ++ y(x) of X with Y such that y(x) = h(x)x for some h(x) E H. For y E Y and g E G , write yg for the member of Y in Hyg. Observe that {y(x)g) = Hy(x)gnY = HxgnY = HxgnY = {y(xg)};thatis y(x)g= y(xg). Also

y(x)g = h(x)xg = h(x)xg(xg)-l(xg)

= h(x)xg(xg)-' h(xg)-' y(xg)

so that

y(x)g(y(x)g)-' = h(x)xg(xg)-lh(xg)-'.

Therefore

= n h ( x ) a ( ~ g ( x g ) - l ) a h ( x g ) - ~ a = n(xg(xg) - l )a xex xex

with the last equality holding as A is abelian and the map x I+ xg is a permutation of X. So the lemma holds.

(37.2) The transfer V is a group homomorphism of G into A.

Proof. Let s, t E G. Observe x ( Z ) = (x5)t. Hence

using the fact that A is abelian and the map x I+ x5 is a permutation on X.

(37.3) Let ( ~ x ~ g j : 0 5 j < ni) , 1 _( i 5 r , be the cycles of g E G on the coset space G / H . Pick X = (xigj: 1 _( i _( r, 0 _( j _(nil. Then

(1) ( g n i ) x ; ' ~ H f o r 1 5 i s r . (2) C;=,ni=lG: HI. (3) g~ = n;, ,ccgni)x;')~.

Transfer 199

Proof. Parts (1) and (2) are immediate from the definitions. By 37.1 we may cal-culate V with respect to this particular choice of coset representatives. By defi-nition of X, (xigj)g=xigi-H1 =(xig1)gfor j < ni-1.Also (xig"j-1)g=xig"jand (xlg"i-1)g =xi. So (3) holds.

(37.4) Let G be a finite group, p a prime, H< G with (p, I G: H p = 1, K < Hwith H/K abelian, and g a p-element in H - K such that g" e e gmK for allintegers m, and all a E G, such that gma E H. Then g V G(l).

Proof. Let A = H/K and a : H A the natural surjection. I'll show gV # 1.Hence g V ker(V), and of course, as GV < A, GV is abelian, so GM < ker(V).

Choose a set X of coset representatives for H in G as in 37.3. By 37.3.1,(g" )x E H, so by hypothesis g" xI ' E g"i K. Hence (g";x )a = g" a = (g,)" .Hence, by 37.3.3, gV = (ga)", where n = E;-1 ni. Finally, by 37.3.2, n =IG : HI, so by hypothesis (p, n) = 1. Hence, as g is a p-element and g V K,also g" V K. Thus 1 # g"a =gV, as desired.

(37.5) Let G be a finite group, p a prime, H < G with (p, I G: H1) = 1, andassume gG n H = gH for all p-elements g in H. Then (OP(G)G()) n H =OP(H)HO).

Proof. Let Go = OP(G)G(l) and K = OP(H)HO). Then certainly K < Go n H,and each p'-element in Go n H is in OP(H) < K, so it remains to showeach p-element g in Go n H is in K. But if m is an integer and a E G withgma E H then by hypothesis gma = gmh, for some h E H, so, as H/K is abelian,gma = gmh E gm K. It follows from 37.4 that if g V K then g V G(l). Then, as allp-elements in Go are in G(1), also g V Go, contrary to the choice of g. So thelemma holds.

Let H < G and S an H-invariant subset of G. Then H is said to control fusionin S if sG n S=sH for each s E S. For example one of the hypotheses in thelast lemma says that H controls fusion of its p-elements.

Let X C H < G. Then X is said to be weakly closed in H with respect to GifXGnH={X}.

(37.6) Let p be a prime, T E Sylp(G), W < T with W weakly closed in Twith respect to G, and D = CG(W). Then NG(W) controls fusion in D.

Proof. Let d E D and g E G, with d9 E D. Then W, W9_' < C(d). By Sy-low's Theorem there is x E CG(d) with U = (W, Wg-`x) a p-group. Let

Transfer 199

Proof. Parts ( 1 ) and (2) are immediate from the definitions. By 37.1 we may calculate V with respect to this particular choice of coset representatives. By definition of X, (x ig j )g = xigj+' = (xigj)g for j < ni - 1. A ~ S O (xigni-')g = xigni and (xigni-')g =xi. SO (3) holds.

(37.4) Let G be a finite group, p a prime, H 5 G with (p, IG : HI) = 1, K H with H / K abelian, and g a p-element in H - K such that gma E gm K for all integers m, and all a E G , such that gma E H. Then g 4 G(').

Proof. Let A = HI K and a : H + A the natural surjection. I'll show g V # 1. Hence g 4 ker(V), and of course, as GV 5 A, GV is abelian, so G(') 5 ker(V).

Choose a set X of coset representatives for H in G as in 37.3. By 37.3.1, (gni)x;l E H, so by hypothesis gnix;l E gni K. Hence (gnix;l)a = gnia = ( g c ~ ) ~ ~ . Hence, by 37.3.3, gV = (ga)", where n = EL=, ni. Finally, by 37.3.2, n = IG : HI, so by hypothesis (p, n ) = 1. Hence, as g is a p-element and g 4 K , also gn 4 K. Thus 1 # gna = g V , as desired.

(37.5) Let G be a finite group, p a prime, H 5 G with (p, IG : HI) = 1, and assume g G n H = g H for all p-elements g in H. Then (oP(G)G(')) n H = OP(H)H( ' ) .

Proof. Let Go = O~(G)G( ' ) and K = OP(H)H( ' ) . Then certainly K 5 Go fl H , and each p'-element in Go n H is in OP(H) 5 K, so it remains to show each p-element g in Go n H is in K. But if m is an integer and a E G with gma E H then by hypothesis gma = gmh, for some h E H, so, as H / K is abelian, gma = gmh E gm K. It follows from 37.4 that if g 4 K then g 4 G('). Then, as all p-elements in Go are in G('), also g 4 Go, contrary to the choice of g. So the lemma holds.

Let H 5 G and S an H-invariant subset of G. Then H is said to controlfusion in S if sG n S = sH for each s E S. For example one of the hypotheses in the last lemma says that H controls fusion of its p-elements.

Let X E H 5 G. Then X is said to be weakly closed in H with respect to G if xG n H = { X ) .

(37.6) Let p be a prime, T E Sylp(G), W 5 T with W weakly closed in T with respect to G , and D = CG(W). Then NG(W) controls fusion in D.

Proof. Let d E D and g E G, with dg E D. Then W , ~ 8 - ' 5 C(d). By Sy- low's Theorem there is x E CG(d) with U = ( W , wg-lX) a p-group. Let

U < S E Sylp(G). As W is weakly closed in T, (W) = We n s = (Wg-'x).Thus h = x-lg E NG(W) and dg = dh.

(37.7) Assume T is an abelian Sylow p-group of G. Then T n O' (G) _[T, NG(T)].

Proof. Let H = NG (T). By the Schur-Zassenhaus Theorem there is a com-plement X to T in H. Then K = X [T, X] < XT = H and H/K = T/[T, X] isan abelian p-group, so K =OP(H)HM. Next G = TOP(G), so G/OP(G)T/(T n OP(G)) is abelian and hence OP(G)=OP(G)GM. Certainly T isweakly closed in itself, so, by 37.6, H controls fusion in CG(T), and hence, asT < CG (T ), H controls fusion in T. But T is the set of p-elements in H soH controls fusion of its p-elements and hence, by 37.5, OP(G) n H = K. AsT n OP(G) and [T, H] are Sylow in OP(G) and K, respectively, it follows thatT n OP(G) = [T, NA(T)].

38 Alperin's Fusion TheoremIn this section G is a finite group, p is a prime, and P is some Sylow p-groupof G. A p-subgroup X of G is said to be a tame intersection of Sylow p-groupsQ and R of G if X = Q n R and NQ(X) and NR(X) are Sylow p-groups ofNG(X ).

The main result of this section is:

(38.1) (Alperin's Fusion Theorem) Let P E Sylp(G), g E G, and A, Ag C P.Then there exists Qi E Sylp(G), 1 < i < n, and xi E NG(P n Qi) such that:

(1) g = x1...xn.

(2) P n Qi is a tame intersection of P and Qi for each i, 1 < i < n.(3) A C P n Q1 and Ax' x, c P n Qi+i for l< i< n.

Alperin's Theorem will follow from Theorem 38.2 and an easy argument. Butfirst some definitions.

For R, Q E Sylp(G) write R -+ Q if there exist Sylow p-groups (Q1:1i < n) of G and elements xi E NG(P n Qi) such that:

(1) P n Qi is a tame intersection of P and Qi for each i with 1 < i < n.(2) P n R < P n Q1 and (P n P n Qt+1 for each i with 1 < i < n.(3) Rx=Q,where x=x1

I'll also write R 4 Q when it's necessary to emphasize the role of the elementx in (3), and say that (Qi, xi: 1 < i < n) accomplish R -+ Q.


U i S E Sylp(G). As W is weakly closed in T, {W) = wG n S = {wg-lx). Thus h =x-lg E NG(W) and dg =dh.

(37.7) Assume T is an abelian Sylow p-group of G. Then T n OP(G) =

[T, Nc(T)I.

Proof. Let H = NG(T). By the Schur-Zassenhaus Theorem there is a complement X to T in H. Then K = X[T, XI XT = H and H/K G T/[T, XI is an abelian p-group, so K = O~(H)H('). Next G = TOP(G), so G/OP(G) Z T/(T n Op(G)) is abelian and hence OP(G) = 0p(G)G('). Certainly T is weakly closed in itself, so, by 37.6, H controls fusion in CG(T), and hence, as T ( CG(T), H controls fusion in T. But T is the set of p-elements in H so H controls fusion of its p-elements and hence, by 37.5, Op(G) i? H = K. As T i? OP(G) and [T, HI are Sylow in OP(G) and K, respectively, it follows that T n OP(G) = [T, NG(T)].

38 Alperin's Fusion Theorem In this section G is a finite group, p is a prime, and P is some Sylow p-group of G. A p-subgroup X of G is said to be a tame intersection of Sylow p-groups Q and R of G if X = Q i? R and NQ(X) and NR(X) are Sylow p-groups of

N G ( ~ ) . The main result of this section is:

(38.1) (Alperin's Fusion Theorem) Let P E Sylp(G), g E G, and A, Ag C P. Then there exists Qi E Sylp(G), 1 5 i 5 n, and xi E Nc(P i l Qi) such that:

(1) g=x1 . . .x,. (2) P n Qi is a tame intersection of P and Qi for each i, 1 ( i 5 n. (3) A E Pi? Ql andAxl...xl E P i l Qi+l for 1 I i < n.

Alperin's Theorem will follow from Theorem 38.2 and an easy argument. But first some definitions.

For R, Q E Sylp(G) write R + Q if there exist Sylow p-groups (Qi: 1 5 i 5 n) of G and elements xi E NG(P i? Qi) such that:

(1) P i? Qi is a tame intersection of P and Qi for each i with 1 5 i 5 n. (2) P n R ( P n Q l a n d ( P n R ) " l . . . X i i P n Q i + l f o r e a c h i w i t h l ( i < n . (3) RX=Q,wherex=xl ... x,.

I'll also write R 5 Q when it's necessary to emphasize the role of the element x in (3), and say that (Qi, xi: 1 5 i 5 n) accomplish R + Q.

Alperin's Fusion Theorem

Theorem 38.2. Q - P for each Q E Sylp(G).

The proof of 38.2 involves several reductions. Observe first that:

(38.3) P P.

Indeed P4 P is accomplished by P, 1.

(38.4) -* is a transitive relation.

201

Proof. Let (R1, yi: 1 < i < m) and (Q;, x; : 1 < i < n) accomplish S -+ Rand R -+ Q, respectively. Then R1, ... , Rm, Q1, ... , Q,,, and yl,... , ym,x1, ... , xn accomplish S -+ Q.

(38.5) If S4P,Qx-+ P, and P n Q < P n S, then Q -+ P.

Proof. By 38.4 it suffices to show Q -+ Qx. Let (S;, x; : 1 I Q n P 1, that S -+ P. ThenQ -+ P.

Proof. By hypothesis there is x E G with R 4 P. Now P n Qx = Rx n Qx =(R n Q)x, so I P n Qx I= I R n Q I> P n Q1. Hence Qx ± P by hypothesis.Now apply 38.5 to complete the proof.

(38.7) Assume P n Q is a tame intersection of P and Q such that S -+ P forall S E Sylp(G) with IS n P) > I Q n PI. Then Q - P.

Proof. By 38.3 we may assume Q 0 P. Thus P n Q < Po = Np(P n Q). Byhypothesis P0 and Qo = NQ(P n Q) are Sylow in M = NG(P n Q), so thereis x E M With Qo = P0. Notice Q -+ Qx is accomplished by (Q, x). FurtherP n Q < P0 < P n Qx, so by hypothesis Qx ± P. Therefore, by 38.4, Q -+ P.

We are now in a position to prove 38.2. Pick a counterexample Q with P n Qof maximal order. By 38.3, P 0 Q, so P n Q P, and hence P n Q <Np(P n Q). Let S E Sylp(G) with Np(P n Q) < Ns(P n Q) E Sylp(NG(PnQ)). As P n Q < Np(P n Q) < P n S, it follows that S -* P by maxima-lity of P n Q. Thus there is x E G with S P.

Alperin's Fusion Theorem

Theorem 38.2. Q + P for each Q E Sylp(G).

The proof of 38.2 involves several reductions. Observe first that:

Indeed P: P is accomplished by P, 1.

(38.4) + is a transitive relation.

Proof. Let (Ri, yi: 1 i ( m) and (Qi,xi: 1 ( i ( n) accomplish S + R and R + Q, respectively. Then R1, . . . , R,, e l , . . . , Q,, and yl, . . . , y,, xl, . . . , x, accomplish S + Q.

(38.5) If S 5 P, Qx + P , and P i l Q ( P i l S, then Q + P.

Proof. By 38.4 it suffices to show Q + QX. Let (Si, xi: 1 5 i 5 n) accomplish S + P . Then (Si, xi: 1 5 i ( n) also accomplish Q + QX.

(38.6) Assume R, Q E Sylp(G) with R + P and P n Q < R i? Q. Assume further,forallS~Syl,(G)withISnPI > IQnPI,thatS+ P.Then Q + P .

Proof. By hypothesis there is x E G with R 5 P . Now P i? Qx = Rx i? Qx = ( R n Q)",soIPn QxI=IRn Ql > I P n Ql.Hence Q x + Pbyhypothesis. Now apply 38.5 to complete the proof.

(38.7) Assume P i? Q is a tame intersection of P and Q such that S + P for all S E Sylp(G) with IS i? PI > lQ n PI. Then Q + P.

Proof. By 38.3 we may assume Q # P . Thus P i? Q < Po = Np(P i? Q). By hypothesis Po and Qo = NQ(P i? Q) are Sylow in M = NG(P i? Q), so there is x E M with Qg = Po. Notice Q + is accomplished by (Q, x). Further Pi? Q < Po 5 Pi? Qx, so by hypothesis Qx + P . Therefore, by 38.4, Q + P .

We are now in a position to prove 38.2. Pick a counterexample Q with P i? Q of maximal order. By 38.3, P # Q, so P n Q # P , and hence P i? Q < Np(P n Q). Let S E Sylp(G) with Np(P n Q) 5 Ns(P n Q) E Sylp(NG ( P n Q)). As P n Q < Np(P i? Q) 5 P n S, it follows that S + P by maximality of P i? Q. Thus there is x E G with S 5 P.

Evidently (P n Q)x < Qx. Also P n Q < S and Sx = P, so (P n Q)x < P.Thus(PnQ)x < PnQx.if(PnQ)x PnQxthen IPnQI < IPnQxl,so,by maximality of I P n Q 1, Qx -+ P. But then Q -+ P by 38.5, contradictingthe choice of Q.

So(PnQ)x=PnQx.Next letT ESylp(G)with NQx(PnQx) < NT(PnQx) E Sylp(NG(P n Qx)). Again P n Qx < NQ=(P n Qx) < T, so P n Qx <T n Qx. Hence if T - P then, by 38.6, Qx -+ P, which we have alreadyobserved to be false. Thus we do not have T -+ P, so, by maximality ofIPnQI,PnQx=PnT.

By choice of T, and as P n Qx = P n T, we have NT (PnT) E Sylp (NG (P nT)). By choice of S, Ns(P n Q) E Sylp(NG(P n Q)) so, as (P n Q)x =P n Qx = P n T and Sx = P, we have Np(P n T) E Sylp(NG(P n T)).Thus P n T is a tame intersection of P and T. But now, by 38.7, T -+ P,contrary to the last paragraph.

This completes the proof of 38.2.

Now for the proof of Alperin's Fusion Theorem. Assume the hypothesis ofAlperin'sTheorem.By38.2, Pg-' -+ P.Let(Q1, x;: 1 < i < n-1)accomplishPg-'-+ P.AsA,AgCP,ACPnPC',so,bydefinition of-+,A c_ PnPg-' < P n Q1 and Axt...xi C (P n Pg-')x,...xi < P n Qi+1 for 1 < i <n - 1. Also, setting x =x1 ... xn_1, Pg-'x = P, so xn =x-1g E NG(P) andg = xxn =x1 ... xn. Finally let Qn = P and observe Ax,...x-' = Agxn' <

P = P n Qn, so the theorem holds.

39 Normal p-complementsIn this section p is a prime and G is a finite group. A normal p-complementfor G is a normal Hall p'-subgroup of G; that is a normal p-complement is anormal complement to a Sylow p-subgroup of G.

(39.1) (Burnside Normal p-Complement Theorem) If a Sylow p-subgroup ofG is in the center of its normalizer then G possesses a normal p-complement.

Proof. This is immediate from 37.7.

(39.2) If p is the smallest prime divisor of the order of G and G has cyclicSylow p-groups, then G has a normal p-complement.

Proof. Let P E Sylp(G). By hypothesis P is cyclic. As P is abelian and Sylowin G, AutG(P) is a p'-group, so, by 23.3, IAutG(P)I divides p - 1. Hence, as


Evidently ( P fl Q)X 5 QX. Also P n Q 5 S and SX = P , so ( P fl Q)X 5 P . Thus(PnQ)x 5 P n Q x . I f ( P n Q ) x # P n Q x t h e n l P n Q l < lPnQxl,so, by maximality of I P n Ql, QX + P. But then Q + P by 38.5, contradicting the choice of Q.

So ( P n Q)X = P n Qx. Next let T E Sylp(G) with N p ( P n QX) 5 NT(P n Qx) E Sylp(NG(P n Qx)). Again P n QX < Np(P n QX) ( T, so P n Qx < T n Qx. Hence if T + P then, by 38.6, QX + P , which we have already observed to be false. Thus we do not have T + P , so, by maximality of I P n Q I , P n Q ~ = P n T.

By choiceof T, andas P n QX = P n T, wehave NT(PnT) E S y l p ( N ~ ( P n T)). By choice of S, Ns(P n Q) E Sylp(NG(P n Q)) so, as ( P n Q)X = P n Qx = P n T and Sx = P , we have Np(P n T) E Sylp(N~(P n T)). Thus P n T is a tame intersection of P and T. But now, by 38.7, T + P , contrary to the last paragraph.

This completes the proof of 38.2.

Now for the proof of Alperin's Fusion Theorem. Assume the hypothesis of Alperin7sTheorem. By 38.2, pg- ' + P. Let (Qi, xi: 1 5 i 5 n-1) accomplish p g - ' + P . As A, Ag E P , A P n ~ g - ' , SO, by definition of +, A E P n pg-' 5 P n Ql and AXl...Xl c - ( P n P~- ' )X~ . . .~ I P n Qi+l for 1 i <

n - 1. Also, setting x = xl . . . x,-1, P ~ - ' x = P , so x, = x-lg E NG(P) and - g = xx, = xl . . . x,. Finally let Q, = P and observe Axl...xn-l - ~ g " n - ' - <

PX;' = P = P n Q,, so the theorem holds.

39 Normal p-complements In this section p is a prime and G is a finite group. A normal p-complement for G is a normal Hall p'-subgroup of G; that is a normal p-complement is a normal complement to a Sylow p-subgroup of G.

(39.1) (Burnside Normal p-Complement Theorem) If a Sylow p-subgroup of G is in the center of its normalizer then G possesses a normal p-complement.

Proof. This is immediate from 37.7.

(39.2) If p is the smallest prime divisor of the order of G and G has cyclic Sylow p-groups, then G has a normal p-complement.

Proof. Let P E Sylp(G). By hypothesis P is cyclic. As P is abelian and Sylow in G, AutG(P) is a p'-group, so, by 23.3, lAut~(P)l divides p - 1. Hence, as

Normal p-complements 203

p is the smallest prime divisor of (G1, AutG(P) = 1. Equivalently, P is in thecenter of its normalizer, so 39.1 completes the proof.

G is metacyclic if there exists a normal subgroup H of G such that H andG/H are cyclic.

(39.3) Assume each Sylow group of G is cyclic. Then G is metacyclic.

Proof. Let p be the smallest prime divisor of I G I. By 39.2, G has a normal p-complement H. By induction on the order of G, H is metacyclic. In particularH is solvable, so, as G/H is cyclic, G is solvable.

Let K = F(G). K is nilpotent with cyclic Sylow groups, so K is cyclic.Thus Aut(K) is abelian. However, by 31.10, K = CG(K), so AutG(K) = G/Kis abelian. As G/K has cyclic Sylow groups, G/K is cyclic, so G is metacyclic.

A subgroup H of G is a p-local subgroup if H = NG(P) for some nontrivialp-subgroup P of G.

(39.4) (Frobenius Normal p-Complement Theorem) The following are equiv-alent:

(1) G has a normal p-complement.(2) Each p-local subgroup of G has a normal p-complement.(3) AutG(P) is a p-group for each p-subgroup P of G.

Proof. The implications (1) implies (2) and (2) implies (3) are easy and leftas exercises. Assume G satisfies (3) but not (1), and, subject to this constraint,choose G minimal.

Observe first that G = OP(G). For if not, by minimality of G, OP(G) has anormal p-complement, which is also a normal p-complement for G.

Next let's see that a Sylow p-subgroup P of G controls fusion in P. For letg E G, a E P, and ag E P. Apply Alperin's Theorem to obtain Qi E Sylp(G)and x, E NG (P n Q1) satisfying the conditions of that theorem with A = (a}.In particular g =x1 ... x,,. It will suffice to show ax,... xi E aP for each 1 <i < n. Assume otherwise and let i be a minimal counterexample. Then b =ax1...x'-1 E P n Q1 = U, x, E H = NG(U), P n H E Sylp(H), and ax,...X; = LX(

By hypothesis AutG(U) is a p-group, so H = CG(U)(P n H), as P n(H/

E

Sylp(H). Thus, as b E U, bx' =b` for some t E P n H, so, as b E aP, ax,...x1 _

bx' Ea1.So P controls fusion in P. Hence by 37.5 (OP(G)G(1)) n P =OP(P)PO).

But OP(P) =1 and, as G = OP(G), (OP(G)G(l)) n P = P. Thus P = PM, so,as p-groups are solvable, P = 1. But then G is its own normal p-complement.

Normal p-complements 203

p is the smallest prime divisor of I G I , AutG (P) = 1. Equivalently, P is in the center of its normalizer, so 39.1 completes the proof.

G is metacyclic if there exists a normal subgroup H of G such that H and G/H are cyclic.

(39.3) Assume each Sylow group of G is cyclic. Then G is metacyclic.

Proof. Let p be the smallest prime divisor of I G I. By 39.2, G has a normal p- complement H. By induction on the order of G, H is metacyclic. In particular H is solvable, so, as G/H is cyclic, G is solvable.

Let K = F(G). K is nilpotent with cyclic Sylow groups, so K is cyclic. Thus Aut(K) is abelian. However, by 31.10, K = CG(K), so Au~G(K) = G/K is abelian. As G/K has cyclic Sylow groups, G/K is cyclic, so G is metacyclic.

A subgroup H of G is a p-local subgroup if H = NG(P) for some nontrivial p-subgroup P of G.

(39.4) (Frobenius Normal p-Complement Theorem) The following are equivalent:

(1) G has a normal p-complement. (2) Each p-local subgroup of G has a normal p-complement. (3) AutG(P) is a p-group for each p-subgroup P of G.

Proof. The implications (1) implies (2) and (2) implies (3) are easy and left as exercises. Assume G satisfies (3) but not (I), and, subject to this constraint, choose G minimal.

Observe first that G = OP(G). For if not, by minimality of G, OP(G) has a normal p-complement, which is also a normal p-complement for G.

Next let's see that a Sylow p-subgroup P of G controls fusion in P . For let g E G, a E P , and ag E P. Apply Alperin's Theorem to obtain Qi E Sylp(G) and xi E NG(P n Qi) satisfying the conditions of that theorem with A = (a). In particular g =XI . . . x,. It will suffice to show E a P for each 1 5 i 5 n. Assume otherwise and let i be a minimal counterexample. Then b = axl...x,-l E P n Qi = U, xi E H = NG(U), P n H E Sylp(H), and aX1...Xi = bxl. By hypothesis AutG(U) is a p-group, so H = CG(U)(P fl H), as P fl H E Sylp(H). Thus, as b E U, bXl = bf for some t E P n H, so, as b E up, aX1...Xz = bXr E up.

So P controls fusion in P . Hence by 37.5 (oP(G)G(')) n P = OP(P) P('). But OP(P) = 1 and, as G = OP(G), (OP(G)G(')) n P = P . Thus P = P('), so, as p-groups are solvable, P = 1. But then G is its own normal p-complement.

(39.5) (Thompson Normal p-Complement Theorem) Let p be odd and P ESylp(G). Assume NG(J(P)) and CG(Q1(Z(P))) have normal p-complements.Then G has a normal p-complement.

Proof. Assume otherwise and let G be a minimal counterexample. By theFrobenius p-Complement Theorem there is a p-local subgroup H of G whichpossesses no normal p-complement. By Sylow's Theorem we take Q = P flH E Sylp(H). Choose H with Q maximal subject to these constraints.

Claim P = Q. If not Q < Np(Q), so I NG(C)Ip > I QI for each C char Q.Hence, by maximality of Q, NG(C) has a normal p-complement. So NH(C)also has a normal p-complement. In particular this holds for C = J(Q) andS21(Z(Q)), so the hypotheses of the theorem hold in H. Hence, by minimalityof G, H has a normal p-complement, contrary to the choice of H.

So P E Sylp(H). Hence, as H has no p-complement, H = G by minimalityof G. Moreover if Op, (G) 1 then, by 18.7 and minimality of G, G/Op,(G)has a p-complement, and then G does too. So Op-(G) = 1.

As G = H is a p-local, Op(G) 0 1. Let G* = G/Op(G). Then Op(G*) = 1so NG(J(P*)) and CG(Q1(Z(P*))) are proper subgroups of G containing P,and hence by minimality of G have p-complements. So, by minimality of G,G* has a p-complement. Hence G = O p,p,, p(G). Thus E(G) = Op,p,, p(E(G)).But [E(G), F(G)] = 1 so Op(E(G)) < Z(E(G)), and thus Op,p,(E(G)) =Op(E(G)) x Op,(E(G)). Now Op,(E(G)) < Op,(G) = 1, so E(G) is a p-group.Hence, as E(G) is perfect, E(G) = 1. So F*(G) = F(G)E(G) = O p(G).

Let p 0 r c ir(G) and R E Syl,(G). R < Op,p,(G), so, by a Frattini Argu-ment, Op(G)NG(Z(R)) contains a Sylow p-group of G which we may taketo be P. In particular P acts on Op(G)Z(R) so PZ(R) = K is a subgroupof G. Now if K 0 G then, by minimality of G, Z(R) = Op,(K) a K. Butthen Z(R) < CG(Op(G)) < Op(G) by 31.10, and hence Z(R) = 1, acontradiction.

So G = PZ(R). In particular G is solvable and R = Z(R) is abelian. But nowby Thompson Factorization, 32.6, G = G 1 G2 where G 1 = NG (J (P)) and G 2 =CG(S21(Z(P)). By hypothesis G, = Op,(G,)P. As Op,(G,) < CG(Op(G))Op (G), Op- (G,) = 1, so G, = P.Thus G = G 1 G2 = P, contradicting the choiceof G as a counterexample.

The next result does not involve normal p-complements, but it has the sameflavor.

(39.6) (Baer-Suzuki Theorem) Let X be a p-subgroup of G. Then either X <Op(G) or there exists g c G with (X, Xg) not a p-group.

204 Transfer andfusion

(39.5) (Thompson Normal p-Complement Theorem) Let p be odd and P E

Syl,(G). Assume NG(J(P)) and CG(Q1 (Z(P))) have normal p-complements. Then G has a normal p-complement.

Proof. Assume otherwise and let G be a minimal counterexample. By the Frobenius p-Complement Theorem there is a p-local subgroup H of G which possesses no normal p-complement. By Sylow's Theorem we take Q = P n H E Syl,(H). Choose H with Q maximal subject to these constraints.

Claim P = Q. If not Q < Np(Q), so ING(C)lp > 1 Ql for each C char Q. Hence, by maximality of Q, NG(C) has a normal p-complement. So NH(C) also has a normal p-complement. In particular this holds for C = J(Q) and Ql (Z(Q)), so the hypotheses of the theorem hold in H . Hence, by minimality of G, H has a normal p-complement, contrary to the choice of H.

So P E Syl,(H). Hence, as H has no p-complement, H = G by minimality of G. Moreover if O,t(G) # 1 then, by 18.7 and minimality of G, G/O,/ (G) has a p-complement, and then G does too. So O,i(G) = 1.

As G = H is a p-local, O,(G) # 1. Let G* = G/O,(G). Then O,(G*) = 1 so NG(J(P*)) and CG(Q1 (Z(P*))) are proper subgroups of G containing P , and hence by minimality of G have p-complements. So, by rninimality of G, G* has a p-complement. Hence G = O,,,/,,(G). Thus E(G) = O,,,~,,(E(G)). But [E(G), F(G)] = 1 so O,(E(G)) ( Z(E(G)), and thus O,,,t(E(G)) = 0, (E(G)) x O,!(E(G)). Now O,t(E(G)) ( O,t(G) = 1, so E(G) is a p-group. Hence, as E(G) is perfect, E(G) = 1. So F*(G) = F(G)E(G) = O,(G).

Let p # r E n(G) and R E Syl,(G). R ( O,,,I(G), so, by a Frattini Argu- ment, Op(G)NG(Z(R)) contains a Sylow p-group of G which we may take to be P . In particular P acts on O,(G)Z(R) so PZ(R) = K is a subgroup of G. Now if K # G then, by minimality of G, Z(R) = O,!(K) 5 K. But then Z(R) ( CG(Op(G)) ( O,(G) by 31.10, and hence Z(R) = 1, a contradiction.

So G = PZ(R). In particular G is solvable and R = Z(R) is abelian. But now by Thompson Factorization, 32.6, G = G 1 G2 where GI = NG(J(P)) and G2 = CG(Ql (Z(P)). By hypothesis Gi = Opf(Gi)P. As O,t(Gi) ( CG(Op(G)) ( O,(G), OPf(Gi)= l ,soGi = P.ThusG=G1G2 = P,contradictingthechoice of G as a counterexample.

The next result does not involve normal p-complements, but it has the same flavor.

(39.6) (Baer-Suzuki Theorem) Let X be a p-subgroup of G. Then either X ( O,(G) or there exists g E G with (X, X g ) not a p-group.

Semiregular action 205

Proof. Assume the theorem is false and let P E Sylp(G) and A = XG flP. (Xc) < G, so, as X Op(G), (Xc) is not a p-group. Thus A 0 XG. ForY E XG - A, (Y, A) is not a p-group, by Sylow's Theorem. Let F be of maxi-mal order subject to r C A and (F, Y) a p-group for some Y E Xc - A. ThenF 0 A, but on the other hand as G is a counter example to the theorem, atleast r is nonempty. Let Q = (F, Y). By maximality of r, r = A fl Q. Henceby Exercise 11.4 we may take Y < N(F) and there is Z E No(F) - F. AsX Op(G), NG(F) 0 G, so, by induction on the order of G, (XG fl NG(F)) isa p-group. But now r C F U {Z} = F' and (I", Y) is a p-group, contradictingthe maximality of F.

40 Semiregular actionIn this section G is a nontrivial finite group and A is a nontrivial group of auto-morphisms of G which acts semiregularly on G. Recall this means CG (a) = 1for each a E A#.

(40.1) 1 G I = 1 mod CAI. In particular (IAA, 1GI)=1.

Proof. As A is semiregular, each orbit of A on G# is of order I Al.

We will wish to apply coprime action, 18.6, to the representation of A on G.Thus, until 40.7, assume either G or A is solvable. In 40.7 we prove G isnilpotent, at which point we see the assumption was unnecessary.

(40.2) A is semiregular on each A-invariant subgroup and factor group of G.

Proof. The first remark is trivial and the second follows from coprime action,18.7.

(40.3) For each p E ir(G), there is a unique A-invariant Sylow p-subgroupof G.

Proof. By coprime action, 18.7, the set A of A-invariant Sylow p-groups isnonempty and CG(A) is transitive on A. So as CG(A) = 1, the lemma follows.

(40.4) For each a E A, the map g H [g, a] is a permutation of G.

Proof. [g, a] =g-lga, So [g, a] _ [h, a] if and only if hg-1 E CG(a). So, asCG(a) = 1, the commutator map is an injection, and hence also a bijection asG is finite.


Proof. Assume the theorem is false and let P E Syl,(G) and A = xG n P. (xG) r! G, so, as X $O,(G), (xG) is not a p-group. Thus A # xG. For Y E xG - A, (Y, A) is not a p-group, by Sylow's Theorem. Let r be of maximal order subject to r s A and ( r , Y) a p-group for some Y E xG - A. Then r # A, but on the other hand as G is a counter example to the theorem, at least r is nonempty. Let Q = ( r , Y) . By maximality of r , r = A n Q. Hence by Exercise 11.4 we may take Y 5 N(r ) and there is Z E NA(r) - r . As X $ O,(G), NG(r) # G, so, by induction on the order of G, (xG fl NG(r)) is a p-group. But now r c r U {Z} = r1 and (rl, Y) is a p-group, contradicting the maximality of r .

40 Semiregular action In this section G is a nontrivial finite group and A is a nontrivial group of automorphisms of G which acts semiregularly on G. Recall this means CG(a) = 1 for each a E A'.

(40.1) IGI 1 mod IAl. In particular (IAl, IGI)= 1.

Proof. As A is semiregular, each orbit of A on G' is of order j A 1.

We will wish to apply coprime action, 18.6, to the representation of A on G. Thus, until 40.7, assume either G or A is solvable. In 40.7 we prove G is nilpotent, at which point we see the assumption was unnecessary.

(40.2) A is semiregular on each A-invariant subgroup and factor group of G.

Proof. The first remark is trivial and the second follows from coprime action, 18.7.

(40.3) For each p E n(G), there is a unique A-invariant Sylow p-subgroup of G.

Proof. By coprime action, 18.7, the set A of A-invariant Sylow p-groups is nonempty and CG(A) is transitive on A. So as CG(A) = 1, the lemma follows.

(40.4) For each a E A, the map g H [ g , a] is a permutation of G.

Proof. [ g , a] = ,g-lga, So [ g , a] = [ h , a] if and only if hg-' E CG(a). So, as CG(a) = 1, the commutator map is an injection, and hence also a bijection as G is finite.

(40.5) If A is of even order there is a unique involution t in A, g` = g-' forg E G, and G is abelian.

Proof. As I A I is even there is an involution t E A. Let g E G. By 40.4, g = [h, t]for some h E G. Thus g` _ (h-1h`)` = h-`h = g-1. Therefore, for each x E G,xr = x-1, so xg = (xg)-` = (g-lx_1)t = g-`x-1 = gx. So G is abelian.Finally, ifs is any involution in A, then I've shown s inverts G, so st E CA(G)=1. Thus t is unique.

(40.6) Let p, q E 7r(A). Then(1) If p is odd then Sylow p-groups of A are cyclic.(2) Sylow 2-groups of A are cyclic or quaternion.(3) Subgroups of A of order pq are cyclic.(4) Subgroups of A of odd order are metacyclic.

Proof. By 40.2 and 40.3 we may assume G is an r-group for some prime r,and indeed replacing G by G/c1(G) we may assume G is elementary abelian.Observe that, for B < A, (CG(b): b E B#) G as CG(b) = 1 for each b E B#.So, by Exercise 8.1, mp(A)= 1. Hence Exercise 8.4 implies (1) and (2). Part(3) follows from 27.18, while (1) and 39.3 imply (4).

(40.7) G is nilpotent.

Proof. Let G be a minimal counterexample and a an element of A of primeorder. By induction on the order of A we may take A = (a). In particular A issolvable, so all lemmas in this section apply.

Suppose q E 7r(G) and 1 Q is an A-invariant normal elementary abelianq-subgroup of G. By 40.2 and minimality of G, G/Q is nilpotent so RQ < Gfor each Sylow r-group R of G. In particular if r = q then R < G. But asG is a counter example there exists r E 7r(G) with Sylow r-groups of G notnormal. By symmetry between r and q, Or(G) = 1. As RQ a G, CR(Q) <Or(G) = 1, so Z(R) is faithful on Q. By 40.3 we may choose R to be a-invariant; then a is faithful on Z(R). But now, as CQ(a) =1, 36.2 supplies acontradiction.

So F(G) = 1. Now if H is a proper A-invariant normal subgroup of G then,by minimality of G, H is nilpotent, so, as F(G) = 1, H = 1.

As G is not nilpotent, G is not a 2-group so there is an odd prime p E 7r(G).Let P be an A-invariant Sylow p-group of G, G1= NG(J(P)) and G2 =CG(S21(Z(P))). As F(G) = 1, G; is a proper A-invariant subgroup of G, so G,

206 Transfer andfision

(40.5) If A is of even order there is a unique involution t in A, gt = g-l for g E G, and G is abelian.

Proof. As IAl is even there is an involution t E A. Let g E G. By 40.4, g = [h, t ] for some h E G. Thus gt = (h-'ht)* = h-'h =g-l. Therefore, for each x E G, xt = X-l, SO xg = (xg)-' = (g-lx-l)t = g-tx-t = gx. So G is abelian. Finally, if s is any involution in A, then I've shown s inverts G, so st E CA(G) = 1. Thus t is unique.

(40.6) Let p, q E n(A). Then (1) If p is odd then Sylow p-groups of A are cyclic. (2) Sylow 2-groups of A are cyclic or quaternion. (3) Subgroups of A of order pq are cyclic. (4) Subgroups of A of odd order are metacyclic.

Proof. By 40.2 and 40.3 we may assume G is an r-group for some prime r , and indeed replacing G by G/@(G) we may assume G is elementary abelian. Observe that, for B ( A, (CG(b): b E B') # G as CG(b) = 1 for each b E B'. So, by Exercise 8.1, m,(A) = 1. Hence Exercise 8.4 implies (1) and (2). Part (3) follows from 27.18, while (1) and 39.3 imply (4).

(40.7) G is nilpotent.

Proof. Let G be a minimal counterexample and a an element of A of prime order. By induction on the order of A we may take A = (a). In particular A is solvable, so all lemmas in this section apply.

Suppose q E n(G) and 1 # Q is an A-invariant normal elementary abelian q-subgroup of G. By 40.2 and minimality of G, G/Q is nilpotent so RQ 9 G for each Sylow r-group R of G. In particular if r = q then R G. But as G is a counter example there exists r E n(G) with Sylow r-groups of G not normal. By symmetry between r and q, O,(G) = 1. As RQ 9 G, CR(Q) 5 O,(G) = 1, so Z(R) is faithful on Q. By 40.3 we may choose R to be a- invariant; then a is faithful on Z(R). But now, as CQ(a) = 1, 36.2 supplies a contradiction.

So F (G) = 1. Now if H is a proper A-invariant normal subgroup of G then, by minimality of G, H is nilpotent, so, as F(G) = 1, H = 1.

As G is not nilpotent, G is not a 2-group so there is an odd prime p E n(G). Let P be an A-invariant Sylow p-group of G, G1 = NG(J(P)) and G2 = CG(n1(Z(P))). As F(G) = 1, Gi is a proper A-invariant subgroup of G, so Gi

is nilpotent by minimality of G. In particular G; has a normal p-complement,so, by the Thompson Normal p-Complement Theorem, G has a normal p-complement. This is impossible as G possesses no nontrivial proper A-invariantnormal subgroups.

Recall from section 35 that a Frobenius group is a finite group H which isthe semidirect product of a nontrivial group K by a nontrivial group B with Bsemiregular on K. K and B are the Frobenius kernel and Frobenius complementof H, respectively. Notice that, by 40.7:

(40.8) Frobenius kernels are nilpotent.

Notice that the lemmas in this section also give lots of information aboutFrobenius complements.

Remarks. Transfer and fusion are basic tools in the study of finite groups. Onecan begin to see the power of these tools in some of the lemmas and exercisesin this chapter, but just barely.

The proof of the Baer-Suzuki Theorem essentially comes from Alperinand Lyons [AL]. Alperin's Fusion Theorem is (surprise) due to Alperin [Al].Thompson was the first to prove the nilpotence of Frobenius kernels in histhesis.

Exercises for chapter 131. Let G be a finite group, p a prime, H < G with (IG : H1, p) =1, and g a

p-element in H. g is extremal in H if ICH(g)I p = ICc(g)1p(1) Represent G on G/H by right multiplication and let Hx be a fixed

point of g on G/H. Prove the orbit HxCG(g) of CG(g) on G/H is oforder prime to p if and only if gx ' is extremal in H.

(2) Assume g is of order p, K < H with H/K abelian, g E H - K, g isextremal in H, and each H-conjugate of g extremal in H is containedin gK. Prove g Gel).

2. Let G be a finite group with G = 02 (G) and let T E Sy12(G). Assume T isdihedral, semidihedral, or l2n wr12, and prove G has one conjugacy classof involutions.

3. Let G be a finite group with G =02 (G). Prove that if m2(G) > 2 thenm2(CG(x)) > 2 for each involution x in G. (Hint: Let T E Sy12(G). Provethere is E4 = U < T and x G fl CT(U) is nonempty for each involution xof G.)


is nilpotent by minimality of G. In particular Gi has a normal p-complement, so, by the Thompson Normal p-Complement Theorem, G has a normal p- complement. This is impossible as G possesses no nontrivial proper A-invariant normal subgroups.

Recall from section 35 that a Frobenius group is a finite group H which is the semidirect product of a nontrivial group K by a nontrivial group B with B semiregular on K. K and B are the Frobenius kernel and Frobenius complement of H , respectively. Notice that, by 40.7:

(40.8) Frobenius kernels are nilpotent.

Notice that the lemmas in this section also give lots of information about Frobenius complements.

Remarks. Transfer and fusion are basic tools in the study of finite groups. One can begin to see the power of these tools in some of the lemmas and exercises in this chapter, but just barely.

The proof of the Baer-Suzuki Theorem essentially comes from Alperin and Lyons [AL]. Alperin's Fusion Theorem is (surprise) due to Alperin [All. Thompson was the first to prove the nilpotence of Frobenius kernels in his thesis.

Exercises for chapter 13 1. Let G be a finite group, p aprime, H 5 G with (IG: HI, p ) = l , and g a

p-element in H. g is extrernal in H if ICH(g)lp = ICG(g)lp. (1) Represent G on G/H by right multiplication and let Hx be a fixed

point of g on G/H. Prove the orbit HxCG(g) of CG(g) on G/H is of order prime to p if and only if gX-' is extremal in H .

(2) Assume g is of order p , K H with H/K abelian, g E H - K, g is extremal in H, and each H-conjugate of g extremal in H is contained in gK. Prove g 4 G(').

2. Let G be a finite group with G = o~(G) and let T E Sy12(G). Assume T is dihedral, semidihedral, or Z2" wrZ2, and prove G has one conjugacy class of involutions.

3. Let G be a finite group with G = 0 2 ( ~ ) . Prove that if m2(G) > 2 then m2(CG(x)) > 2 for each involution x in G. (Hint: Let T E Sy12(G). Prove there is E4 Z U T and xG f l CT(U) is nonempty for each involution x of G.)

4. Let G be a finite group, p a prime, and assume a Sylow p-subgroup of Gis the modular group Mp., of order p" # 8. Prove(1) OP(G) has cyclic Sylow p-subgroups.(2) If p = 2 then G has a normal 2-complement.

5. Let G be a finite group with quaternion Sylow 2-subgroups. Prove that eitherG has a normal 2-complement or there exists K < H < G with H/KSL2(3).

6. Let G be a group of order 60. Prove either G is solvable or G is isomorphicto the alternating group A5 of degree 5. (Hint: Let T E Sy12(G). Show T a Gor G has a normal 2-complement or JG : NG(T)I = 5 and kerNG(T)(G) = 1.)


4. Let G be a finite group, p a prime, and assume a Sylow p-subgroup of G is the modular group MpJZ of order pn # 8. Prove (1) Op(G) has cyclic Sylow p-subgroups. (2) If p = 2 then G has a normal 2-complement.

5. Let G be a finite group with quaternion Sylow 2-subgroups. Prove that either G has a normal 2-complement or there exists K 9 H 5 G with H / K 2

sL2(3). 6. Let G be a group of order 60. Prove either G is solvable or G is isomorphic

to the alternating group Ag of degree 5. (Hint: Let T E Syl,(G). Show T 9 G or G has a normal 2-complement or (G : Nc(T)( = 5 and ker~ , (~ , (G) = 1.)

14

The geometry of groups of Lie type

Chapters 4 and 7 introduced geometries preserved by the classical groups.Chapter 14 considers these geometries (and related geometries preserved byCoxeter groups) in detail, and uses the representations of the classical groupson their geometries to establish various group theoretical results.

For example we'll see that the finite classical groups Ln(q), Un(q), PSpn(q),and PS2n(q) are simple, with a few exceptions when n and q are small. AlsoLn(F) and PSpn(F) are simple for infinite fields F, as are Un(F) and PQn(F)under suitable restrictions on F. If F is finite of characteristic p, it will developthat the stabilizer B of a maximal flag of the geometry of a classical group Gover F is the normalizer of a Sylow p-group of G, and the subgroups of Gcontaining B are precisely the stabilizers of flags fixed by B. These subgroupsand their conjugates are the parabolic subgroups of G. We say B is the Borelgroup of G.

It also turns out that to each classical group G there is associated a Coxetergroup called the Weyl group of G. The Weyl groups of the classical groups areof type A, Cn, or Dn. The structure of G is controlled to a large extent by thatof its Weyl group: see for example lemma 43.7 and Exercise 14.6.

41 ComplexesBefore beginning this section the reader may wish to review the discussionof geometries in section 3. Section 41 is devoted to a related class of objects:complexes. A complex is a pair -6' = (F, -') where F is a geometry over somefinite index set I and -6' is a collection of flags of F of type I. The members of6' are called chambers. Subflags of chambers are called simplices. Simplicesof corank 1 are called walls. A complex is thin if each wall is contained inexactly two chambers. A complex is thick if each wall is contained in at leastthree chambers.

Define the chamber graph of 6' to be the graph on 6' obtained by joiningchambers which have a common wall. A path in the chamber graph is called agallery. A complex is said to be connected if its chamber graph is connected. Aconnected complex (F, -6') in which every flag of F of rank 1 or 2 is a simplexis called a chamber complex.

A morphism a: (F, f) -* (A, 9) of complexes is a morphism F - A ofgeometries with -6'a contained in 9. A subcomplex of (F, -2) is a complex

The geometry of groups of Lie type

Chapters 4 and 7 introduced geometries preserved by the classical groups. Chapter 14 considers these geometries (and related geometries preserved by Coxeter groups) in detail, and uses the representations of the classical groups on their geometries to establish various group theoretical results.

For example we'll see that the finite classical groups L,(q), U,(q), PSp,(q), and PQE(q) are simple, with a few exceptions when n and q are small. Also L,(F) and PSp,(F) are simple for infinite fields F , as are U,(F) and PQ,(F) under suitable restrictions on F . If F is finite of characteristic p, it will develop that the stabilizer B of a maximal flag of the geometry of a classical group G over F is the normalizer of a Sylow p-group of G, and the subgroups of G containing B are precisely the stabilizers of flags fixed by B. These subgroups and their conjugates are the parabolic subgroups of G . We say B is the Bore1 group of G .

It also turns out that to each classical group G there is associated a Coxeter group called the Weyl group of G . The Weyl groups of the classical groups are of type A,, C,, or D,. The structure of G is controlled to a large extent by that of its Weyl group: see for example lemma 43.7 and Exercise 14.6.

41 Complexes Before beginning this section the reader may wish to review the discussion of geometries in section 3. Section 41 is devoted to a related class of objects: complexes. A complex is a pair 6' = ( r , 6') where r is a geometry over some finite index set I and 6' is a collection of flags of r of type I. The members of 67 are called chambers. Subflags of chambers are called simplices. Simplices of corank 1 are called walls. A complex is thin if each wall is contained in exactly two chambers. A complex is thick if each wall is contained in at least three chambers.

Define the chamber graph of 67 to be the graph on 6' obtained by joining chambers which have a common wall. A path in the chamber graph is called a gallery. A complex is said to be connected if its chamber graph is connected. A connected complex ( r , 6') in which every flag of r of rank 1 or 2 is a simplex is called a chamber complex.

A morphism a : ( r , 6') -+ (A, !%) of complexes is a morphism r + A of geometries with 6'a contained in !%. A subcomplex of ( r , 6') is a complex

210 The geometry of groups of Lie type

(0, 5) with 0 a subgeometry of F and _T _ t' fl o the set of chamberscontained in A.

The notions of complex, building, and Tits system (which are the subjectof this chapter) come from Tits [Ti]. However the definition of a complexI've just given is somewhat less general than that of Tits [Ti] in that undermy definition there is a type function defined on simplices inherited from theassociated geometry. However Tits shows how to associate type functions toCoxeter complexes and buildings, so in the end the class of object consideredis the same.

The treatment of complexes, buildings, and Tits systems given here is ex-tracted from Tits [Ti] and Bourbaki [Bo], modulo remarks above.

Let G be a group and -IF= (G;: i E 1) a family of subgroups of G. Thecoset geometry F(G, T) determined by G and is defined in section 3. Let-t'(G, 9) be the complex on F(G, _,F) whose chamber set (also denoted by- '(G, ST)) consists of the flags SI,,, X E G, where SI,x = (G,x: i E fl.

(41.1) Let G be a group, = (G1: i E I) a family of subgroups of G, F =F(G, ST), and -' = t'(G, ST). Then

(1) G is represented as a group of automorphisms of -6 by right multiplica-tion on the cosets in F.

(2) G is transitive on the simplices of -t' of type J for each subset J of I.In particular G is transitive on chambers.

(3) G,I is the stabilizer of the simplex Sj of type J.(4) The walls of chambers of -6 are the conjugates of Si', i E 1, under G.

Si, is contained in exactly IGp: GI ( chambers.(5) Every flag in r of rank 1 or 2 is a simplex of -t'.

The proof is straightforward; the notation is explained in section 3.

(41.2) Let C be a chamber in a chamber complex -0 and a a morphism of -6fixing C. Then

(1) a fixes each simplex contained in C.(2) If -0 is thin and a is a bijection on e, then a = 1.

Proof. a is a morphism of geometries so it preserves type. Hence as C containsa unique simplex of each type, (1) holds.

To prove (2) it suffices to show a fixes each chamber adjacent to C, sinceic is connected. Let D be such a chamber; then D fl c = W is a wall of C,so a fixes W by (1). As -t' is thin, D and C are the only chambers containingW, so, as W = Wet = Da fl Ca, (Ca, Da) = (C, D). Now, as C = Ca, alsoD = Da.


(A, GB) with A a subgeometry of r and GB = 6' r l A the set of chambers contained in A.

The notions of complex, building, and Tits system (which are the subject of this chapter) come from Tits [Ti]. However the definition of a complex I've just given is somewhat less general than that of Tits [Ti] in that under my definition there is a type function defined on simplices inherited from the associated geometry. However Tits shows how to associate type functions to Coxeter complexes and buildings, so in the end the class of object considered is the same.

The treatment of complexes, buildings, and Tits systems given here is ex- tracted from Tits [Ti] and Bourbaki [Bo], modulo remarks above.

Let G be a group and F= (Gi: i E I ) a family of subgroups of G. The coset geometry r(G, g) determined by G and 9 is defined in section 3. Let 6'(G, 9) be the complex on r(G, F ) whose chamber set (also denoted by &(G, F ) ) consists of the flags SI,,, x E G, where SI,, = (Gix: i E I).

(41.1) Let G be a group, F = (Gi: i E I ) a family of subgroups of G, r = r(G, F ) , and f3 = 6'(G, F ) . Then

(1) G is represented as a group of automorphisms of B by right multiplication on the cosets in r.

(2) G is transitive on the simplices of 6' of type J for each subset J of I . In particular G is transitive on chambers.

(3) G is the stabilizer of the simplex S j of type J . (4) The walls of chambers of 6' are the conjugates of Si,, i E I , under G.

Sit is contained in exactly I Gi,: GI I chambers. (5) Every flag in r of rank 1 or 2 is a simplex of 6'.

The proof is straightforward; the notation is explained in section 3.

(41.2) Let C be a chamber in a chamber complex 6' and a a morphism of B fixing C. Then

(1) a fixes each simplex contained in C. (2) If 6' is thin and a is a bijection on 6', then a = 1.

Proof. a is a morphism of geometries so it preserves type. Hence as C contains a unique simplex of each type, (1) holds.

To prove (2) it suffices to show a fixes each chamber adjacent to C , since f3 is connected. Let D be such a chamber; then D n C = W is a wall of C, so a fixes W by (1). As 6' is thin, D and C are the only chambers containing W , so, as W = W a = Da r l Ca, (Ca, Da) = (C , D). Now, as C = C a , also D = Da.

Complexes 211

Let 6' be a thin chamber complex. A folding of -t' is an idempotent morphismwhose fibres on chambers are all of order 2. -t' is a Coxeter complex if for eachpair C, D of adjacent chambers there exists a folding mapping C to D.

In the remainder of this section assume = (F, -') is a Coxeter complex.

(41.3) Let 0 be a folding of t'. Then(1) 0 induces a bijection 0: (P - eo) -> eo.(2) 0 fixes each member of r o and i ' j .

Proof. Part (2) is just a restatement of the hypothesis that 4) is idempotent.Let C E -6'0. By hypothesis there is a unique chamber D distinct from C withDo = C. By (2), D is not in eo.

For C, D in -' let d(C, D) be the distance between C and D in the chambergraph.

(41.4) Let 0 be a folding of t', C in eo, and D in t' - eo. Then(1) If C = Co, ... , C = D is a gallery, there exists 0 < i < n with

Ci E -6'0 and Ci+1 eo(2) If C is adjacent to D then Do = C.(3) Let Ci = Ci if Ci is not in eo and C = 1(Ci) - (Ci } if Ci is in i ' .

Then Ci is adjacent to C1+1.

Proof. Part (1) is clear. Assume W = C fl D is a wall. By 41.3, 0 fixes C andW. Thus W = WO c_ Do, so, as is thin, Do = C or D. As D is not in-00, Dip = C, so (2) holds.

Part (3) is clear if neither Ci nor Ci+1 is in -00, so let Ci be in eo. LetU = Ci fl Ci+1, v = O-1(U) fl C, and E the chamber through V distinct fromCi'.Then U=VcpcEO,so EO=C,or Ci+1.If EO=Ci+1then E=C+1is adjacent to Cl. If EO = Ci then, as 0-1(Ci) = (C, Ci}, we have E = Ci.Hence V = VO = U by 41.2.1, so C = Ci+1. Therefore Ci+10 = Ci Ci+1,so Ci+1 e o, and hence (Ci+i )' = Ci+i Thus C' = Ci+1 is adjacent to Ci+1 =

Ci+1

(41.5) Let C and C' be distinct adjacent chambers and let 0 and O' be foldingswith C'O = C and Cep' = C. Then

(1) If D is in eo then d(C', D) = d(C, D) + 1.(2) If D is not in eo then d(C', D) = d(C, D) - 1.(3) eo' _ e - eo.

Proof. As C is adjacent to C', Id(C, D) - d(C', D)J < 1 for each D E P.

Complexes 21 1

Let 6' be a thin chamber complex. A folding of 6' is an idempotent morphism whose fibres on chambers are all of order 2.6' is a Coxeter complex if for each pair C , D of adjacent chambers there exists a folding mapping C to D.

In the remainder of this section assume 6' = (r, 6') is a Coxeter complex.

(41.3) Let 4 be a folding of 6'. Then (1) 4 induces a bijection 4: (6' - 6'4) --+ 6'4. (2) 4 fixes each member of r4 and &?4.

Proof. Part (2) is just a restatement of the hypothesis that 4 is idempotent. Let C E -64. By hypothesis there is a unique chamber D distinct from C with D@ = C . By (2), D is not in 6'4.

For C , D in 6' let d(C, D ) be the distance between C and D in the chamber graph.

(41.4) Let 4 be a folding of 6', C in 6'4, and D in 6' - 6'4. Then (1) If C = Co, . . . , C,, = D is a gallery, there exists 0 5 i < n with

Ci E 6'4 and Ci+l $6 '4 . (2) If C is adjacent to D then D@ = C . (3) Let Cf = Ci if Ci is not in 6'4 and Cf = @-'(Ci) - (Ci} if Ci is in 6'4.

Then C; is adjacent to C;+l.

Proof. Part (1) is clear. Assume W = C rl D is a wall. By 41.3,4 fixes C and W . Thus W = W @ E D4, so, as 6' is thin, D@ = C or D. As D is not in 6'4, D@ = C , so (2) holds. Part (3) is clear if neither Ci nor Ci+l is in 6'4, so let Ci be in 6'4. Let

U = Ci r l Ci+l, V = 4-'(u) r l C;, and E the chamber through V distinct from Cf. Then U = V @ E E 4 , so E@ = Ci or Ci+1. If E@ = Ci+1 then E = Cf+, is adjacent to Cf. If E@ = Ci then, as 4-'(ci) = (Cf, C i} , we have E = Ci. Hence V = V @ = U by41.2.1, SO Cf = Ci+1. Therefore Ci+1@ = Ci # Ci+17 so Ci+l $! 6'4, and hence (Ci+1)' = Ci+1. Thus C,! = Ci+1 is adjacent to Ci+1 =

Cf+l.

(41.5) Let C and C' be distinct adjacent chambers and let 4 and 4' be foldings with C'4 = C and C4' = C'. Then

(1) If D is in 6'4 then d(C1, D ) = d(C, D ) + 1. (2) If D is not in 6'4 then d(C1, D ) = d(C, D ) - 1. (3) -e4l= e - 64 .

Proof. As C is adjacent to C', Id(C, D) - d(C1, D)I _< 1 for each D E 6.

Assume D is in-00 and C' = Co, ... , C, = D is a galleryoflength d(C', D).Then (Cio:0 < i < n) is a gallery between C = C'O and Do = D as 0 isa morphism. By 41.4 there is 0 i < n with CiO = C;+1 = Ci+10. HenceC = CO, ... , C;_10, Ci+10, ... , CnO = D is a gallery of length at mostn -1 from C to D. Therefore d(C, D) < n = d(C', D), so (1) follows fromthe first paragraph of the proof.

Similarly if D 0 -00 let C = Co, ... , Cn = D be a gallery of length d(C, D).Define C' as in 41.4.3. By 41.4.3, C' = Co, ... , Cn = D is a gallery, whileby 41.4.1 and 41.4.2 there is 0 < i < n with C; in ,°¢, C;+1 not in -6'0 andC;+1O = C. Hence C= C,+1 = Ci+i, so, as in the preceding paragraph,d(C', D) < n and then (2) holds.

Finally let D E '. If D is in 6' ', then, by (1) applied to 0' and 0, D is notin e o. So io' c_ -C - e o. If D is not in 6' then, by (2) applied to 0 and 0',D is in i/'. Thus i' - eo c io'. So (3) is established.

Foldings 0 and 0' are defined to be opposite if e o' = .6' - 60.

(41.6) Let 0 and 0' be opposite foldings and define a = a(O, 0'): r -* r byva = v/ if v E ro' and va = vO' if v E ro. Then a is an automorphism of- of order 2 whose orbits on -0 are the fibres of 0.

Proof. By hypothesis 6' = 6' U -00. By definition of chamber complex, eachmember of r is a simplex, so r = ro U ro'. By 41.3.2, 0 and 0' agree onr¢ fl ro', so a is well defined. By 41.3.1, a is bijective on '. If u and v areincident members of r, then as i' is a chamber complex there is C in -' withu and v in C. Then ua, va c Ca c e, so ua * va. Hence a is morphism. By41.4.1 there exist adjacent chambers C and D with C in io and D not in eo.By 41.4.2, DO = C and CO' = D. Then Ca 2 = Cq5'q5 = Do = C. Hence, by41.2.2, a2 = 1. Therefore a-1 = a is a morphism, so a is an automorphism oforder 2.

Complexes 213

there exist foldings ¢ and 0' of 6 with CO = C' and C'O' = C. By 41.5, ¢and 0' are opposite, so by 41.6 they determine an involutory automorphisma(O, 0') of 6, called a reflection through C fl C. Let S denote the set of allreflections through C fl C' as C' varies over A(C), and let W be the subgroupof Aut(-) generated by S. It will develop that (W, S) is a Coxeter system andW = Aut(o).

For W E W let 1(w) be the length of w with respect to the generating set S.

(41.7) Let w = sl ... Sn E W with Si E S. Then(1) C, CSn, CSn_1Sn, ... , CS1... Sn is a gallery from C to Cw.(2) W is transitive on -0.(3) d(C, Cw) = 1(w).

Proof. Let S E S. By definition of S there exists C' E A(C) and foldingsthrough c fl C' with s = a(0, 0'), and, by 41.6, (C, C') is a cycle of s.

Let w; = sn_i+1... sn. We've just seen C is adjacent to Csn_;+1 so Cw; _CSn_i+l Wj_1 is adjacent to Cwi_,. Hence C, Cw 1, ... , Cwn = Cw is a galleryfrom C to Cw of length n, so d(C, Cw) < n. In particular d(C, Cw) < 1(w).

Conversely let C = CO, C1,..., C,n be a gallery of length d(C, Cm) = m. I'llshow there exist r, E S, 1 < i < m, such that Ck = CUk, where Uk =rm-k+1 ... rm. Notice that this establishes the transitivity of W on -0. Pro-ceed by induction on m; the case m = 1 has been handled in the first paragraphof this proof, so take m > 1. By induction Ck = Cuk fork < m. Then Cm isadjacent to Cm-1 = Cum_, so Cm(um_1)-1 is adjacent to C. Hence there isr1 E S with Cr, = Cm(um-1) 1, so Cum = Crlum-1 = Cm.

Finally let u = r1... rm and assume Cm = Cw. Then Cu = Cm = Cw souw-1 fixes C and hence, by 41.2, u = w. So 1(w) = 1(u) < m = d(C, Cw),completing the proof.

(41.8) (1) W = Aut(6').(2) W is regular on e.

The element a(0, rp') of 41.6 is called a reflection and is said to be a reflection(3) 1 SI = 111; that is there exists a unique reflection through each wall of C.

through c fl C' for each pair C, C' of adjacent chambers with C E eo andC' 0'q.

Fix a chamber C in i. is defined over some finite index set I ; let m = Ifor the moment. Notice each of the m subsets of I of order m - 1 determinesa wall of C, and, as i' is thin, each wall X determines a unique chamber C'with C' fl c = X. Let A(C) denote the set of these chambers; then A(C) isthe set of chambers distinct from C and adjacent to C in the chamber graph.Further A(C) is of order m. As -' is a Coxeter complex, for each C' in A(C)

Proof. By 41.7, W is transitive on 6', so G = Aut(i') = WGc where GC isthe stabilizer in G of C. But, by 41.2.2, Gc = 1, so (1) and (2) hold. For eachC' E A(C) there is S E S with cycle (C, C'), so BSI IA(C)I _ III. Further ift is a member of S with cycle (C, C') then St E We =1, So S = t.

(41.9) (W, S) is a Coxeter system.

Complexes 213

there exist foldings 0 and 0' of 6 with Co = C' and C'O' = C. By 41.5, 0and 0' are opposite, so by 41.6 they determine an involutory automorphisma(0, 0') of 6, called a reflection through C n C. Let S denote the set of allreflections through C n C' as C' varies over A(C), and let W be the subgroupof Aut(-') generated by S. It will develop that (W, S) is a Coxeter system andW = Aut(o).

For W E W let 1(w) be the length of w with respect to the generating set S.

(41.7) Let w = sl ... Sn E W with si E S. Then(1) C, Csn, CSn-Is n. ... . Csl ... sn is a gallery from C to Cw.(2) W is transitive on -0.(3) d(C, Cw) = 1(w).

Proof. Let S E S. By definition of S there exists C' E A(C) and foldingsthrough c n C' with s = a(0, 4'), and, by 41.6, (C, C') is a cycle of s.

Let wi = Sn-i+l ... sn. We've just seen C is adjacent to Csn_i+l so Cwi =Csn-i+l wi -1 is adjacent to Cw i_ 1 . Hence C, Cw 1, ... , Cwn = Cw is a galleryfrom C to Cw of length n, so d(C, Cw) < n. In particular d(C, Cw) < 1(w).

Conversely let C = Co, C1, ... , Cbe a gallery of length d(C, Cm) = m. I'llshow there exist ri E S, 1 < i < m, such that Ck = CUk, where Uk =rm_k+i ... rm. Notice that this establishes the transitivity of W on P. Pro-ceed by induction on m; the case m = 1 has been handled in the first paragraphof this proof, so take m > 1. By induction Ck = Cuk fork < m. Then Cm isadjacent to Cm-1 = Cum_1 so Cm(um_1)-1 is adjacent to C. Hence there isr1 E S with Cr, = Cm(um_1)-1, so Cum = Crium-1 = Cm.

Finally let u = r1 ... rm and assume Cm = Cw. Then Cu = Cm = Cw souw-1 fixes C and hence, by 41.2, u = w. So 1(w) = 1(u) < m = d(C, Cw),completing the proof.

(41.8) (1) W = Aut(6 ).(2) W is regular on e.(3) I SI = I 11; that is there exists a unique reflection through each wall of C.

Proof. By 41.7, W is transitive on ', so G = Aut(o) = WGc where Ge isthe stabilizer in G of C. But, by 41.2.2, Gc = 1, so (1) and (2) hold. For eachC' E 0(C) there is s E S with cycle (C, C'), so ISI ? I A(C)I = III. Further ift is a member of S with cycle (C, C') then St E We = 1, so s = t.


Complexes

there exist foldings 4 and 4' of fi? with C 4 = C' and C'4' = C. By 41.5, 4 and 4' are opposite, so by 41.6 they determine an involutory automorphism a ( 4 , 4 ' ) of fi?, called a reflection through C n C'. Let S denote the set of all reflections through C n C' as C' varies over A(C), and let W be the subgroup of Aut(6) generated by S. It will develop that (W , S ) is a Coxeter system and W = Aut(fi?).

For w E W let l (w) be the length of w with respect to the generating set S.

(41.7) Let w = sl . . . s, E W with si E S. Then (1) C , Cs,, CS,-~S,, . . . , Csl . . . s, is a gallery from C to Cw. (2) W is transitive on fi?. (3) d (C, C w ) = l(w).

Proof. Lets E S . By definition of S there exists C' E A(C) and foldings 4,4' through C f' C' with s = a ( 4 , 40, and, by 41.6, (C , C') is a cycle of s.

Let w, = s,-i+l . . . s,. We've just seen C is adjacent to CS,-,+~ so Cw, = is adjacent to CwiVl . Hence C , C w l , . . . , Cw, = Cw is agallery

from C to Cw of length n, so d(C, C w ) 5 n. In particular d(C, C w ) 5 l(w). Conversely let C = Co, C 1 , . . . , C,,, be agallery of length d(C, C,) = m. I'll

show there exist r, E S , 1 5 i 5 m, such that Ck = Cuk, where uk= rm-k+l . . . r,. Notice that this establishes the transitivity of W on fi?. Pro- ceed by induction on m; the case m = 1 has been handled in the first paragraph of this proof, so take m > 1. By induction Ck = Cuk for k < m. Then C, is adjacent to CmP1 = CumPl so C,(U,-~)-' is adjacent to C. Hence there is rl E S with Crl = C,(U,-~)-', so Cum = Crlu,,-1 = C,.

Finally let u = rl . . . r, and assume C, = Cw. Then Cu = C, = Cw so uw-' fixes C and hence, by 41.2, u = w. So l (w ) = l (u) 5 m = d(C, Cw) , completing the proof.

(41.8) ( 1 ) W = Aut(8). (2) W is regular on fi?. (3) I SI = I I I; that is there exists a unique reflection through each wall of C.

Proof. By 41.7, W is transitive on 8, so G = Aut(&) = WGc where Gc is the stabilizer in G of C. But, by 41.2.2, Gc = 1, so (1) and (2) hold. For each C' E A(C) there is s E S with cycle (C, C'), so IS1 2 IA(C)l = 111. Further if t is a member of S with cycle (C, C') then st E Wc = 1, so s = t .

(41.9) ( W , S ) is a Coxeter system.

Proof. It suffices to establish the exchange condition of 29.4. So let w =S1 ... sn E W with l (w) =n ands E S with l (w s) < l (w ). Let Ck = CSn_k+l ... Sn

forl <k <n.By41.7,

--9=(C=Co, C1, .. .,Cn=Cw)

is a gallery and d(C, Cw) = 1(w) > l(ws) = d(C, Cws). Let 0 be a foldingwith Cso = C.Nowd(Cs, Cw) = d(C, Cws) < d(C, Cw), so, by 41.5, CwCo. Then by 41.4 applied tog there exists i, 1 < i < n with Ci in Co, Ci+1 notin Co, and Ci+1/. = Ci. By definition of s, Cis = Ci+1, SO CSn-i+l ... SnS =Ci s = Ci+1 = Csn_i ... sn, and hence, by 41.8.2, Sn-i+1 . . SnS = Sn_i ... Sn, SO

the exchange condition is verified.

(41.10) Let (G, R) be a Coxeter system, R = (r1: i E I), Gi = (ri : j 0 i), and= R) = (Gi : i E I). Then P(G, J) is a Coxeter complex, G =

Aut(-'(G, 1F, )), and R is the set of reflections through the walls of the chamberC = (Gi:i E I).

Proof. Adopt the notation of section 3. By 29.13.3, Gi, = (r1) (recall i'=I - {i}) and GI = 1. So, by 41.1.4, _T = -'(G, J) is thin. Also G = (R) =(Gi,: i E I), so, by Exercise 14.3, _T is a chamber complex. By 41.1.2, G istransitive on - and, by 41.1.3, GI is the stabilizer of the chamber C, so, asGI = 1, G is regular on -T. In particular G is faithful on -9 so, by 41.1.1, Gis a subgroup of Aut(_9). Also G is transitive on _q so it remains only to showfor each wall Si, of C that there is a folding 0 through Si, with C in -90.

Let r = ri and define 0: G -+ G by

go = g if l(gr) = l(g) + 1

go = gr if 1(gr) = 1(g) - 1.

As G, = 1 we can identify G with _T via g H Cg and regard 0 as a functionfrom _T into -9. Claim that if D, D' E -9 and D fl D' is a wall of type j' thenDo fl D'o is also a wall of type j'. Now D = Cg and D' = Crjg for someg E G, so we must show Cgo fl C (r j g)o is a wall of type j', so it suffices to show(rjg)4 =rj(g4)foreachg E G.Thisisclearifl(rjgr)-l(rjg)=l(gr)-1(g).So, by symmetry between rjg and g, we may assume l(gr) = l(g) - 1 andl(rjgr)=l(rjg) + 1. Thus go=gr and (rjg)o=rjg. Let g=sn...si withl(g) = n and sk = rik. As l (gr) < l (g), the Exchange Condition says sn ... s1=sii-1 ... sir for some m. Thus g = S, ... Sm+lSm-1 . . sir, so without lossr = si, Hence go = gr = Sn ... S2. Next rjgr = rjs, ... s2 is of length atmost n = 1(g) so l(rjg) = l(rjgr) - 1 < 1(g). Then, by the Exchange Con-dition, r j Sn . . . Sk+i = Sn ... Sk for some k. If k 0 1 we may take sn = r.


Proof. It suffices to establish the exchange condition of 29.4. So let w = sl . . . s, E W withl(w) = n ands E S withl(ws) 5 l(w). LetCk = C~n-k+l . . . Sn

for 1 5 k 5 n. By 41.7,

-LC7 = (C = Co, C1,. . . , Cn = Cw)

is a gallery and d(C, Cw) = l(w) 2 l(ws) = d(C, Cws). Let @ be a folding withCs@ = C.Nowd(Cs, Cw) = d(C, Cws) 5 d(C, Cw),so, by41.5,Cw $ &@.Then by 41.4 applied to $? thereexists i, 1 5 i 5 n with Ci in&@, Ci+l not in &@, and Ci+i@ = Ci. By definition of s, Cis = Ci+l, SO CS,-~+~ . . . S,S = Cis =Ci+l = CS,-~. . . sn,andhence,by41.8.2,sn-i+l.. . S,S = s,-i .. . S,,SO the exchange condition is verified.

(41.10) Let (G, R) be a Coxeter system, R = (ri: i E I), Gi = (ri: j # i ) , and F = F(G, R) = (Gi: i E I). Then &(G, 9 ) is a Coxeter complex, G = Aut(&(G, F ) ) , and R is the set of reflections through the walls of the chamber C = (Gi: i E I}.

Proof. Adopt the notation of section 3. By 29.13.3, Git = (ri) (recall i '= I - {i}) and GI = 1. So, by 41.1.4, L3 = B(G, 9 ) is thin. Also G = (R) = (Gi,: i E I ) , SO, by Exercise 14.3, L3 is a chamber complex. By 41.1.2, G is transitive on g- and, by 41.1.3, GI is the stabilizer of the chamber C, so, as GI = 1, G is regular on L3. In particular G is faithful on g so, by 4 1.1.1, G is a subgroup of Aut(g). Also G is transitive on 5B so it remains only to show for each wall Sit of C that there is a folding @ through Sit with C in g@.

Let r = ri and define @: G + G by

As GI = 1 we can identify G with L3 via g H Cg and regard @ as a function from L3 into g. Claim that if D, D' E g and D n D' is a wall of type j' then D@ n Dl@ is also a wall of type j'. Now D = Cg and D' = Crjg for some g E G, so wemust show Cg@ fl C(r;g)@ is a wall of type j', so it suffices to show (rig)@ = rj(g@) for eachg E G. This is clear if 2(rjgr) - 2(rjg) = 2(gr) - 2 (g). So, by symmetry between rjg and g, we may assume l(gr) = l(g) - 1 and l(rjgr) = l(rjg) + 1. Thus g@ = gr and (rjg)@ = rig. Let g = s, . . . sl with l(g) = n and sk = ri,. As l(gr) 5 l(g), the Exchange Condition says sm . . . sl = sm-1 . . . slr for some m. Thus g = s, . . . sm+lsm-1 . . . slr, so without loss r = s1, Hence g@ = gr = s, . . . s2. Next r j gr = r j s, . . . s2 is of length at most n = l(g) so l(r;g) = l(rjgr) - 1 5 l(g). Then, by the Exchange Con- dition, rjs, . . . sk+l = s, . . . sk for some k. If k # 1 we may take s, = r;.

Buildings 215

But then r j gr = sn_ 1 ... s2 is of length n - 2 <n-1 = l (r j g ). So k = landr j Sn ... S2 = Sn ... Si = g. Thus (r j g)o = r j g = Sn ... S2 = go, as desired.

So the claim is established. By 29.13.3,

Gk' = (G{k,j}' i E k')

for each k E I, if III > 2. Hence by Exercise 14.8, the map Gkg H Gk(g0)is a morphism of the complex (F(G, (). -T). I also write this map as 0.

Next observe that for each g E G, g02 = go. So 0 is an idempotent mor-phism. Also if Cg E -90 then 0-1(Cg) = (Cg, Cgr} so 0 is a folding. FinallyC and Cr are the chambers through Si, and Cro = C so 0 is a folding throughSi'. By construction r is the reflection through Sip; that is the fibres of 0 on -9are the orbits of r on -9. So R is the set of reflection through the walls of Cand hence Aut(-T) = (R) = G.

(41.11) The map (G, R) H '(G, -14-(G, R)) is a bijection between the set ofall Coxeter systems and the set of all Coxeter complexes (up to isomorphism).4-(G, R) is the family of maximal parabolics defined in 41.12 and the inverseof the correspondence is 6 i-+ (Aut(o), R) where R is the set of reflectionsthrough the walls of some fixed chamber of -C.

Proof. If (G, R) is a Coxeter system then by 41.10, (G, R)cp = -6(G, R))is a Coxeter complex, while if 6 is a Coxeter complex then by 41.8 and 41.9,6'* _ (Aut(o), R) is a Coxeter system. By Exercise 14.2.3, Picp = -6', whileby 41.10, (G, R)cpi = (G, R), so cp and 1 are inverses of each other and thelemma holds.

42 BuildingsA building is a thick chamber complex 0 _ (F, ') together with a set sad ofsubcomplexes of 0, called apartments, such that the following axioms aresatisfied:

(B 1) The apartments are thin chamber complexes.(B2) Each pair of chambers of 0 is contained in an apartment.(B3) If A and B are simplices of 0 contained in apartments E and E', then

there exists an isomorphism of E with E' which is the identity on A U B.

In sections 13 and 22 a geometry r was associated to each classical groupG (occasionally subject to restrictions on the field), and it was shown that Gis flag transitive on F. Now -6'(F) is a thick chamber complex and in Exercise14.5 a set of apartments sa is defined which admits the transitive action ofG, and it is shown that (6'(F), s1) is a building admitting G as a group ofautomorphisms. This representation is then used to study G in section 43.

Buildings 215

But then r jgr = s,-1 . . . s2 is of length n - 2 < n - 1 = l(rjg). So k = 1 and rjs, . . . s2 = S, . . . s1 = g. Thus (rig)@ = rjg = s, . . . s2 = g4, as desired.

So the claim is established. By 29.13.3,

Gkl = (G{k,jY: j E k')

for each k E I, if I Il > 2. Hence by Exercise 14.8, the map Gkg H Gk(g4) is a morphism of the complex (r(G, 9 ) . g ) . I also write this map as 4.

Next observe that for each g E G, g42 = g4. So 4 is an idempotent morphism. Also if Cg E g 4 then 4-'(Cg) = (Cg, Cgr] so 4 is a folding. Finally C and C r are the chambers through Sit and Cr4 = C so 4 is a folding through Sij. By construction r is the reflection through $ 1 ; that is the fibres of 4 on g are the orbits of r on g. So R is the set of reflection through the walls of C and hence Aut(g) = (R) = G.

(41.11) The map (G, R) H &(G, F (G , R)) is a bijection between the set of all Coxeter systems and the set of all Coxeter complexes (up to isomorphism). F (G , R) is the family of maximal parabolics defined in 41.12 and the inverse of the correspondence is & H (Aut(&), R) where R is the set of reflections through the walls of some fixed chamber of 8.

Proof. If (G, R) is a Coxeter system then by 41.10, (G, R)p = 8 ( G , F (G , R)) is a Coxeter complex, while if & is a Coxeter complex then by 41.8 and 41.9, &@ = (Aut(&), R) is a Coxeter system. By Exercise 14.2.3, &@p Z 8 , while by 41.10, (G, R)p@ = (G, R), so p and @ are inverses of each other and the lemma holds.

42 Buildings A building is a thick chamber complex 93 = ( r , &) together with a set d of subcomplexes of 93, called apartments, such that the following axioms are satisfied:

(B 1) The apartments are thin chamber complexes. (B2) Each pair of chambers of 93 is contained in an apartment. (B3) If A and B are simplices of 93 contained in apartments E and E', then

there exists an isomorphism of E with E' which is the identity on A U B.

In sections 13 and 22 a geometry r was associated to each classical group G (occasionally subject to restrictions on the field), and it was shown that G is flag transitive on r. Now B ( r ) is a thick chamber complex and in Exercise 14.5 a set of apartments d is defined which admits the transitive action of G, and it is shown that (&(r), d) is a building admitting G as a group of automorphisms. This representation is then used to study G in section 43.

But before all that let's take a closer look at buildings. So for the rest of thissection let A = (F, ') be a building with apartment set a.

(42.1) Each pair of apartments is isomorphic.

Proof. Let E i = 1, 2, be apartments and pick chambers Ci E E. By axiomB2, there is an apartment E containing C1 and C2, and by axiom B3, E1E=E2.

Given a pair (E, C) with E an apartment and C a chamber in E define a mapp = p(E, C): F - E by vp = vo, where C, v are contained in some apartmentE' and 0: E'--> E is an isomorphism which is the identity on C. There are anumber of points to be made here. First by B2 there does indeed exist anapartment E' containing C and v, and by B3 the map 0 exists. Moreover ifv, C are contained in an apartment 0 and 1/f: 0 - E is an isomorphism trivial onC then by B2 there exists an isomorphism a: E' 0 trivial on C U {v). Then(0-1)uVf = ,B E Aut(E) is trivial on C, so, by 41.2.2, ,B = 1 . Thus a = 0,so v4 = vat = v . What all this shows is that p(E, C) is independent of thechoice of E' and 0, so in particular p(E, C) is well defined. As a matter of factit shows there exists a unique isomorphism 0: E' -> E trivial on C. Let's seenext that:

(42.2) p = p(E, C) is a morphism of onto E with p-1(C) = C. Moreoverif E' is an apartment containing C then p: E' E is the unique isomorphismof E' with E trivial on C. In particular p is trivial on E.

I've already made the last observation of 42.2. The remaining parts of thelemma are an easy consequence of this observation and the building axioms.

Given simplices A and B of .M define d(A, B) to be the minimal integer nsuch that there exists a gallery 7 = (Ci: 0 < i < n) of A with A contained inCo and B contained in Cn. Then C is said to be a gallery between A and B oflength n.

(42.3) (Rainy'Day Lemma) Let E be an apartment, C a chamber of E, andX a simplex of E. Then E contains every gallery of A between X and C oflength d(X, C).

Proof. Let 9 = (Ce:0 < i < n) be such a gallery and assume ' is not con-tained in E. Then there exists an i with C; not contained in E and Ci+icontained in E. Then w = Ci f1 CZ+i is a wall of C,+1 so there is a chamberT 0 CZ+1 of E with W c T. Let p = p(E, T). As p is a morphism trivial


But before all that let's take a closer look at buildings. So for the rest of this section let B = (r , 6') be a building with apartment set sf.

(42.1) Each pair of apartments is isomorphic.

Proof. Let X i , i = 1,2, be apartments and pick chambers Ci E C, . By axiom I

B2, there is an apartment C containing C1 and C2, and by axiom B3, C1 %

C % C2.

Given a pair ( C , C ) with C an apartment and C a chamber in C define a map p = p(C, C): r 4 C by up = v4 , where C , v are contained in some apartment C' and 4: C' + C is an isomorphism which is the identity on C. There are a number of points to be made here. First by B2 there does indeed exist an apartment C' containing C and v, and by B3 the map 4 exists. Moreover if v , C are contained in an apartment 0 and $: 0 4 C is an isomorphism trivial on C then by B2 there exists an isomorphism a: C' + 0 trivial on C U {v) . Then (4-')a$ = ,!? E Aut(C) is trivial on C, so, by 41.2.2, ,!? = 1. Thus a$ = 4, so v4 = va$ = v$. What all this shows is that p(C, C ) is independent of the choice of C' and 4, so in particular p(C, C ) is well defined. As a matter of fact it shows there exists a unique isomorphism 4: C' + C trivial on C. Let's see next that:

(42.2) p = p(C , C ) is a morphism of B onto C with p-' (c) = C . Moreover if C' is an apartment containing C then p: C' + C is the unique isomorphism of C' with C trivial on C. In particular p is trivial on C .

I've already made the last observation of 42.2. The remaining parts of the lemma are an easy consequence of this observation and the building axioms.

Given simplices A and B of 93 define d(A , B ) to be the minimal integer n such that there exists a gallery 9 = (Ci: 0 5 i 5 n ) of B with A contained in Co and B contained in C,. Then 9 is said to be a gallery between A and B of length n .

(42.3) (Rainy'Day Lemma) Let C be an apartment, C a chamber of C , and X a simplex of C. Then C contains every gallery of B between X and C of length d(X, C).

Proof. Let 9 = (C,: 0 5 i 5 n ) be such a gallery and assume 9 is not contained in C . Then there exists an i with C, not contained in C and C,+1 contained in C. Then W = C, n C,+' is a wall of Ci+' so there is a chamber T # C,+, of C with W E T . Let p = p(C, T ) . As p is a morphism trivial

Buildings 217

on E, Cip is a chamber of E containing W, so, as E is thin, C; p = T orC,+1. As p-1(T) = T, C,p = C,+1. But qp is a gallery between X = Xpand Cp = C of length n = d(X, C), so, as C, p = Ci+1 = Ci+lp, c o p ,--- ,C1_1 p, Ci+l p, . .., is a gallery between X and C of length n - 1, contra-dicting n = d(X, C).

(42.4) Let C and D be chambers, X a subset of C, and E an apartment con-taining C. Let p = p(E, C) and q a gallery of length d(X, D) between X andD in A. Then d(Xp, Dp) = d(X, D) and qp is a gallery of length d(Xp, Dp)between X = Xp and Dp.

Proof. By B2, C U D is contained in an apartment E' and, by 42.3, C iscontained in V. By 42.2, p: E' - E is an isomorphism so the lemma holds.

(42.5) Each apartment is a Coxeter complex.

Proof. Let C and C' be adjacent chambers in an apartment E. We must showthere exist opposite foldings 0 and 0' of E through B = C f1 C' with C'O = Cand CO' = C'. For the first time the hypothesis that A is thick is used. Namelyas 2B is thick there is a chamber C* distinct from C and C' through B. LetE1 be an apartment containing C and C*, p, = p(E1, C), P2 = p(E, C'), and0 = pipe: E E. As the composition of morphisms trivial on C, 0 is a mor-phism trivial on C. Moreover applying 42.4 to p, and p2 we obtain:

(42.5.1) If 9 is a gallery of length d(B, T) = n between B and a chamber Tof E, then d(B, TO) = n and 9O is a gallery of length n between B and TO.

Next C'O is a chamber containing B, so C't = C or C'. As pl : E E1 is an

isomorphism trivial on C, C'p1 = C*. Then C'O = C* p2 0 C' as (p2)-1(C')C'. So C'q5 = C.

Similarly there is a morphism 0' of E trivial on C', with CO' = C', andsatisfying 42.5.1.

Let D be a chamber in E and (C,:0 < i < n) = C a gallery of lengthd(B, D) in E from B to D. Notice Co = C or C'. Claim

(42.5.2) If Co = C then

(i) 0 and 0'0 are trivial on D, and(ii) Do' 0 D.

Assume otherwise and choose a counterexample with n minimal. As CO' = C',C'O = C, and 0 is trivial on C, it follows that n > 0. Now (Ci: 0 < i < n) is

Buildings 217

on C , Cip is a chamber of C containing W, so, as C is thin, Cip = T or Ci+1. As p - l ( ~ ) = T , Cip = Ci+1. But 9 p is a gallery between X = Xp and Cp = C of length n = d(X , C) , so, as Cip = Ci+1 = c i + l ~ , COP, ..., Ci-1 p, Ci+1 p, . . . , C,,p is a gallery between X and C of length n - 1, contradicting n = d(X , C).

(42.4) Let C and D be chambers, X a subset of C , and C an apartment containing C. Let p = p(C, C ) and 9 a gallery of length d(X , D) between X and D in B . Then d(Xp , Dp) = d(X , D ) and 9p is a gallery of length d(Xp , Dp) between X = Xp and Dp.

ProoJ By B2, C U D is contained in an apartment C' and, by 42.3, 9 is contained in C'. By 42.2, p: C' + C is an isomorphism so the lemma holds.

(42.5) Each apartment is a Coxeter complex.

ProoJ Let C and C' be adjacent chambers in an apartment C. We must show there exist opposite foldings 4 and 4' of C through B = C n C' with C'4 = C and C4' = C'. For the first time the hypothesis that B is thick is used. Namely as B is thick there is a chamber C* distinct from C and C' through B. Let C1 be an apartment containing C and C*, pl = p(C1, C ) , p:! = p(C, C'), and 4 = pl p2: C + C . As the composition of morphisms trivial on C , 4 is a morphism trivial on C. Moreover applying 42.4 to pl and pz we obtain:

(42.5.1) If 9 is a gallery of length d(B , T ) = n between B and a chamber T of C , then d(B , T 4 ) = n and 94 is a gallery of length n between B and T 4 .

Next C'4 is a chamber containing B , so C'4 = C or C'. As pl : C + C1 is an isomorphism trivial on C , C1pl = C*. Then C'4 = C*p2 # C' as (,3:!)-' (c') = C'. So C14 = C.

Similarly there is a morphism 4' of C trivial on C', with C4' = C', and satisfying 42.5.1.

Let D be a chamber in C and (Ci: 0 5 i 5 n) = 9 a gallery of length d(B , D) in C from B to D. Notice Co = C or C'. Claim

(42.5.2) If Co = C then

(i) 4 and 4'4 are trivial on D, and (ii) D4' # D.

Assume otherwise and choose acounterexample withn minimal. As C4' = C', C'4 = C , and 4 is trivial on C , it follows that n > 0. Now (Ci: 0 5 i < n) is

a gallery from B to Cii_1 -= E of length d(B, E) = n - 1, so (i) and (ii) holdfor E by minimality of n. Let A = D fl E. As A = Aq c Do, Do = D or E.By 42.5.1, n = d(B, DO), so, as d(B, E) = n - 1, it follows that DO = D.Similarly, as 0 and 0' both satisfy 42.5.1, so does 0'0, and then as 0'0 is trivialon E the same argument shows DO'O = D.

So (i) is established. Notice -9' = 6/'0' is gallery of length n = d(B, Do')with Coo' = CO' = C', so, by symmetry between C and C', 0' is trivial onDo'.Thus, if D=Do',A=Aq'cDflEq',so Eq'=Dsince Eq'#E.This is impossible as n - 1 = d(B, E) = d(B, E/') while d(B, D) = n. Thiscompletes the proof of 42.5.2.

Now, by 42.5.2, either Co = C and D = Do, or Co = C', in which caseCoo = C and, applying 42.5.2 to qq5, we get (DO)o = Do. In either caseDq = Dq52, so 0 is idempotent. Also if D = Do then by 42.5.2, Co # C',so Co = C, and then again by 42.5.2, D Do' and D = Dq'q. Moreover ifT# D is a chamber of E with To = D then T 0 To, so T = To' =Too'.Thus T = Do' and so {D, Do'} is the fibre of D under 0. Hence 0 is a foldingof E through B with CO = C, and 0' is the opposite folding. The proof of42.5 is complete.

43 BN-pairs and Tits systemsA Tits system is a quadruple (G, B, N, S) such that G is a group, B and Nare subgroups of G, S is a finite collection of cosets of B fl N in N, and thefollowing axioms are satisfied:

(BNl) G=(B,N)andH=BIN<N.(BN2) W =N/H is generated by S, and the members of S are involutions in

W.

(BN3) For each s E S and w E W, sBw c_ BwB U BswB.(BN4) For each s E S, B 0 BS.

By convention, wB = nB and B' = B'1, for n a representative of the cosetw = Hn E W. As H < N and H < B, nB and B" are independent of thechoice of coset representative n, so the notation is well defined. Axioms BN3and BN4 should be read subject to this convention.

B and N are also said to be a (B, N) pair for G.

(43.1) Let 3 = (I', t) be a building with apartment set .1 and assume G is agroup of automorphisms of -M transitive on

cZ={(C,E):CE 6,E E W,CCE}.


a gallery from B to CnV1 = E of length d(B, E) = n - 1, so (i) and (ii) hold for E by minimality of n. Let A = D i l E. As A = Aq5 c Dq5, Dq5 = D or E. By 42.5.1, n = d(B, Dq5), so, as d(B, E) = n - 1, it follows that Dq5 = D. Similarly, as q5 and 4' both satisfy 42.5.1, so does 4'4, and then as 4'4 is trivial on E the same argument shows Dq5'q5 = D.

So (i) is established. Notice g' = Hq5' is gallery of length n = d(B, Dq5') with Coq5' = Cq5' = C', so, by symmetry between C and C', 4' is trivial on Dq5'. Thus, if D = Dd', A = Aq5' c D n Eq5', so Eq5' = D since Eq5' # E. This is impossible as n - 1 = d(B, E) = d(B, Eq5') while d(B, D ) = n. This completes the proof of 42.5.2.

Now, by 42.5.2, either Co = C and D = Dq5, or Co = C', in which case .

Coq5 = C and, applying 42.5.2 to gq5, we get (Dq5)q5 = Dq5. In either case Dq5 = Dq52, SO q5 is idempotent. Also if D = Dq5 then by 42.5.2, Co # C', so Co = C , and then again by 42.5.2, D # Dq5' and D = Dq5'4. Moreover if T # D is a chamber of C with Tq5 = D then T # Tq5, so T = Tq5' = Tq5q5'. Thus T = Dq5' and so { D , D@'} is the fibre of D under q5. Hence q5 is a folding of C through B with C'q5 = C , and q5' is the opposite folding. The proof of 42.5 is complete.

43 BN-pairs and Tits systems A Tits system is a quadruple (G, B, N, S) such that G is a group, B and N are subgroups of G, S is a finite collection of cosets of B i l N in N, and the following axioms are satisfied:

(BN1) G = ( B , N ) a n d H = B n N a N . (BN2) W = N/H is generated by S, and the members of S are involutions in

W . (BN3) For each s E S and w E W, sBw E BwB U BswB. (BN4) For each s E S, B # BS.

By convention, w B = nB and BW = Bn, for n a representative of the coset w = Hn E W. As H <I N and H 5 B, nB and Bn are independent of the choice of coset representative n, so the notation is well defined. Axioms BN3 and BN4 should be read subject to this convention.

B and N are also said to be a (B, N)-pair for G.

(43.1) Let 93 = (r, B) be a building with apartment set d and assume G is a group of automorphisms of 93 transitive on

S2 = { ( C , C ) : C E 8, C E d, C C C } .

BN-pairs and Tits systems 219

Let (C, E) E 0, B = Gc, and N = NG(E). Then(1) The representation of N on E maps N surjectively onto Aut(E) with

kernel H = B fl N.(2) (W, S) is a Coxeter system, where W = N/H and S is the set of reflections

in W through the walls of C.(3) (G, B, N, S) is a Tits system.

Proof. By 42.5, E is a Coxeter complex, while by hypothesis N is transitiveon the chambers of E, so (1) follows from 41.8. Then (2) follows from (1) and41.9.

Lets E S and w E W. Then C and Cs are adjacent and E = C fl Cs is a wallof C. Each b E B fixes E, so E C_ Csb and Ew CCsbw. Let 6' = (Ci: 0 <i < n) be a gallery of length d(C, Ew) from C to Ew, and E' an apart-ment containing C and Csbw. By the Rainy Day Lemma, 42.3, c E fl V.Let a E G with (C, E')a = (C, E). Then Ca = C, so a E B. Ci and C arethe only chambers of E or E' containing C fl CI so, as a fixes C, it alsofixes C1. Proceeding by induction on k, a fixes Ck for each 0 < k < n. Thus afixes Ew CC,,. But Ew C Csbw C V. So Ew = Ewa C Csbwa C E'a = E,so Csbwa = Cw or Csw. Hence sbw e BwB or BswB, so BN3 is estab-lished.

As 3 is thick, there exists a chamber D through E distinct from C and Cs.Let 9 be an apartment containing C and D, and let g e G with (C, E)g = (C, 0).Then g E B and C and Csg are the chambers through E in Eg = 0, so Csg =D # Cs. Thus sgs B, so BN4 is established and the proof is complete.

It will develop later in this section that the converse of 43.1 also holds; that iseach Tits system defines a building.

Notice that by 43.1 and Exercise 14.5 the classical groups possess a BN-pair.The BN-pair structure will be used to establish various facts about the classicalgroups.

In the remainder of this section assume (G, B, N, S) is a Tits system andlet H = B fl N and W = N/H. We say B is the Borel subgroup of G, H isthe Cartan subgroup of G, and W is the Weyl group of G. Let l be the lengthfunction on W defined by the generating set S.

(43.2) If u, w e W with BwB = BuB then u = w.

Proof. Let m = 1(w) < 1(u) and induct on m. If m = 0 then w = 1, so u eB fl N = H; that is u = 1 (remember the convention on M. So take m > 0.Then w sv with s E S and 1(v) = m - 1. Now svB c BuB by hypothesis,so vB c sBuB c BuB U BsuB by BN3. Then BvB = BuB or BsuB, so, by

BN-pairs and Tits systems

Let (C, C) E 52, B = Gc, and N = NG(C). Then (1) The representation of N on Z maps N surjectively onto Aut(Z) with

kernel H = B i l N. (2) (W, S) is a Coxeter system, where W = N/H and S is the set of reflections

in W through the walls of C. (3) (G, B, N, S) is a Tits system.

Proof. By 42.5, C is a Coxeter complex, while by hypothesis N is transitive on the chambers of C, so (1) follows from 41.8. Then (2) follows from (1) and 41.9.

Lets E S and w E W. Then C and Cs are adjacent and E = C i l Cs is a wall of C. Each b E B fixes E, so E E Csb and Ew Csbw. Let L+9 = (Ci: 0 5 i 5 n) be a gallery of length d ( C , Ew) from C to Ew, and C' an apartment containing C and Csbw. By the Rainy Day Lemma, 42.3, L+9 E Z i l C'. Let a E G with (C, Z')a = (C, C). Then Ca = C, so a E B. C1 and C are the only chambers of C or Z' containing C n C1 so, as a fixes C , it also fixes C1. Proceeding by induction on k, a fixes Ck for each 0 5 k 5 n. Thus a fixes Ew C,. But Ew Csbw E C'. So Ew = Ewa c Csbwa c C'a = C, so Csbwa = Cw or Csw. Hence sbw EBWB or BswB, so BN3 is established.

As 93 is thick, there exists a chamber D through E distinct from C and Cs. Let 8 be an apartment containing C and D, and let g E G with (C, C)g = (C, 8). Then g E B and C and Csg are the chambers through E in Cg = 8, so Csg = D # Cs. Thus sgs $ B, so BN4 is established and the proof is complete.

It will develop later in this section that the converse of 43.1 also holds; that is each Tits system defines a building.

Notice that by 43.1 and Exercise 14.5 the classical groups possess a BN-pair. The BN-pair structure will be used to establish various facts about the classical groups.

In the remainder of this section assume (G, B, N, S) is a Tits system and let H = B i l N and W = NIH. We say B is the Bore1 subgroup of G, H is the Cartan subgroup of G, and W is the Weyl group of G. Let 1 be the length function on W defined by the generating set S.

(43.2) If u , w E W with BwB = BuB then u = w.

Proof. Let m = l(w) 5 l(u) and induct on m. If m = 0 then w = 1, so u E

B i l N = H; that is u = 1 (remember the convention on W). So take m > 0. Then w = sv with s E S and l(v) = m - 1. Now svB BuB by hypothesis, so vB E sBuB 2 BuB U BsuB by BN3. Then BvB = BuB or BsuB, so, by

induction on m, v = u or su. As 1(v) = m - 1 and m < 1(u), v $ u, so v = suand hence w = sv = u as desired.

(43.3) Let W E W and S E S.(1) If l(sw) > 1(w) then sBw C BswB.(2) If l(sw) < l(w) then sBw fl BwB is nonempty.

Proof. First the proof of (1), which is again by induction on m = 1(w). If m = 0then w = 1 and the result is trivial, so take m > 0. Then w = ur, with r E S and1(u) = m - 1. If l(su) < m - 1 then l(sw) = l(sur) < l(su) + 1 < m, con-trary to hypothesis, so l(su) > m - 1 = 1(u). Then, by induction on m, sBu CBsuB. Hence sBw = sBur C BsuBr C_ BsuB U BsurB by BN3. Also, byBN3, sBw C BwB U BswB. If BwB = BsuB then, by 43.2, w = su. Henceu = sw so m - 1 = 1(u) = l(sw) > 1(w) = in, a contradiction. Similarly ifBwB = BsurB then w = Sur = sw, a contradiction. It follows that sBw CBswB, as desired.

So (1) is established and it's on to (2). By BN3 and BN4, BS fl BsB isnonempty, so sBsu fl BsBu is nonempty for each u E W. Take u = sw. Thenl(su) = 1(w) > l(sw) = 1(u), so, by (1), sBu C BsuB = BwB. Therefore, assBsu fl BsBu is nonempty and w = su, (2) is established.

(43.4) Let w = si ... Sn E W with Si E S. Then(1) If n = 1(w) then siB C (B, B'_1

) D BwB for 1 < i < n.(2) For each u E W, BwBuB c UiEO Bsi, ... SiruB, where A consists of

the sequences i = i l , ... , i t with i j E { 1 , ... , n j and i t < i2 < < i,..(3) If So E S with 1 (sow) < 1(w) = n then there exists 1 < k < n with

SOS1 ... Sk-1 = S1 ... Sk.

Proof. Part (2) follows from BN3 by induction on n. Next part (1).Let wi =si ... s1w for 1 < i < n, n = 1(w), and wo = w. By hypothesisl(wi+i) < 1(wi) so, by 43.3.2, si+iBwi fl BwiB is nonempty. Hence

si+,B C BwiBwi-1B C (B, B"''.

By induction on i, sj E (B, B") = X for each j < i, so si+i c (B, B-1_') < X.Hence w = sl ... sn c X and then BwB C X. So (1) holds.

Similarly if l(sow) < 1(w) then, by 43.3.2, SOB C BwBw-1B so, by (2),BsOB = Bxw-1B for X = Sil ...sir and some i c A. Hence, by 43.2, so =xw-1. But l(xw-1) > l(w-1) - 1(x) > n - r, so, as l(so) = 1, r = n - 1 andx = S1 ... Sk-1Sk+1 . . . Sn for some 1 < k < n. Therefore (3) holds.



induction onm, v = u orsu. As l ( v ) = m - 1 andm 5 l(u) , v # u, so v = su and hence w = sv = u as desired.

(43.3) Let w E Wand s E S. ( 1 ) If l(sw) 2 l (w) then sBw 2 BswB. (2) If l(sw) 5 l(w) then sBw f' BwB is nonempty.

Prooj First the proof of ( I ) , which is again by induction on m = l(w). If m = 0 then w = 1 and the result is trivial, so take m > 0. Then w = ur, with r E S and l (u) = m - 1. If l (su) < m - 1 then l ( sw) = l (sur) 5 l(su) + 1 < m, contrary to hypothesis, so Z(su) > m - 1 = l(u). Then, by induction on m, sBu s .

BsuB. Hence sBw = sBur 2 BsuBr 2 BsuB U BsurB by BN3. Also, by BN3, sBw 2 BwB U BswB. If BwB = BsuB then, by 43.2, w = su. Hence u = sw so m - 1 = l (u ) = l (sw) 2 l (w) = m, a contradiction. Similarly if BwB = BsurB then w = sur = sw, a contradiction. It follows that sBw 5 BswB, as desired.

So ( 1 ) is established and it's on to (2). By BN3 and BN4, BS f' BsB is nonempty, so sBsu n BsBu is nonempty for each u E W. Take u = sw. Then l(su) = l (w ) 2 l ( sw) = l(u), so, by ( I ) , sBu 2 BsuB = BwB. Therefore, as sBsu n BsBu is nonempty and w = su, (2) is established.

(43.4) Let w = sl . . . s, E W with si E S. Then ( 1 ) If n = l (w) then siB 2 ( B , B"-') 2 BwB for 1 5 i 5 n. (2) For each u E W, Bw BuB 5 Ui,, Bsil . . . siruB, where A consists of

the sequences i = i l , . . . , i, with i j E ( 1 , . . . , n} and il < i2 < . . . < i,. (3) If so E S with l(s0w) 5 l (w) = n then there exists 1 5 k 5 n with

S0Sl . . . Sk-1 = S1 . . . S k .

Prooj Part ( 2 ) follows from BN3 by induction on n. Next part ( 1 ) . Let wi=si .. .slw for 15 i 5 n , n = l (w) , and wo = w. By hypothesis l(wi+l) < l (wi) so, by 43.3.2, si+lBwi f' BwiB is nonempty. Hence

-1 By induction on i , sj E ( B , B"-') = X for each j 5 i , so si+l E ( B , B W ~ ) 5 X . Hence w = sl . . . s, E X and then Bw B 5 X . So ( 1 ) holds.

Similarly if l(s0w) 5 l (w) then, by 43.3.2, soB 2 B w ~ w - ' B so, by (2), BsoB = ~ x w - ' ~ for x = si, . . .sir and some i E A. Hence, by 43.2, so = xw-'. But l ( xwP1) > l (w- ' ) - l ( x ) 2 n - r , so, as l(so) = 1 , r = n - 1 and x = sl . . . sk-lsk+l . . . S, for some 1 5 k 5 n. Therefore (3) holds.

(43.5) (W, S ) is a Coxeter system.

Proof. This is immediate from 43.4.3 and 29.4.

(43.6) S is the set of w E W* such that B U Bw B is a group.

221

Proof. By BN3, B U Bs B is a group for S E S. Conversely if w = s1 ... S" E Wwith Si E S and n = l(w) > 0 then, by 43.4.1, Si E (B, w), so if B U BwB is agroup then, by 43.2, si = w for each i.

Let S = (Si: i E I) and for J C I let Pj = (B, sj: j E J). The conjugates ofthe subgroups Pj, J C I, are called parabolic subgroups of G. Recall Wj _(sj: j E J) is a parabolic of W.

(43.7) (1) Pi = BWjB. In particular G = BNB.(2) The map J H Pj is a bijection of the power set of I with the set of all

subgroups of G containing B.

(3) PJUK = (Pr, PK)-(4) PJnK = Pj n PK.(5) If g E G with B9 < PJ then g E PJ. In particular NG(Pj) = Pi is con-

jugate in G to PK only if J = K.(6) Wr=WnPj.

Proof. To prove (1) it suffices to show uBw C BWjB for each u, w E Wi.Proceeding by induction on l(u) it suffices to show this for u = sj, j E J. Butthis is just BN3. So (1) is established.

LetB < X < G.By(1),X = UYEYByBforsomeY C W.Lety = r1 ... r EY with ri E S and 1(y) = n. By 43.4, ri E (B, By-) < X. Let J be the set ofj E I such that sj = ri for some y E Y and some such expression for y. I'veshown Pj < X, while by definition certainly X < Pj. Thus X = PJ.

Suppose Pi = PK; to complete the proof of (2) it remains to show K = J.By (1) and 43.2, Wj = W n P, = W n PK = WK. Hence, by 29.13.4, J = K.So (2) and (6) are established.

Evidently the bijection of (1) preserves inclusion, and hence also least upperbounds and greatest lower bounds. Hence (3) and (4) hold.

Let B9 < Pi.g-1 E BwBforsome wE W, and, by 43.4.1,w c (B, B") =(B, B9) < Pi. So g E (B, g) = (B, w) < Pi.

Exercises 4.9 and 7.8 show that, if G is a classical group over a finite fieldof characteristic p, then the Cartan group H is a Hall p'-group of the Borelgroup B and B possesses a normal p-complement U with U E Syl p (G). FurtherB = NG (U), so the parabolics are the subgroups of G containing the normalizer


Proof. This is immediate from 43.4.3 and 29.4.

(43.6) S is the set of w E W' such that B U Bw B is a group.

Proof. By BN3, B U BsB is a group fors E S . Conversely if w = sl . . . sn E W withsi E Sand n = l(w) > 0 then, by 43.4.1, si E (B, w), so if B U BwB is a group then, by 43.2, si = w for each i.

Let S = (si: i E I ) and for J 5 I let P j = (B, sj: j E J). The conjugates of the subgroups P j , J 5 I , are called parabolic subgroups of G. Recall Wj = (sj: j E J) is a parabolic of W.

(43.7) (1) P j = BWjB. In particular G = BNB. (2) The map J H P j is a bijection of the power set of I with the set of all

subgroups of G containing B.

(3) PJUK = (PJ, PK). (4) P J ~ K = PJ f' PK (5) If g E G with Bg ( P j then g E Pj . In particular NG(PJ) = P j is con-

jugate in G to PK only if J = K. (6) wj = w n Pj .

Proof. To prove (1) it suffices to show uBw 2 BWjB for each u, w E Wj. Proceeding by induction on l(u) it suffices to show this for u = sj , j E J. But this is just BN3. So (1) is established.

LetB I X I. G.By(l),X=Uy,yByBforsomeYSW.Lety = r l . . . rnE

Y with ri E S and l(y) = n. By 43.4, ri E (B, BY-') ( X . Let J be the set of j E I such that s j = ri for some y E Y and some such expression for y. I've shown P j I. X , while by definition certainly X I. Pj . Thus X = Pj.

Suppose P j = PK; to complete the proof of (2) it remains to show K = J . By (1)and43.2, Wj = W n P J = W n P K = WK.Hence,by29.13.4, J = K . So (2) and (6) are established.

Evidently the bijection of (1) preserves inclusion, and hence also least upper bounds and greatest lower bounds. Hence (3) and (4) hold.

Let Bg 5 pJ. g-l E Bw B for some w E W, and, by 43.4.1, w E (B, fIW-') = (B, Bg) I Pj. SO g E (B, g) = (B, W ) I. Pj.

Exercises 4.9 and 7.8 show that, if G is a classical group over a finite field of characteristic p, then the Cartan group H is a Hall p'-group of the Bore1 group B and B possesses a normal p-complement U with U E Syl,(G). Further B = NG(U), so the parabolics are the subgroups of G containing the normalizer


of a Sylow p-group of G. Indeed it can be shown that the maximal parabolicsP;', i E I, are the maximal subgroups of G containing the Sylow p-group U,and further every p-local is contained in some maximal parabolic (cf. section47).

Some more notation. For i E I let G; = Pi, (recall for J C I that J' = I - J)and let _ (G1: i E I) be the set of maximal parabolics containing B. Formthe geometry F(G, _1F), the subgeometry E = {G;w: i E I, w E W}, and thecomplex A = -'(G, OT). Let C be the chamber {G1: i E I); then ={Cg: g E G} is the set of chambers of 3, and we can also regard E as thecomplex (E, fl E). Let szl = {Eg: g E G}, so that a is a collection ofsubcomplexes of A. Write 3 (G, B, N, S) for the pair (A, sa?). Finally letUl = W', 6' = (U1: i E I) and form the complex -'(W, d').

(43.8) (1) The map G1 w i-± UU w is an isomorphism of the complex (E, E me)with '(W, e).

(2) G is represented as a group of automorphisms of 3 by right multipli-cation; the kernel of this representation is kerB (G).

(3) A(G, B, N, S) is a building.(4) G is transitive on S2 = {(D, 9): D E', 9 E s1, DC 9}(5) B = Gc and (BE)N = NE(E) are the stabilizers in G of C and E,

respectively. NE = Aut(E) - W and H = NE.

Proof. By 29.13.3, WJnK = win WK for J, K C I. Thus Uj = Wj,, whereUj = n j EJ Ui. In particular, as (W, S) is a Coxeter system, UI = WO = 1and Up = W, = (si) is of order 2, so -'(W, 1) is a thin chamber complex by41.1.4 and Exercise 14.3.

Next 43.7.6 says the map 7r: Gi w i-+ U; w of (1) is well defined, after which7r is evidently a bijection and 7r-1 is a morphism of geometries. So to prove (1)it remains to show that if Gkw fl G1 v is nonempty then Ukw fl U1 v is too. ButGku fl Gj v is nonempty if and only if 1 E Gkuv-1Gj, in which case Ukw fl UU vis nonempty by the following observation, which is an easy consequence of43.2 and 43.4.2:

(43.9) For each J, K C I and each w E W, (Gj WGK) fl W = Uj WUK.

Hence (1) holds and therefore by the first paragraph of this proof, satisfies-04

B1.Notice (2) is immediate from 41.1.1. Exercise 14.3 and 43.7 show is-04

a chamber complex. To show is thick, by 41.1.4 and 43.7 we must show-04

JP,: BI > 2 for each i c I. But, by 43.6, Pi = B U Bs, B so I P, : B I = 2 if andonly if si E NG(B). Thus BN4 says is thick.-04


of a Sylow p-group of G. Indeed it can be shown that the maximal parabolics Pit, i E I , are the maximal subgroups of G containing the Sylow p-group U , and further every p-local is contained in some maximal parabolic (cf. section 47).

Some more notation. For i E I let Gi = Pi, (recall for J c I that J' = I - J ) and let F = (Gi: i E I ) be the set of maximal parabolics containing B. Form the geometry r (G , F ) , the subgeometry Z = {Giw: i E I , w E W), and the complex G8 = &(G, 9). Let C be the chamber {Gi: i E 1); then & = {Cg: g E G) is the set of chambers of 93, and we can also regard Z as the complex (Z, & n Z). Let d = {Zg: g E G), so that d is a collection of subcomplexes of 93. Write B(G, B, N, S) for the pair ( a , d ) . Finally let .

Ui = W~J, 8 = (Ui : i E I ) and form the complex &(W, 8).

(43.8) (1) The map Gi w H Ui w is an isomorphism of the complex (Z , Z n 8) with&(W, 8) .

(2) G is represented as a group of automorphisms of % by right multiplication; the kernel of this representation is kerB (G).

(3) B ( G , B, N, S) is a building. (4) G is transitive on C2 = {(D, 0): D E &, 0 E d , D 2 0). (5) B = Gc and (Bc)N = NG(Z) are the stabilizers in G of C and Z,

respectively. N' = Aut(Z) S W and H = Nc.

Proof. By 29.13.3, WJnK = Wj n WK for J, K C I. Thus UJ = Wj!, where UJ = njEJ Uj. In particular, as (W, S) is a Coxeter system, Ul = WD = 1 and Uit = Wi = (q) is of order 2, so &(W, 8 ) is a thin chamber complex by 41.1.4 and Exercise 14.3.

Next 43.7.6 says the map n : Giw H Uiw of (1) is well defined, after which n is evidently a bijection and n-' is a morphism of geometries. So to prove (1) it remains to show that if Gkw n G j v is nonempty then Ukw n Uj v is too. But Gku n Gj v is nonempty if and only if 1 E G ~ u v - ' G ~ , in which case Ukw n Uj v is nonempty by the following observation, which is an easy consequence of 43.2 and 43.4.2:

(43.9) For each J , K E I and each w E W, (GJwGK) n w = UJWUK.

Hence (1) holds and therefore by the first paragraph of this proof, 93 satisfies B1.

Notice (2) is immediate from 41.1.1. Exercise 14.3 and 43.7 show 93 is a chamber complex. To show 93 is thick, by 41.1.4 and 43.7 we must show IPi:BI >2foreachi E I.But,by43.6, Pi = BUBsiBsoIPi:BI =2i fand only if si E NG(B). Thus BN4 says 93 is thick.

For x E G and J C I let Cj,x = {Gjx: j E J} be the simplex of type J inCx. Let Cj,x and CK,y be simplices. Then there exist a, b E B and n E N withxy-1 = anb. Now CK,y = CK,by and Cj,x = CJ,Rby are in Eby, so satisfiesB2.

Suppose Ci,x, CK,y c E fl Eg. Then we can choose x, y E N and there ex-ist n, m E N with Cj,x = CJ,ng and CK,y = CK,mg. Notice h = mgy-1 E G.Define u = nm-1 and v = xy-1. Then (Gj)uh = (Gj)ngy-1 = (Gj)xy-1=(Gj)v, so u c (Gj)v(GK). So, by 43.9, u = rvs for some r E Gi n N ands E GKfN. Setl = y-1 shy. ThenGjxl = Gj vyl = Gj vshy = Gtr-1uhy =Gang = Gjx, and GKyl = GKShy = GKhy = GK y, and El = Eyl = Eshy =Ehy = Emg = Eg. So El = Eg, (Cj,x)l = Cj,x, and (CK,y)l = CK,y. That isl induces an isomorphism of E with Eg trivial on Cj,x and CK,y. Therefore 3satisfies B3. This completes the proof of (3).

As W is transitive on the chambers of -6(W, 6' ), (1) says N is transitive on-' fl E. So, as G is transitive on c1 by construction, (4) holds. Also NG(E) =NNB(E) as NB(E) is the stabilizer of C in NG(E). As 6' is thin, NB(E) = BEby 41.2, so (5) holds.

(43.10) Let G* = G/kerB(G). Then (G*, B*, N*, S*) is a Tits system.

Proof. This is clear.

Because of 43.10 it does little harm to assume kerB (G) = 1. By 43.8 this hasthe effect of insuring that G is faithful on its building.

Recall that the Coxeter system (W, S) is irreducible if the graph of its Coxeterdiagram is connected.

(43.11) Assume (W, S) is an irreducible Coxeter system and kerB(G) = 1.Then

(1) If X < G then G = XB, and(2) if G is perfect and B is solvable then G is simple.

Proof. Let X < G. Then XB < G so, by 43.7, XB = Pj for some J C I. LetJo = {i E I: (Bsi B) fl x : Q}. If i E JO then si c Bsi B c XB = Pj so i E Jby 43.7.6. Conversely, as Pj = XB, X intersects each coset of B in Pj non-trivially so J CJO. Thus J = JO.

Claim J = I. If not as (W, S) is irreducible there exists i E I - J and j E Jwith [si, sj] 1. Therefore, as (si, sj) is dihedral, l (si sjsi) > l (sjsi) > l (si ),so, by 43.3.1, Bsi Bsi Bsi B = Bsi sjsi B. As X < G and J = JO it follows thatsi sjsi E W fl Pj = Wj. But then si sjsi E Wj fl W{i,J } = W3 = (sj), contra-dicting [si, sj] : 1.


For x E G and J c I let Cj,, = {Gjx: j E J ) be the simplex of type J in Cx. Let Cj,, and C K , ~ be simplices. Then there exist a , b E B and n E N with xy-' = anb. Now C K , ~ = CK,by and C j , = Cj,nby are in Cby, so 93 satisfies B2.

Suppose Cj,,, C K , ~ C C n Cg. Then we can choose x, y E N and there exist n, m E N with Cj,, = Cj,,, and C K , ~ = C K , ~ ~ . Notice h = mgy-' E GK. Define u = nm-' and v = xy-'. Then (Gj)uh = ( ~ j ) n ~ y - ' = ( G ~ ) X ~ - ' = (Gj)v, so u E (G~)u(GK). SO, by 43.9, u = rvs for some r E G j n N and s E GKnN.Set l = y-'shy.ThenGJxl = Gjvyl = Gjvshy = GJr-'uhy = GJng = Gjx, and GKyl = GKshy = GKhy = GKy, and Cl = Cyl= Cshy = Chy = Cmg= Cg.So Cl = Cg,(CJ,,)l =CJ,,,and(C~,,)1 = C ~ , ~ . T h a t i s 1 induces an isomorphism of C with Cg trivial on Cj,, and C K , ~ . Therefore 93 satisfies B3. This completes the proof of (3).

As W is transitive on the chambers of 6(W, &), (1) says N is transitive on fi? n C. So, as G is transitive on d by construction, (4) holds. Also NG(C) = NNB(C) as NB (C) is the stabilizer of C in NG(C). As &is thin, NB(C) = BE by 41.2, so (5) holds.

(43.10) Let G* = G/kerB(G). Then (G*, B*, N*, S*) is a Tits system.

Proof. This is clear.

Because of 43.10 it does little harm to assume kerB(G) = 1. By 43.8 this has the effect of insuring that G is faithful on its building.

Recall that the Coxeter system (W, S) is irreducible if the graph of its Coxeter diagram is connected.

(43.11) Assume (W, S) is an irreducible Coxeter system and kerB(G) = 1. Then

(1) I fXgGthenG=XB,and (2) if G is perfect and B is solvable then G is simple.

Proof. Let X 9 G. Then XB < G so, by 43.7, XB = P j for some J E I. Let Jo = {i E I: (Bsi B) n X # 4). If i E JO then si E Bsi B E XB = P j so i E J by 43.7.6. Conversely, as P j = XB, X intersects each coset of B in P j nontrivially so J c Jo. Thus J = Jo.

Claim J = I. If not as (W, S) is irreducible there exists i E I - J and j E J with [si, sj] # 1. Therefore, as (si, sj) is dihedral, l(sisjsi) > l(sjsi) > l(si), so, by 43.3.1, BsiBsjBsiB = BsisjsiB. As X G and J = Jo it follows that sisjsi E W n PJ = Wj. But then sisjsi E Wj n Wgi,jJ = Wj = (sj), contradicting [si , sj] # 1.

Thus I = J, so XB = Pj = PI = G. Hence (1) is established and of course(1) implies (2).

(43.12) Let F be a field and 1 < n and integer. Then(1) SLn(F) is quasisimple and Ln(F) is simple unless (n, IFI) _ (2, 2) or

(2, 3).(2) Spn(F) is quasisimple and PSpn(F) is simple unless (n, IFI) _ (2, 2),

(2, 3), or (4, 2).(3) If F is finite then SUn(F) is quasisimple and UU(F) is simple unless

(n, IFI) = (2, 4), (2, 9) or (3, 4).(4) If F is finite or algebraically closed, and n > 6, then Stn (F) is quasisim-

ple and PS2n(F) is simple.

Proof. Let G = SLn(F), Spn(F), SUn(F), or 01 (F), in the respective case.By Exercise 14.5, parts 1, 2, and 6, and 43.1, G possesses a BN-pair (B, N).Observe next that B is solvable. This follows from Exercise 4.9 and Exercise7.8. Indeed these exercises show B is the semidirect product of a nilpotentgroup by a solvable group. In applying Exercise 7.8 observe n - 2m < 2 asZ1 has no singular points; thus O(Z1, f) is solvable by Exercise 7.2.

Next, by Exercise 14.5.5, kerB(G) = Z(G). Further, except in the specialcases listed in the lemma, G is perfect. This follows from 13.8, 22.3.4, 22.4,and Exercise 7.6. Finally, to complete the proof, 43.11.2 says G/Z(G) is sim-ple, since by Exercise 14.5.3 the Coxeter system of the Weyl group of G isirreducible.

Remarks. The material in chapter 14 comes from Tits [Ti] and Bourbaki [Bo].Already in this chapter we begin to see the power of the Tits system buildingapproach to the study of groups of Lie type. The proof of the simplicity ofvarious classical groups in 43.12 probably provides the best example. Furtherresults are established in section 47, where groups with a BN-pair generatedby root subgroups and satisfying a weak version of the Chevalley commutatorrelations are investigated. These extra axioms facilitate the proof of a numberof interesting results.

In [Ti], Tits classifies all buildings of rank at least 3 with a finite Weyl group,and hence also all groups with a Tits system of rank at least 3 and finite Weylgroup. The rank of a building is I I I while the rank of a Tits system (G, B, N, S)is ISI.

Exercises for chapter 141. If ci = (A, a) is a complex let E (sxl) be the geometry whose objects of

type i are the simplices of sa( of type i and whose edges are the simplicesof sl of rank 2. If F is a geometry let 6(I') be the complex (I', -'(I'))


Thus I = J, so XB = PJ = PI = G. Hence (1) is established and of course (1) implies (2).

(43.12) Let F be a field and 1 < n and integer. Then (1) SL,(F) is quasisimple and L,(F) is simple unless (n, I FI) = (2,2) or

(2,3). (2) Sp,(F) is quasisimple and PSp,(F) is simple unless (n, I FI) = (2,2),

(2,3), or (4,2). (3) If F is finite then SU,(F) is quasisimple and U,(F) is simple unless

(n, IF11 = (2,419 (279) or (3,4>. (4) If F is finite or algebraically closed, and n > 6, then Qi(F) is quasisim-

ple and PQi(F) is simple.

Proof. Let G = SL,(F), Sp,(F), SU,(F), or Qi(F), in the respective case. By Exercise 14.5, parts 1, 2, and 6, and 43.1, G possesses a BN-pair (B, N). Observe next that B is solvable. This follows from Exercise 4.9 and Exercise 7.8. Indeed these exercises show B is the semidirect product of a nilpotent group by a solvable group. In applying Exercise 7.8 observe n - 2m 5 2 as 2' has no singular points; thus O(Z', f ) is solvable by Exercise 7.2.

Next, by Exercise 14.5.5, ker~(G) = Z(G). Further, except in the special cases listed in the lemma, G is perfect. This follows from 13.8, 22.3.4, 22.4, and Exercise 7.6. Finally, to complete the proof, 43.1 1.2 says G/Z(G) is simple, since by Exercise 14.5.3 the Coxeter system of the Weyl group of G is irreducible.

Remarks. The material in chapter 14 comes from Tits [Ti] and Bourbaki [Bo]. Already in this chapter we begin to see the power of the Tits system building approach to the study of groups of Lie type. The proof of the simplicity of various classical groups in 43.12 probably provides the best example. Further results are established in section 47, where groups with a BN-pair generated by root subgroups and satisfying a weak version of the Chevalley commutator relations are investigated. These extra axioms facilitate the proof of a number of interesting results.

In [Ti], Tits classifies all buildings of rank at least 3 with a finite Weyl group, and hence also all groups with a Tits system of rank at least 3 and finite Weyl group. The rank of a building is 111 while the rank of aTits system (G, B, N, S) is IS/.

Exercises for chapter 14 1. If sf = ( A , d ) is a complex let C ( d ) be the geometry whose objects of

type i are the simplices of d of type i and whose edges are the simplices of d of rank 2. If r is a geometry let B ( r ) be the complex ( r , &(r))

where -6(I') is the set of flags of F of type I. Prove(1) The inclusion map 7r is an injective morphism of E(sa') into A; 7r is

an isomorphism if and only if each rank 2 flag of A is a simplex ofsxl;7r is an isomorphism of-6(E(sy1)) with if and only if every flagof A is a simplex of c(.

(2) 7r is an isomorphism of E(6(I')) with F if and only if every rank 2flag of r' is contained in a flag of type I.

(3) For each simplex T of a. let .(T = {C - T : T < C E s-I and ET =E(AT, aT). Prove the map 7rT: ET -+ AT is an isomorphism of ATwith ET for each simplex T of , 1 if and only if every flag of A is asimplex of s..

(4) Aut(F) = Aut(6(F)).2. Let i6 = (I', -6) be a Coxeter complex over I = (1, ... , n 1, let W =

Aut(6), C in 6, and for J C I let J' = I - J and Tj the subflag ofC of type J. For i c I let ri be the reflection through the wall Tip, letR = {ri, ... , and Rj = (ri: i E J'). Prove(1) If J C I then (ET, , -6T f) is a Coxeter complex and WT, = Aut(ET,, 6Tf )

(in the notation of Exercise 14.1).(2) WT, = (Rj) and if I # J then (WT,, Rj) is a Coxeter system.(3) Let Ti = (vi 1, Wi = WT, and = 9,-(W, R) = (Wi: i E I). Prove

the map

viwI-Wiw iEI,wEW

is an isomorphism of -6 with 6(W,3. Let G be a group and 97= (Gi : i c I) a family of subgroups of G. Prove the

complex 6(G, is a chamber complex if and only if G = (G1': i E I),where i' = I - {i 1. Prove 6(G, ) = 6(I'(G, 3T)) if and only if G is flagtrasitive on I'(G, 9-).

4. Let -6' = (F, -6) be a Coxeter complex and W = Aut(6). Prove(1) Let A and B be simplices. Prove A U B is a simplex if and only if A U B

is contained in F0 or Fo' for each pair 0, 0' of opposite foldings.(2) Let Ai, 1 < i < 3, be simplices such that Ai U Aj is a simplex for

each i, j. Then U3=1 Ai is a simplex.(3) Every flag of r' is a simplex.(4) W is flag transitive on F.(5) Prove (in the notation of Exercise 14.1) that (IT,, 6T) is a Coxeter

complex for each J C I.(6) W = Aut(I').

5. Let F be a field, V a finite dimensional vector space over F, and assumeone of the following holds:(A) F is the projective geometry on V, Y = {(xi): 1 < i < n) for some

basis X = (xi: 1 < i < m) of V, m = dim(V), and G = GL(V).


where &(r ) is the set of flags of r of type I. Prove (1) The inclusion map n is an injective morphism of C ( d ) into A; n is

an isomorphism if and only if each rank 2 flag of A is a simplex of d, n is an isomorphism of &(C(d)) with d i f and only if every flag of A is a simplex of d.

(2) n is an isomorphism of C(&(r)) with r if and only if every rank 2 flag of r is contained in a flag of type I .

(3) Foreachsimplex T o f d l e t d , = {C - T: T 5 C E d ] and CT = C(AT, d T ) . Prove the map n T : CT + AT is an isomorphism of AT with CT for each simplex T of d if and only if every flag of A is a simplex of d.

(4) Aut(r) = Aut(&(r)). 2. Let B = ( r , &) be a Coxeter complex over I = (1, . . . , n], let W =

Aut(8), C in B , and for J E I let J' = I - J and Tj the subflag of C of type J. For i E I let r, be the reflection through the wall T,I, let R = {r l , . . . , r,], and R j = (r,: i E J']. Prove (1) If J c I then (Cz , g z ) is a Coxetercomplex and WE = Aut(CTI,

(in the notation of Exercise 14.1). (2) WE = (Rj) and if I # J then (WE, R j) is a Coxeter system. (3) Let T, = (v,], W, = Wz, and 9 = 9(W, R) = (W,: i E I). Prove

the map

is an isomorphism of 8 with 8(W, 9 ) . 3. Let G be a group and F = (Gi : i E I ) a family of subgroups of G. Prove the

complex &(G, 9 ) is a chamber complex if and only if G = (G,!: i E I ) , where i' = I - {i 1. Prove B(G, 9) = B(r(G, 9 ) ) if and only if G is flag trasitive on r (G , 9 ) .

4. Let & = ( r , &) be a Coxeter complex and W = Aut(8). Prove (1) Let A and B be simplices. Prove A U B is a simplex if and only if A U B

is contained in r4 or r4' for each pair 4 , 4 ' of opposite foldings. (2) Let A,, 1 5 i 5 3, be simplices such that A, U A, is a simplex for

each i, j. Then uLl Ai is a simplex. (3) Every flag of r is a simplex. (4) W is flag transitive on r. (5) Prove (in the notation of Exercise 14.1) that (rfi, B E ) is a Coxeter

complex for each J c I. (6) W = Aut(r).

5. Let F be a field, V a finite dimensional vector space over F, and assume one of the following holds: (A) r is the projective geometry on V, Y = {(xi): 1 5 i 5 n] for some

basis X = (xi: 1 I: i 5 m) of V, m = dim(V), and G = GL(V).

(C) (V, f) is a symplectic or unitary space and G = O(V, f) or (V, Q)is an orthogonal space and G = O(V, Q). If (V, Q) is orthogonalassume it is not hyperbolic. 0 < m is the Witt index of the space andr its polar geometry. X = (xi: I < i < 2m) is a hyperbolic basis fora maximal hyperbolic subspace of V and Y= ((x): x E X}.

(D) (V, Q) is a 2m-dimensional hyperbolic orthogonal space and G isthe subgroup of O(V, Q) preserving the equivalence relation of 22.8,F is the oriflamme geometry of (V, Q), X = (xi: 1 < i < 2m) is ahyperbolic basis of V, and Y = ((x) : x E X }.In each case let A = -'(F) _ (F, e) be the complex on r definedin Exercise 14.1, let Ey be the subgeometry consisting of the objectsof r generated by members of Y, identify Ey with the subcomplex(Ey, -6 fl Ey), and let . ' = ((Ey)g: g E G). Prove

(1) (A a) is a building.(2) G is transitive on SZ = ((C, E): C E -6, E E sz'}.(3) The Weyl group of A is of type A,,,, C,,,, orDm in case A, C, and D,

respectively.(4) Let C be a chamber in Ey, B = Gc, and N = NG(Ey). Prove G is

the semidirect product of a nilpotent group U by H = B fl N, and His abelian if A or D holds, or if (V, f) is symplectic or unitary; or ifF is finite or algebraically closed.

(5) kerB(G) = kerH(G) is the group of scalar transformations of V in G.(6) SL(V) and Sp(V) are transitive on Q and if F is finite or algebraically

closed then S2(V, Q) and SU(V) are transitive on Q.6. Let (G, B, N, S) be a Tits system with Weyl group W = N/H and S =

(Si: i E I). Prove(1) For each J, K C I, the map

(PJ)w(PK) H (WJ)w(WK)is a bijection of the set of orbits of the parabolic WK on the coset spaceB/PJ with the orbits of the parabolic WK on the coset space W/WJ.

(2) Assume the Coxeter diagram A of W is one of the following:

1 2 n-1 nAn o--o ... p_...p

1 2 n-2 n-1 nCn o- o p------ cz::=

Dn


(C) (V, f ) is a symplectic or unitary space and G = O(V, f ) or (V, Q) is an orthogonal space and G = O(V, Q). If (V, Q) is orthogonal assume it is not hyperbolic. 0 < m is the Witt index of the space and r its polar geometry. X = (xi: 1 5 i 5 2m) is a hyperbolic basis for a maximal hyperbolic subspace of V and Y = ((x):x E X).

(D) (V, Q) is a 2m-dimensional hyperbolic orthogonal space and G is the subgroup of O(V, Q) preserving the equivalence relation of 22.8, r is the oriflamme geometry of (V, Q), X = (xi: 1 5 i 5 2m) is a hyperbolic basis of V, and Y = ((x): x E XI. In each case let 6% = 8 ( r ) = ( r , 8 ) be the complex on r defined in Exercise 14.1, let Cy be the subgeometry consisting of the objects of r generated by members of Y, identify Xy with the subcomplex (Cy, 6 il Xy), and let d = ((Cy)g: g E G). Prove

(1) ( B d) is a building. (2) G is transitive on !2 = ((C, X): C E 6, Z E d l . (3) The Weyl group of B is of type A,, C,, orD, in case A, C, and D,

respectively. (4) Let C be a chamber in Xy, B = Gc, and N = NG(Xy). Prove G is

the semidirect product of a nilpotent group U by H = B i l N, and H is abelian if A or D holds, or if (V, f ) is symplectic or unitary, or if F is finite or algebraically closed.

(5) kerB(G) = kerH(G) is the group of scalar transformations of V in G. (6) SL(V) and Sp(V) are transitive on !2 and if F is finite or algebraically

closed then Q(V, Q) and SU(V) are transitive on Q. 6. Let (G, B, N, S) be a Tits system with Weyl group W = N/H and S =

(si : i E I). Prove (1) For each J, K E I , the map

( P J ) W ( ~ K ) H (WJ)W(WK) is a bijection of the set of orbits of the parabolic WK on the coset space B/PJ with the orbits of the parabolic WK on the coset space W/ Wj.

(2) Assume the Coxeter diagram A of W is one of the following:

Prove:(a) If A is of type An then G is 2-transitive on G/PI,, rank 3 on G/P2,

if n > 2, and rank 6 on P{t,n}' if n > 1.(c) If A is of type Cn and n > 1, then G is rank n + 1 on G/Pn,, rank

3 on G/P1,, and rank 6 on G/P2 if n > 2.(d) If A is of type Dn, n > 4, then G is rank 3 on G/Pl, and rank 8

or 7 on G/P2, for n = 4 or n > 4, respectively.7. (1) Every flag in a building is a simplex.

(2) If (G, B, N, S) is a Tits system then G is flag transitive on its building9 (G, B, N, S).

(3) If A = (F, e) is a building then Aut(A) = Aut(F).8. Let (I', -') be a chamber complex and for x E F let ex = fl F.

(1) If I I I > 2 assume for each x E F that (Fx, -ex) is a connected complex.Assume 0: -.' H -' is a function such that, for all C, D E i' and alli E I, if CflDis awall of type i'then CO flDo is also awall oftype P. For X E F define xa to be the element of CO of the same typeas x, where x E C E -e. Prove a: (F, ') --> (F, ') is a well-definedmorphism.

(2) Assume III > 2, F = F(G, -12-), and -' = C(G, -12-) for some groupG and some family 9 = (G;: i E I) of subgroups. Assume for eachi E I that G;, = (Gy,j),: j E i'). Prove (I'x, -fix) is a connectedcomplex.

9. Let T = (G, B, N, S) be a Tits system with finite Weyl group W = N/H.Let 3 _ 1-213 (T) be the building of T, C the chamber of A fixed by B,and E _ (xw: x E C, W E W) the apartment of A stabilized by N (cf. thediscussion before 43.8). Let be a root system for W and it a simplesystem for W with S = (r,,: a E 7r) (cf. 29.12 and 30.1). Let < be thepartial order on W defined in Exercise 10.6. Prove(1) If u, w E W with u < w then there exists a gallery (C,: 0 < i < n) of

length n = 1(w) from C to Cw with C/(u) = Cu.(2) Ifs E S and W E W with l(sw) < 1(w) then B fl BS" BW C BSw.(3) Ifs ES and wEW with 1(w) <l(ws)then BSflB'<BandB"'fl

BSB C B.(4) If u,wE Wwith u<wthen BflBW <Bu.(5) Define T to be saturated if H= nwEw Bw. Prove the following are

equivalent:(a) T is saturated.(b) H is the pointwise stabilizer in G of E.(c) B fl Bw0 = H, where wo is the element of W of maximal lengthin the alphabet S (cf. Exercise 10.5).

(6) Let f, = nwEw Bw, N = FIN, and S = (Ns: S E S). Prove (G, B,N, S) is a saturated Tits system.


Prove: (a) If A is of type A, then G is 2-transitive on G/Pl!, rank 3 on G/P21

if n > 2, and rank 6 on Ptl,ny if n > 1. (c) If A is of type C, and n > 1, then G is rank n + 1 on G/ P,!, rank

3 on G/P1!, and rank 6 on G/P2! if n > 2. (d) If A is of type D,, n 2 4, then G is rank 3 on G/Pl1 and rank 8

or 7 on G/ P2t for n = 4 or n > 4, respectively. 7. (1) Every flag in a building is a simplex.

(2) If (G, B , N, S) is a Tits system then G is flag transitive on its building 98 (G, B , N, S).

(3) If 98 = ( r , 8 ) is a building then Aut(98) = Aut(r). 8. Let (r, 8 ) be a chamber complex and for x E r let 8, = B i l r,.

(1) If 11 I > 2 assume for each x E r that (r, , 8 , ) is a connected complex. Assume @: 8 H B is a function such that, for all C, D E 8 and all i E I , if C n D is a wall of type i' then C@ i l D@ is also a wall of type i'. For x E r define xcr to be the element of C@ of the same type as x, where x E C E 8. Prove cr: ( r , 8 ) + ( r , 8 ) is a well-defined morphism.

(2) Assume 11 I > 2, r = r (G, 9), and 8 = C(G, 9) for some group G and some family 9 = (Gi: i E I ) of subgroups. Assume for each i E I that Git = (G{i,,)~: j E it). Prove (r,, 8 , ) is a connected complex.

9. Let T = (G, B , N, S) be aTits system with finite Weyl group W = N/H. Let 98 = 98 (T) be the building of T, C the chamber of 98 fixed by B , and X = (XW: x E C, w E W) the apartment of 98 stabilized by N (cf. the discussion before 43.8). Let @ be a root system for W and n a simple system for W with S = (r,: cr E n) (cf. 29.12 and 30.1). Let 5 be the partial order on W defined in Exercise 10.6. Prove (1) If u, w E W with u 5 w then there exists a gallery (Ci: 0 I i I n) of

length n = l(w) from C to Cw with C1(,) = Cu. (2) If s E S and w E W with l(sw) < l(w) then B i l BSWBW C B S W . (3) If s E S and w E W with l(w) 5 l(ws) then BS i l BW i B and Bw i l

B S B C B . (4) I fu ,w E W w i t h u i w t h e n B n B W ( B U . (5) Define T to be saturated if H = nw,, B w . Prove the following are

(a) T is saturated. (b) H is the pointwise stabilizer in G of X. (c) B n BWO = H, where wo is the element of W of maximal length in the alphabet S (cf. Exercise 10.5).

B ~ , IV = AN, and S = (IVs:s E s). Prove (G, B , s a saturated Tits system.

(Hints: Use 41.7 in (1). To prove (2) observe B fl BSS BW C NB(Csw fl Cw) and use the Rainy Day Lemma and (1) to prove NB(Csw fl Cw) < BSW. Use (2) to prove (3) and (4) and use (4) andExercise 10.6 to prove (5).)

10. (Richen [Ri]) Assume the hypothesis of Exercise 14.9 and assume furtherthat T is saturated. Let a E jr and s = ra E S. For W E W define B," =B fl Bw0w-' , where w0 is the maximal element of W. For ,8 E jr defineBp = Br, . Prove(1) If w E W with aw > 0 then Ba < Bw '(2) If W E W with aw < 0 then Ba fl Bw-' = H.(3) If W E W with 1(w) < l(ws) then B = (B fl B'')(B fl B').(4) B = Ba(B f1 BS) with BS f1 Ba = H.(5) Ba H.(6) For w E W, Ba < Bw ' if and only if aw > 0.(7) Let A = {(Ba)w: a E 7r, w E W). Prove the map (Ba)' H aw is a

well-defined permutation equivalence of the representations of W onA and (D. In particular we may define By = (Ba)' for each y E (D,where y = aw.

(8) If w E W with aw < 0 then B", = Ba(B,S,,,)S and Ba fl (B,r",)S = H.(9) Let w0 = r1 ... r with n = l(wo) _ I(D+I and rl = ra,, a; E jr. Let

wi = riri_1 ... ri. Prove B = BaiBa2wt ...Ba w = H. Further the map i H aiwi_1 is a bijection ofj i+1nj with 4)+.

(Hints: In (1) use Exercise 10.6.4 to show w-1 < wos and then appealto Exercise 14.9.4. In (2) use parts (2) and (5) of Exercise 14.9 toconclude Bw-'S fl BS < B and B fl Bw0 = H. In (3) use 43.3.1 toget B C BSBw and then appeal to Exercise 14.9.3. Use (4) and BN4to prove (5). In (7) use (6) to show for ,8 E Jr, (Ba)w = Bp if and onlyif aw = P. In (8) use (1) to show Ba < Bw. Then use (4) to showBw = Ba(B fl BS) and use Exercise 14.9.3 to show Bsw < B. Use (8)to prove (9).)


(Hints: Use 41.7 in (1). To prove (2) observe B n BSWBW E NB (Csw n C w ) and use the Rainy Day Lemma and (1) to prove NB (Csw n C w ) 5 BSW. Use (2 ) to prove (3 ) and (4) and use (4) and Exercise 10.6 to prove (3.)

10. (Richen [Ri]) Assume the hypothesis of Exercise 14.9 and assume further that T is saturated. Let a E n and s = r, E S. For w E W define Bw =

B n B W O W - ' , where wo is the maximal element of W. For B E n define Bg = B,, . Prove (1) If w E W with aw > 0 then B, 5 B W - I .

(2) If w E W with a w < 0 then B, n B W - I = H . (3) I f w E W with l (w) 5 l(ws) then B = ( B fl BS)(B fl BW). (4) B = B,(B n B" with BW B, = H .

(5) B, # H . (6) For w E W, B, 5 B ~ - ' if and only if aw > 0. (7) Let A = {(B,)": a E n, w E W). Prove the map (B,)W H aw is a

well-defined permutation equivalence of the representations of W on A and @. In particular we may define B, = (Ba)W for each y E @,

where y = aw . (8) I f w E W with a w < 0 then Bw = B,(Bsw)S and B, n (Bsw)S = H . (9) Let wo = rl . . . r, with n = l(wo) = I @+ 1 and ri = raL , ai E n. Let

wi =riri-1 ... rl. Prove B = Ba1Ba,,, ... B, , , ,,-, with Ba,w,-,fl n;=,+, BaJwJ-, = H . Further the map i H ai w,-1 is a bijection of ( 1 , . . . , n) with @+. (Hints: In ( 1 ) use Exercise 10.6.4 to show w -' < wos and then appeal to Exercise 14.9.4. In (2) use parts (2) and (5 ) of Exercise 14.9 to conclude B ~ - ~ ~ n BS _( B and B n BWO = H . In (3 ) use 43.3.1 to get B BSBW and then appeal to Exercise 14.9.3. Use (4) and BN4 to prove (5). In (7) use (6) to show for B E n, (Ba)W = Bg if and only if aw = B. In (8) use ( 1 ) to show B, 5 B,. Then use (4 ) to show Bw = B,(B fl Bs ) and use Exercise 14.9.3 to show Biw 5 B. Use (8 ) to prove (9).)

15

Signalizerfunctors

Let r be a prime, G a finite group, and A an abelian r-subgroup of G. AnA-signalizer functor on G is a map 0 from A# into the set of A-invariantr'-subgroups of G such that, for each a, b E A#, 0(a):!:- CG(a) and 0(a) flCG(b) <9(b). The signalizer functor 0 is said to be complete if there is anA-invariant r'-subgroup 0(G) such that 0(a) = CB(G)(a) for each a E A#.

Notice that one way to construct an A-signalizer functor is to select someA-invariant r'-subgroup X of G and define 0(a) = Cx(a) for a E A#. By con-struction this signalizer functor is complete. If m(A) > 3 it turns out that this isthe only way to construct signalizer functors. That is, if m(A) > 3 then everyA-signalizer functor is complete. This result is called the Signalizer FunctorTheorem. It's one of the fundamental theorems in the classification of the finitesimple groups. Unfortunately the proof of the Signalizer Functor Theorem isbeyond the scope of this book. However, chapter 15 does contain a proof of aspecial case: the so-called Solvable 2-Signalizer Functor Theorem. It turns outthat the Solvable Signalizer Functor Theorem suffices for many applicationsof signalizer functors. -

An A-signalizer functor 0 on G is said to be solvable if 0(a) is solvablefor each a E A#. We say 0 is solvably complete if 0 is complete and 0(G) issolvable. The main result of chapter 15 is:

Solvable 2-Signalizer Functor Theorem. Let A be an abelian 2-subgroup ofa finite group G with m(A) > 3. Then each solvable A-signalizer functor on Gis solvably complete.

Chapter 16 contains a discussion of the Classification Theorem which illus-trates how the Signalizer Functor Theorem is used. Observe that the conditionthat m(A) > 3 is necessary in the Signalizer Functor Theorem by Exercise 15.1.

44 Solvable signalizer functorsIn section 44, r is a prime, G is a finite group, A is an abelian r-subgroup ofG of r-rank at least 3, and 0 is a solvable A-signalizer functor on G.

Let Q(G) denote the set of solvable A-invariant r'-subgroups X of G withCx(a) = X fl 0(a) for each a e A#. For A < H < G, let Q(H) = H fl Q(G) and

0(H) = (H fl 0(a): a E A#).

Signalizer functors

Let r be a prime, G a finite group, and A an abelian r-subgroup of G. An A-signalizer functor on G is a map 8 from A' into the set of A-invariant rl-subgroups of G such that, for each a , b E A', 8(a) I: CG(a) and 8(a) n CG(b) I: 8(b). The signalizer functor 8 is said to be complete if there is an A-invariant rl-subgroup 8(G) such that 8(a) = Ce(G)(a) for each a E A'.

Notice that one way to construct an A-signalizer functor is to select some A-invariant rl-subgroup X of G and define 8(a) = Cx(a) for a E A'. By construction this signalizer functor is complete. If m(A) 2 3 it turns out that this is the only way to construct signalizer functors. That is, if m(A) 2 3 then every A-signalizer functor is complete. This result is called the Signalizer Functor Theorem. It's one of the fundamental theorems in the classification of the finite simple groups. Unfortunately the proof of the Signalizer Functor Theorem is beyond the scope of this book. However, chapter 15 does contain a proof of a special case: the so-called Solvable 2-Signalizer Functor Theorem. It turns out that the Solvable Signalizer Functor Theorem suffices for many applications of signalizer functors.

An A-signalizer functor 8 on G is said to be solvable if 8(a) is solvable for each a E A'. We say 8 is solvably complete if 8 is complete and 8(G) is solvable. The main result of chapter 15 is:

Solvable ZSignalizer Functor Theorem. Let A be an abelian Zsubgroup of a finite group G with m(A) 2 3. Then each solvable A-signalizer functor on G is solvably complete.

Chapter 16 contains a discussion of the Classification Theorem which illustrates how the Signalizer Functor Theorem is used. Observe that the condition that m(A) 2 3 is necessary in the Signalizer Functor Theorem by Exercise 15.1.

44 Solvable signalizer functors In section 44, r is a prime, G is a finite group, A is an abelian r-subgroup of G of r-rank at least 3, and 8 is a solvable A-signalizer functor on G.

Let Q(G) denote the set of solvable A-invariant rl-subgroups X of G with Cx(a) = X n O(a) for each a E A'. For A 5 H 5 G, let Q(H) = H fl Q(G) and

230 Signalizerfunctors

If 7r is a set of primes then Q(H, 7r) denotes the set of 7r-groups in Q(H). WriteQ*(H) and Q*(H, 7r) for the maximal members of Q(H) and Q(H, 7r), respec-tively, under the partial order of inclusion. For X E Q(G), write Qx(G, 7r) forthe set of H E Q(G, n) with X < H, and define

n(0)= U 7r(0(a)).aEA#

(44.1) If X E Q(G) and Y is an A-invariant subgroup of X then Y E Q(G).

(44.2) If X, Y E Q(G) with X < NG (Y) then XYE Q(G).

Proof. This follows from Exercise 6.1.

(44.3) Let H E Q(G) with H a G, set G* = G/H, and define 9*(a*) = B(a)*for a* E A*#. Then

(1) 8* is an A*-signalizer functor.(2) Q(G*) = QH(G)* = {X *: X E Q(G) and H < X1.(3) Q(G*) = Q(G)* = {X*: X E Q(G)}.

Proof. Let a, b E A#. To prove (1) we must show 9*(a*) fl 9*(b*).

By coprime action, 18.7, CG- (b*) = CG(b)*, so it suffices to show CHOW (b) <H9(b). But by Exercise 6.1, CHO(a)(b)=CH(b)CB(a)(b), so as CB(a)(b) <9(b),(1) is established.

Let X E QH(G). Then X is a solvable A-invariant r'-group, so its imageX* has the same properties. Further for a E A#, Cx (a) < 0 (a), and by coprimeaction, Cx{(a*) = Cx(a)*, so Cx.(a*) < 0(a)* = 9*(a*). That is, X* E Q(G*).Conversely, suppose Y* E Q(G*) and let Y be the full preimage of Y* in G;to complete the proof of (2), we must show Y E Q(G). As Y/H = Y* and Hare solvable r'-groups, so is Y by 9.3. As Y* E Q(G*), Cy(a*) <9*(a*), soCy(a) <9(a)H. But H E Q(G), so CH(a) <0(a), and then

Cy(a) < CG(a) fl 0(a)H = 0(a)CH(a) < 0(a)

by the modular property of groups 1.14. So (2) is established.By 44.2, HX E QH(G) for each X E Q(G), so QH(G)* = Q(G)*. Hence (2)

implies (3).

(44.4) For each 10 B < A, 9(CG(B)) E Q(G).

Proof. Recall 9(CG(B)) _ (CG(B) fl 9(a): a E A#). But for b E B#, CG(B) fl0(a) < CG(b) fl 0(a) < 0(b), so 9 (CG (B )) < 0(b), and hence the lemma followsfrom 44.1.

230 Signalizer functors

If rr is a set of primes then &(H, n ) denotes the set of n-groups in &(H). Write &*(H) and &*(H, n ) for the maximal members of &(H) and Q(H, n), respectively, under the partial order of inclusion. For X E &(G), write &x(G, n ) for the set of H E Q(G, n ) with X ( H , and define

(44.1) If X E &(G) and Y is an A-invariant subgroup of X then Y E &(G).

(44.2) If X, Y E &(G) with X I N c ( Y ) then XYE &(GI.

Proof. This follows from Exercise 6.1.

(44.3) Let H E &(G) with H 5 G , set G* = GIH, and define O*(a*) = O(a)* for a* E A*'. Then

(1) e* is an A*-signalizer functor. (2) &(G*) = QH(G)* = {X*: X E &(G) and H 5 X}. (3) &(G*) = &(G)* = {X*: X E Q(G)}.

Proof. Let a , b E A'. To prove ( 1 ) we must show O*(a*) f l Cc*(b*) 5 O*(b*). By coprime action, 18.7, CG*(b*) = CG(b)*, SO it suffices to show C~s( , ) (b) 5 HO(b). But by Exercise 6.1, CHsca)(b) = CH(b)Ce(a)(b), SO as Ce(a)(b) I O(b), ( 1 ) is established.

Let X E QH(G) . Then X is a solvable A-invariant r'-group, so its image X* has the same properties. Further for a E A', Cx(a) i O(a), and by coprime . action, Cx*(a*) = Cx(a)*, so Cx*(a*) 5 O(a)* =O*(a*). That is, X* E &(G*). Conversely, suppose Y* E &(G*) and let Y be the full preimage of Y* in G ; to complete the proof of (2), we must show Y E &(G). As Y I H = Y* and H are solvable r'-groups, so is Y by 9.3. As Y* E &(G*), Cr*(a*) _( O*(a*), so Cy(a) (B (a )H . But H E Q(G), so CH(a) 5 $(a), and then

by the modular property of groups 1.14. So (2) is established. By 44.2, HX E QH(G) for each X E &(G), so &H(G)* = &(G)*. Hence (2)

implies (3).

(44.4) For each 1 # B 5 A, O(Cc(B)) f &(GI.

Proof. Recall O(CG(B)) = (CG(B) f l @(a): a E A'). But for b E B', Cc (B) f l

@(a) i Cc(b) flO(a) ( @(b), so @(CG(B)) 5 O(b), and hence the lemma follows from 44.1.

Solvable signalizer functors 231

(44.5) For each X E Q(G) and Tr C.7r(9), Cx(A) <0(CG(A)) and Cx(A) istransitive on Q*(X,.7r). Indeed, Q*(X,.7r) is the set of A-invariant Hall .7r-subgroups of X.

Proof. Let T be the set of A-invariant yr-subgroups of X and T* the set of maxi-mal members of T under inclusion. By 44.1, T = Q(X, .7r), so T * = Q* (X, .7r).By Exercise 6.2, T* is the set of A-invariant Hall yr-subgroups of X and Cx (A)is transitive on T*. Finally, as X E Q(G), we have Cx(A) <0(a) for eacha E A#, so Cx(A) < 0(CG(A)), completing the proof.

(44.6) Let P E 7r (0). Then(1) For each a E A# there is a unique maximal B(CG(A))-invariant member

0(a) of Q(6(a), p').(2) A is an A-signalizer functor.

Proof. By 44.4, Y = B(CG(A)) < X = 9(a). Let R be the set of Y-invariantmembers of Q(X, p') and S = Q*(X, p'). Then for R E R, R <_ S E S, so asY is transitive on S by 45.5, R is contained in each member of S. ThereforeM = (R) < S, so M is the unique maximal member of R by 44.1. This estab-lishes (1). Next for b E B#, 0(a) fl CG(b) = CM(b) <9(b) as B is a signalizerfunctor. Also CM(b) is an A-invariant p'-group, so CM(b) < 0(b), completingthe proof of (2).

(44.7) (Transitivity Theorem) B(CG(A)) is transitive on Q*(G, p) for eachp E 7r (B).

Proof. The proof is left as Exercise 15.2.

(44.8) Let X E Q(G). Then(1) If B is a noncyclic elementary abelian subgroup of A then

X = (B(C(D)) f1 X: I B: D) =r).

(2) If1#T<Athen

0(CG(T)) = (0(CG(B)): T < B < A and B is noncyclic).

(3) X =0(X).

Proof. If 1 T < A then by Exercise 6.5,

X = (Cx(S): T/S is cyclic).

Solvable signalizerfinctors 23 1

(44.5) For each X E &(G) and n E: n (0), Cx (A) 5 O (CG (A)) and Cx (A) is transitive on &*(X, n). Indeed, &*(X, n ) is the set of A-invariant Hall n- subgroups of X.

Proof. Let T be the set of A-invariant n-subgroups of X and T* the set of maximal members of T under inclusion. By 44.1, T = &(X, n), so T* = &*(X, n). By Exercise 6.2, T* is the set of A-invariant Hall n-subgroups of X and Cx(A) is transitive on T*. Finally, as X E &(G), we have Cx(A) 5 O(a) for each a E A', so Cx (A) ( O (CG (A)), completing the proof.

(44.6) Let p E n(O). Then (1) For each a E A# there is a unique maximal O(CG(A))-invariant member

A(a> of &(O(a), PI). (2) A is an A-signalizer functor.

Proof. By 44.4, Y = O(CG(A)) 5 X = O(a). Let R be the set of Y-invariant members of &(X, p') and S = &*(X, p'). Then for R E R , R ( S E S, so as Y is transitive on S by 45.5, R is contained in each member of S. Therefore M = (R) 5 S, so M is the unique maximal member of R by 44.1. This establishes (1). Next for b E B', A(a) f l CG(b) = CM(b) 5 O(b) as O is a signalizer functor. Also CM(b) is an A-invariant pl-group, so CM(b) 5 A(b), completing the proof of (2).

, (44.7) (Transitivity Theorem) O(CG(A)) is transitive on &*(G, p) for each

P E n(@.

Proof. The proof is left as Exercise 15.2.

(44.8) Let X E &(G). Then (1) If B is a noncyclic elementary abelian subgroup of A then

(2) If 1 # T 5 A then

O(CG(T)) = (O(CG(B)): T 5 B 5 A and B is noncyclic).

(3) X =O(X).

Proof. If 1 # T I A then by Exercise 6.5,

X = (Cx(S): T/S is cyclic).

Further if S 01 then CX(S) <0(CG(S)) by 44.5, so (1) follows. Similarly by44.4, Y =9(CG(T)) E Q(G) and A= A/T is abelian and acts on Y, so byExercise 6.5,

Y = (Cy(B): Alb is cyclic)

with Cy(B) = 9(CG(B)). If T is noncyclic then (2) is trivial, so we may assumeT is cyclic. Then as m(A) > 3, B is noncyclic, so (2) holds.

For H < G, 9(H) < H by definition of 9(H), while X <9(X) by (1).

This about exhausts the results on solvable signalizer functors which canbe established outside of an inductive setting. To proceed further we mustwork inside a minimal counterexample to the Solvable 2-Signalizer FunctorTheorem.

So in the remainder of chapter 15 assume (A, G, 9) is a counterexampleto the Solvable 2-Signalizer Functor Theorem with (G( + In(9)1 minimal. Inparticular now r = 2. Let I = 9(CG (A)).

(44.9) (1) If A < H < G then 9(H) E Q(G).(2) G = A(9(b): b E B#) for each E4-subgroup B of A.(3) If 10 H E Q(G) then G 0 NG(H).

Proof. In each case we use the minimality of the counterexample. Under thehypotheses of (1), 9H(a) = 9(a) fl H is an A-signalizer functor on H, so (1)follows from minimality of I G 1.

Let B be a 4-subgroup of A and K = (9(b): b c B#). If G 0 KA then by (1),Y = 9(KA) E Q(G). But now for a E A#,

9(a) = (9(a) fl 9(b): b E B#) <Cy(a) = (9(b) fl Cy(a): b E B#) <6(a)

by 44.8.1, so 9 is solvably complete, contrary to the choice of (A, G, 9) as acounterexample. Thus (2) holds.

Suppose 1 * H E Q(G) and H < G. Define G* and 0* as in 44.3. By 44.3.1,9* is an A*-signalizer functor, so by minimality of J G 1, 9 * is solvably complete.

Thus there is a unique maximal member X* of Q(G*). By 44.3.2, X E Q(G).As 6(a)*=9*(a*) <X*, 9(a) < X, so 9(a)=CX(a) as X E Q(G). Therefore9 is solvably complete, contrary to the choice of (A, G, 9) as a counterexample.

(44.10) Let lr c n(9), M E Q*(G, 7r), and 10 X E Q(G) with X < M. Then(1) M is a Hall n-subgroup of 9(NG(X)).(2) Op(M) <Opn(9(NG(X)) for each p E7r.(3) If7r=7r(9),sothatMEQ*(G),then9(NG(X))=M. HenceifX<YE

Q(G), then Ny(X) <M.


Further if S # 1 then Cx (S) 5 8(CG(S)) by 44.5, so (1) follows. Similarly by 44.4, Y = 8(CG(T)) E Q(G) and A = AIT is abelian and acts on Y, so by Exercise 6.5,

Y = (Cy (B): A/B is cyclic)

with Cy (B) = 8(CG(B)). If T is noncyclic then (2) is trivial, so we may assume T is cyclic. Then as m(A) > 3, B is noncyclic, so (2) holds.

For H 5 G, 8(H) j H by definition of 8(H), while X 5 8(X) by (1).

This about exhausts the results on solvable signalizer functors which can be established outside of an inductive setting. To proceed further we must work inside a minimal counterexample to the Solvable 2-Signalizer Functor Theorem.

So in the remainder of chapter 15 assume (A, G, 8) is a counterexample to the Solvable 2-Signalizer Functor Theorem with IG I + In (8)/ minimal. In particular now r = 2. Let I = 8(CG(A)).

(44.9) (1) If A 5 H < G then 8(H) E Q(G). (2) G = A(8(b): b E B') for each E4-subgroup B of A. (3) If 1 # H E Q(G) then G # NG(H).

Proo$ In each case we use the minimality of the counterexample. Under the hypotheses of (I), OH(a) = 8(a) n H is an A-signalizer functor on H, so (1) follows from minimality of (GI.

Let B be a 4-subgroup of A and K = (8(b): b E B'). If G # KA then by (I), Y = 8(KA) E Q(G). But now for a E A',

by 44.8.1, so 8 is solvably complete, contrary to the choice of ( A , G, 8) as a counterexample. Thus (2) holds.

Suppose 1 # H E Q(G) and H a G. Define G* and 8* as in 44.3. By 44.3.1, 8* is an A*-signalizer functor, so by minimality of IG1,8* is solvably complete. Thus there is a unique maximal member X* of Q(G*). By 44.3.2, X E Q(G). As 8(a)* = 8*(a*) 5 X*, 8(a) 5 X, so 8(a) = Cx(a) as X E Q(G). Therefore 8 is solvably complete, contrary to the choice of (A, G, 8) as a counterexample.

(44.10) Let n 2 n(8), M E Q*(G, n), and 1 # X E Q(G) with X 9 M. Then (1) M is a Hall n-subgroup of 8(NG(X)). (2) 0, (M) j Opn (8 (NG (X)) for each p E n . (3) If n = n(8), so that M E Q*(G), then 8(NG(X)) = M. Hence if X j Y E

Q(G), then Ny (X) 5 M.

Solvable signalizerfunctors 233

Proof. As X < M, M < H = 9(NG(X)) by 44.8.3. By 44.9.3, G 0 NG(X), soH E Q(G) by 44.9.1. As M E Q*(G, n), M E Q*(H, n), so (1) follows from44.5. Part (2) follows from (1) and Exercise 11.1. If M E Q*(G) then H = Mby maximality of M, so (3) holds.

(44.11) CA(P) = 1 for each p E7r(0) and each P E Q*(G, p).

Proof. Suppose T = CA(P) 0 1. Then by 44.7, Q*(G, p) c CG(T), so, byExercise 11.1.3, [H, T] <Op-(H) for each H E Q(G).

Form the signalizer functor A of 44.6 with respect to p. By minimality of17r(0)1, A is solvably complete. Hence K = O(G) E Q(G). Observe

[6(a), TI Op'(9(a)) < O(a) = CK(a)

by paragraph one of this proof. Recall that, by 24.5, [9(a), T, T] = [9(a), T],so

[CK(a), Ti _ [9(a), T].

Then by 8.5.6, [CK(a), T] < 9(a).Let S be the set of noncyclic subgroups of A containing T. By Exercise 8.9,

[K, Ti = ([CK(b), Ti: b c B#)

for each B E S. Therefore .

0(CG(B)) = n 9(b)bEB#

acts on [K, T]. Hence 9 (CG(T )) acts on [K, T] by 44.8.2, so by 44.2 and 44.4,H = [K, T]9(CG(T)) E Q(G). But by 24.4, 9(a) <9(CG(T))[9(a), T], so, bythe second paragraph of this proof, 0(a) < H for each a E A#. This contradicts44.9.2.

(44.12) For each lr C 7r(6), a c A#, and M E Q*(G, 7r), [a, F(M)] 0 1.

Proof. Assume otherwise. Then by Exercise 11.1.2, a centralizes M.Let p c n (F(M)) and Op (M) = X < P E Q*(G, p). By 44.11, [a, P] 0 1, so

by the Thompson A x B Lemma, [Np (X), a] 0 1. But Np (X) is contained in aHall n-subgroup of H = 0(NG(X)), so by 44.10.1 and 44.5, Np(X)h < M forsome h E CH(A). This is impossible, as a centralizes M but [Np(X)h, a] 0 1.

For 7r C r(0) and a E A#, let M(a, r) denote those M E Q*(G, r) suchthat CM(a) E Q*(CG(a), n). Recall that for Jr a set of primes and p E 7r, p"7r'U{p}.


Proof. AS X <I M, M 5 H = 8(NG(X)) by 44.8.3. By 44.9.3, G # NG(X), SO

H E Q(G) by 44.9.1. As M E Q*(G, n) , M E Q*(H, n), so (1) follows from 44.5. Part (2) follows from (1) and Exercise 11.1. If M E Q*(G) then H = M by maximality of M, so (3) holds.

(44.11) CA(P) = 1 for each p E n(8) and each P E Q*(G, p).

Proof. Suppose T = CA(P) # 1. Then by 44.7, Q*(G, p) c Cc(T), so, by Exercise 11.1.3, [H, TI 5 Opf(H) for each H E Q(G).

Form the signalizer functor A of 44.6 with respect to p. By minimality of ln(8)1, A is solvably complete. Hence K = A(G) E Q(G). Observe

by paragraph one of this proof. Recall that, by 24.5, [8(a), T, TI = [8(a), TI, SO

Then by 8.5.6, [CK(a), TI <I 8(a). Let S be the set of noncyclic subgroups of A containing T. By Exercise 8.9,

[K, TI = ([CK(b), TI: b E B')

for each B E S. Therefore

b€B#

acts on [K, TI. HenceB(CG(T)) acts on [K, TI by 44.8.2, so by 44.2 and 44.4, H = [K, T]~(CG(T)) E Q(G). But by 24.4,8(a) 58(C~(T))[8(a) , TI, so, by the second paragraph of this proof, O(a) 5 H for each a E A'. This contradicts 44.9.2.

(44.12) Foreachn En(@, a EA', and M E Q*(G, n), [a, F(M)]# 1.

Proof. Assume otherwise. Then by Exercise 11.1.2, a centralizes M. LetpEn(F(M))andO,(M)=X 5 P E Q*(G,p).By44.11,[~, PI# 1,so

by the Thompson A x B Lemma, [Np(X), a ] # 1. But Np (X) is contained in a Hall n-subgroup of H = 8(NG(X)), so by 44.10.1 and 44.5, N P ( X ) ~ 5 M for some h E CH(A). This is impossible, as a centralizes M but [Np(xlh, a ] # 1.

For n c n ( 8 ) and a E A', let M ( a , n ) denote those M E Q*(G, n ) such that CM(a) E Q*(CG(a), n). Recall that for n a set of primes and p E n , pK =

n ' u {PI.

(44.13) Let jr C- 7r p c 7r, a E A#, M E M(a, n), V E Q(Op(M)), and N EQ(G) with N = (V, CN(a)). Then V <Opn(N).

Proof. Assume V ¢ O pn (N). By 31.20.2, there is a 4-subgroup B of A withCv(B) ¢ Opn((CN(a), Cv(B))), so replacing V by Cv(B), we may assume[V, B] = 1. Next by 31.20.3 there is b c= B# with V¢ Op- ((CN((a, b) ), V)), sowe may assume N <0(b). However as Y = Cm(a) E Q* (CG (a), 7r), Cy(b) EQ*(Co(b)(a), yr). Further Cy(b) acts on CoP(M)(b) and Cv(b) <Cop(M)(b), so31.20.1 supplies a contradiction, completing the proof.

For.7r c .7r(O) and a E A#, let U(a, 7r) consist of the those U E Q(G,.7r) suchthat [U, a] 1 and U is invariant under some member of Q*(0(a), 7r). LetU*(a, 7r) be the minimal members of U(a, 7r) under inclusion. Given p c 1r,let V(p,.7r) consist of those V E Q(G, p) such that V for eachH E Qv(G).

(44.14) For each.7r c 7r (0) and a c A#:(1) U(a, Ir) # 0.(2) For U E U*(a, pr), U = [U, a] is a p-group for some prime p.(3) Let Y E Q*(O(a), 7r) with Y < NG(U) and M E QJ (G, ir). Then U <

Op(M) and U centralizes CF(M)(a).(4) U E V(p, 7r).

Proof. Let Y E Q*(O(a), 7r) and Y < X E Q*(G, n). By 44.12, [a, F(X)] # 1,so X E U(a, 7r), and (1) is established. Let U E U*(a,.7r); without loss Y actson U. By Exercise 11.1.2, [Op(U), a] * I for some prime p, and by 24.5,[Op(U), a, a] _ [Op(U), a], so (2) follows from minimality of U.

Let M E Q I (G, 7r). Then Y = CM(a), so U is CM(a)-invariant. Thus U <Op(M) by 36.3. Hence U centralizes OP(F(M)), while by the ThompsonLemma and minimality of U, U centralizes COP(M) (a). This establishes (3).

Finally let H E Qu(G) and N= (CH(a), U). By 44.13, U <Opn(N), soU = [U, a] < [Op" (N), a] < Op" (H) by 36.3, establishing (4).

(44.15) Let p c 7r C7r (0), B a hyperplane of A, P E Q* (G, p), and for b E B#,let M(b) E M(b, 7r). Then

(1) If F(M(b)) =Op(M(b)) and P fl M(b) E Q*(M(b), p) for each b E B#,then

Z(P) < n Op(M(b))bEB#


(44.13) Let n E n ( 8 ) , p E n, a E A', M E M ( a , n), V E Q(Op(M)) , and N E

Q(G) with N = ( V , CN(a) ) . Then V 5 O,n(N).

Proo$ Assume V $ OPT ( N ) . By 3 1.20.2, there is a 4-subgroup B of A with C v ( B ) $ Opn((CN(a) , C v ( B ) ) ) , SO replacing V by Cv(B) , we may assume [ V , B] = 1. Next by 31.20.3 there is b E B' with V $ O P n ( ( C ~ ( ( a , b ) ) , V ) ) , SO

we may assume N 5 8(b). However as Y = C M ( U ) E Q*(Cc(a), n), CY (b) E Q*(Ce(b)(a), n). Further C Y ( ~ ) acts on C O , ( M ) ( ~ ) and Cv(b) i C o , ( ~ ) ( b ) , so 3 1.20.1 supplies a contradiction, completing the proof.

For n g n ( 8 ) and a E A', let U(a, n) consist of the those U E Q(G, n ) such that [U, a ] # 1 and U is invariant under some member of Q*(8(a), n). Let U*(a, n ) be the minimal members of U ( a , n ) under inclusion. Given p E n, let V ( p , n ) consist of those V E Q(G, p) such that V _ ( O P r ( H ) for each H E Qv(G).

(44.14) For each n E: n ( 8 ) and a E A':

(1) U(a , n ) # 0. (2) For U E U*(a, n), U = [U, a ] is a p-group for some prime p. (3) Let Y E Q*(O(a), n) with Y 5 Nc(U) and M E Q*,,(G, n). Then U 5

O,(M) and U centralizes C F ( ~ ) ( U ) .

(4) U E V ( p , n).

Proof. Let Y E Q*(O(a), n ) and Y X E Q*(G, n). By 44.12, [a, F ( X ) ] # 1, so X E U ( a , n), and (1) is established. Let U E U*(a, n); without loss Y acts on U . By Exercise 11.1.2, [Op(U), a] # 1 for some prime p, and by 24.5, [O,(U), a , a ] = [O,(U), a ] , so (2) follows from minimality of U .

Let M E QTuy(G, n). Then Y = CM(a) , so U is CM(~)-invariant. Thus U 5 O,(M) by 36.3. Hence U centralizes Op(F(M)), while by the Thompson Lemma and minimality of U , U centralizes Co,(M)(a). This establishes (3).

Finally let H E Qu(G) and N = (CH(U) , U ) . By 44.13, U Opn(N), SO

U = [U, a ] 5 [O,r ( N ) , a ] 5 Opn ( H ) by 36.3, establishing (4).

(44.15) Let p E n E n ( 8 ) , B a hyperplane of A, P E Q*(G, p), and forb E B', let M(b) E M ( b , n). Then

(1) If F(M(b))=O,(M(b)) and P n M(b) E P ( M ( b ) , p) for each b E B', then

(2) We have

Q n Op(M(b))) c V(p, n).bEB#(

Proof. Assume the hypotheses of (1) and let Q(b) = Op(M(b)) and N(b) _O(NG(Q(b))). By 44.10.1, M(b) is a Hall 7r-subgroup of N(b), so as P(b) = P nM(b) E Sylp(M(b)) by the hypotheses of (1), P(b) E Sylp(N(b)). Thus Q(b) <P(b) < P, so Z(P) <B(CG(Q(b))) < N(b). Hence Z(P) < P(b), so as CM(b)(F(M(b))) < F(M(b)), (1) holds. Let

1#XEQ(n Op(M(b))bEB#

and N E Qx (G). By 44.13, X < Op" ((CN (b), X)) for each b E B#, so by 31.20.3,X < O pn (N), establishing (2).

(44.16) Let M E Q*(G, 7r), a c,r(F(M)), and X < F(M) with.7r(X) =a andZ(O.(F(M))) < X. Then for each N E Qx(G,.7r):

(1) [OP(X), Op(N)] =1 for each p E a.(2) If 7r(F(N)) c a thence = 7r(F(N)) and [Op(X), OP(F(N))] =1 foreach

pEa.(3) If Tr(F(N)) c a and laI > 1 then X < F(N).

Proof. For p E a, let XP = Op(X) and Zp = Z(O p(M)). By 44.10.2,

Xq <Ogr(O(NG(Zp))) for each p, q E a,

so Xq <Oq(NN(Zp)). Thus OP(X) < Op, (NN (Zp)), so by 31.15, OP(X) <Op,(N) <CN(Op(N)), establishing (1).

Suppose 7r(F(N)) c a. Then by (1), 1 # Op(X) < CN(Op(F(N)), so by31.10, p E 7r(F(N)), establishing (2). Assume in addition that loll > 1 butOp(X) ¢ F(N). Pick p # q E a and let H = CN(Op(F(N))) and K = HZq.By (2), X < H, so as H < N, we have X, ¢ O p(H). But if Zq < Oq(K),then by 31.14, X p <Op(NH(Zq)) = Op(H), SO Zq ¢Oq(K). However,F(H) = Op(N)Z(F(N)) and [Zq,Op(N)] = 1, so [Zq, H] <CH(Op(N))=CH(F(H)) = Z(F(H)). As [Zq, OP(F(H))] = 1, we have Zq <Oq(ZgZ(F(H))), so [Zq, H] <Oq(H), contradicting Zq ¢ Oq(K). This completes theproof of (3).

(44.17) Let 7r c.7r(0), H E Q* (G, 7r), X, L E Q(G, ir), and CF(H) (X) < X <F(H) n L. Then

(1) [Op(X), Op(L)] = 1 for each p E.7r(F(H)).(2) rr(F(L) n H) 97r(F(H))

Solvable signalizer functors

(2) We have

Proof Assume the hypotheses of (1) and let Q(b) = O,(M(b)) and N(b) = 0(NG(Q(b))).By44.10.1, M(b)isaHalln-subgroupofN(b),soas P(b) = Pn M(b) E Syl,(M(b)) by the hypotheses of (I), P(b) E Syl,(N(b)). Thus Q(b) 5 P(b) ( P , so Z(P) 5 0(c~(Q(b))) 5 N(b). Hence Z(P) i P(b), so as CM(~) (F(M(b))) ( F(M(b)), (1) holds. Let

and N E Qx(G).By 44.13,X ( Opz((CN(b), fo for each b E so so by 31.20.3, X ( Opn (N), establishing (2).

(44.16) Let M E P ( G , n), a E n(F(M)), and X ( F(M) with n(X) = a and Z(O,(F(M))) ( X. Then for each N E Qx(G, n):

(1) [OP(X), O,(N)] = 1 for each p E a.

(2) Ifn(F(N)) E. a thena = n(F(N)) and [O,(X), OP(F(N))] = 1 foreach p E a.

(3) If n(F(N)) E a and la1 > 1 then X (F(N).

Proof. For p E a, let Xp = Op(X) and Z, = Z(O,(M)). By 44.10.2,

X, I Oqn(O(N~(Zp))) for each p , q E a,

SO X, ( Oq(N~(Zp)). Thus OP(X) ( O,J(NN(Z,)), SO by 31.15, OP(X) ( O,l(N) ( CN(Op(N)), establishing (1).

Suppose n(F(N)) E a. Then by (I), 1 # O,(X) 5 CN(OP(F(N)), so by 31.10, p E n(F(N)), establishing (2). Assume in addition that la1 > 1 but O,(X) $ F(N). Pick p # q E a and let H = CN(OP(F(N))) and K = HZ,. By (2), X i H, so as H 9 N, we have X, $ O,(H). But if Z, ( O,(K), then by 31.14, X, 5 Op(NH(Z,)) = O,(H), so Z, $ O,(K). However, F(H) = Op(N)Z(F(N)) and [Z,, Op(N)l = 1, so [Z,, HI i CH(O,(N>) = C,y(F(H))=Z(F(H)). As [Z,,OP(F(H))] = 1, we have Z, iO,(Z,Z (F(H))), so [Z,, HI 5 O,(H), contradicting Z, $ O,(K). This completes the proof of (3).

(44.17) Let n E n(0), H E Q*(G, n), X, L E Q(G, n), and CF(~) (X) ( X 5 F(H) n L. Then

(1) [OP(X), O,(L)] = 1 for each p E n(F(H)). (2) n(F(L) n H) E n(F(H)).

(3) If7r(F(L))Cn(F(H))then 7r(F(L))=n(F(H))andeither In(F(H))I =1 or X < F(L).

(4) Suppose 1 V E Q(Op(L) fl O,(H)) with NH(V) E Q*(NG(V), 7r)and L E Q*(G, 7r). Then 7r(F(H))=7r(F(L)), and either 7r(F(H)) = {p} orX < F(L), L E HI , and L = H if 7r = 7r(9).

Proof. Part (1) follows from 44.16.1, and (3) follows from 44.16.2 and 44.16.3.As

[Op(L) fl H, OP(X)] =1 by (1) and CF(H)(X) <X,

(2) is a consequence of the A x B Lemma and 31.10.Assume the hypotheses of (4) and let Y = NF(W (V ). By hypothesis, NH (V) E

Q*(NG(V),7r), so by 44.5, Y < H` for some i E NI(V). Thus (L, Y, H`)satisfies the hypotheses of (H, X, L). Also, by Exercise 1 1 . 1 , O p(X) < O p(B (NG (V ))), so X < F (NL(V )), and hence OP(X) < F(L) by 31.14.2. Therefore7r(F(H)) C 7r(F(L)), so applying (3) to (L, Y, H'), we conclude 7r(F(H)) =7r(F(L)) and either Y < F(H') or 7r(F(H)) = {p}. So to complete the proofof (4) we may assume I7r(F(H))l > 1 and Y < F(H`). By (3), X < F(L). AsOP(Y)=OP(F(L)) <OP(F(H1)), IOP(F(L))I < IOP(F(H))J. As (Z(OP(L)),L, H`, Y) satisfy the hypotheses of (V, H, L, X), by symmetry, JOP(F(H))I <JOP(F(L))I, so OP(F(L))=OP(Y)=O"(F(H`)). Then by 44.10.1 and 44.5,H' is conjugate to L in NI(OP(Y)), so L E HI. Finally, if 7r =7r(0) thenL = 0(NG(OP(Y))) = H' = H by 44.10.3.

(44.18) Let 7r C 7r(9), U E V(p, 7r), and H, L E Q* (G, 70 with NL(U) EQ*(NG(U), yr)

Then(1) If F(L)=OP(L) then F(H)=OP(H).(2) If q E Jr, V E V(q, Jr), NH (V) E Q*(NG(V), jr), and [U, v] < U fl V

then either(i) p = q and F(H) and F(L) are p-groups, or(ii) L E HI and if 7r=7r(O)then H=L.

Proof. As U E V(p, 7r), U <OP(L) fl O,(H). As NL(U) E Q*(NG(U), 7r),X = NF(H) (U) < L` for some i E NI(U), and we may assume i = 1. Thus wehave the hypotheses of 44.17. If F*(L) = OP(L) then 7r(F(L)) C 7r(F(H)),so (1) holds by 44.17.3. Assume the hypotheses of (2). Then V < Oq(H) be-cause V E V(p, 7r), so as [V, U] < U, we have V < X < L and then V < Oq(L)because V E V(q, 7r). So we have the hypotheses of 44.17.4, so that lemmacompletes the proof of (2).

For 7r C 7r(0) let 7-1(7r) consist of those H E Q(G, 7r) such that Q*(H, p) CQ*(G, p) for each p E 7r.


(3) Ifn(F(L)) 2 n(F(H))thenn(F(L)) =n(F(H))andeither In(F(H))I = 1 or X 5 F(L).

(4) Suppose 1 # V E &(Op(L) n Op(H)) with NH(V) E P(NG(V), n ) and L E Q*(G, n). Then n(F(H)) =n(F(L)), and either n(F(H)) = (p} or X_( F(L), L E HI, a n d L = H i fn=n(0 ) .

Proof. Part (1) follows from 44.16.1, and (3) follows from 44.16.2 and44.16.3. As

[Op(L) n H, OP(X)I = 1 by (1) and CF(H)(X) _( X,

(2) is a consequence of the A x B Lemma and 31.10. Assume the hypotheses of (4) andlet Y = NF(~)(V). By hypothesis, NH(V) E

P(NG(V), n), so by 44.5, Y _( H' for some i E NI(V). Thus (L, Y, H i ) satisfies the hypotheses of (H, X, L). Also, by Exercise 11.1, Op(X) _( OPn (O(NG(V))), SOX 5 F(NL(V)), and henceOP(X) _( F(L) by 31.14.2. Therefore n(F(H)) 2 n(F(L)), so applying (3) to (L, Y, H'), we conclude n(F(H)) = n(F(L)) and either Y 5 F(H') or n(F(H)) = (p}. So to complete the proof of (4) we may assume In(F(H))I > 1 and Y _( F(H'). By (3), X ( F(L). As

OP(Y) =OP(F(L)) i OP(F(Hi)), IOP(F(L))I 5 IOP(F(H))I. As (Z(Op(L)), L, H', Y) satisfy the hypotheses of (V, H, L, X), by symmetry, IOP(F(H))I 5 IOp(F(L))I, so Op(F(L)) = OP(Y) = OP(F(H')). Then by 44.10.1 and 44.5, H' is conjugate to L in NI(O"(Y)), SO L E HI. Finally, if n =n(0) then L = O(NG(OP(Y))) = H' = H by 44.10.3.

(44.18) Let n 2 n(0), U E V(p, n), and H, L E Pu(G, n ) with NL(U) E

P(NG(U), n). Then (1) If F(L) = Op(L) then F(H) = Op(H). (2) If q E n , V E V(q, n), NH(V) E P(NG(V), n), and [U, Vl 5 U n V

then either (i) p = q and F(H) and F(L) are p-groups, or

(ii) L E H' and if n = n(0) then H = L.

Proof. As U E V(p, n), U _( Op(L) n Op(H). As NL(U) E P(NG(U), n), X = NF(~) (U) ( L' for some i E NI(U), and we may assume i = 1. Thus we have the hypotheses of 44.17. If F*(L) = Op(L) then n (F(L)) 2 n(F(H)), so (1) holds by 44.17.3. Assume the hypotheses of (2). Then V ( Oq(H) because V E V(p, n), so as [V, U] _( U, we have V 5 X ( L and then V _( Oq(L) because V E V(q, n). So we have the hypotheses of 44.17.4, so that lemma completes the proof of (2).

For n 2 n(0) let X(n) consist of those H E &(G, n ) such that P ( H , p) 2 &*(G, p) for each p E n .

(44.19) If B is a hyperplane of A and H E M(b, 7r) for each b E B# thenH E 7-l (7r ).

Proof. Let p E 7r, Q E Sylp(H), and Q P E Q*(G, p). As H E M(b, 7r), wehave CH(b) E Q*(CG(b), 7r), so CQ(b)=CP(b). Thus P = Q by 44.8.1.

(44.20) Let 7r C 7r (0). Then either(1) R(7r) 0 0, or(2) There exists aprime p such that F*(M) = O p(M) for each ME M (a, 7r)

and each a E A#.

Proof. Let a E A#. By 44.14.1 there is U E U*(a, 7r), and by 44.14.2, U is ap-group for some prime p. By minimality of U and Exercise 8.9, CA(U) = Bis a hyperplane of A. Thus U < M(b) E M(b, 7r) for each b E B#. By 44.14.4,U E V (p, 7r), so U < O p(M(b)). By 44.12 there is U(b) E U*(b, 7r) with U(b) <F(M(b)), and by 44.14.3, [U, U(b)] =1.

Let H E Q*(G, 7r) with NH(U) E Q*(NG(U), 7r). If F(H) is a p-group thenby 44.18.1, so is F*(M) for each M E Q* (G, 7r). In particular F(M(b)) _Op(M(b)). Further, for C E A# and M E M(c, 7r), from the proof of 44.14there is U(c) E U(a, 7r) with U(c) < M. By symmetry there is a hyperplaneB(c) of A centralizing U(c), and as B n B(c) $ 1, we have U(c) < M(b)` forsome i E I. As F(M(b)) is a p-group, so is F(H(c)) for H(c) E Q*(G, 7r)with NH(c) (U (0) E Q*(NG(U(c)), 7r) by 44.18.2. Hence F*(M)=Op(M), asU(c) <M. Thus (2) holds in this case, so we may assume F*(H) is not a p-group. Hence by 44.18.2, M(b) E HI for each b E B#. Thus (1) holds by 44.19.

(44.21) There is a prime p such that for each a E A# and each M E M(a, 7r (9)),F(M) = O p(M).

Proof. If not, by 44.20 there is H E7-l(7r(9)). But then for p E7r(0) and P ESylp(H), P E Q*(G, p), so for a E A#, Cp(a) E Sylp(9(a)), and hence 9(a) _(Cp(a): p E 7E (0)) < H, contrary to 44.9.2.

In the remainder of this section pick p as in 44.21 and let 7r = 7r (9) -(p). If H E 7-1(7r), pick q E7r(F(H)), Q E Q*(G, q), and B a hyperplane of Awith z = Z(Q) n Oq(H) n CH(B) 0 1. As H E R(rr), H E M(b, ir) for eachbEB#. In this case let M(b)=Hfor bEB#.

If 7-l(7r) = 0 then by 44.20 there is a prime q such that F*(M) = Oq (M) foreach M E M(a, 7r) and each a E A#. Pick Q E Q*(G, q), B a hyperplane of Awith Z = Cz(Q)(B) 01, and for b E B#, M(b) E M(b, 7r) with Q n M(b) E Q*(M(b), q). By 44.15.1, Z <Oq(M(b)).


(44.19) If B is a hyperplane of A and H E M ( b , n ) for each b E B# then H E IFt(n).

Proof. Let p E n , Q E Sylp(H), and Q 5 P E P ( G , p). As H E M ( b , n), we have CH(b) E p ( C ~ ( b ) , n), SO CQ(b) = Cp(b). Thus P = Q by 44.8.1.

(44.20) Let n c n (0). Then either (1) IFt(n) # 0, or (2) There exists aprime p such that F*(M) = O p ( M ) for each M E M (a , n )

and each a E A'.

Proof. Let a E A'. By 44.14.1 there is U E U*(a, n), and by 44.14.2, U is a p-group for some prime p. By minimality of U and Exercise 8.9, CA(U) = B is a hyperplane of A. Thus U 5 M(b) E M ( b , n ) for each b E B#. By 44.14.4, U E V(p, n),so U 5 Op(M(b)).By 44.12thereis U(b) E U * ( ~ , n ) with U(b) 5 F(M(b)), and by 44.14.3, [U, U(b)] = 1.

Let H E P ( G , n ) with NH(U) E Q(NG(U), n). If F ( H ) is a p-group then by 44.18.1, so is F*(M) for each M E Q*,(G, n). In particular F(M(b)) = O,(M(b)). Further, for c E A' and M E M ( c , n), from the proof of 44.14 there is U(c) E U(a , n ) with U(c) 5 M. By symmetry there is a hyperplane B(c) of A centralizing U(c), and as B n B(c) # 1, we have U(c) 5 ~ ( b ) ' for some i E I. As F(M(b)) is a p-group, so is F(H(c)) for H(c) E P ( G , n ) with NHCc)(U(c)) E Q*(NG(U(C)), n ) by 44.18.2. Hence F*(M) = O,(M), as U(c) 5 M. Thus (2) holds in this case, so we may assume F*(H) is not a p- group. Hence by 44.18.2, M(b) E H' for each b E B'. Thus (1) holds by 44.19.

(44.21) There is a prime p such that for each a E A# and each M E M (a , n ( O ) ) , F (M) = O,(M).

Proof. If not, by 44.20 there is H E IFt(n(0)). But then for p E n ( 0 ) and P E

Syl,(H), P E P ( G , p), so for a E A#, Cp(a ) E Sylp(O(a)), and hence 0(a) = ( C p (a): p E n (0 ) ) 5 H , contrary to 44.9.2.

In the remainder of this section pick p as in 44.21 and let n = n ( 0 ) - (p) . If H E IFt(n), pick q E n ( F ( H ) ) , Q E P ( G , q) , and B a hyperplane of A with Z = Z ( Q ) n Oq(H) n C H ( B ) # 1. As H E IFt(n), H E M ( b , n ) for each b E B'. In this case let M(b) = H for b E B'.

If X ( n ) = 0 then by 44.20 there is a prime q such that F*(M) = O,(M) for each M E M ( a , n ) and each a E A'. Pick Q E Q*(G, q) , B a hyperplane of A with Z = C Z ( ~ ) ( B ) # 1, and forb E B', M(b) E M ( b , n ) with Q n M(b) E P (M(b), q) . By 44.15.1, Z 5 Oq(M(b)).

So in any case, Z < Oq (M(b)) for each b E B#. Hence by 44.15.2, Z E V(q, 7r).For a E A#, pick Ma E M(a, 7r(9)) and let P E Q*(G, p). As I < Ma and

I is transitive on Q*(G, p), P n Ma E Q*(Ma, p). Thus Z(P) <Op(MQ) by44.15.1 and 44.21. Therefore

Z(P) < Po = n Op(Mb).bEB"

Next, (Z, I) < 0(b) < Mb for each b E B#, so (Z, I) acts on P0. Let U = (ZI ),and )/V the set of W E Q(G) with I < NG (W), W = Op (W) U and Po < Op (W).

(44.22) Po E V(p, 7r(0)), and F(M) =Op(M) for each M E Q* (G).

Proof. First Po E V(p, n-(9)) by 44.15.2. Let L E Q*(G) with 0(NG(PO)) < L.As Po E V(P, 7r(9)), Po <Op(L). If 10Or(L) for some prime r p, then asB is noncyclic, 1 Co,{L}(b) for some b E B#. But X = CF{M6}(Po) < L, so by

44.17.2, r E 7r(F(Mb)), contrary to 44.21.Thus F(L) =Op(L), so the lemma follows from 44.18.1.

(44.23) Let M, L E Q* (G) and Op(M) < R E Q(M, p) with B(NG(C)) < Lfor each 1 C char R. Then M = L.

Proof. Let X = ROp,F(M). As Q =Op(M) < R, R E Sylp(X), and by 44.22,Q = F(M), so Q = F(X). Thus by Thompson Factorization, 32.6, we haveX = Nx(J(R))Cx(Z(R)), so X < L by hypothesis. As Q =Op(X), Q = Op(L n M). By 44.10.3, 0(NG(C)) < M for each 1 C char Q. In particularNoP(L)(Q) <Op(L n M) = Q, so Op(L) < Q by Exercise 3.2.1. But now(M, Q, L) satisfies the hypotheses of (L, R, M), so by symmetry, Q < Op(L).Therefore Q = O p (L), so M = L by 44.10.3.

(44.24) LetWEW,S=Op(W),Y=OP(W),LEQ*(G)with 0(NG (Y))<L,and M E Q*W(G). Then

(1) Y <Op,q(M)(2) If Op(M) < S then M = L.

Proof. Recall Z E V(q, 7r), so Z < Oqn (M) = O{ p,q} (M) by definition ofV(q, n). By construction Z < Z(Q) for some Q E Q*(G, q), and as I < M,Q n M E Sylq (M). Thus as Z < O{p,q} (M), Z O p,q (M) by 31.10. HenceU = (ZI) <Op,q(M). But W = UOp(W), so y =O'(W) _ (Uw) <Op,q(M),establishing (1).

Suppose R =Op(M) < S and let M* = M/R and X* = F(M*). By (1),Y* <Oq(X*) and as R < S < NG(Y), Y =OP(YR) a Nx(Y*), so Nx(Y*) _


Soin any case, Z ( O,(M(b))foreach b E B'. Hence by 44.15.2, Z E V(q , n). For a E A#, pick Ma E M ( a , n (0 ) ) and let P E Q*(G, p). As I ( Ma and

I is transitive on Q*(G, p), P n Ma E Q*(Ma, p). Thus Z ( P ) 5 Op(Ma) by 44.15.1 and 44.2 1. Therefore

z ( p ) 5 PO = op(Mb). b€BU

Next, ( 2 , I ) ( 8(b) 5 Mb for each b E B#, so ( 2 , I ) acts on PO. Let U = ( z ' } , and Wthe set of W E Q(G) with I 5 Nc(W), W = Op(W)U and Po 5 Op(W).

(44.22) Po E V ( p , n(8) ) , and F(M) = Op(M) for each M E Pp0 (G).

Proof. First Po E V ( p , n ( 8 ) ) by 44.15.2. Let L E Q*(G) with 6(NG(Po)) ( L. As Po E V ( P , n(8)) , Po 5 Op(L). If 1 # O,(L) for some prime r # p, then as B is noncyclic, 1 # C O ~ ( ~ ) ( ~ ) for some b E B'. But X = CF(Mb)(PO) 5 L, SO by 44.17.2, r E n(F(Mb)) , contrary to 44.21.

Thus F(L) = Op(L) , so the lemma follows from 44.18.1.

(44.23) Let M, L E Q>o(G) and Op(M) ( R E Q(M, p) with ~ ( N G ( C ) ) ( L for each 1 # C char R. Then M = L.

Proof. Let X = ROp,F(M). As Q = Op(M) ( R, R E Sylp(X), and by 44.22, Q = F(M), so Q = F(X) . Thus by Thompson Factorization, 32.6, we have X = Nx(J(R))Cx(Z(R)) , so X 5 L by hypothesis. As Q = Op(X) , Q = 0, ( L n M). By 44.10.3, O(NG(C)) 5 M for each 1 # C char Q. In particular No,(~) (Q) 5 Op(L n M ) = Q, so Op(L) 5 Q by Exercise 3.2.1. But now ( M , Q, L ) satisfies the hypotheses of ( L , R, M), so by symmetry, Q 5 Op(L). Therefore Q = Op(L) , so M = L by 44.10.3.

(44.24) Let W E W , S =Op(W), Y = OP(W), L E Q*(G) w i thB(N~(Y) ) 5 L , and M E QT w(G). Then

( 1 ) y 5 Op,q(M). (2) If Op(M) 5 S then M = L.

Proof. Recall Z E V(q , n), so Z ( 0," ( M ) = O1p,ql(M) by definition of V(q , n). By construction Z 5 Z(Q) for some Q E Q*(G, q ) , and as I 5 M , Q n M E Syl,(M). Thus as Z ( OIp,,)(M), Z 5 OP,,(M) by 31.10. Hence U = (2') 5 Op,,(M). But W = UOp(W), so Y = OP(W) = (uW) ( Op,,(M), establishing (1) .

Suppose R = O,(M) 5 S and let M* = M / R and X* = F(M*). By ( I ) , Y* 5 O,(X*) and as R ( S ( NG(Y), Y = OP(YR) 9 Nx(Y*), SO Nx(Y*) =

Nx(Y) < L. Let T = O p(L). Then NT(R) < M by 44.10.3, so

[NT(R)*, Nx,(Y*)] < NT(R)* n Nx,(Y*) = 1.

Therefore by the A x B Lemma, NT(R)* < CM.(X*) < X*, so NT(R) < R andhence T < R. Finally by 44.10.3, 0(NG (C)) < M for each 1 C char R, soM=L by 44.23.

If Mb = Mb' for all b, b' E B#, then the argument establishing 44.21 suppliesa contradiction. Therefore as Mb E Q* UI(G) for each b E B#, IQP0U,(G)l > 1.Pick W E W with S=Op(W) maximal subject to IQ*1(G)I > 1. Set Y =OP(W) and pick M E Q*(G) with 0(NG (S)) <M.

(44.25) For each N E Q* I(G) - {M}, S is the unique maximal WI-invariantmember of Q(N, p).

Proof. Suppose S < So E Q(N, p) with WI < NG (SO). Then S < Nso(S), sowithout loss, S < So. Therefore So < 0(NG(S)) < M. Let WO = WSo. ThenWO E W and M, N E Qw0I(G), contradicting the maximal choice of S.

(44.26) (1) Q*(NG(y)) (G) contains a unique member L.

(2) Qe(NG(s))(G) = {M}.(3) Q* (G) =I M, L}.

Proof. Let L E QB(NG(y))(G) and N E Q** (G) - {M}. By 44.25, 02(N) < S,so by 44.24.2, N = L. Thus (3) holds and as (Q*i (G)) > 1, M L, so (1) and(2) hold.

We are now in a position to obtain a contradiction and hence establish theSolvable 2-Signalizer Functor Theorem. By 44.26.2, B(NG(C)) < M for each1 # C char S, while by 44.25, Op (L) ::S S. Hence M = L by 44.23, contradicting44.26.3 and the choice of W with IQ* (G) I > I.

This completes the proof of the Solvable 2-Signalizer Functor Theorem.

Remarks. Gorenstein introduced the concept of the signalizer functor and ini-tiated the study of these objects [Gor 1]. He was motivated by earlier workof Thompson. Goldschmidt [Gol 1, Gol 2] simplified some of Gorenstein'sdefinitions and proved the Solvable 2-Signalizer Functor Theorem and theSolvable Signalizer Functor Theorem for r odd when Mr (A) > 3. Glauberman[Gl 1] established the general theorem for solvable functors. Bender gave anew short proof of the Solvable 2-Signalizer Functor Theorem [Be 2]; theproof given here is based on his proof, although our proof is longer and more


Nx(Y) 5 L. Let T =Op(L). Then NT(R) 5 M by 44.10.3, so

[NT (R)*, Nx* (Y*)] 5 NT(R)* n Nx* (Y*) = 1.

Therefore by the A x B Lemma, NT(R)* 5 CM*(X*) 5 X*, SO NT(R) 5 R and hence T 5 R. Finally by 44.10.3, B(Nc(C)) I M for each 1 # Cchar R, so M = L by 44.23.

If Mb = Mb, for all b, b' E B', then the argument establishing 44.21 supplies a contradiction. Therefore as Mb E Q*&,y1(G) for each b E B#, ] Q>ouI (G)I > 1. Pick W E W with S = O,(W) maximal subject to IQ*,,(G)l> 1. Set Y = OP(W) and pick M E Q*(G) with B(NG(S)) 5 M .

(44.25) For each N E Q&,(G) - {MJ, S is the unique maximal WZ-invariant member of Q(N, p).

Proof. Suppose S < So E Q(N, p) with WZi NG(So). Then S < Nso(S), so without loss, S 9 So. Therefore So 5 B(NG(S)) 5 M. Let Wo = WSo. Then Wo E W and M , N E PWOI (G), contradicting the maximal choice of S.

(44.26) (1) QZ;(NG(Y))(G) contains a unique member L.

(2) QZ;(NG(S))(G) = (3) PwI(G)= { M , LJ.

Proof. Let L E Q&NG(Y))(G) and N E PwI(G) - {MI . By 44.25, 02(N) 5 S, ! so by 44.24.2, N = L. Thus (3) holds and as I Pw1 (G)I > 1, M # L, so (1) and

(2) hold.

We are now in a position to obtain a contradiction and hence establish the Solvable 2-Signalizer Functor Theorem. By 44.26.2, B(NG(C)) 5 M for each 1 # C char S, while by 44.25, OP(L) 5 S. Hence M = L by 44.23, contradicting 44.26.3 and the choice of W with I Pw1(G)I > 1.

This completes the proof of the Solvable 2-Signalizer Functor Theorem.

Remarks. Gorenstein introduced the concept of the signalizer functor and ini- tiated the study of these objects [Gor 11. He was motivated by earlier work of ?hompson. Goldschmidt [Gol 1, Go1 21 simplified some of Gorenstein's definitions and proved the Solvable 2-Signalizer Functor Theorem and the Solvable Signalizer Functor Theorem for r odd when m,(A) > 3. Glauberman [Gl 11 established the general theorem for solvable functors. Bender gave a new short proof of the Solvable 2-Signalizer Functor Theorem [Be 21; the proof given here is based on his proof, although our proof is longer and more

complicated than Bender's for reasons I'll explain in a moment. Finally,McBride [Mc] proved the Signalizer Functor Theorem subject to the hypothesisthat the composition factors of 9(a) are known simple groups. The Classifica-tion Theorem allows this hypothesis to be removed.

Our proof is longer than Bender's because Bender uses the ZJ-Theorem,whereas we use Thompson Factorization, since the ZJ-Theorem is not provedin this text. There are at least two advantages to this approach: First, the increasein the length of the proof caused by not appealing to the ZJ-Theorem is probablyless than the length of the proof of the ZJ-Theorem, although of course the ZJ-Theorem is of interest in its own right. Second, the change gives some insightinto how Thompson Factorization is used in the literature, and ThompsonFactorization is used often, while the ZJ-Theorem is not.

In the first edition of the book, we gave a proof of the Solvable SignalizerFunctor Theorem for all primes which was similar to the proof given here butstill more complicated. In this edition we have opted for less generality in thehope that a simpler proof will better expose the underlying concepts.

Exercises for chapter 151. Let G = A7, A = ((1, 2)(3, 4), (3, 4)(5, 6)), and 9(a) = 03(CG(a)) for a c

A#. Prove 9 is a solvable A-signalizer functor on G which is not complete.2. Prove lemma 44.7. (Hint: Assume the lemma is false and choose P, Q E

Q*(G, p) not conjugate under 9(CG(A)) with P fl Q maximal subject tothese constraints. Prove P fl Q 1 by observing Cp(a) 1 CQ(a) forsome a E A#, and considering 9(a). Then consider NG(P n Q)/(P fl Q)and use 44.3.)

3. Let A be a noncyclic abelian r-subgroup of a finite group G and 9 an A-signalizer functor on G. For B < A let WB = (9 (b): b E B#) and let W = WA.Prove(1) If B is a noncyclic subgroup of A then W = WB.(2) If 6(a8) = 6(a)9 for each g E G and a E A#, then F2,A(G) < NG(W),

where I'2,A(G) = (NG (B): B < A, m(B) > 2).(3) Assume the hypothesis of (2) and let VIG(A, r') denote the set of A-

invariant r'-subgroups of G. Prove VIG(A, r') c NG(W). If 9 is com-plete, G = F2,A(G) and Or'(G) = 1, prove 9(a) =1 for all a c- A#.

4. Let A be an elementary abelian r-subgroup of a finite group G with m (A) > 3.Prove(1) If a E A# and CG (a) is solvable then CG (a) is balanced for the primer.

(See section 31 for the definition of balance.)(2) If CG(a) is balanced for the prime r for each a c A# then 9 is an A-

signalizer functor on G, where 9(a) =Or,(CG (a)).


complicated than Bender's for reasons I'll explain in a moment. Finally, McBride [Mc] proved the Signalizer Functor Theorem subject to the hypothesis that the composition factors of 0(a) are known simple groups. The Classifica- tion Theorem allows this hypothesis to be removed.

Our proof is longer than Bender's because Bender uses the ZJ-Theorem, whereas we use Thompson Factorization, since the ZJ-Theorem is not proved in this text. There are at least two advantages to this approach: First, the increase in the length of the proof caused by not appealing to the ZJ-Theorem is probably less than the length of the proof of the ZJ-Theorem, although of course the ZJ- Theorem is of interest in its own right. Second, the change gives some insight into how Thompson Factorization is used in the literature, and Thompson Factorization is used often, while the ZJ-Theorem is not.

In the first edition of the book, we gave a proof of the Solvable Signalizer Functor Theorem for all primes which was similar to the proof given here but still more complicated. In this edition we have opted for less generality in the hope that a simpler proof will better expose the underlying concepts.

Exercises for chapter 15 1. Let G = A7, A = ((1,2)(3,4), (3,4)(5,6)), and 0(a) = Os(CG(a)) for a E

A'. Prove 0 is a solvable A-signalizer functor on G which is not complete. 2. Prove lemma 44.7. (Hint: Assume the lemma is false and choose P , Q E

Q(G, p) not conjugate under O(CG(A)) with P n Q maximal subject to these constraints. Prove P n Q # 1 by observing Cp(a) # 1 # CQ(a) for some a E A', and considering 0(a). Then consider NG(P fl Q)/(P fl Q) and use 44.3.)

3. Let A be a noncyclic abelian r-subgroup of a finite group G and 0 an A- signalizer functor on G. For B 5 A let WE = (0(b): b E B') and let W = WA. Prove (1) If B is a noncyclic subgroup of A then W = WE. (2) If 0(as) = 0(a)g for each g E G and a E A', then r 2 , ~ ( G ) 5 NG(W),

where r 2 , ~ ( G ) = (NG(B): B 5 A, m(B) 2 2). (3) Assume the hypothesis of (2) and let MG(A, r') denote the set of A-

invariant rt-subgroups of G. Prove MG(A , rt) 5 NG( W). If 0 is complete, G = r2,A(G) and O,t(G) = 1, prove 8(a) = 1 for all a E A'.

4. Let A be an elementary abelian r -subgroup of a finite group G with m(A) >_ 3. Prove (1) If a E A' and CG(a) is solvable then CG(a) is balanced for the prime r.

(See section 3 1 for the definition of balance.) (2) If CG(a) is balanced for the prime r for each a E A# then 0 is an A-

signalizer functor on G, where @(a) = O,!(CG(a)).

(3) Assume CG(a) is solvable for each a E A# and G = F2,A(G).Prove Or'(CG(a)) < Or'(G) for each a c- A#, and if Or'(G) =1 thenF*(CG(a)) = Or(CG(a)).

5. Let r be a prime, G a finite group, Era = A < G, and 7r a set of primes withrV7r.ForEr2=B <Gdefine

(nO,r(CG(b))bEB#

Assume ac0(,)(B) < O,(CG(c)) for each c c- A# and each hyperplane B ofA. Also either assume the Signalizer Functor Theorem or assume On (CG(c))is solvable for each c e A#. Prove a(B) is a 7r-group for each hyperplane Bof A and a(B) is independent of the choice of B. (Hint: Define

yB(a) = [On(CG(a)), B](O,,(CG(B)) fl C(a))

for EP2 = B < A and a E A#. Prove yB is an A-signalizer functor anda(B) < a(D) for each pair B, D of hyperplanes of A.)

Solvable signalizer functors 24 1

(3) Assume CG(a) is solvable for each a EA' and G = I '~ ,A(G). Prove Or1(CG(a)) 5 Or!(G) for each a E A#, and if O,!(G) = 1 then F*(Cc(a)) = Or(Cc(a)).

5. Let r be a prime, G a finite group, Ers E A 5 G , and n a set of primes with r e n . For E,z Z B 5 G define

Assume U ~ , ( ~ ) ( B ) I: O,(CG(c)) for each c E A# and each hyperplane B of A. Also either assume the Signalizer Functor Theorem or assume 0, (CG(C)) is solvable for each c E A'. Prove a ( B ) is a n-group for each hyperplane B of A and a ( B ) is independent of the choice of B. (Hint: Define

for EP2 =N B 5 A and a E A'. Prove y~ is an A-signalizer functor and a ( B ) 5 a ( D ) for each pair B , D of hyperplanes of A.)

16

Finite simple groups

To my mind the theorem classifying the finite simple groups is the most im-portant result in finite group theory. As I indicated in the preface, the Classifi-cation Theorem is the foundation for a powerful theory of finite groups whichproceeds by reducing suitable group theoretical questions to questions aboutrepresentations of simple groups. The final chapter of this book is devotedprimarily to a discussion of the Classification Theorem and the finite simplegroups themselves.

Sections 45 and 46 introduce two classes of techniques useful in the studyof simple groups. Section 45 investigates consequences of the fact that eachpair of involutions generates a dihedral group. The two principal results ofthe section are the Thompson Order Formula and the Brauer-Fowler The-orem. The Thompson Order Formula supplies a formula for the order of afinite group with at least two conjugacy classes of involutions in terms ofthe fusion of those involutions in the centralizers of involutions. The Brauer-Fowler Theorem shows that there are at most a finite number of finite simplegroups possessing an involution whose centralizer is isomorphic to any givengroup.

Section 46 considers the commuting graph on the set of elementary abelianp-subgroups of p-rank at least k in a group G. The determination of the groupsfor which this graph is disconnected for small k plays a crucial role in theClassification Theorem.

Section 47 contains a brief description of the finite simple groups. The groupsof Lie type are described as groups with a split BN-pair and generated by rootgroups satisfying a weak version of the Chevalley commutator relations. Thelast portion of section 47 explores consequences of these axioms. In particularthe existence of Levi complements is derived, it is shown that the maximalparabolics are the maximal overgroups of Sylowp-groups of groups in charac-teristicp, and the Borel-Tits Theorem is (essentially) proved for finite groups ofLie type. This last result says that in a finite group of Lie type and characteristicp, each p-local is contained in a maximal parabolic.

Finally section 48 consists of a sketchy outline of the Classification Theorem.This discussion provides a nice illustration of many of the techniques developedin earlier chapters.


To my mind the theorem classifying the finite simple groups is the most important result in finite group theory. As I indicated in the preface, the Classifi- cation Theorem is the foundation for a powerful theory of finite groups which proceeds by reducing suitable group theoretical questions to questions about representations of simple groups. The final chapter of this book is devoted primarily to a discussion of the Classification Theorem and the finite simple groups themselves.

Sections 45 and 46 introduce two classes of techniques useful in the study of simple groups. Section 45 investigates consequences of the fact that each pair of involutions generates a dihedral group. The two principal results of the section are the Thompson Order Formula and the Brauer-Fowler The- orem. The Thompson Order Formula supplies a formula for the order of a finite group with at least two conjugacy classes of involutions in terms of the fusion of those involutions in the centralizers of involutions. The Brauer- Fowler Theorem shows that there are at most a finite number of finite simple groups possessing an involution whose centralizer is isomorphic to any given group.

Section 46 considers the commuting graph on the set of elementary abelian p-subgroups of p-rank at least k in a group G. The determination of the groups for which this graph is disconnected for small k plays a crucial role in the Classification Theorem.

Section 47 contains a brief description of the finite simple groups. The groups of Lie type are described as groups with a split BN-pair and generated by root groups satisfying a weak version of the Chevalley commutator relations. The last portion of section 47 explores consequences of these axioms. In particular the existence of Levi complements is derived, it is shown that the maximal parabolics are the maximal overgroups of Sylow p-groups of groups in characteristic~, and the Borel-Tits Theoremis (essentially) proved for finite groups of Lie type. This last result says that in a finite group of Lie type and characteristic p, each p-local is contained in a maximal parabolic.

Finally section 48 consists of a sketchy outline of the Classification Theorem. This discussion provides a nice illustration of many of the techniques developed in earlier chapters.

Involutions in finite groups 243

45 Involutions in finite groupsSection 45 seeks to exploit the following property of involutions established inExercise 10.1:

(45.1) Let x and y be distinct involutions of a group G. Then (x, y) is a dihedralgroup of order 21xyl.

Throughout section 45, G will be assumed to be a finite group. To begin let'slook more closely at 45.1:

(45.2) Let x and y be distinct involutions in G, n = IxyI, and D = (x, y). Then(1) Each element in D - (xy) is an involution.(2) If n is odd then D is transitive on its involutions, so in particular x is

conjugate to y in D.(3) If n is even then each involution in D is conjugate to exactly one of x, y,

or z, where z is the unique involution in (xy). Further z E Z(D).(4) If n is even and z is the involution in (xy) then xz is conjugate to x in D

if and only if n - 0 mod 4.

Proof. Let u = xy and U = (u). Then uX = u-1 so VX = v-1 for each v E U.Then (vx)2 = V VX = vv-1 =1, so (1) holds. Further for w E U, (vx)'= vxw _vw-1wXx = vw-2x,soxD = {v2x:v E U}. In particular if n is odd then XDD - U and U contains no involutions, so (2) holds.

So take n even and let z be the involution in U. Then

D - U {V2x: V E U} U {V2ux: V E U}

with ux = y, so D - U = xD U yD and (3) holds. Finally zx E XD preciselywhen z is a square in U; that is when n = 0 mod 4.

(45.3) Let G be of even order with Z(G) = 1, let m be the number of involutionsin G, and n = Gm. Then G possesses a proper subgroup of index at most2n2.

Proof. Let I be the set of involutions of G, R the set of elements of G invertedby a member of I, and (xi : 0 < i < k) a set of representatives for the conjugacyclasses of Gin R. Pick xo = 1 and let mi = IxG I, let Bi be the set of pairs (u, v)with u, v E I and uv = xi, and let bi = I Bi I.

Observe first that if u, v E I then either u = v and uv = 1 or (u, v) is di-hedral. In either case u inverts uv, so uv E R. Thus counting I x I in two

Involutions inJinite groups

45 Involutions in finite groups Section 45 seeks to exploit the following property of involutions established in Exercise 10.1 :

(45.1) Let x and y be distinct involutions of a group G. Then (x, y) is a dihedral group of order 21xy 1 .

Throughout section 45, G will be assumed to be a finite group. To begin let's look more closely at 45.1:

(45.2) Let x and y be distinct involutions in G, n = Ixyl, and D = (x, y). Then (1) Each element in D - (xy) is an involution. (2) If n is odd then D is transitive on its involutions, so in particular x is

conjugate to y in D. (3) If n is even then each involution in D is conjugate to exactly one of x, y,

or z , where z is the unique involution in (xy). Further z E Z(D). (4) If n is even and z is the involution in (xy) then xz is conjugate to x in D

if and only if n = 0 mod 4.

Proof. Let u = xy and U = (u). Then uX = u-' so vX = v-' for each v E U . Then ( v x ) ~ = vvX = vv-' = 1, so (1) holds. Further for w E U , ( V X ) ~ = vx =

VW-'wXx = V W - ~ X , soxD = (v2x: u E U ] . Inparticularifn is oddthenxD = D - U and U contains no involutions, so (2) holds.

So take n even and let z be the involution in U . Then

with ux = y, so D - U = xD U y and (3) holds. Finally zx E x precisely when z is a square in U ; that is when n 0 mod 4.

(45.3) Let G be of even order with Z(G) = 1, let m be the number of involutions in G, and n = IGllm. Then G possesses a proper subgroup of index at most 2n2.

Proof. Let I be the set of involutions of G, R the set of elements of G inverted by a member of I, and (xi : 0 5 i 5 k) a set of representatives for the conjugacy classes of G in R. Pick xo = 1 and let mi = I, let Bi be the set of pairs (u, v) with u, v E I and uv =xi , andlet bi = IBil.

Observe first that if u, v E I then either u = v and uv = 1 or (u, v) is dihedral. In either case u inverts uv, so uv E R. Thus counting I x I in two

244 Finite simple groups

different ways, we obtain:

(a)

k

m2=II x71 =k mibi.

i-0

Moreover there is an involution ti inverting xi and if (u, v) E Bi then u invertsxi and v = uxi. Hence the map (u, v) -r u is an inj ection of Bi into ti CG (Xi) sobi < I tiCG(xi)) = I CG(xi)I. Also mi = IG : CG(xi)I, so mibi < G. If i = 0we can be more precise: mo = 1 and bo = m. Combining these remarks with(a) gives:

(b) m2 <m+kIGl.

Let s be the minimal index of a proper subgroup of G. If i > 0 thenby hypothesis xi V Z(G), so mi = I G : CG(xi)I > s. Hence IGI > Ek

o ml >1 + ks, which I record as:

(c) k < (IGI - 1)/s.

Combining (b) and (c) gives:

(d) m2 < (IGI(IGI -1)/s)+m.

Then, as n = IGI /m, it follows from (d) that s < n(n - m-1)/(1 - m-'), and,as m > 2, s < 2n2.

(45.4) Let G be a finite simple group of even order, t an involution in G, andn = I CG(t)I. Then IGI < (2n2)!.

Proof. By 45.3, G possesses a proper subgroup H 9f index at most 2no,where no = I G I /m and in is the number of involutions in G. Then m > I tG I =(G : CG(t)1, so no < IGI/IG : CG(t)1 = n.

Represent G as a permutation group on the coset space G/H. Then k =I G/HI < 2n2, and as G is simple the representation is faithful. So G is isomor-phic to a subgroup of the symmetric group Sk, and hence IGI < k! < (2n2)!.

As an immediate corollary to 45.4 we have the following theorem of Brauerand Fowler:

(45.5) (Brauer-Fowler [BF]) Let H be a finite group. Then there exists atmost a finite number of finite simple groups G with an involution t such thatCG(t) = H.

Recall that the Odd Order Theorem of Feit and Thompson says that nonabelianfinite simple groups are of even order and hence possess involutions. TheBrauer-Fowler Theorem suggests it is possible to classify finite simple groups


different ways, we obtain:

Moreover there is an involution ti inverting xi and if (u, v) E Bi then u inverts xi and v = uxi. Hence the map (u, v) H u is an injection of Bi into ti CG(X~) SO

bi i ItiCG(xi)l = ICG(xi)l. Also mi = IG: Cc(xi)l, so mibi i IGI. If i = 0 we can be more precise: mo = 1 and bo = m. Combining these remarks with (a) gives:

(b) m2 5 m + klGI.

Let s be the minimal index of a proper subgroup of G. If i > 0 then by hypothesis xi 6 Z(G), so mi = IG : CG(xi)l L S . Hence (GI 1 ~ f = ~ mi 3 1 + ks, which I record as:

Combining (b) and (c) gives:

( 4 m2 5 (lGI(IGI - l)/s) + m.

Then, as n = IGllm, it follows from (d) that s 5 n(n - mP1)/(l - m-I), and, as m 2 2, s 5 2n2.

(45.4) Let G be a finite simple group of even order, t an involution in G, and n = lCG(t)l. Then IG1 ( (2n2)!.

Proof. By 45.3, G possesses a proper subgroup H,gf index at most 2 4 , where no = I G l/rn and rn is the number of involutions in G. Then m q ItG 1 =

IG : CG(t)l, SO no i IGJIJG: CG(t)J = n. Represent G as a permutation group on the coset space G/H. Then k =

IG/HI 5 2n2, and as G is simple the representation is faithful. So G is isomorphic to a subgroup of the symmetric group Sk, and hence JGI ( k! 5 (2n2)!.

As an immediate corollary to 45.4 we have the following theorem of Brauer and Fowler:

(45.5) (Brauer-Fowler [BF]) Let H be a finite group. Then there exists at most a finite number of finite simple groups G with an involution t such that CG(t) H.

Recall that the Odd Order Theorem of Feit and Thompson says that nonabelian finite simple groups are of even order and hence possess involutions. The Brauer-Fowler Theorem suggests it is possible to classify finite simple groups

Connected groups 245

in terms of the centralizers of involutions. This approach will be discussed inmore detail in section 48.

If G has more than one class of involutions it is possible to make a muchmore precise statement than 45.5:

(45.6) (Thompson Order Formula) Assume G has k > 2 conjugacy classes ofinvolutions x 1 < i < k, and define ni to be the number of ordered pairs(u,v)with u Ex1 ,v Ex2,andxi c (uv). Then

k

IGI = ICG(xi)IICG(x2)I ni/ICG(xi)Ii=1

Proof. Let I be the set of involutions in G and Q = xG x x2 . Then

(*) ICI = J X G J J X G J1 2 = IG : CG(xi)I IG : CG(x2)I

For (u, v) E Q, u vG, so by 45.2 there is a unique involution z(u, v) in(uv). Hence Q is partitioned by the subsets QZ consisting of those pairs (u, v)with z = z(u, v). Thus IQI = >-+ZEI IQz1. Further IQ,I = IQx, I for z c xP, so

Y IQzl = IxGIIQxiI = IG:CG(xi)IniZExG

Therefore

k

(**)101

= IG:CG(xi)Ini.i=1

Of course the lemma follows from (*) and (**).

Notice that the integer ni can be calculated if x g fl CG (xi ), j = 1, 2, is knownfor each i. Hence the order of G is determined by the fusion of xl and x2 inCG(xi), 1 < i < k. In particular the order of G can be determined from thefusion of involutions in local subgroups.

46 Connected groupsIn this section G is a finite group and p is a prime.

If Q is a collection of subgroups of G permuted by G via conjugation, define5(Q) to be the graph on Q obtained by joining A to B if [A, B] = 1. Evidently

G is represented as a group of automorphisms of 5(Q) via conjugation.

(46.1) Let A be a G-invariant collection of subgroups of G and H < G. Thenthe following are equivalent:

(1) H controls fusion in H fl A and NG(X) < H for each X E H fl A.(2) H fl Hg fl A is empty for g E G - H.


in terms of the centralizers of involutions. This approach will be discussed in more detail in section 48.

If G has more than one class of involutions it is possible to make a much more precise statement than 45.5:

(45.6) (Thompson Order Formula) Assume G has k 2 2 conjugacy classes of involutions xf , 1 5 i 5 k, and define ni to be the number of ordered pairs (u, v) with u E x f , u E xf , and xi E (uv). Then

ProoJ Let I be the set of involutions in G and Q = xf x x:. Then

For (u, v) E a, u $ vG, so by 45.2 there is a unique involution z(u, v) in (uv). Hence Q is partitioned by the subsets Q, consisting of those pairs (u, v) with z = z(u, v). Thus 1Q1 = C,,, lQzl. Further 1Q,1 = 1Q,1 for z E xc, so

Therefore k

' Of course the lemma follows from (*) and (**).

Notice that the integer ni can be calculated if xy n CG(xi), j = 1,2, is known for each i. Hence the order of G is determined by the fusion of xl and x2 in CG(xi), 1 5 i 5 k. In particular the order of G can be determined from the fusion of involutions in local subgroups.

46 Connected groups In this section G is a finite group and p is a prime.

If Q is a collection of subgroups of G permuted by G via conjugation, define g (Q) to be the graph on Q obtained by joining A to B if [A, B] = 1. Evidently G is represented as a group of automorphisms of g (Q) via conjugation.

(46.1) Let A be a G-invariant collection of subgroups of G and H 5 G. Then the following are equivalent:

(1) H controls fusion in H n A and NG(X) _( H for each X E H n A. (2 ) H r l Hg n A is empty for g E G - H.

(3) The members of H fl A fix a unique point in the permutation representationof G on G/H by right multiplication.

(4) H fl A is the union of a set F of connected components of 2(0), andrfl Fg is emptyforg c G-H.

Proof. Assume (1) and let g c G with H fl Hg fl A nonempty. Then there isX E H fl A with Xg < H so, as H controls fusion in H fl A, Xgh = X forsome h E H. Then gh E NG(X) < H, so g E H. Thus (1) implies (2).

Assume (2) and consider the representation of G on G/H. Then X E H fl Afixes Hg if and only if X < Hg, which holds precisely when g c H by (2).Thus (2) implies (3).

Assume (3). Then, for A E H fl A, {H} = Fix(A), so NG(A) < H, as H isthe stabilizer in G of the coset H. In particular if B E A is incident to A in1(0) then B < C(A) < H, so B E H fl A. Thus H fl A is the union of someset F of connected components of 2(0). Further if A E 8 E F and 8g E F thenAg < H, so {H} = Fix(Ag) = {Hg} and hence g E H. So (3) implies (4).

Finally assume (4). If X E H fl A and g E G with X g< H, then X E 9 E I'andXg E 8' E I',so8' = 8g E I'nFg.Hence, by(4),g E H.So(4)implies(1).

If k is a positive integer write Sk (G) for the set of all elementary abelian p-subgroups of G of p-rank at least k. G is said to be k-connected for the primep if 0(61k(G)) is connected.

(46.2) Let G be a p-group. Then(1) G is 1-connected for the prime p.(2) If m(G) > 2 then there is Ep2 = U< G and X E S' (G) is in the same

connected component of 0(S' (G)) as U precisely when m(CG(X)) > 2.(3) If p = 2 and G has a normal E8-subgroup then G is 2-connected for

the prime 2.(4) If p = 3 and m(G) > 3, then G is 2-connected for the prime 3.

Proof. Part (1) follows from the fact that Z(G) ,- 1. Assume m(G) > 2;by Exercise 8.4 there is Ep2 = U< G. G/CG(U) SL2(p) and SL2(p) is ofp-rank 1. So if A E 63(G) then m(CA(U)) > 2, so m(CA(U)U) > 2. Hencem(CG(U)) > 2. Also, if U ,- E E S' (G) is in the same connected componentas U, then there is E D E °z (G) adjacent to E and m(CG(E)) > m(DE) >3. Thus to prove (2) it remains to show that each A E 613 (G) is in the connectedcomponent of U. But A, CA(U), CA(U)U, U is a path in _T(6'2P(G)), so theproof of (2) is complete.

Assume p = 2 and E8 - Va G. Let E E 6122(G); to prove (3) we must ex-hibit a path from E to V. Fore c E, m (CV (e)) > 2 by Exercise 9.8, and we mayassume there is e c E - V, so E, (e, CV (E)), Cv (e), V is a path in 2(d0z (G)).


(3) The members of H i l A fix a unique point in the permutation representation of G on GIH by right multiplication.

(4) H i l A is the union of a set r of connected components of LB(A), and r n rgisemptyforg E G - H.

Proof. Assume (1) and let g E G with H i l Hg i l A nonempty. Then there is X E H n A with Xg 5 H so, as H controls fusion in H i l A, xgh = X for some h E H. Then gh E NG(X) 5 H, so g E H. Thus (1) implies (2).

Assume (2) and consider the representation of G on GI H. Then X E H i l A fixes Hg if and only if X 5 Hg, which holds precisely when g E H by (2). Thus (2) implies (3).

Assume (3). Then, for A E H i l A, {H) = Fix(A), so NG(A) 5 H, as H is the stabilizer in G of the coset H . In particular if B E A is incident to A in LB(A) then B 5 C(A) 5 H, so B E H i l A. Thus H i l A is the union of some set r of connected components of g(A). Further if A E 6' E r and 6'g E r then Ag ( H, so {H) = Fix(A8) = {Hg) and hence g E H. So (3) implies (4).

Finally assume (4). If X E H i l A and g E G with Xg 5 H, then X E 6' E r andXg E 6" E r , SOB' = 6'8 E r n r g . Hence, by (4), g E H. So (4)implies (1).

If k is a positive integer write &l(G) for the set of all elementary abelian p- subgroups of G of p-rank at least k. G is said to be k-connected for the prime p if g(&j!(G)) is connected.

(46.2) Let G be ap-group. Then (1) G is 1-connected for the prime p. (2) If m(G) > 2 then there is Epz E U: G and X E &[(G) is in the same

connected component of g(&,P(G)) as u precisely when m(CG(X)) > 2. (3) If p = 2 and G has a normal E8-subgroup then G is 2-connected for

the prime 2. (4) If p = 3 and m(G) > 3, then G is 2-connected for the prime 3.

Proof. Part (1) follows from the fact that Z(G) # 1. Assume m(G) > 2; by Exercise 8.4 there is Ep2 E U 9 G. G/CG(U) ( SLz(p) and SL2(p) is of p-rank 1. So if A E &;(G) then m(CA(U)) 2 2, so m(CA(U)U) > 2. Hence m(CG(U)) > 2. Also, if u # E E &,P(G) is in the same connected component as U, then there is E # D E 8; (G) adjacent to E and m(CG (E)) 2 m (D E) 2 3. Thus to prove (2) it remains to show that each A E &f(G) is in the connected component of U. But A , CA(U), CA(U)U, U is a path in LB(&;(G)), SO the proof of (2) is complete.

Assume p = 2 and E8 E V 9 G. Let E E &$(G); to prove (3) we must exhibit apathfrom E to V. For e E E, m(Cv(e)) > 2 by Exercise 9.8, and we may assume there is e E E - V, so E, (e, Cv(E)), Cv(e), V is apathin g(&;(~) ) .

Similarly under the hypothesis of (4) there is E81 = V a G by Exercise 8.11,after which we can use Exercise 9.8 and the argument of the last paragraph toestablish (4).

Given a p-group P acting as a group of automorphisms of G, and given apositive integer k, define

I'P,k(G) = (NG(X): X < P, m(X) > k)

I'P k(G) = (NG(X): X < P, m(X) > k, m(XCP(X)) > k).

The following observation is an easy consequence of Exercise 3.2.1:

(46.3) Let H < G, P E Sylp(H), and k a positive integer. Then either of thefollowing implies that P E Sylp(G):

(1) m(P) > kandFP,k(G) < H(2) m(P) > k and I'P k(G) < H.

(46.4) Let H < G, P E Sylp(H), and m(P) > k. Then the following areequivalent:

(1) FP,k(G) < H.(2) H controls fusion in olk (H) and NG(X) < H for each X E 6k (H).(3) m p(H fl Hg) < k for each g E G - H.(4) Each member of &'k (G) fixes a unique point in the permutation represen-

tation of G by right multiplication on G/H.

Proof. Parts (2), (3), and (4) are equivalent by 46.1, except in (4), °k (G)should be gk(H). This weaker version of (4) evidently implies (1) and (1)and 46.3 show P E Sylp(G), from which the strong version of (4) follows. Itremains to show (1) implies (2).

Assume FP,k(G) < H and let X E °k (H). It suffices to show that if g E Gwith Xg < H, then g E H. By Sylow's Theorem we may assume (X, Xg) <P. By 46.3, P E Sylp(G). By Alperin's Fusion Theorem there exists Pi ESylp(G), 1 k, so g; E NG(P f Pi) < FP,k(G) < H.Hence g = gl ... g E H, so that (2) holds.

If P E Sy1p(G) then FP,k(G) is called the k-generated p-core of G. By Sylowthe k-generatedp-core of G is determined up to conjugacy in G. The hypothesisof 46.4 are equivalent to the assertion that H contains the k-generated p-core ofG, and hence, if H is a proper subgroup of G, thatp-core is a proper subgroupof G. By 46.1, if the k-generatedp-core is proper then G is k-disconnected for


Similarly under the hypothesis of (4) there is Esl V 9 G by Exercise 8.1 1, after which we can use Exercise 9.8 and the argument of the last paragraph to establish (4).

Given a p-group P acting as a group of automorphisms of G, and given a positive integer k, define

~ P , L ( G ) = (NG(X): X 5 P , m(X) > k)

r:,,(~) = (NG(X): X 5 P , m(X) > k, m(XCp(X)) > k).

The following observation is an easy consequence of Exercise 3.2.1:

(46.3) Let H ( G, P E Sylp(H), and k a positive integer. Then either of the following implies that P E Sylp(G):

(1) m(P) > kandrp,k(G) 5 H (2) m(P) > k and r $ . k ( ~ ) 5 H.

(46.4) Let H 5 G, P E Sylp(H), and m(P) > k. Then the following are equivalent:

(1) ~ P , L ( G ) 5 H. (2) H controls fusion in &i(H) and NG(X) ( H for each X E &:(H). (3) mp(H i l Hg) < kforeachg E G - H. (4) Each member of &f(G) fixes a unique point in the permutation represen-

tation of G by right multiplication on GIH.

Prooj Parts (2), (3), and (4) are equivalent by 46.1, except in (4), &:(G) should be &:(H). This weaker version of (4) evidently implies (1) and (1) and 46.3 show P E Sylp(G), from which the strong version of (4) follows. It remains to show (1) implies (2).

Assume rP,L(G) 5 H and let X E &;(H). It suffices to show that if g E G with Xg 5 H, then g E H. By Sylow's Theorem we may assume (X, Xg) 5 P . By 46.3, P E Sylp(G). By Alperin's Fusion Theorem there exists Pi E Sylp(G), 1 5 i 5 n, and gi E NG(P fl Pi) with g = gl . . . g,, X 5 PI, and Xg'...g' 5 I' fl Pi. Then m(P fl Pi) 2 k, SO gi E NG(P fl Pi) 5 rP,L(G) 5 H. Hence g = gl . . . g, E H, so that (2) holds.

If P E Sylp(G) then rP,L(G) is called the k-generatedp-core of G. By Sylow the k-generatedp-core of G is determined up to conjugacy in G. The hypothesis of 46.4 are equivalent to the assertion that H contains the k-generatedp-core of G, and hence, if H is a proper subgroup of G, thatp-core is a proper subgroup of G. By 46.1, if the k-generatedp-core is proper then G is k-disconnected for

the prime p, although the converse need not be true. However in the followingspecial case the converse is valid:

(46.5) Let P E Sylp(G) and assume P is k-connected. Then(1) ek (1, p,k(G)) is a connected component of _q((ffk (G)) and FP,k(G) is the

stabilizer in G of that connected component.(2) G is k-disconnected for the prime p if and only if G has a proper k-

generated p-core.

Proof. Evidently (1) implies (2). Let H = i,P,k(G). By 46.1 and 46.4 ek (H) isthe union of connected components of _T((ffk (G)), while, as P is k-connected,(9k (P) is contained in some component A and of course H < NG(A). Henceek (H) C A by Sylow's Theorem. So A = ,gk (H) and H = NG(A) by 46.1.4.That is (1) holds.

If H is a proper subgroup of G satisfying any of the equivalent conditions of46.4 with k = 1, then we say H is a strongly p-embedded subgroup of G. Asa direct consequence of 46.5 and 46.2.1 we have:

(46.6) G possesses a strongly p-embedded subgroup if and only if G is1-disconnected for the prime p.

(46.7) Let mp(G) > 2, P E Sy1P(G), and let (92(G)° denote the set of sub-groups X E o' (G) with mp(XCG(X)) > 2. Then

(1) e2 (G)° is the set of points of e2 (G) which are not isolated in (o'2 (G)).(2) Sz (I'P 2(G))° is a connected component of 0(ff2 (G)) and rP 2(G) is

the stabilizer in G of this connected component.(3) 1((ffz (G)°) is connected if and only if G = F , 2(G).

Proof. Part (1) is trivial, as is (3) given (2). So it remains to establish (2).By 46.2.2, e (P)° is contained in a connected component A of ?(ff2 (G)).Thus H = rP 2(G) < NG(A). Let F = ez(H)°. As ff2(P)° c A and H actson A, I' C A by Sylow's Theorem. If A 0 17 there is x E F, Y E A - 17 withX and Y adjacent in 1(A). Without loss X E ff2(P)°, so Y < CG(X) < H.Hence, as mp(CH(Y)) > mp(XY) > 3, Y E F, a contradiction.

So A = F. Thus, to complete the proof of (2), it remains to show that ifX, Xg E I' then g E H. Suppose X, Xg E r. Then NG(X) < H > NG(X9)so there is EP3 = A < CG(X) < H and Ag < H. By Sylow we may takeA, Ag < P. Now apply Alperin's Fusion Theorem as in the proof of 46.4,using 17P,3(G) < I'P 2(G) < H, to conclude g E H.


the prime p, although the converse need not be true. However in the following special case the converse is valid:

(46.5) Let P E Sylp(G) and assume P is k-connected. Then (1) &;(rp,k(G)) is a connected component of LB(&:(G)) and rp,k(G) is the

stabilizer in G of that connected component. (2) G is k-disconnected for the prime p if and only if G has a proper k-

generated p-core.

Proof. Evidently (1) implies (2). Let H = ~ P , ~ ( G ) . By 46.1 and 46.4 &:(H) is the union of connected components of LB(&;(G)), while, as P is k-connected, &[(P) is contained in some component A and of course H <_ NG(A). Hence &:(H) E A by Sylow's Theorem. So A = &:(HI and H = NG(A) by 46.1.4. That is (1) holds.

If H is a proper subgroup of G satisfying any of the equivalent conditions of 46.4 with k = 1, then we say H is a strongly p-embedded subgroup of G. As a direct consequence of 46.5 and 46.2.1 we have:

(46.6) G possesses a strongly p-embedded subgroup if and only if G is 1-disconnected for the prime p.

(46.7) Let mp(G) > 2, P E Sylp(G), and let &;(G)' denote the set of subgroups X E &;(G) with 9(XCG(X)) > 2. Then

(1) &:(G)' is the set of points of &:(G) which are not isolated in LB(&:(G)). (2) &;(ri,,(G))' is a connected component of LB(&;(G)) and r i , 2 ( ~ ) is

the stabilizer in G of this connected component. (3) LB(&;(G)') is connected if and only if G = I';,,(G).

Proof. Part (1) is trivial, as is (3) given (2). So it remains to establish (2). By 46.2.2, &;(P)' is contained in a connected component A of LB(&;(G)). Thus H = r;,,(G) 5 NG(A). Let r = &:(H)'. AS &;(P)' E A and H acts on A, r E A by Sylow's Theorem. If A # r there is x E r , Y E A - r with X and Y adjacent in LB(A). Without loss X E &;(P)', so Y 5 CG(X) ( H . Hence, as mp(CH(Y)) 2 mp(XY) 2 3, Y E r , a contradiction.

So A = r. Thus, to complete the proof of (2), it remains to show that if X, Xg E r then g E H. Suppose X, Xg E r . Then NG(X) 5 H 2 NG(Xg) so there is Ep3 A i: CG(X) i: H and A8 i: H. By Sylow we may take A, Ag i: P. Now apply Alperin's Fusion Theorem as in the proof of 46.4, using ~ P , ~ ( G ) 5 I';,2(~) 5 H, to conclude g E H.

The finite simple groups 249

Lemma 46.7 says that if m p(G) > 2 and P E Sylp(G), then, neglecting theisolated points of 1((g2 (G)), G is 2-connected for the prime p precisely when

G = F 2(G)These observations are very useful in conjunction with the Signalizer Functor

Theorem, as the next two lemmas and Exercise 16.1 indicate.

(46.8) Let 17 be the set of elements a of G of order p with Mp(CG (a)) > 2 andlet 9 be a map from 17 into the set of p'-subgroups of G such that, for all a, b E Fwith [a, b] = 1 and all g E G, 9(ag) = 9(a)9 and 9(a) fl CG(b) < 9(b). LetP E Sylp(G), assume G = F , 2(G), Op-(G) = 1, and either

(1) 9(a) is solvable for each a E 17, or(2) the Signalizer Functor Theorem holds on G.

Then 9(a) = 1 for each a E F.

Proof. For A E (P3 (G), 9 is an A-signalizer functor by hypothesis. ForB E (ffz (G)° define WB = (9(b): b E B#). Then there exists A E 6'3 (G) withB < A and, by Exercise 15.3, WB = WA and FA,2(G) < NG(WA). In partic-ular if B and D are distinct members of S2 (G) adjacent in 1(ez (G)) thenBD E 6'3 (G) so WB = WBD = WD Thus, as F0° 2(G) = G, 46.7 says WB = Wis independent of the choice of B E (9z (G)°. Thus G = rP 2(G) < NG(W) asNG(B) < FA,2(G) < NG(W). But, by (1) and the Solvable Signalizer Func-tor Theorem or by (2), W is a p'-group, so W < Op,(G). As Op,(G) = 1 byhypothesis, and as 9(a) < W for each a E 17, the lemma is established.

(46.9) Let 17 be the set of elements a of G of order p with mp(CG (a)) > 2, letP E Sylp(G), and assume F 2(G) = G, Op,(G) = 1, CG(a) is balanced forthe prime p for each a E 17, and either

(1) Op'(CG (a)) is solvable for each a E F, or(2) the Signalizer Functor Theorem holds on G.

Then Op'(CG(a)) = 1 for each a E 1.

Proof. For a E F let 9(a) = Opr(CG(a)). By Exercise 15.4.2, 9(a) fl CG(b) <9(b) for each a, b E 17 with [a, b] = 1. Hence 46.8 completes the proof.

47 The finite simple groupsSection 47 describes a list .7C of finite simple groups. The Classification The-orem says that any finite simple group is isomorphic to a member of X. Theproof of the Classification Theorem is far beyond the scope of this book, butthere is a brief outline of that proof in the final section of this chapter.


Lemma 46.7 says that if mp(G) > 2 and P E Sylp(G), then, neglecting the isolated points of LB(&:(G)), G is 2-connected for the prime p precisely when G = ~ O , , , ( G ) .

These observations are very useful in conjunction with the Signalizer Functor Theorem, as the next two lemmas and Exercise 16.1 indicate.

(46.8) Let r be the set of elements a of G of order p with mp(CG(a)) > 2 and let 8 be a map from r into the set of p'-subgroups of G such that, for all a , b E r with [a, b] = 1 and all g E G, B(ag) = 8(a)g and B(a) f l CG(b) ( 8(b). Let P E Sylp(G), assume G = I';,,(G), Op!(G) = 1 , and either

( 1 ) 8(a) is solvable for each a E r , or (2) the Signalizer Functor Theorem holds on G.

Then 8(a) = 1 for each a E r.

Proof. For A E d':(G), 8 is an A-signalizer functor by hypothesis. For B E &;(G)' define WE = (8(b): b E B'). Then there exists A E &:(G) with B 5 A and, by Exercise 15.3, WE = WA and rA,,(G) ( NG(WA). In particular if B and D are distinct members of &:(G) adjacent in LB(&;(G)) then BD E &f(G) so WB = WBD = WD.Thus, as I';,,(G) = G,46.7 says WB = W is independent of the choice of B E &;(G)'. Thus G = r ! , , ( ~ ) ( NG(W) as NG(B) ( rA,,(G) ( NG(W). But, by (1) and the Solvable Signalizer Func- tor Theorem or by (2), W is a p'-group, so W ( Opf(G). As Opf (G) = 1 by hypothesis, and as 8(a) ( W for each a E r , the lemma is established.

(46.9) Let r be the set of elements a of G of order p with mp(CG(a)) > 2, let P E Sylp(G), and assume I';,,(G) = G , Op!(G) = 1, CG(a) is balanced for the prime p for each a E r , and either

(1) OPt(CG(a)) is solvable for each a E r , or (2) the Signalizer Functor Theorem holds on G.

Then Opt(CG(a)) = 1 for each a E r .

Proof. For a E r let 8(a) = Op~(CG(a)) . By Exercise 15.4.2,8(a) f l CG(b) ( 8(b) for each a , b E r with [a, b] = 1. Hence 46.8 completes the proof.

47 The finite simple groups Section 47 describes a list YC of finite simple groups. The Classification The- orem says that any finite simple group is isomorphic to a member of YC. The proof of the Classification Theorem is far beyond the scope of this book, but there is a brief outline of that proof in the final section of this chapter.

. C can be described as the following collection of groups:

(1) The groups of prime order.(2) The alternating groups An of degree n > 5.(3) The finite simple groups of Lie type listed in Table 16.1.(4) The 26 sporadic simple groups listed in Table 16.3.

The groups of prime order are the abelian simple groups (cf. 8.4.1). Theyare the simplest of the simple groups.

By 15.16, the alternating groups An, n > 5, are simple. The permutationrepresentation of An of degree n is an excellent tool for investigating the group,and, as in 15.3 and Exercise 5.3, to determine its conjugacy classes. Lemma33.15 describes the covering group and Schur multiplier of An. Exercise 16.2says S,,, unless n = 6.

The groups of Lie type are the analogues of the semisimple Lie groups. Thisclass of groups is extremely interesting; for one thing by some measure mostfinite simple groups are of Lie type. We've already encountered examples ofthese groups; the classical groups are of Lie type.

The groups of Lie type can be described in terms of various representations:as groups of automorphisms of certain Lie algebras or fixed points of suitableautomorphisms of such groups; as fixed points of suitable endomorphisms ofsemisimple algebraic groups; as groups of automorphisms of buildings withfinite Weyl groups; as groups with a BN-pair and a finite Weyl group. I'll takea modified version of this last point of view here to avoid complications.

A finite group of Lie type satisfies the following conditions:

(Ll) G possesses a Tits system (G, B, N, S) with a finite Weyl group W =N/(B fl N). The Lie rank of G is the integer I S!.

(L2) H = B fl N possesses a normal complement U in B.(L3) There is a root system E for W and a simple system r for E with

S = (sa: a E jr) where sa is the reflection through a. Let E+ be thepositive system of rr. There exists an injection a H Ua of E into the set ofsubgroups of G such that for each a E E and w E W, H < NG(Ua) and(Ua)' = U. Each member of U can be written uniquely as a product

with ua E Ua, in some fixed ordering of E+ respecting height.(L4) The root groups (Ua: a E E+) satisfy

R E E+,[Ua, UU] C (Z(Ua) fl Z(UU)) f Uia+lp, a,/where the product is over all roots i a + j,8 with i and j positive integers.

We will be most interested in groups which also satisfy:

(L5) Let w0 be the unique member of W of maximal length in W (cf. Exercise10.3). Then G = (U, U'O), B fl B'O = H, and W is irreducible.


X can be described as the following collection of groups:

(1) The groups of prime order. (2) The alternating groups A, of degree n > 5. (3) The finite simple groups of Lie type listed in Table 16.1. (4) The 26 sporadic simple groups listed in Table 16.3.

The groups of prime order are the abelian simple groups (cf. 8.4.1). They are the simplest of the simple groups.

By 15.16, the alternating groups A,, n 2 5, are simple. The permutation representation of A, of degree n is an excellent tool for investigating the group, and, as in 15.3 and Exercise 5.3, to determine its conjugacy classes. Lemma 33.15 describes the covering group and Schur multiplier of A,. Exercise 16.2 says Aut(A,) = S,, unless n = 6.

The groups of Lie type are the analogues of the semisimple Lie groups. This class of groups is extremely interesting; for one thing by some measure most finite simple groups are of Lie type. We've already encountered examples of these groups; the classical groups are of Lie type.

The groups of Lie type can be described in tenns of various representations: as groups of automorphisms of certain Lie algebras or fixed points of suitable automorphisms of such groups; as fixed points of suitable endomorphisms of semisimple algebraic groups; as groups of automorphisms of buildings with finite Weyl groups; as groups with a BN-pair and a finite Weyl group. I'll take a modified version of this last point of view here to avoid complications.

A finite group of Lie type satisfies the following conditions:

(Ll) G possesses a Tits system (G, B, N, S) with a finite Weyl group W = N/(B f l N). The Lie rank of G is the integer JSJ.

(L2) H = B n N possesses a normal complement U in B. (L3) There is a root system Z for W and a simple system n for Z with

S = (s,: a E n ) where s, is the reflection through a. Let E+ be the positive system of n . There exists an injection a F-+ U, of Z into the set of subgroups of G such that for each a E Z and w E W, H 5 NG(U,) and (U,), = U,,. Each member of U can be written uniquely as a product l-IaEC+u,, with u, E U,, in some fixed ordering of Z + respecting height.

(LA) The root groups (U,: a E Z +) satisfy

where the product is over all roots ia + jB with i and j positive integers.

We will be most interested in groups which also satisfy:

(L5) Let wo be the unique member of W of maximal length in W (cf. Exercise 10.3). Then G = (U, UWo), B n BWO = H, and W is irreducible.


Some observations about these axioms are in order. Define a Tits systemT = (G, B, N, S) to be saturated if fl w B' = H. By Exercise 14.9.5, T issaturated if and only if B fl B"'0 = H. So the condition u fl U"'0 = H in L5can be replaced by the hypothesis that T is saturated.

Next, from Exercise 10.3, E+wo = E-. Hence, by L3, 47.1.1, and 30.7,Ua = U fl U"'0,° for a E r. Moreover the proof of 47.1 depends on Ll-L4 butnot L5. By 30.9 each y E E can be written y = aw for some a E Tr, W E W.So, by L3, U, = UaN, _ (Ua)"' = (U fl UwOS°)'. Hence L1-L4 determine theroot groups Uy, y E E uniquely.

Conversely if T is a saturated Tits system satisfying L2 then by Exercise14.10 we can define Uy = (U n U'0'°)', where y = aw, a E r, and w E W.Then, by Exercise 14.10 parts (7) and (9), L3 is satisfied. Thus L3 can bedispensed with if we assume Ll and L2 with T saturated. Equivalently L1, L2,and L5 imply L3.

The properties L1 and L2 essentially characterize the finite groups of Lietype (cf. [FS], [HKS], and [Ti]). There are some rather trivial examples ofgroups of Lie rank 1 satisfying Ll-L5 which are not of Lie type. The finitegroups of Lie type satisfying L5 and possessing a trivial center are listed inTable 16.1. Column 1 lists G. The parameters q and n are an arbitrary primepower and positive integer, respectively, unless some restriction is listed ex-plicitly. Column 2 lists the order of a certain central extension G of G. UsuallyG is simple, G is the universal covering group of G, and d(G) is the order ofthe Schur multiplier of G. In any event IGI = I61/d(G). Finally, column 4lists the Dynkin diagram of the root system E of G. Table 16.2 explains thenotation An, B, etc. The nodes of the Dynkin diagram are indexed by the setr of simple roots. Distinct nodes a and ,B are joined by an edge of weight2(a, ,8)(,8, a)/(a, a)(,8, ,B) = map - 2. It turns out Isasp I = man, so, if we ne-glect the arrows in the diagram, the Dynkin diagram of E is just the Coxeterdiagram of the Weyl group W of E or G, as defined in section 29. By 30.9, eachmember of E is conjugate to a member of r under W, and it turns out that thereare one or two orbits of W on E depending on whether the Dynkin diagramhas no multiple bonds or has multiple bonds, respectively. Indeed if a and ,Bare joined by an edge of weight 1 then a is conjugate to ,B under (sa, so). If thediagram has multiple bonds, roots in different orbits are of different lengths,and hence are called long or short roots. The arrow in the diagram points tothe short root in the pair joined.

Let q be a power of the prime p. p is called the characteristic of G. Recallthat G can be defined in terms of one of several representations. Consider forthe moment the representation of G as a group of automorphisms of a Liealgebra L. L is a Lie algebra over a field of characteristic p obtained froma simple Lie algebra L' over the complex numbers. L' has an associated root

The$nite simple groups 25 1

Some observations about these axioms are in order. Define a Tits system T = ( G , B, N , S ) to be saturated if ow,, BW = H . By Exercise 14.9.5, T is saturated if and only if B n BWO = H . So the condition U n UWO = H in L5 can be replaced by the hypothesis that T is saturated.

Next, from Exercise 10.3, Zf wo = Z-. Hence, by L3, 47.1.1, and 30.7, U, = U f l U wOSa for a E n . Moreover the proof of 47.1 depends on L 1-LA but not L5. By 30.9 each y E Z can be written y = a w for some a E n, w E W . So, by L3, U, = U,, = (U,)W = (U n UwOS~)W. Hence L1-LA determine the root groups U,, y E Z uniquely.

Conversely if T is a saturated Tits system satisfying L2 then by Exercise 14.10 we can define U, = (U n UWOSa)W, where y = a w , a E n, and w E W. Then, by Exercise 14.10 parts (7) and (9), L3 is satisfied. Thus L3 can be dispensed with if we assume L1 and L2 with T saturated. Equivalently L1, L2, and L5 imply L3.

The properties L1 and L2 essentially characterize the finite groups of Lie type (cf. [FS], [HKS], and [Ti]). There are some rather trivial examples of groups of Lie rank 1 satisfying L1-L5 which are not of Lie type. The finite groups of Lie type satisfying L5 and possessing a trivial center are listed in Table 16.1. Column 1 lists G. The parameters q and n are an arbitrary prime power and positive integer, respectively, unless some restriction is listed explicitly. Column 2 lists the order of a certain central extension of G . Usually G is simple, e is the universal covering group of G , and d ( G ) is the order of the Schur multiplier of G . In any event ]GI = ~ e l / d ( ~ ) . Finally, column 4 lists the Dynkin diagram of the root system Z of G . Table 16.2 explains the notation A,, B,,, etc. The nodes of the Dynkin diagram are indexed by the set n of simple roots. Distinct nodes a and B are joined by an edge of weight 2(a, B ) ( B , a ) / ( a , a)(B, B ) = r n , ~ - 2. It turns out Is,sSI = rn4 , so, if we ne- glect the arrows in the diagram, the Dynkin diagram of Z is just the Coxeter diagram of the Weyl group W of Z or G , as defined in section 29. By 30.9, each member of Z is conjugate to a member of n under W, and it turns out that there are one or two orbits of W on Z depending on whether the Dynkin diagram has no multiple bonds or has multiple bonds, respectively. Indeed if a and B are joined by an edge of weight 1 then a is conjugate to B under (s,, s S ) . If the diagram has multiple bonds, roots in different orbits are of different lengths, and hence are called long or short roots. The arrow in the diagram points to the short root in the pair joined.

Let q be a power of the prime p. p is called the characteristic of G . Recall that G can be defined in terms of one of several representations. Consider for the moment the representation of G as a group of automorphisms of a Lie algebra L . L is a Lie algebra over a field of characteristic p obtained from a simple Lie algebra L' over the complex numbers. L' has an associated root


Table 16.1 Finite groups of Lie type with L5 and trivial center

G IGId(G) = IGI d(G) E

An(q)

Bn(q), q odd

Cn(q)

Dn(q), n > 2

E6(q)

E7(q)

E8(q)

F4(q)

G2(q)

2An(q)

2B2(q), q = 22,n+1

2Dn(q), n >

3D4(q)

2E6(q)

2F4(q), q = 22m+I

2G2(q), q = 32m+1

Jr

qn(n+l)/2 rl(gi+l - 1)i=1

q"2fl(g2, - 1)

i=I

q"2 fl(g2i - 1)i=1

n-1

qn(n-l)(qn - 1) (g2i - 1)i=1

g3fi(g12 - 1)(q9 - 1)(q8 - 1)

(q6 - 1)(q5 - 1)(q2 - 1)

g63(g18 - 1)(g14 - 1)

(812 - 1)(g10 - 1)(q8 - 1)(q6 - 1)(q2-1)g120(g3° - 1)(g24 - 1)

(q20 - 1)(g18 - 1)(814 - 1)(g12 - 1)

(q8 - 1)(q2 - 1)

g24(g12 - 1)(q8 - 1)(q6 - 1)(q2 - 1)

q6(q6 - 1)(q2 - 1)n

qn(n+1)/2 rl(gi+l -(-1)i+l)i=1

q2(q2 + 1)(q - 1)n-1

qn(n-l)(q" + 1) rl (q2i - 1)i=1

g12(g8 + q4 + 1)(q6 - 1)

(q2 - 1)g36(g12 - 1)(q9 + 1)(q8 - 1)

(q6 - 1)(q5 + 1)(q2 - 1)

gl2(g6 + 1)(q4 - 1)

(q3 + 1)(q - 1)q3(q3 + 1)(q - 1)

(n+1,q-1) An

(2, q - 1) Bn

(2, q - 1) Cn

(4, qn - 1) Dn

(3, q - 1) E6

(2, q - 1) E7

1 E8

1 F4

1 G2

(n + 1, q + 1) C[n+l/2]

1 Al

(4, q" + 1) Cn-1

1

(3, q + 1)C2

F4

1 dihedral 16

1

system E' with Dynkin diagram V. If G is of type A, B, C, D, F, or G then G iscalled an ordinary Chevalley group and E = V. On the other hand the groupsof type 'X are called twisted Chevalley groups. mX (q) is the fixed points on anordinary Chevalley group of type X(qe) of a suitable automorphism of orderm, and in this case E is not equal to V.

It turns out also that U is a Sylow p-subgroup of G, where p is the charac-teristic of G (cf. 47.3).


Table 16.1 Finite groups of Lie type with L5 and trivial center

G IGld(G) = I ~ I d(G) Z

dihedral 16

system C' with Dynkin diagram C'. If G is of type A, B, C, D, E or G then G is called an ordinary Chevalley group and C = C'. On the other hand the groups of type "X are called twisted Chevalley groups. "X(q) is the fixed points on an ordinary Chevalley group of type X(qe) of a suitable automorphism of order m, and in this case X is not equal to XI.

It turns out also that U is a Sylow p-subgroup of G, where p is the characteristic of G (cf. 47.3).

Table 16.2 Some Dynkin diagrams

A.1 2 n-I no--- ... -

B

C

1 2 n-2 n- I n

1 2 n-2 n-1 n

n-11 2 -<

n

1 2 3 5 n-I nEn 0-0

F4

4

1 2 3 4o ate-c1 2

G2 o<Em4

Dihedral 161 2

^ 6 ^

The classical groups are groups of Lie type; namely An(q) = Ln+1(q),

Bn(q) = P22n+1(q), for q odd, Cn(q) = PSP2n(q), Dn(q) = PQ2n(q),2An(q) = Un+1(q), and 2Dn(q) = PS22n(q). Also the groups 2B2(q) were firstdiscovered as permutation groups by M. Suzuki and hence are called Suzukigroups and sometimes denoted by Sz(q).

There are some isomorphisms among the groups of Lie type, and of groupsof Lie type with alternating groups. Also some centerless groups of Lie typeare not simple. Here's the list of such exceptions; we've already seen a numberof them:

A1(q) = B1(q) - C1(q) 2A1(q) - L2(q),

B2(q) - C2(q) PSP4(q),

D2(q) - PS24 (q) = L2(q) x L2(q),

2D2(q) = PS24 -(q) --- L2(q 2),

D3(q) - A3(q), 2D3(q)2A3(q).

In particular recall L2(2) and L2(3) are not simple, and of course neither isL2(q) x L2(q). Also C2(2) - S6, and 2A2(2) - PSU3(2) are not simple. If

Thejnite simple groups

Table 16.2 Some Dynkin diagrams

1 2 Dihedral 16 -

6

The classical groups are groups of Lie type; namely An(q) = Ln+l(q),

Bn(q) = PQ2n+l(q), for q odd, Cn(q) = PSp2n(q), Dn(q) = f'filn(q), 2 ~ n ( q ) = Un+l(q), and Dn(q) = P!2,(q). Also the groups B2(q) were first discovered as permutation groups by M. Suzuki and hence are called Suzuki groups and sometimes denoted by Sz(q).

There are some isomorphisms among the groups of Lie type, and of groups of Lie type with alternating groups. Also some centerless groups of Lie type are not simple. Here's the list of such exceptions; we've already seen a number of them:

In particular recall L2(2) and L2(3) are not simple, and of course neither is L2(q) x L2(q). Also C2(2) % S6, and 2 ~ 2 ( 2 ) % Psu3(2) are not simple. If


G = G2(2) or 2F4(2) then IG : G(I)) = 2 with G(I) simple. (G2(2))(1) = U3(3).(2F4(2))(1) is called the Tits group. 2B2(2) is a Frobenius group of order 20,and hence solvable. 12G2(3) : 2G2(3)(I)I = 3 with 2G2(3)(1 = L2(8). All othercenterless groups of Lie type are simple. L2(4) = L2(5) - A5, L2(9) = A6,L4(2) - A8, U4(2) = PSp4(3), and L3(2) = L2(7). This exhausts the iso-morphisms among groups of Lie type and between groups of Lie type andalternating groups.

Later in this section we will explore further consequences of the propertiesLl-L5, using the theory of BN-pairs developed in chapter 14. But first let'stake a look at the sporadic simple groups.

The sporadic simple groups and their orders are listed in Table 16.3. Atpresent there is no nice class of representations available to describe thesporadic groups in a uniform manner, although some recent work on the

Table 16.3 The sporadic simple groups

Notation Name Order

M11 Mathieu 24 .32.5.11Mil 26.33.5.11M22 27 .32.5.7.11M23 27.32.5.7.11.23M24 210.33.5.7.11.23J1 Janko 23.3.5.7.11.19J2 =HJ Hall-Janko f. 33 .52 .7J3 =HJM Higman-Janko-McKay 2.35.5.17.19J4 Janko 221 .33 .5 .7 . 113 .23-29-31-37-43HS Higman-Sims 29.32.53.7.11Mc McLaughlinSz Suzuki 213.37.52.7.11.13Ly = LyS Lyons-Sims 28 . 37 .56 .7-11-31-37-67He = HHM Held-Higman-McKay 210 33 .52 .73 . 17Ru Rudvalis 214.33.53.7.13.29O'N = O'NS O'Nan-Sims 29.34.5.73.11.19.31Co3 = .3 Conway 210.37.53.7.11.23Cot = .2 218 36 .53 .7-11-23Col = .1 221 .39 .54 .72 . 11 .13.23M(22) = F22 Fischer 217.39.52.7.11.13M(23) = F23 218.313.52.7.11.13. 17.23M(24)' = F24 221.316.52.73.11.13.17.23.29F3 = E Thompson 215.310.53.72.13.19.31F5 = D Harada-Norton 214.36.56.7.11.19F2 = B Baby monster 241 .313.56.72.11.13.17

.19 .23-31-47

F1 =M Monster 246 . 320 59 .76 . 112. 133 . 17-19-23-29-31-41-47-59-71


G = G2(2) or ' ~ ~ ( 2 ) then IG : G(')I = 2 with G(') simple. ( ~ ~ ( 2 ) ) ( ' ) E U3(3). (2~4(2))(1) is called the Tits group. '~ '(2) is a Frobenius group of order 20, and hence solvable. I2G2(3) : 2G2(3)(1)1 = 3 with ' ~ ~ ( 3 ) ( ' ) =" L2(8). All other centerless groups of Lie type are simple. L2(4) E L2(5) E As, L2(9) E Ag,

L4(2) E A8, U4(2) E PSp4(3), and L3(2) =" L2(7). This exhausts the isomorphisms among groups of Lie type and between groups of Lie type and alternating groups.

Later in this section we will explore further consequences of the properties El-L5, using the theory of BN-pairs developed in chapter 14. But first let's take a look at the sporadic simple groups.

The sporadic simple groups and their orders are listed in Table 16.3. At present there is no nice class of representations available to describe the sporadic groups in a uniform manner, although some recent work on the

Table 16.3 The sporadic simple groups

Notation Name Order

MII Mathieu 2 4 . 3 2 . 5 . 1 1 M12 2 6 . 3 3 . 5 . 1 1 MZZ 2 7 . 3 2 . 5 . 7 . 1 1 M7.3 27 . 3 2 . S . 7 . 1 1 . 2 3 M24 210 .33 .S .7 .11 .23 J1 Janko 2 3 . 3 . 5 . 7 . 1 1 . 1 9 J2 =HJ Hall-Janko 27 . 33 .52 .7 J3 =HJM Higman-Janko-McKay z7 . 35 . S . 17 . 19 J4 Janko 2". 33 . 5 . 7 . l13 . 23 .29 .31 .37 .43 HS Higman-Sims 2 9 . 3 2 . ~ 3 . 7 . 1 1 Mc McLaughlin 2 7 . 3 6 . S 3 . 7 . 1 1 Sz Suzuki 213 .37 .s2 . 7 . 1 1 . 1 3 Ly = LyS Lyons-Sims 2 8 . 3 7 . S 6 . 7 . 1 1 . 3 1 . 3 7 . 6 7 He = HHM Held-Higman-McKay 21° . 33 .S2 . 73 , 17 Ru Rudvalis 214 . 33 . S3 . 7 . 13 .29 O'N = O'NS O'Nan-Sims 2 9 . 3 4 . 5 . 7 3 . 1 1 . 1 9 . 3 1 Cog = .3 Conway 21° .37 .S3 . 7 . 11 .23 Coz = .2 2 1 8 . 3 6 . S 3 . 7 . 11 .23 Col = .l 2" . 3 9 . S 4 . 7 2 . 1 1 . 1 3 . 2 3 M(22) = Fz2 Fischer 2 1 7 . 3 9 - ~ 2 ~ 7 . 1 1 . 1 3 M(23) = FZ3 218 .313 .s2 . 7 . 11 . 13. 17 + 2 3 M(24)' = FZ4 2" . 316 .S2 . 7 3 . 11. 13. 17.23 .29 F3 = E Thompson 215 . 310.53 . 72 . 13. 19.31 F5 = D Harada-Norton 214. 36 . s 6 . 7 . 1 1 . 1 9 F2 = B Baby monster 241 .313 .s6 .72 . 11 . 13 . 17

.19 .23 .31 .47 F1 = M Monster 246. 320 .59 .76. 112. 133. 17

.19 .23 .29 .31 .41 . 47 .59 .71

representations of the sporadic groups on geometries may eventually lead tosuch a description. Instead the sporadic groups were discovered in variousways. The Mathieu groups all have multiply transitive permutation represen-tations; specifically M is 3, 4, or 5-transitive of degree n. J2, Mc, HS, Sz, Ru,M(22), M(23), and M(24) all have rank-3 permutation representations. Henceeach is a group of automorphisms of a strongly regular graph via the con-struction of section 16. The Conway groups act on the Leech lattice, a certaindiscrete integer submodule of 24-dimensional Euclidean space. The other spo-radic groups were discovered through the study of centralizers of involutions(cf. section 48).

In the remainder of this section assume G satisfies the conditions L1-L5.We'll see that these conditions together with the theory in chapter 14 have anumber of interesting consequences.

A subset 0 of E is closed if, for each a, ,B E 0 with a + P E E, we havea + P E A. A subset of r of 0 is an ideal of 0 if, whenever a E F, ,B E 0,and a +,B E E, then a +,B E r. Define Uo = (Ua: a E 0).

(47.1) Let 0 be a closed subset of E+ and r an ideal of A. Then(1) Uo = 11aEAUa, with the product in any order, and each element in Uo

can be written uniquely as a product 11aEoua, ua E 0 (for any fixed orderingof 0).

(2) Urd Uo.(3) H < NG (UA).

Proof. H acts on Ua by L3, so (3) holds. Property L4 says Ur4 U. Recallthe definition of the height function h from section 30, and let a E 0 withh(a) minimal. Observe 0' = 0 - {a} is an ideal of A. Thus Uo-4 Uo, soUo = Ua UA,, and then, by induction on the order of 0, U = r1,,, UPwith respect to some ordering of A. Now L3 and Exercise 16.3 complete theproof.

(47.2) Let W E W, n E N with Hn = w, and 0 =E+ fl E-w. Then 0 isa closed subset of E and each element of BwB can be written uniquely in theform bnu, b E B, u E U.

Proof. Observe E+ and E- are closed subsets of E, the image of a closedsubset under an element of W is closed, and the intersection of closed sub-sets is closed. So 0 is closed. Similarly r = E+ - 0 = E+ fl E+w is closed.By 47.1, U = UrUA so BwB = BwHUrUA = B((Ur)' ')wUo = BwUoas (Ur)'"-' = Urw-[ < U. Suppose bnu = any, a, b E B, u, v E U. Then

Thejnite simple groups 255

representations of the sporadic groups on geometries may eventually lead to such a description. Instead the sporadic groups were discovered in various ways. The Mathieu groups all have multiply transitive permutation representations; specifically M, is 3,4, or 5-transitive of degree n. Jz, Mc, HS, Sz, Ru, M(22), M(23), and M(24) all have rank-3 permutation representations. Hence each is a group of automorphisms of a strongly regular graph via the construction of section 16. The Conway groups act on the Leech lattice, a certain discrete integer submodule of 24-dimensional Euclidean space. The other sporadic groups were discovered through the study of centralizers of involutions (cf. section 48).

In the remainder of this section assume G satisfies the conditions L1-L5. We'll see that these conditions together with the theory in chapter 14 have a number of interesting consequences.

A subset A of C is closed if, for each a , ,!? E A with a + ,!? E C, we have a + ,!? E A. A subset of r of A is an ideal of A if, whenever a E r, ,!? E A, a n d a + p E C,thena+,!? E r.DefineUA = (U,:a E A).

(47.1) Let A be a closed subset of C+ and r an ideal of A. Then (1) UA = naeAU,, with the product in any order, and each element in UA

can be written uniquely as a product lIaEAu,, u, E A (for any fixed ordering of A).

(2) u r UA (3) H i N G ( ~ A ) .

Proof. H acts on U, by L3, so (3) holds. Property L4 says Ur l] UA. Recall the definition of the height function h from section 30, and let a E A with h(a) minimal. Observe A' = A - {a} is an ideal of A. Thus UAl: UA, so UA = U,UAf, and then, by induction on the order of A, U = nSEAUg with respect to some ordering of A. Now L3 and Exercise 16.3 complete the proof.

(47.2) Let w E W,n E N with Hn = w, and A = C+ i l C-w. Then A is a closed subset of C and each element of BwB can be written uniquely in the form bnu, b E B, u E UA.

Proof. Observe C+ and C- are closed subsets of C, the image of a closed subset under an element of W is closed, and the intersection of closed subsets is closed. So A is closed. Similarly r = C+ - A = C+ n C+w is closed. By 47.1, U = Ur UA SO BwB =I: BwHUr UA B((u~)w-')w UA =: BwUA as (u~)w- ' = UrW-l 5 U. Suppose bnu = anv, a , b E B, u, v E UA. Then

a-1b = (vu-1)n-' E B n (UA)w-' < B n U'O as Aw-1 C E- = E+wo, byExercise 10.3. But by L5, B fl B'O = H, so B fl Uw° = 1. Hence a-1b = 1,so a = b and u = v.

If G is a finite group of Lie type in characteristic p satisfying L5 then U ESylp(B). Thus the next result shows U E Sylp(G).

(47.3) If G is finite and U E Sylp(B) then U E Sylp(G).

Proof. For w E W let A(w) = E+ fl E-w. Then, by 47.2,

IGI = IBI E WA(w)lWEW

As U E Syl p(B), I B I p= I U I and I UA(w) l- 0 mod p if UA(w) # 1. But UA(w) = 1if and only if A(w) = 0, and, by 30.12, this happens precisely when w = 1.Hence IGI - IBlmod(pIUI), so U E Sylp(G).

Recall from section 43 that, for J C rr, SJ = (s1: j E J) and further that Wj =(Si) and Pi = (B, Wj) are parabolics of W and G, respectively. Let EJ bethe set of roots spanned by J and let *J = E+ - E J. Observe *J is an ideal ofE+ and let Vi = U*,. Similarly Ei is a closed subset of E and set UJ = UE+.Finally set LJ = (WJ, UJ).

(47.4) (1) Pi = NG(VJ)(2) LJ is a complement to Vj in Pj.(3) (Li, HUj, N, Si) is a Tits system satisfying L1-L4 with respect to the

root system EJ and the simple system J.

Proof. By 30.20, EJ is a root system with simple system J and Weyl group WJ.As Yf J is an ideal of E+, Vi :!l U and U = UJ Vi by 47.1. Then PJ = (U, WJ) _(LJ, VJ). Finally, for sj E Si and a E 1frj, asj = a - 2(a, j)j/(j, j) E 1/rj,because a has a positive projection on 7r - J, and hence as j does too. ThusU." = U.,, < Vj, so Wj < NG(Vi). Therefore Pj = (U, Wj) < NN(Vi).So, by 43.7, NG(VJ) = PK for some K C 7r with J C K. If k E K - J thenk E 1/rj and ksk = -k, so U_k = Ukk < V. But then U_k < Uw0 fl u = 1.So K = J and hence (1) holds.

Let A = HUJ,Claim LJ = A Wj A. It suffices to show w Ujw' C_ AWJA =Y for all w, w' E Wj. Then by induction on the length of w', it suffices toshow w Ujs j C Y for each W E Wj and j E J. Let A = EJ - {j}. By30.7, As j C E , so Uo = UAsj C Uj. Thus as Uj = U j UA, it suffices toshow wUJs.. C Y. Similarly, if l(wsj) > 1(w), then by 30.10, jw-1 > 0, so

25 6 Finite simple groups

a-'b =(vu-')"-I E B n (U~)W- ' 5 B n UWo as AW-' & C- = C+wo, by Exercise 10.3. But by L5, B i l BWO = H, so B i l UWO = 1. Hence a-'b = 1, s o a = b a n d u = v .

If G is a finite group of Lie type in characteristic p satisfying L5 then U E Syl,(B). Thus the next result shows U E Syl,(G).

(47.3) If G is finite and U E Syl,(B) then U E Syl,(G).

Proof. For w E W let A(w) = C+ i l C-w. Then, by 47.2,

AsU ~ S y l , ( B ) , ~ B ~ , = ~ U ~ a n d ~ U ~ ( ~ ) ~ O r n o d p i f U ~ ( ~ ~ # l .ButU~( , )= 1 if and only if A(w) = 0, and, by 30.12, this happens precisely when w = 1. Hence [GI = IBlmod(plUI), so U E Syl,(G).

Recall from section 43 that, for J & n, S j = (sj: j E J) and further that Wj = (Sj) and P j = (B, Wj) are parabolics of W and G, respectively. Let C be the set of roots spanned by J and let $rJ = C+ - C j. Observe $rJ is an ideal of C+ and let Vj = U$, . Similarly Cf is a closed subset of C and set UJ = U,;. Finally set L j = (Wj, UJ).

(47.4) (1) P j = NG(Vj). (2) L is a complement to Vj in P j . (3) (L j , HUj , N, Sj) is a Tits system satisfying L1-L4 with respect to the

root system C j and the simple system J.

Proof. By 30.20, C is aroot system withsimplesystem J and Weyl group Wj. As$rJisanidealofC+, VjGUandU = UJVJby47.1.ThenPJ = (U, Wj) = (Lj , Vj). Finally, for s j E S j and a E $rJ, a s j = a - 2(a, j ) j / ( j , j ) E 1+9j, because a has a positive projection on n - J, and hence a s j does too. Thus U: = U,, 5 VJ, so WJ 5 NG(Vj). Therefore P j = (U, Wj) _( NG(Vj). So, by 43.7, NG(Vj) = PK for some K & n with J & K. If k E K - J then k E $rJ and ksk = -k, SO UPk = U: ( Vj. But then U-k 5 UWO n U = 1. So K = J and hence (1) holds.

LetA = HUJ,ClaimLJ = AWJA.ItsufficestoshowwUJwl & AWjA = Y for all w, w1 E Wj. Then by induction on the length of wl, it suffices to show wUJsj & Y for each w E Wj and j E J. Let A = C j - {j). By 30.7, Asj c Cf, so U: = UA,, & UJ. Thus as UJ = UjUA, it suffices to show wUJSI & Y. Similarly, if l(wsj) > l(w), then by 30.10, jw-' > 0, so

Uj'-' = Uj", < Uj and hence w Ujsj = Uj'-'wsj c Y. So assume l(wsj) <l(w). Then by induction on l(w), ws j Uj sj c Y. Now if AWjA is a group thenUJ' c AWjA, so

wUjsj = wsjUjj c ws1AWWA = wsjA U wsjAsjA C Y.

Thus it suffices to show AWj A is a group.

By Exercise 10.3, s" = si for some i E 7t, and wo is an involution. Let X =Pi". I'll show X n Pj = AWjA, which will show AWjA is a group and henceestablish the claim. By 43.7, Pj = B U Bs j B = V j T, where T = AWj A. Sim-ilarlyX = BWOUBWOsjBWO = BWOUSjBW0SiBW'0,and, assj E X,sjX = X,soX = s j B'OU BWOSJ BWO. Let Q = E - - {- j }. Then UWO = U_ j Uu so Uw°S; =

Uj Uc2 by 30.7, and hence BWOS; B'O = Uj BwO. Next B n s j Bw0 = (Bwo ns jwoB)wo and, as wo 0 s jwo, Bw0B n Bs jwoB = 0 by 43.2. So B nsj B'O =0. Hence BnX = BnB'Os;BIO = BnU;B"'0 = Uj(BnBw0) = UjH = A.Assj,ACX,TCX,soXnPj=TVjnX=T(VjnX)c_T(BnX)=TA = T. Thus the claim is at last established.

It's now easy to see that (LJ, A, N, Sj) is a Tits system. The only one of theBN-pair axioms which is not evident is BN3. But, for s E Sj, w E Wj,

sAw C (BwB u BswB) n AWj A

= AwA U AswA

with the equality following from 47.2 and 43.2. Similarly it is clear that L2-L4inherit to L j. So (3) holds.

We saw Vj< Pj = (L j, Vj), so Pj = L jVj. By 43.2 and 47. 1, Vj n L j =V. n A = 1, establishing (2).

The subgroup Lj is called a Levi factor of the parabolic Pj, and Vj is theunipotent radical of Pj.

(47.5) (1) HUj n u,'- = 1, where xo is the element of Wj of maximal length.

(2) CH(UJ) = kerHU,(LJ)(3) If G is finite, Z(G) = 1, and U E Syl p(B) for some prime p, then Vj =

Op(Pj) = F*(Pj) for each J C Tr.

Proof. Let A = HUj and L = L j. By Exercise 10.3, (Ei )xo = EJ soUJ° n B < U"'° n B = 1. Thus (1) holds. Let D = kerA(L). Then D <AnAxO=Hby(1).So[D,Uj]<HnUj=1,andhence D<CH(Uj)=E. I'll prove E < D by induction on the Lie rank 1 of L. Without loss ofgenerality, J = I. (I'll only use (1), not L5.) As U< B, E = CH(U)< H, soE 1 then, by induction on 1, E < kerHU; (Li) for each i ETr.

The jinite simple groups 257

- I

Uw J = U,,-I 5 UJ and hence wU;s; = U;-'WS; G Y. So assumel(wsj) <

l(w). Then by induction on l(w), ws,U;s, C Y. Now ifAW;A is a group then U: G AW; A, so

Thus it suffices to show AW; A is a group. By Exercise 10.3, s y = si for some i E n, and wo is an involution. Let X =

PTO. I'll show X i l Pi = AW; A, which will show AW; A is a group and hence establish the claim. By 43.7, Pi = B U Bs; B = VjT, where T = A W,A. Sim- ilarly X = BwO UBWOs;BWO = BWO UsjBWoSj BWo, and, ass; E X, sjX = X, so X = s;BWOUBWoS~BWO.Let i-2 = C--{-j).ThenUWO = U_;UQ soUwOSj = UjUQ by 30.7, and hence BWoSj BWO = U; BWO. Next B i l s j BWO = (Bwo i l

sjwoB)wo and, as wo # sjwo, BwoBnBsjwoB = @by 43.2. So BnsiBWO = @.Hence B n X = BnBWOsjBWO = BnUjBWO = U;(BnBWO) = UjH = A. ASS^, A c X, T c X, S O X ~ P; = TV, n x = T(V; n x ) c T(B n x ) = TA = T. Thus the claim is at last established.

It's now easy to see that (L j , A, N, Sj) is a Tits system. The only one of the BN-pair axioms which is not evident is BN3. But, for s E Sj, w E Wj,

SAW G (BWB u BSWB) n AWjA

= AwA U AswA

with the equality following from 47.2 and 43.2. Similarly it is clear that L2-L4 inherit to L j . So (3) holds.

W e s a w V J g P J = ( L J , V J ) , s o P J = L J V J . B y 4 3 . 2 a n d 4 7 . 1 , V J n L J = Vj i l A = 1, establishing (2).

The subgroup L is called a Levi factor of the parabolic P j , and Vj is the unipotent radical of Pj.

(47.5) (1) HUj i l Up = 1, where xo is the element of Wj of maximal length.

(2) CH(UJ) = ~ ~ ~ H u ~ ( L J > . (3) If G is finite, Z(G) = 1, and U E Sylp(B) for some prime p, then Vj =

Op(Pj) = F*(Pj) for each J c n.

Proof. Let A = HUj and L = L j. By Exercise 10.3, (Cf)xo = C J, so Up i l B 5 UWo i l B = 1. Thus (1) holds. Let D = kerA(L). Then D 5 A i l AXO = H by (1). So [D, UJ] 5 H n UJ = 1, and hence D 5 CH(Uj) = E. I'll prove E 5 D by induction on the Lie rank 1 of L. Without loss of generality, J = I. (I'll only use (I), not L5.) As U<1 B, E = CH(U): H , so E a B = UH. If 1 > 1 then, by induction on I, E 5 kerHu, (Li) for each i E n.

Thus E < C(Uf) for each x e Li, so L, = (H, UiL') < NG(E). ThereforeG = (Li, B: i e n) < NG(E), so E < D. So take l = 1, and consider theaction of G on BG by conjugation. By BN3, G = B U BwB, so B is transitiveon BG - {B). By 47.2, BwB = BwU, so BG - {B) = BW'U. Hence, as E < Bwand U < C(E), E< ngEG Bg = D. So (2) is established.

Let X = CG(Vj). By 47.4.1, X < Pj. Let Y=BX and for 1 < m E Z defineQ. = {a E *j: h(a) > m). Observe 52,E is an ideal in E+, so V = Vim < B.Indeed by L4, U,, Vm+1 < B for each a E Qm. So also UaVm+14 Y. Now forY E Y fl Wj, UU = Uay so UcgyVm+1 = and hence by 47.1.1, ay = a.Therefore y fixes each member of '/ij. Now if y = sj for some j e J thenVri c C(E)(y). Hence, by Exercise 10.4, y centralizes (E), a contradiction.Therefore Sj fl Y is empty. However, by 43.7, Y = PK for some K C J so asSiflY=1,K=0andY=B.

So X < B. Next, as Lj is a complement to Vj in Pj, XVj = Vj(XVj fl Lj)and, as XV j < B, XVV l Lj < ker,l(Lj) = CH(UJ). Assume the hypothesis of(3). Then H is a p'-group and Ho = XVj fl Lj induces inner automorphismson the p-group Vi, so [Ho, Vj] = 1. Thus Ho = CH(UJVJ) = CH(U) _kerB(G). But, as G = (UG), kerB(G) = Z(G) = 1.

I've shown CG(VJ) < Vi under the hypothesis of (3). Thus to complete theproof of (3) it remains only to observe that, by (2), Vj = Oy(Pi).

(47.6) UG = UwEw(Uw)U

Proof. By 43.7, G = nWEwBwU, so the remark holds.

(47.7) If U < X = (UG fl X) < G then H < NI(X) and HX = Pj for someJ c .r. In particular (P,-{r}: i E .r) is the set of maximal subgroups of Gcontaining U.

Proof. By 47.6, X = (Ud: d E D) for some D C W. So, as H normalizesUw for each w E W, H normalizes X. So HX < G and, as B = HU < HX,HX = Pj for some J c n, by 43.7.1.

Let M be a maximal subgroup of G containing U and X = (UG fl M).Then X < M, so by maximality of M either M = NG (X) or X < G. In the lattercase G = (U, Uw0) = X, a contradiction. So HX < NI(X) = M and henceM = Pj for some J c 7r. By maximality of M, J = 7r - {i) for some i E 7r.

If G is a finite group of Lie type in characteristic p then, as a consequenceof a theorem of Borel and Tits [BT], each p-local subgroup of G is containedin a maximal parabolic. This result can be derived using L1-L5 together with

25 8 Finite simple groups

Thus E ( C(U,") for each x E Li, so Li = (H, u:') 5 Nc(E). Therefore G = (Li, B: i E n) ( NG(E), so E ( D. So take 1 = 1, and consider the action of G on B' by conjugation. By BN3, G = B U BwB, so B is transitive on B' - {B). By 47.2, BwB = BwU, so B' - {B) = B~ ' . Hence, as E ( Bw and U ( C(E), E ( n,,, Bg = D. SO (2) is established.

LetX = CG(VJ).By47.4.1,X 5 PJ.LetY=BXandforl ( m E Zdefine a,,, = {a E $j: h(a) > m). Observe a, is an ideal in C+, so V,,, = VQm B. Indeed by L4, U, V,+l 9 B for each a E M,. So also U, V,+l <I Y. Now for y E Y ~ w ~ , u ; =U,y~~U,yVm+l =U,Vm+landhenceby47.1.1,ay =a . Therefore y fixes each member of $j. Now if y = s j for some j E J then $j G Hence, by Exercise 10.4, y centralizes (C), a contradiction. Therefore S j i l Y is empty. However, by 43.7, Y = PK for some K G J so as S j n Y = l , K = I Z I a n d Y = B .

So X 5 B. Next, as L is a complement to Vj in Pj , XVj = Vj(XVj i l L J) and, as XVj 5 B, XVj i l L 5 kerA(L j) = CH(Uj). Assume the hypothesis of (3). Then H is a p'-group and Ho = XVj i l L induces inner automorphisms on the p-group VJ, so [Ho, Vj] = 1. Thus Ho = CH(UJVj) = CH(U) = kerB(G). But, as G = (u'), kerB(G) = Z(G) = 1.

I've shown CG(Vj) ( Vj under the hypothesis of (3). Thus to complete the proof of (3) it remains only to observe that, by (2), Vj = O,(Pj).

ProoJ By 43.7, G = nWEwBwU, so the remark holds.

(47.7) If U ( X = (u' i l X) ( G then H ( NG(X) and HX = P j for some J G n. In particular (P,+): i E n ) is the set of maximal subgroups of G containing U.

ProoJ By 47.6, X = ( u d : d E D) for some D G W. So, as H normalizes Uw for each w E W, H normalizes X. So HX 5 G and, as B = HU 5 HX, HX = P j for some J G n , by 43.7.1.

Let M be a maximal subgroup of G containing U and X = (u' i l M). Then X 9 M, so by maximality of M either M = NG(X) or X 9 G. In the latter case G = (U, UwO) = X, a contradiction. So HX ( NG(X) = M and hence M = P j for some J G n. By maximality of M, J = n - {i) for some i E n.

If G is a finite group of Lie type in characteristic p then, as a consequence of a theorem of Bore1 and Tits [BT], each p-local subgroup of G is contained in a maximal parabolic. This result can be derived using L1-L5 together with

three extra properties of finite groups of Lie type listed as hypotheses in thenext lemma:

(47.8) Assume the following hold:

(a) G is a finite and U E Sylp(B).(b) Let I' be the set of nontrivial p-subgroups R of G with R = Op(NG(R)).

Assume for each R E F that R = P fl Q for some P, Q E Sylp(NG(R)).(c) U=(Ui:iE7r).

Then the following hold:

(1) (NG(R): R E F) is the set of proper parabolic subgroups of G.(2) If Q is a nontrivial p-subgroup of G then there exists a proper parabolic

M of G with NG(Q) < M and Q :S Op (M).(3) F*(X) = Op(X) for each p-local X of G, if Z(G) = 1.

Proof. Let R E F and M = NG(R). By hypothesis there are P, Q E Sylp(M)with R = P fl Q. By 47.3, U E Sylp(G), so without loss P < U. Similarly,by 47.6, Q < U" for some w E W, U E U, and replacing R by R"-` we mayassume Q < U'. Observe R = U fl U'. For if not, R < U fl U"' fl M <P fl Q = R, a contradiction.

Let A = E+ fl E+w and S2 = E- fl E+w. Then U' = UoUQ with Uo <R, so R = Uo(U fl UQ) c Uo(U fl U"'°) = Uo. That is R = U. I'll showUT < M for each i E n and some vi E U. Then, by (c) and Exercise 8.12,U < M. Also H < N(Uo) < M, so M is a parabolic by 43.7, and (1) holds.

Let i E 7r; it remains to show Ui < M. If not U; R, so i A. Hence -i E(E+)w by Exercise 10.5, so U_i < U'. By 47.5, U fl US, = U*< (U, US! ),where ,k = E+ - {i). So U_i < USi < NG(U fl USi) and of course U_1 <U"' < NG(Uw), so U_i acts on U fl U"' fl Usi = U* fl Uo = U. = R,as A c ,. That is U_i < M.

Observe next that, as P E Sylp(M), P = U fl M. If P < U,, then even P =U,, fl M so, as U_i acts on U,,, U_; acts on P. But then, as P E Sylp(M),U_i < P, contradicting U fl Ulo = 1. So P U*. Consider the parabolic Pi.By 43.7, Bs; is a maximal subgroup of Pi, while, as P U*, Mn Pi BS1.

Hence, as Bsi = HU_iU* < (M n Pi)U*, it follows that Pi = (M n Pi)U,,.Therefore siv E M for some v e U, so Ui" = (U-i)'1° < P

This completes the proof of (1). Arguing by induction on the order of Q, thereexists R e F with Q < R and NG(Q) < NG(R), so (1) implies (2). Finally (2),47.5.3, and 31.16 imply (3).

TheJinite simple groups 259

three extra properties of finite groups of Lie type listed as hypotheses in the next lemma:

(47.8) Assume the following hold:

(a) G is a finite and U E Sylp(B). (b) Let r be the set of nontrivial p-subgroups R of G with R = Op(NG(R)).

Assume for each R E r that R = P i l Q for some P, Q E SylP(NG(R)). (c) U = (Ui: i E n) .

Then the following hold:

(1) (NG(R): R E r ) is the set of proper parabolic subgroups of G. (2) If Q is a nontrivial p-subgroup of G then there exists a proper parabolic

M of G with NG(Q) 5 M and Q 5 Op(M). (3) F * ( X ) = O,(X) for each p-local X of G, if Z(G) = 1.

Proof. Let R E r and M = NG(R). By hypothesis there are P, Q E Sylp(M) with R = P n Q. By 47.3, U E Sylp(G), so without loss P 5 U. Similarly,

by 47.6, Q 5 Uwu for some w E W, u E U, and replacing R by R"-' we may assume Q 5 UW. Observe R = U n UW. For if not, R < U n UW n M 5 P i l Q = R, a contradiction.

Let A = Z+ n Z+w and '2 = Z- f l Z+w . Then Uw = UAUn with UA 5 R, so R = UA(U n U,) E UA(U n UWo) = UA. That is R = UA. I'll show Ulu' 5 M for each i E n and some v, E U. Then, by (c) and Exercise 8.12, U 5 M. Also H 5 N(UA) 5 M, so M is a parabolic by 43.7, and (I) holds.

Let i E n; it remains to show U, 5 M. If not U, -$ R, so i & A. Hence -i E (Z+)w by Exercise 10.5, so U-, 5 Uw. By 47.5, U i l US' = U+<1 (U, US'), where I) = Z+ - {i). So U-, 5 USa 5 NG(U i l US,) and of course U-, 5 Uw 5 NG(Uw), so U-, acts on U n Uw n US' = U+ n UA = UA = R, as A G I). That is U-, < M.

Observe next that, as P E SylP(M), P = U i l M. If P 5 U+ then even P = U+ i l M so, as U-, acts on U+, U-, acts on P. But then, as P E Sylp(M), U-, 5 P , contradicting U n UwO = 1. So P -$ U+. Consider the parabolic P, . By 43.7, BSl is a maximal subgroup of P,, while, as P -$ U+, M i l P, -$ BSi. Hence, as BSl = HU-, U+ 5 (M i l P,)U+, it follows that PI = (M i l P,)U+. Therefore s,v E M for some v E U, so U: = (U-,)SIU 5 P.

This completes the proof of (1). Arguing by induction on the order of Q, there exists R E r with Q 5 R and NG(Q) 5 NG(R), so (1) implies (2). Finally (2), 47.5.3, and 31.16 imply (3).


48 An outline of the Classification TheoremLet . 9 C be the list of finite simple groups appearing in section 47. Section 48provides a brief outline of the Classification Theorem, which asserts:

Classification Theorem. Every finite simple group is isomorphic to a memberof X.

The usual procedure for classifying a collection of objects is to associate toeach member of the collection some family of invariants, prove that each objectis uniquely determined by its invariants, and determine which sets of invariantsactually correspond to objects. The invariants used to classify the finite simplegroups are certain local subgroups of the group, usually the normalizers ofsuitable subgroups of prime order, particularly centralizers of involutions.

A rationale for this approach can be obtained from the Odd Order Theoremof Feit and Thompson and the Brauer-Fowler Theorem (Theorem 45.5).

Odd Order Theorem. (Feit-Thompson [FT]) Groups of odd order aresolvable.

The Odd Order Theorem says that every nonabelians simple group G is of evenorder and hence possesses an involution t. The Brauer-Fowler Theorem saysthere is only a finite number of finite simple groups Go possessing an involutionto with CG0(to) = CG(t). In practice, with a small number of exceptions, Gis the unique simple group with such a centralizer. Even in the exceptionalcases at most three simple groups possess the same centralizer (e.g. L5(2),M24, and He all possess an involution with centralizer L3(2)/Dg). Exercise16.6 illustrates how the isomorphism type of a simple group can be recoveredfrom the isomorphism type of the centralizers of involutions.

So centralizers of involutions provide a set of invariants upon which to basethe Classification Theorem. For various reasons it turns out to be better toenlarge this set of invariants to include suitable normalizers of subgroups ofodd prime order.

To be more precise, define a standard subgroup of a group G for the prime pto be a subgroup H of G such that H = CG(x) for some element x of order p,H has a unique component L, and CH(L) has cyclic Sylow p-groups. Standardsubgroups provide the principal set of invariants for the classification of thefinite simple groups.

However certain small groups either possess no standard subgroup or cannotbe effectively characterized in terms of this invariant. Such groups are charac-terized by other methods. Hence we have our first partition of the simple groupsfor purposes of the classification: the partition into generic groups and small


48 An outline of the Classification Theorem Let K be the list of finite simple groups appearing in section 47. Section 48 provides a brief outline of the Classification Theorem, which asserts:

Classification Theorem. Every finite simple group is isomorphic to a member of K.

The usual procedure for classifying a collection of objects is to associate to each member of the collection some family of invariants, prove that each object is uniquely determined by its invariants, and determine which sets of invariants actually correspond to objects. The invariants used to classify the finite simple groups are certain local subgroups of the group, usually the normalizers of suitable subgroups of prime order, particularly centralizers of involutions.

A rationale for this approach can be obtained from the Odd Order Theorem of Feit and Thompson and the Brauer-Fowler Theorem (Theorem 45.5).

Odd Order Theorem. (Feit-Thompson [FT]) Groups of odd order are solvable.

The Odd Order Theorem says that every nonabelians simple group G is of even order and hence possesses an involution t. The Brauer-Fowler Theorem says there is only a finite number of finite simple groups Go possessing an involution to with CGo(tO) Z CG(t). In practice, with a small number of exceptions, G is the unique simple group with such a centralizer. Even in the exceptional cases at most three simple groups possess the same centralizer (e.g. L5(2), MZ4, and He all possess an involution with centralizer L ~ ( ~ ) / D ; ) . Exercise 16.6 illustrates how the isomorphism type of a simple group can be recovered from the isomorphism type of the centralizers of involutions.

So centralizers of involutions provide a set of invariants upon which to base the Classification-Theorem. For various reasons it turns out to be better to enlarge this set of invariants to include suitable normalizers of subgroups of odd prime order.

To be more precise, define a standard subgroup of a group G for the prime p to be a subgroup H of G such that H = CG(x) for some element x of order p, H has a unique component L, and CH(L) has cyclic Sylow p-groups. Standard subgroups provide the principal set of invariants for the classification of the finite simple groups.

However certain small groups either possess no standard subgroup or cannot be effectively characterized in terms of this invariant. Such groups are characterized by other methods. Hence we have our first partition of the simple groups for purposes of the classification: the partition into generic groups and small

An outline of the Classification Theorem 261

groups. I'll define the appropriate measure of size in a moment; but before thatanother partition.

When possible, we'd like to characterize a simple group G in terms of astandard subgroup for the prime 2. But often G possesses no such subgroup.G is said to be of characteristic p-type if F*(H) = Op(H) for each p-localsubgroup H of G. For example, if G is of Lie type and characteristic p, we sawin 47.8 that G is of characteristic p-type. In particular if G is of characteristic 2-type then it possesses no standard subgroup for the prime 2. Our second partitionof the simple groups is the partition into groups of even and odd characteristic,where by definition G is of even characteristic if G is of characteristic 2-typeand G is of odd characteristic otherwise.

Define the 2-local p-rank of G to be

m2, p(G) = max{m p(H): His a 2-local of G},

and define

e(G) = max{m2,p(G): p odd}.

In a group of Lie type and characteristic 2, e(G) is a good approximation ofthe Lie rank.

A group G of even characteristic is small if e(G) < 2 and generic otherwise.A group of odd characteristic is small if G is 2-disconnected for the prime 2and generic otherwise. Thus we have a four part partition of the finite simplegroups for purposes of the classification.

The proof of the Classification Theorem proceeds by induction on the orderof the simple group to be classified. Thus we consider a minimal counterexample G to the Classification Theorem; that is G is a finite simple groupof minimal order subject to G ¢ X. Define a finite group H to be a .7C-group if every simple section of H is in .7C (a section of H is a factor groupA/B, where B < A < H). Observe that every proper subgroup of our minimalcounterexample is a X -group. This property will be used repeatedly.

If G is a finite simple group with m2(G) < 2, a moment's thought shows Gto be 2-disconnected for the prime 2. If m2(G) > 2 then, by 46.7, neglectingisolated points in the graph -T(ez(G)), G is 2-disconnected for the prime 2precisely when G $ I' , 2(G), where P E Sy12(G). As a matter of fact, if G issimple and m2(G) > 2, it can be shown that G is 2-disconnected for the prime2 if and only if G has a proper 2-generated 2-core. Thus to classify the smallsimple groups of odd characteristic it suffices to prove the following two results:

Theorem 48.1. If G is a nonabelian finite simple group with m2(G) < 2 theneither:

(1) a Sylow 2-group of G is dihedral, semidihedral, or Z2» wr Z2, and GL2(q), L3 (q), U3 (q), q odd, or M11, or

(2) G U3 (4).

An outline of the ClassiJication Theorem 26 1

groups. I'll define the appropriate measure of size in a moment; but before that another partition.

When possible, we'd like to characterize a simple group G in terms of a standard subgroup for the prime 2. But often G possesses no such subgroup. G is said to be of characteristic p-type if F*(H) = O,(H) for each p-local subgroup H of G. For example, if G is of Lie type and characteristic p, we saw in 47.8 that G is of characteristic p-type. In particular if G is of characteristic 2- type then it possesses no standard subgroup for the prime 2. Our second partition of the simple groups is the partition into groups of even and odd characteristic, where by definition G is of even characteristic if G is of characteristic 2-type and G is of odd characteristic otherwise.

Define the 2-local p-rank of G to be

m2,,(G) = max{mp(H): His a 2-local of G),

and define

e(G) = max{m2,,(G): p odd).

In a group of Lie type and characteristic 2, e(G) is a good approximation of the Lie rank.

A group G of even characteristic is small if e(G) 5 2 and generic otherwise. A group of odd characteristic is small if G is z-disconnected for the prime 2 and generic otherwise. Thus we have a four part partition of the finite simple groups for purposes of the classification.

The proof of the Classification Theorem proceeds by induction on the order of the simple group to be classified. Thus we consider a minimal counter example G to the Classification Theorem; that is G is a finite simple group of minimal order subject to G @ X. Define a finite group H to be a X- group if every simple section of H is in X (a section of H is a factor group A/ B, where B 5 A 5 H). Observe that every proper subgroup of our minimal counterexample is a X-group. This property will be used repeatedly.

If G is a finite simple group with m2(G) 5 2, a moment's thought shows G to be z-disconnected for the prime 2. If m2(G) > 2 then, by 46.7, neglecting isolated points in the graph g(8?j(~)), G is 2-disconnected for the prime 2 precisely when G # r:,2(G), where P E Sy12(G). As a matter of fact, if G is simple and m2(G) > 2, it can be shown that G is Zdisconnected for the prime 2 if and only if G has a proper 2-generated 2-core. Thus to classify the small simple groups of odd characteristic it suffices to prove the following two results:

Theorem 48.1. If G is a nonabelian finite simple group with m2(G) 5 2 then either:

(1) a Sylow 2-group of G is dihedral, semidihedral, or Z2,, wr 22, and G "=

L2(4)7 L3(4), U3(4), q odd, or M11, or (2) G 2 U3(4).


Theorem 48.2. Let G be a nonabelian finite simple group with m2(G) > 2and assume G has a proper 2-generated 2-core. Then either G is a group ofLie type of characteristic 2 and Lie rank 1 (i.e. L2(2"), U3(2"), or Sz(2")) orG= J1.

In brief, Theorem 48.1 is proved by using local theory to restrict the subgroupstructure of our minimal counterexample G. At this point there is either enoughinformation available to conclude G E . C or G possesses many TI-sets. TI-setscan be exploited using character theory (along the lines of 35.22 or Frobenius'Theorem) to derive a contradiction or to restrict the structure of G further andshow G E X.

It's interesting to note that character theory plays an important role in theproof of Theorem 48.1, but is rarely used in the remainder of the classification.In essence character theory is used to deal with small groups such as the groupsof Lie rank 1 and some groups of Lie rank 2. The generic groups of higherdimension can be identified using geometric or quasigeometric techniques.

The local theory used in the proof of Theorem 48.1 is of two sorts. Transferand fusion techniques are used to pin down the structure of a Sylow 2-groupof G and the fusion of 2-elements; see for example Exercise 13.2. One suchtool, used sparingly in the proof of Theorem 48.1 but frequently in later stagesof the classification, is Glauberman's Z*-Theorem:

Glauberman Z*-Theorem. [G13]. Let G be a finite group and t an involutionin G such that t is weakly closed in CG(t). Then t* E Z(G*), where G* _G/OT(G).

Recall a subset S of G is weakly closed in a subgroup H of G (with respectto G) if SG fl H = {S}. The proof of the Z* -Theorem uses modular charactertheory and is beyond the scope of this book.

The second kind of local theory used in the proof of Theorem 48.1 involvesan analysis of subgroups of G of odd order using signalizer functor theory orsome variant of that theory. Notice the Odd Order Theorem is one step in theproof of Theorem 48.1, since groups of odd order are of 2-rank 0. When Feitand Thompson proved the Odd Order Theorem signalizer functor theory didnot exist; instead they generated their own techniques, which in time evolvedinto signalizer functor theory. I'll illustrate the signalizer functor approach inthe generic case a little later.

The techniques used to establish Theorem 48.2 are rather different. The proofdepends heavily on the fact that each pair of involutions generate a dihedralgroup. This observation makes possible a number of combinatorial and grouptheoretic arguments of the flavor of sections 45 and 46. Exercises 16.5 and


Theorem 48.2. Let G be a nonabelian finite simple group with rnz(G) > 2 and assume G has a proper 2-generated 2-core. Then either G is a group of Lie type of characteristic 2 and Lie rank 1 (i.e. L2(2"), U3(2"), or Sz(2")) or G Z J 1 .

In brief, Theorem 48.1 is proved by using local theory to restrict the subgroup structure of our minimal counterexample G. At this point there is either enough information available to conclude G E 3% or G possesses many TI-sets. TI-sets can be exploited using character theory (along the lines of 35.22 or Frobenius' Theorem) to derive a contradiction or to restrict the structure of G further and show G E 3%.

It's interesting to note that character theory plays an important role in the proof of Theorem 48.1, but is rarely used in the remainder of the classification. In essence character theory is used to deal with small groups such as the groups of Lie rank 1 and some groups of Lie rank 2. The generic groups of higher dimension can be identified using geometric or quasigeometric techniques.

The local theory used in the proof of Theorem 48.1 is of two sorts. Transfer and fusion techniques are used to pin down the structure of a Sylow 2-group of G and the fusion of 2-elements; see for example Exercise 13.2. One such tool, used sparingly in the proof of Theorem 48.1 but frequently in later stages of the classification, is Glauberman's Z*-Theorem:

Glauberman Z*-Theorem. [GI 31. Let G be a finite group and t an involution in G such that t is weakly closed in CG(t). Then t* E Z(G*), where G* = G/Oz4G).

Recall a subset S of G is weakly closed in a subgroup H of G (with respect to G) if S' n H = {S). The proof of the Z*-Theorem uses modular character theory and is beyond the scope of this book.

The second kind of local theory used in the proof of Theorem 48.1 involves an analysis of subgroups of G of odd order using signalizer functor theory or some variant of that theory. Notice the Odd Order Theorem is one step in the proof of Theorem 48.1, since groups of odd order are of 2-rank 0. When Feit and Thompson proved the Odd Order Theorem signalizer functor theory did not exist; instead they generated their own techniques, which in time evolved into signalizer functor theory. I'll illustrate the signalizer functor approach in the generic case a little later.

The techniques used to establish Theorem 48.2 are rather different. The proof depends heavily on the fact that each pair of involutions generate a dihedral group. This observation makes possible a number of combinatorial and group theoretic arguments of the flavor of sections 45 and 46. Exercises 16.5 and


16.6 illustrate some of these arguments. Indeed Exercise 16.6 establishes avery special case of Theorem 48.2. It would be nice to have the analogue ofTheorem 48.2 for odd primes, but as nothing in particular can be said aboutgroups generated by a pair of elements of odd prime order, a different approachis required.

Let's turn next to the generic groups of odd characteristic. Observe that byExercise 16.1 a generic group G of odd characteristic possesses an involutiont such that O2',E(Cc(t)) :O2'(Cc(t)). I encourage you to retrace the steps inthe proof of this exercise; the proof provides a good illustration of signalizerfunctor theory. Similar arguments reappear in the proof of Theorem 48.1 andin the analysis of the generic groups of even characteristic.

This brings us to the following fundamental property of finite groups:

Bp-Property. Let p be a prime, G a finite group with Op,(G) = 1, and x anelement of order p in G. Then Op',E(CG(x)) =Op'(CG(x))E(CG(x)).

The verification of the Bp-Property is one of the most critical and difficult stepsin the classification. Only the B2-Property is established directly; for odd p,the Bp-Property follows only as a corollary to the Classification by inspectionof the groups in X. (Notice that by Exercise 16.4, to verify the Bp-Propertyit suffices to consider the case where F*(G) is simple.)

Observe next that the B2-Property follows from the following result:

Unbalanced Group Theorem. Let G be a finite group with F*(G) simplewhich is unbalanced for the prime 2. Then F*(G) is a group of Lie type andodd characteristic, A2n+i, L3(4), or He.

For to verify the B2-Property it suffices to assume F*(G) is simple by Exercise16.5. If O2'(Cc(t)) = 1 for each involution t in G, then the B2-Property istrivially satisfied, so we may assume G is unbalanced. Hence, by the Unbal-anced Group Theorem, F*(G) E X. But, by inspection of the local structureof Aut(L) for L E X, if F*(G) E X then G satisfies the B2-Property.

Suppose for the moment that the Unbalanced Group Theorem, and hencealso the B2-conjecture, is established. The B2-conjecture makes possible ma-nipulations which prove:

Component Theorem. (Aschbacher [As 1]) Let G be a finite group withF*(G) simple satisfying the B2-Property and possessing an involution t suchthat 02', E (CG (t)) O2' (CG (t)). The G possesses a standard subgroup for theprime 2.

An outline of the Class$cation Theorem 263

16.6 illustrate some of these arguments. Indeed Exercise 16.6 establishes a very special case of Theorem 48.2. It would be nice to have the analogue of Theorem 48.2 for odd primes, but as nothing in particular can be said about groups generated by a pair of elements of odd prime order, a different approach is required.

Let's turn next to the generic groups of odd characteristic. Observe that by Exercise 16.1 a generic group G of odd characteristic possesses an involution t such that OZ',E(CG(t)) # OZ'(CG(t)). I encourage you to retrace the steps in the proof of this exercise; the proof provides a good illustration of signalizer functor theory. Similar arguments reappear in the proof of Theorem 48.1 and in the analysis of the generic groups of even characteristic.

This brings us to the following fundamental property of finite groups:

Bp-Property. Let p be a prime, G a finite group with Opf(G) = 1, and x an element of order p in G. Then O p f , ~ (CG (x)) =Opt (CG (x))E(CG(X)).

The verification of the Bp-Property is one of the most critical and difficult steps in the classification. Only the Bz-Property is established directly; for odd p, the Bp-Property follows only as a corollary to the Classification by inspection of the groups in X. (Notice that by Exercise 16.4, to verify the Bp-Property it suffices to consider the case where F*(G) is simple.)

Observe next that the Bz-Property follows from the following result:

Unbalanced Group Theorem. Let G be a finite group with F*(G) simple which is unbalanced for the prime 2. Then F*(G) is a group of Lie type and odd characteristic, AZ*+~, L3(4), or He.

For to verify the Bz-Property it suffices to assume F*(G) is simple by Exercise 16.5. If Ozl(CG(t)) = 1 for each involution t in G, then the B2-Property is trivially satisfied, so we may assume G is unbalanced. Hence, by the Unbal- anced Group Theorem, F*(G) E YC. But, by inspection of the local structure of Aut(L) for L E X, if F*(G) E 3% then G satisfies the Bz-Property.

suppose for the moment that the Unbalanced Group Theorem, and hence also the Bz-conjecture, is established. The Bz-conjecture makes possible ma- nipulations which prove:

Component Theorem. (Aschbacher [As 11) Let G be a finite group with F*(G) simple satisfying the B2-Property and possessing an involution t such that 02',E(CG(t)) # OZ'(CG(t)). The G possesses a standard subgroup for the prime 2.

Actually the definition of a standard subgroup has to be relaxed a little to makethe Component Theorem correct as stated above, but the spirit is accurate.Notice that at this stage we have associated to each generic group of oddcharacteristic the desired set of invariants: its collection of standard subgroupsfor the prime 2. It remains to characterize simple groups via these invariants.Thus we must consider:

Standard form problem for (L, r): Determine all finite groups G possessinga standard subgroup H for the prime r with E(H) = L.

If G is our minimal counterexample and H a standard subgroup of G, then His a .X-group, so E(H)/Z(E(H)) E X. As we know the universal coveringgroup of each member of X, we know E(H), and to prove G e X it remainsto treat the standard form problem for each perfect central extension of eachmember of X.

How does one retrieve the structure of G from that of the standard subgroupH? Let L = E(H). Then H/CG(H) < Aut(L), so we have good control overHI CG(H), while as CG(H) has cyclic Sylow p-groups we have good controlof CG(H). Then analysis of fusion of p-elements of H gives us a conjugate Hgof H such that the intersection H fl Hg is large. Results similar to Theorem 48.2allow us to conclude G = (H, Hg). This information can be used to define arepresentation of G on a subgroup geometry along the lines of the constructionin section 3, or to obtain a presentation of G. For example we might representG on a building or show it possesses a BN-pair. If so, the machinery in chapter14 becomes available to identify G as a member of X.

It remains to prove the Unbalanced Group Theorem. By the Odd Order The-orem O2 (CG(t)) is solvable for each involution t in G, so by 46.9 and Exercise13.3, if G is not balanced for the prime 2, also CG(t)* = CG(t)/O2'(CG(t))is not balanced for the prime 2 for some involution t in G. Hence by 31.19there is a component L* of C(t)* and X* < N(L*) such that is notbalanced for the prime 2. By induction on the order of G, L*/Z(L*) is one ofthe groups listed in the conclusion of the Unbalanced Group Theorem. Thispiece of information is critical; together with some deep local theory it can beused to produce a standard subgroup for the prime 2, reducing us to a previouscase.

The groups of odd characteristic have been treated; let us turn next to thegroups of even characteristic. If G is a generic group of even characteristic weseek to produce a standard subgroup for some odd prime p. More precisely pis a prime in the set o-(G) where

or(G) = {p: p odd, m2,p(G) > 31.


Actually the definition of a standard subgroup has to be relaxed a little to make the Component Theorem correct as stated above, but the spirit is accurate. Notice that at this stage we have associated to each generic group of odd characteristic the desired set of invariants: its collection of standard subgroups for the prime 2. It remains to characterize simple groups via these invariants. Thus we must consider:

Standard form problem for (L, r): Determine all finite groups G possessing a standard subgroup H for the prime r with E(H) Z L.

If G is our minimal counterexample and H a standard subgroup of G, then H is a K-group, so E(H)/Z(E(H)) E YC. As we know the universal covering group of each member of 3%, we know E(H), and to prove G E YC it remains to treat the standard form problem for each perfect central extension of each member of x.

How does one retrieve the structure of G from that of the standard subgroup H? Let L = E(H). Then H/CG(H) 5 Aut(L), so we have good control over H/CG(H), while as CG(H) has cyclic Sylow p-groups we have good control of CG(H). Then analysis of fusion of p-elements of H gives us a conjugate Hg of H such that the intersection H fl Hg is large. Results similar to Theorem 48.2 allow us to conclude G = (H, Hg). This information can be used to define a representation of G on a subgroup geometry along the lines of the construction in section 3, or to obtain a presentation of G. For example we might represent G on a building or show it possesses a BN-pair. If so, the machinery in chapter 14 becomes available to identify G as a member of 3%.

It remains to prove the Unbalanced Group Theorem. By the Odd Order The- orem O2,(CG(t)) is solvable for each involution t in G, so by 46.9 and Exercise 13.3, if G is not balanced for the prime 2, also CG(~)* = CG(t)/o2l(C~(t)) is not balanced for the prime 2 for some involution t in G. Hence by 31.19 there is a component L* of C(t)* and X* 5 N(L*) such that Autp(L*) is not balanced for the prime 2. By induction on the order of G, L*/Z(L*) is one of the groups listed in the conclusion of the Unbalanced Group Theorem. This piece of information is critical; together with some deep local theory it can be used to produce a standard subgroup for the prime 2, reducing us to a previous case.

The groups of odd characteristic have been treated; let us turn next to the groups of even characteristic. If G is a generic group of even characteristic we seek to produce a standard subgroup for some odd prime p. More precisely p is a prime in the set a(G) where

a(G) = {p: p odd, rn~,,(G) Z 31.


The actual definition of a(G) is a little more complicated, but again the defini-tion above is in the right spirit. Using signalizer functor theory and other localgroup theoretic techniques one shows:

Theorem 48.3. (Trichotomy Theorem [As 2], [GL]) let G be a minimal counter-example to the Classification Theorem and assume G is generic of even charac-teristic. Then one of the following holds:

(1) G possesses a standard subgroup for some p e a(G).(2) There exists an involution t in G such that F*(CG(t)) is a 2-group of

symplectic type.(3) G is in the uniqueness case.

G is in the uniqueness case if, for each p e a(G), G possesses a stronglyp-embedded maximal 2-local subgroup. Recall that a p-group is of symplectictype if it possesses no noncyclic characteristic abelian subgroups, and thatgroups of symplectic type are described completely in chapter 8.

I've already discussed briefly how one deals with standard subgroups. In case2 of Theorem 48.3, the structure of F*(C(t)) = Q is determined from chapter8, as is Aut(Q) (cf. Exercise 8.5). With this information and some work onecan recover C(t) and then proceed as though C(t) were a standard subgroup.

The arguments used to deal with the uniqueness case and the small groupsof even characteristic are quite similar. They involve factoring 2-locals as theproduct of normalizers of certain subgroups of a Sylow 2-group of the local.The Thompson Factorization, discussed in section 32, is the prototype of suchfactorizations. The proof of the Thompson Normal p-Complement Theorem(39.5) and of the Solvable Signalizer Functor Theorem give some indicationof how such factorizations can be used.

Remarks. See Gorenstein's series of books [Gor 2, Gor 3] for a more detailedoutline of the proof of the Classification Theorem and a more complete discus-sion of the sporadic simple groups. In particular [Gor 2, Gor 3] contain explicitreferences to the articles which, taken together, supply a proof of Theorems48.1 and 48.2 and the Unbalanced Group Theorem.

Carter [Ca] and Steinberg [St] are good places to learn about groups of Lietype.

Exercises for chapter 161. Let G be a finite group with P E Sy12(G) and G = 02(G). Assume G =

T , 2(G), m2(G) > 2, and 02',E(CG(t)) = O2(CG(t)) for each involutiont in G. Assume O2,(G) = 1. Prove G is of characteristic 2-type; that isF*(H) = 02(H) for each 2-local subgroup H of G.

An outline of the Class$cation Theorem

The actual definition of n(G) is a little more complicated, but again the definition above is in the right spirit. Using signalizer functor theory and other local group theoretic techniques one shows:

Theorem 48.3. (Trichotomy Theorem [As 21, [GL]) let G be a minimal counterexample to the Classification Theorem and assume G is generic of even characteristic. Then one of the following holds:

(1) G possesses a standard subgroup for some p E n(G). (2) There exists an involution t in G such that F*(Cc(t)) is a 2-group of

symplectic type. (3) G is in the uniqueness case.

G is in the uniqueness case if, for each p E n(G), G possesses a strongly p-embedded maximal 2-local subgroup. Recall that a p-group is of symplectic type if it possesses no noncyclic characteristic abelian subgroups, and that groups of symplectic type are described completely in chapter 8.

I've already discussed briefly how one deals with standard subgroups. In case 2 of Theorem 48.3, the structure of F*(C(t)) = Q is determined from chapter 8, as is Aut(Q) (cf. Exercise 8.5). With this information and some work one can recover C(t) and then proceed as though C(t) were a standard subgroup.

The arguments used to deal with the uniqueness case and the small groups of even characteristic are quite similar. They involve factoring 2-locals as the product of normalizers of certain subgroups of a Sylow 2-group of the local. The Thompson Factorization, discussed in section 32, is the prototype of such factorizations. The proof of the Thompson Normal p-Complement Theorem (39.5) and of the Solvable Signalizer Functor Theorem give some indication of how such factorizations can be used.

Remarks. See Gorenstein's series of books [Gor 2, Gor 31 for a more detailed outline of the proof of the Classification Theorem and a more complete discussion of the sporadic simple groups. In particular [Gor 2, Gor 31 contain explicit references to the articles which, taken together, supply a proof of Theorems 48.1 and 48.2 and the Unbalanced Group Theorem.

Carter [Ca] and Steinberg [St] are good places to learn about groups of Lie

type.

Exercises for chapter 16 1. Let G be a finite group with P E Sy12(G) and G = o~(G). Assume G =

rOp,2(~), m;?(G) > 2, and 0 2 1 , ~ ( C ~ ( t ) ) = O2(CG(t)) for each involution t in G. Assume Or(G) = 1. Prove G is of characteristic 2-type; that is F*(H) = 02(H) for each 2-local subgroup H of G.

2. Let G = A,, be the alternating group on n > 5 letters and A = Aut(G).Prove:(1) A=S,, if n#6.(2) A/G=E4ifn=6.

3. Let (Gi: 1 < i < n) be a family of subgroups of a group G such that(1) Each element of G can be written uniquely as a product

91 ... gn, with gi E Gi, and

(2) GmGm+1 ... Gn < G for each m, 1 < m <n.Then, for each permutation a of 11, ... , n), each element of G can be writtenuniquely as a product gia ... gnu, gi c G1.

4. Let G be a finite group and p a prime. Assume OP,(G) = 1. AssumeAutH(L) satisfies the BP-Property for each component L of G and eachsubgroup H of G with L < H < NG(L). Finally assume each componentof G satisfies the Schreier conjecture. Prove G satisfies the BP-Property.

5. Let H be strongly 2-embedded in G and represent G by right multiplicationon X = G/H. Let I be the set of involutions of G, t E I fl H, u c I - H,D=HfH",m=IIfHI,andJ={dED:du=d-1}.Prove:(1) G is transitive on I.(2) H is transitive on I fl H.(3) D is of odd order.(4) uJ=uDfI.(5) CG (j) is of odd order for each j c J#.(6) Distinct involutions in uD are in distinct cosets of CG(t).(7) For each x, y E X with x # y there are exactly m involutions in G with

cycle (x, y).(8) D is transitive on I fl H.(Hint: For (1) observe each member of I is conjugate to some s c I - H;then use 45.2 to prove s is conjugate to tin (s, t). Use 46.4 in (2) and (3)and 45.2 in (4). In (5) set Y = CG (j) (u) and observe that if t E CG (j) thenH fl Y is strongly embedded in Y; now appeal to (1) for a contradiction.Derive (6) from (5). To prove (7), let 0 be the set of triples (i, x, y) withi E I and (x, y) a cycle in i on G/H. Count IQ1 in two ways, using (6) toconclude there are at most m involutions with cycle (x, y).)

6. Let G be a finite group with noncyclic Sylow 2-group T. Assume CG(t)is an elementary abelian 2-group for each involution t of G and T is notnormal in G. Let q = IT 1, H = NG(T ), X = G/H, u an involution inG - H, and D = H fl H". Let F = GF(q) and Y = F U too). RegardL2(q) = G* as the group of all permutations of Y of the form

0(a,b,c,d):y i-+ (ay+b)/(cy+d) a,b,c,d c F,ad - be # 0


2. Let G = A, be the alternating group on n L 5 letters and A = Aut(G). Prove: (1) A = S, if n # 6. (2) AIG E E4 i fn = 6.

3. Let (Gi: 1 I i 5 n) be a family of subgroups of a group G such that (1) Each element of G can be written uniquely as a product

gi ...g,, withgi E Gi, and

(2) GrnGrn+'.. . G,s G for each rn, 1 5 rn 5 n. Then, for each permutation a of 11, . . . , n}, each element of G can be written uniquely as a product gl, . . . gnu, gi E Gi.

4. Let G be a finite group and p a prime. Assume Oe,(G) = 1. Assume AutH(L) satisfies the Be-Property for each component L of G and each subgroup H of G with L 5 H 5 NG(L). Finally assume each component of G satisfies the Schreier conjecture. Prove G satisfies the B,-Property.

5. Let H be strongly 2-embedded in G and represent G by right multiplication on X = GIH. Let I be the set of involutions of G, t E I n H, u E I - H, D = H n HU,rn = II n HI,and J = {d E D:dU =d-'].Prove: (1) G is transitive on I. (2) H is transitive on I n H. (3) D is of odd order. (4) U J = U D ~ I . (5) CG(j) is of odd order for each j E J'. (6) Distinct involutions in uD are in distinct cosets of CG(t). (7) For each x, y E X with x # y there are exactly rn involutions in G with

cycle (x, Y). (8) D is transitive on I n H. (Hint: For (1) observe each member of I is conjugate to some s E I - H; then use 45.2 to prove s is conjugate to t in (s , t). Use 46.4 in (2) and (3) and 45.2 in (4). In (5) set Y = CG ( j ) ( u ) and observe that if t E CG ( j ) then H n Y is strongly embedded in Y; now appeal to (1) for a contradiction. Derive (6) from (5). To prove (7), let S2 be the set of triples (i, x, y) with i E I and (x, y) a cycle in i on GIH. Count 1 S21 in two ways, using (6) to conclude there are at most rn involutions with cycle (x, y).)

6. Let G be a finite group with noncyclic Sylow 2-group T. Assume CG(t) is an elementary abelian 2-group for each involution t of G and T is not normal in G. Let q = [TI, H = NG(T), X = GIH, u an involution in G - H, and D = H n H U . Let F = GF(q) and Y = F U {oo]. Regard Lz(q) = G* as the group of all permutations of Y of the form

@(a, b, c, d): y H (ay + b)/(cy + d) a , b, c, d E F , a d - bc # 0

An outline of the Classification Theorem

as in Exercise 4.10. Let

H*={4(a,b,1,1):aEF#,bEF).

267

Prove:(1) T is elementary abelian.(2) H is strongly 2-embedded in G.(3) D is inverted by u, I H I = q (q - 1), and D is a complement to T in H.(4) NG(D) = D(u) and {H, Hu} is the fixed point set of D on X.(5) G is 3-transitive on X with only the identity fixing 3 or more points. T

is regular on X - {H}.(6) There is an isomorphism n: H -+ H* and a bijection a: X -+ Y such

that Ha = oo, (Hu)a = 0, and (xa)h7r = (xh)a for all x E X andh E H.

(7) There exists v c uD with (act-1)v = a-la-1 for all a E F#.(8) n extends to isomorphism of G and G* = L2 (q) with vn = 4(0, 1, 1, 0).(Hints: Use Exercise 16.5 in the proof of (3) and (4). In (4) show NG (D) _E(u) where E = 02'(NG(D)) and E(u)/D is regular on the fixed point set0 of D on X. Let r be the set of triples (i, x, y) such that i c uE and (x, y)is a cycle of i on A. Count r in two ways to get I 0 I = 2. Use (4) andPhillip Hall's Theorem, 18.5, to get T regular on X - {H}. Then complete(5) using 15.11. For (6) use (5) and the observation that, as D is regular onT#, D = EndGF(2)D(T)# and EndGF(2)D(T) = F.)

An outline of the ClassiJication Theorem

as in Exercise 4.10. Let

H* = {#(a, b, 1 , l ) : a E F', b E F) .

Prove: (1) T is elementary abelian. (2) H is strongly 2-embedded in G. (3) Disinvertedbyu,~H~=q(q-1),andDisacomplementtoTinH. (4) NG(D) = D(u) and {H, Hu} is the fixed point set of D on X. (5) G is 3-transitive on X with only the identity fixing 3 or more points. T

is regular on X - {HI. (6) There is an isomorphism n : H + H* and a bijection a: X + Y such

that H a = oo, (Hu)a = 0, and (xa)hn = (xh)a for all x E X and h E H.

(7) There exists v E uD with (aa-')v = a-la-' for all a E F'. (8) n extends to isomorphism of G and G* = L2(q) with vn = #(0,1,1,0). (Hints: Use Exercise 16.5 in the proof of (3) and (4). In (4) show NG(D) = E(u) where E = 02t(NG(D)) and E(u)/D is regular on the fixed point set A of D on X. Let r be the set of triples (i, x, y) such that i E uE and (x, y) is a cycle of i on A. Count r in two ways to get lAl = 2. Use (4) and Phillip Hall's Theorem, 18.5, to get T regular on X - {HI. Then complete (5) using 15.1 1. For (6) use (5) and the observation that, as D is regular on T', D = ~ n d ~ ~ ( 2 ) D(T)' and EndGF(2)D(T) = F.)

Appendix

Solutions to selected exercisesChapter 3, Exercise 5. First, as a is of order n and (a) is transitive on A ={Gi : 1 < i < n} of order n, (a) is regular on 0, so renumbering if necessarywe may take Gia=Gi+1, with the subscripts read modulo n. Thus a`-1:G1 Gi is an isomorphism.

Next, by definition of "central product" in 11.1,

G

n

= fi1Yi:YiEGij

with yi y, = y j yi for i $ j, so fli yi is independent of the order of the factors.Now as ai-1: G1 Gi is an isomorphism,

n

(*) G = lFIxjai_1:XiEG1I.i=1

Define 7r: G1 G by x7r = fli xac-1. Then

x7ra = (flxat_1) a = FI XU'= X7r,i i

as the product is independent of the order of its factors and indices are readmodulo n. Thus G17r < CG(a). Similarly

(xy)7r = fl(xy)a1-1= fxai-1yai-1 = T7xai-1 f yai-1 =x7r y7r,i i

ii

so 7r is a homomorphism. Further, if x E ker(7r) then 1= fli xai-1, so

1

x = (flxa') E G1 r1 G2G3 ... Gn < Z(G1),i>1

so ker(7r) < Z(G1).Claim CG(a) = G17rZ, where Z = CZ(G)(a). We just saw G17r < CG(a), so

G17rZ < CG(a). Suppose g E CG(a). By (*), g = fli xiai-1 for some xi E G1.Then

x1u-1 = flxia`-1 =g =gai+1 = fxiai+' =x-iu.ii i

Appendix

Solutions to selected exercises Chapter 3, Exercise 5. First, as a is of order n and (a) is transitive on A = {Gi: 1 ( i ( n} of order n, (a) is regular on A, so renumbering if necessary we may take Gia = Gi+1, with the subscripts read modulo n. Thus a'-': G1 Gi is an isomorphism.

Next, by definition of "central product" in 11.1,

with yi yj = y, yi for i # j, so ni yi is independent of the order of the factors. Now as a'-': G1 + Gi is an isomorphism,

Define n: G1 + G by x n = ni xui-'. Then

as the product is independent of the order of its factors and indices are read modulo n. Thus G 1 n ( CG (a). Similarly

so n is a homomorphism. Further, if x E ker(n) then 1 = ni xai--', so

so ker(n) ( Z(G1). Claim CG(a) = GlnZ , where Z = Cz(G)(ff). We just saw G i n ( CG(a), SO

G l n Z I CG(ff). Suppose g E CG(ff). BY (*), g = ni xiai-' for some xi E GI. Then

270 Appendix

for each j, where uj = fi 0 -i xia`+j E G2G3 Gn, so

xlxI = u u-1 E G1 fl G2G3 Gn < Z(G1)

and hence xj=xlzj for some zjE Z(G1). Therefore g =xiJrz, where z=fi zjaj-1 E Z(G). As g, x17r E CG(a), also z =g(xiJr)-1 E CZ(G)(a), com-pleting the proof that CG((X) = G1nCZ(G)(01).

Let K = Glrr and Z = CZ(G)(a). As 7r: G1 -+ K is a surjective homomor-phism and G1 is perfect, K is perfect and as Z is abelian Z(1) = 1. Thus

CG((X)(1) = (KZ)(1) = K(1)Z(1) = K.

Thus it remains to show that NAut(G)(G1) f1 CAut(G)(G1). LetL = G1 and M = CG(L). As G2 ... Gn < M, G = LM. Let 0 E NAnt(G)(L) f1CAut(G)(K). Then ,B acts on M. For x E L, y =x-1 x7r E M, so as ,B acts onM, y-1 yfi E M. Now

y-1 yN=x-17r x x-1,B x7rf=(x .x-1P)x" EL

as ,B centralizes xir E K and L < G. Thus (x x-1,8)x'7 = y-1 y,B EL flM = Z(G), so x x-1,B E Z(L), so [,B, L, L] =1. Hence [,B, L] 1 by 8.9,so indeed ,B E CAut(G)(G1)

Chapter 4, Exercise 7. (1) Let f (x, y) _ 1:j 'j ai,1xi y3 E V = f [x, y] andg E G. By definition

.f (g7r) = E ai,i (xg)` (yg)j ,i,i

so as xg = ax + by and yg = cx + dy for some a, b, c, d E F, we have f (g7r) E

V. It is easy to check that gir preserves addition and scalar multiplication,so g7r E EndF(V). For h E G,

.f (gh)ir _ .f (x(gh), y(gh)) = .f ((xg)h, (yg)h) _ .f (xg, yg)(h7r) _ (.f (g7r))(h7r),

so (gh)ir = girhir, and hence 7r: G -+ GL(V) is a representation.(2) Observe Vn has basis Bn = {xi yn-i: 0 < i < n}. Further, (xi yn-i)(gir) _

(ax + by)i (cx + dy)n-i with

(ax + by)k(k)

= ()a1xi(by)k_J E Vk,

ij

so (xiyn-i)(gir) E Vn. Thus G acts on V, of dimension n + 1.(3) For i E Z, let i be the residue of i modulo p; that is 0 r}.

270 Appendix

for each j, where uj = ni+-j xiai+j E G2G3 . . . Gn, SO

and hence xj = xlzj for some z j E Z(G1). Therefore g =xlnz , where z = nj zjaj-l E Z(G). As g, x ln E Cc(a) , also z =g(xln)- l E Cz(c)(a) , completing the proof that CG(a) = G l n C ~ ( ~ ) ( a ) .

Let K = G l n and Z = Cz(G)(a). As n: G1 + K is a surjective homomorphism and G1 is perfect, K is perfect and as Z is abelian z(') = 1. Thus

Thus it remains to show that NAut(G)(Gl) n C A ~ ~ ( G ) ( K ) 4 C A ~ ~ ( G ) ( G ~ ) . Let L = G I and M = CG(L). AS G 2 . . . G,, 5 M , G = LM. Let B E NAnqc)(L) n CAut(G)(K). Then p acts on M. For x E L , y =x-' . x n E M , so as p acts on M, y-l . yp E M . Now

as B centralizes x n E K and L 9 G. Thus ( x . x - ' ~ ) ~ ~ = y-l . yp E L n M = Z(G), so x . x-'B E Z(L) , so [B, L , L ] = 1. Hence [ p , L] = 1 by 8.9, so indeed p E CAut(~)(G1).

Chapter 4, Exercise 7. ( 1 ) Let f ( x , y) = Ci,j ai,jxiyj E V = f [ x , y] and g E G. By definition

soasxg = ax+byandyg = ex+dyforsomea, b , c ,d E F,wehave f ( g n ) ~ V . It is easy to check that gn preserves addition and scalar multiplication, so gn E EndF(V). For h E G ,

so (gh)n = gn hn , and hence n: G + GL(V) is a representation. . .

(2) Observe V, has basis B, = {xlyn-" : 0 5 i _( n). Further, (xiyn- ' ) (pr) = (ax + by)' (ex + dy),-' with

so (xiyn-')(gn) E V,. Thus G acts on V, of dimension n + 1. (3) For i E Z, let 7 be the residue of i modulo p; that is 0 5 i r ] .

Appendix 271

Then 0 U Vn as xPyn_P

E U but xP-1 yn-P+l U. Claim U is G-invariant,so that G is not irreducible on V,,. It suffices to show (x' yn-i )g7r E U for all iwith 1 = s < r. To do so, we will show that all monomials in x1 g7r and yj g7rare of the form xt yi_t with t < J. Hence as n - i = r - s, all monomials in(x' yn-' )g7r = x i g7r yn-i g7r are of the form Xt yn-t with t < s + (r - s) = r.

Let j = v, so that j = up + v and xig7r = (xng7r)Px"g7r. As (f + h)P =fP + hP for all f, g E V, all monomials in (xng7r)P are of the formxPtyP(j-t)so

we can assume j = v. Then

_ (xgn)' = (ax + by)' = k1 (i)()k(by)J_k,=O

k)

establishing the claim.Finally assume n < p; it remains to show G is irreducible on Vn. By hy-

pothesis, X = {x, y} is a basis for U. Identify g E G with its matrix MX(g) anddefine

ga =

Then

(0a),

ha=(1a 10

/

S={ga:aEF} = F=T={ha:aEF}.

Let Mi=(yjxn-j:0< j <i) for -1 1.

Proof. xha = x and yha = ax + y, so

[y'xn-j, ha] _ (y' x'-' )ha - n-i

j-1' (I)yai_kXn_k ajyj 1xn-j+1 modMj_2

k=O

andifa0O jthen aj00.

Corollary. If i > 0 and w E M, - M_1 then [w, T] = Mi -1.

Proof. The proof is by induction on i. If i = 0 then w E (x n) < Cv (T),so [w, T] = 0 = M-1. Assume the result holds at i -1. By the Lemma, [w, h] EMi-1 - Mi-2 for some h E T, so by the induction hypothesis, Mi_2 =[w, h, T] < [w, T], so [w, T] = Mi_1 as dim(Mi-1/Mi-2) =1.

Appendix 27 1

Then 0 # U # Vn as xpyn-p E U but X P - l yn-p+' 6 U . Claim U is G-invariant, so that G is not irreducible on Vn. It suffices to show (xiyn-')gn E U for all i with i = s 5 r . To do so, we will show that all monomials in x j g n and y j g n - are of the form xtyj-' with i 5 7. Hence as n - i = r - s , all monomials in (,i yn-i )gn =xignyn- 'gn are of the form xtyn-' with t 5 s + ( r - s ) = r .

Let J= v, so that j = u p + v and x j g n = (xUgn)Pxugn. As (f + h)P = f P + hP for all f, g E V , all monomials in (xugn)P are of the form xptyp(j-'), so we can assume j = v. Then

i x j g n = ( x g n ) j = (ax + by)j = c (L) (ax)*(by)j-*,

k=O

establishing the claim. Finally assume n < p; it remains to show G is irreducible on Vn. By hy-

pothesis, X = { x , y ] is a basis for U . Identify g E G with its matrix Mx(g) and define

Then

Let Mi = ( y j x n - j : O i j i i ) for -1 ( i i n , so that Mn= Vn, Mo= ( x n ) , and MW1 =O.

Lemma. T acts on Mi for all i and [Mi/Mi-2, TI = Mi-l/Mi-2 for i > 1.

Prooj: xh, = x and yh, = ax + y, so

Corollary. If i > O and w E Mi - Mi-1 then [w, TI =Mi-1.

Prooj: The proof is by induction on i . If i = O then w E ( x n ) 5 C"(T) , so [w , TI = 0 = M - l . Assume the result holds at i - 1. By the Lemma, [w , h ] E Mipl - Mi-2 for some h E T , so by the induction hypothesis, Mi-2 = [ w , h , TI 5 [ w , T I , SO [w , TI = Mi-i as dim(Mi-l/Mi-2) = 1.

272 Appendix

Now SL(U) = H = (T, S) and we will show H is irreducible on Vn. Let0 W be an FH-submodule of Vn. By the Corollary and symmetry betweenT and S, yn E [W, S] or W = (yn), so in either case yn E W. Hence by theCorollary, Mn-1 = [W, T], so Vn = (yn, Mn-1) < W.

(4) Supposeg c ker(nn).Thenxn = xng7r = (xg7r)n,soxg7r = Xx forsomenth root of unity A. Similarly yg r = µy. Then xyn-1 = (xyn-1)g7r = ),µn-1

xyn 1, so It.

Chapter 4, Exercise 10. (1) and (2): First observe that if I is the identitymatrix in G then z'(I) = z for all z c 17, so 0(I) =1 is the identity of themonoid S of all functions from r into F.

Next, if A = (at,1) and B = (b;,1) are in G then for z c F with al,2z+a2,2 0:

z(P(A)(P(B) =Ca1,1z + a2,11

(P(B)(a 1,2Z + a2,2 /J)

(a1,1z + a2,1)bi,I + (al,2Z + a2,2)b2,1

(al,iz + a2,1)b1,2 + (al,2Z + a2,2)b2,2

- (a1,ib1,1 +a1,2b2,1)z +a2,ib1,1 +a2,2b2,1

(ai,1bi,2 + a1,2b2,2)z + a2,1b1,2 + a2,2b2,2

If a1,2z + a2,2 = 0 then zo(A) = oo, so

zO(A)O(B)= oo0(B)=bi,i/b1,2

On the otherhand as ai,2z + a2,2 = 0 and det(A) 0, a1,2 0, SO Z = -a2,2/ai,2and

z.P(AB)a2,1b1,i - ai,1a2,2b1,1/a1,2 - bi,i(ai,2a2,i - a1,1a2,2) =b1,1/b1,2.

a2,ib1,2 - ai,ia2,2b1,2/a1,2-

b1,2(a1,2a2,1 - a1,ia2,2)

Finally, ooo(A)O(B) _ (a1,1/a1,2)0(B) and if a1,2 0 then

b1,1a1,1/a1,2 + b2,1 _ b1,1a1,1 + b2,1a1,2(a1,1/a1,2)(P(B) =

b1,2a1,1/a1,2 + b2,2 b12a1,1 + b2,2a1,2 = oo(P(AB).

On the other hand if a1,2 = 0 then a1,1 :0 and a1, i /ai,2 = oo, so

bi,iai,i +b2,1a1,2oo(P(A)(P(B) _ oo(P(B) = b1,1/b1,2 -

b1 za1 1 + b2,2a1,2= oo-P(AB).

Thus we have verified that 0: G -+ S is a monoid homomorphism. Henceas G is a group and 0(I) =1, G* _ O(G) is a subgroup of S and 0: G -+ G* isa surjective group homomorphism. As G* is a subgroup of S, each O(A) E G*is invertible in S and hence (1) holds. Further to complete the proof of (2)it remains to show ker(o) is the group of scalar matrices. So let A E ker(o).

272 Appendix

Now SL(U) = H = ( T , S ) and we will show H is irreducible on Vn. Let 0 # W be an FH-submodule of Vn. By the Corollary and symmetry between T and S, yn E [W, S] or W = ( y n ) , so in either case yn E W. Hence by the Corollary, Mn-1 = [ W , TI, so Vn = (yn , MnP1) < W .

(4) Suppose g E ker(nn). Then xn = xngn = (xgn)", so xgn = hx for some nth root of unity h. Similarly ygn = py. Then xyn-' = ( x ~ - ' ) ~ n = hpn-' xYn-l, SO h = pl-" = p.

Chapter 4, Exercise 10. (1) and (2): First observe that if I is the identity matrix in G then zq5(I) = z for all z E r, so @ ( I ) = 1 is the identity of the monoid S of all functions from r into r.

Next, if A = (ai, j ) and B = (b;, j ) are in G then for z E F with al,2z+a2,2 # 0:

If al,2z + a2,2 = 0 then zq5(A) = m, so

On the other hand as al.2z + a2,2 = 0 and det(A) # 0, a1,2 # 0, so z = -a2,2/a1,2 and

Finally, mq5(A)q5(B) = (a131/al,2)q5(B) and if al,2 # 0 then

On the other hand if al-2 = 0 then al,l # 0 and al , 1 /al ,2 = m, so

Thus we have verified that q5: G + S is a monoid homomorphism. Hence as G is a group and @ ( I ) = 1, G* = q5(G) is a subgroup of S and q5: G + G* is a surjective group homomorphism. As G* is a subgroup of S, each @(A) E G* is invertible in S and hence (1) holds. Further to complete the proof of (2) it remains to show ker(q5) is the group of scalar matrices. So let A E ker(q5).

Appendix 273

Thus zo(A) = z for all z E F. In particular oo = ooo(A) = a1,1 /ai,2, so a1,2 =O a1,1. Also 0=O (A)=a2,1/a2,2, so a2,1 =O a2,2. Finally

a1,1 +a2,11 1/a2,2,

a1,2 + a2,2

so a 1,1 = a2,2 and hence A = a 1,11 is scalar.(3) By construction a: Q -+ F is a bijection.Let H be the stabilizer in G of Fx1 E Q. By 13.5, G is 2-transitive on S2, so

G is primitive on Q by 15.14. Hence H is maximal in G by 5.19, so G = (H, t),where t c G interchanges x1 and x2. Thus to show (t)g)a = for each

E G, it suffices to show (t)g)a = (coa)0(g) for each g c H and g = t.Each h E H is the product of a scalar matrix and a matrix

_ (a 0g b 1

aEF#, bEF,

so it suffices to show (t)g)a = (wa)o(g). But 0(g): z v--> az + b, so g fixes Fx1and 0(g) fixes oo = (Fx1)a. Further, for co = F(Xx1 + x2), ,X E F,

(wg)a = F((a.1. + b)x1 + x2)a = a), + b = )fi(g) _ (wa)o(g)

Next, t interchanges x1 and x2, so t has cycles (Fx1, Fx2) and (F(ax1 +x2),F(a.-1x1 + x2)), a. E F#, on Q. Then 0(t): z H 1/z, so 0(t) has cycles (oo, 0)and (X, A 1) on F. Hence as (Fx1)a = oo, (Fx2)a = 0, and F(.lx1 + x2)athe proof is complete.

Chapter 5, Exercise 6. (1) Induct on in. By Jordan's Theorem, 15.17, G is2-transitive on X, so the result holds when in = 1. Again by Jordan, Gx isprimitive on X - {x} = X' for X E Y. So as Y= Y - {x} is of order in - 1with Gx,y' = Gy primitive on X' - Y'= X - Y, Gx is in-transitive on X' byinduction on in. Thus G is (m + 1)-transitive on X by 15.12.1.

(2) Let t be a transposition or cycle of length 3 in G and set Y = Fix(t).Then t E Gy and (t) is transitive on X - Y. Thus Gy is primitive on X - Y asIX - YI is prime. Therefore G is (n - 2)-transitive on X by (1), so G containsthe alternating group by 15.12.4.

(3) Let {x} = M(a) rl M(b), y =xa, z =xb, and A = {x, y, z}. Now

Fix(a) rl Fix(b) = Fix((a, b)) C Fix([a, b]).

If V E M(a) - {x, y} then v, va-1 E M(a) - {x} C Fix(b), so

v[a, b] = va-lb-lab = va-lab = vb = v.

Similarly if v c M(b) - {x, z} then v, vb-1 E Fix(a), so

v[a, b] = va-lb-lab = vb-lab = vb-1b = v.

Appendix 273

Thus zq5(A) = z for all z E r. In particular m = mq5(A) = a1,1/a1,2r SO al,2 = 0 # al,l. Also 0 = Oq5(A) = a2,1/a2,2, so a2.1 = 0 # a2,2. Finally

so al.1 = a 2 , ~ and hence A = al,l I is scalar. (3) By construction a: C2 + r is a bijection. Let H be the stabilizer in G of Fxl E C2. By 13.5, G is Ztransitive on C2, so

GisprimitiveonC2 by 15.14.Hence H ismaximalinG by5.19,soG = (H, t), where t E G interchanges xl and x2. Thus to show (wg)a = (wa)$(g) for each g E c, it suffices to show (wg)a = (wa)@(g) for each g E H and g = t.

Each h E H is the product of a scalar matrix and a matrix

so it suffices to show (wg)a = (wa)@(g). But q5(g): z H az + b, so g fixes Fxl and q5(g) fixes m = (Fx1)a. Further, for w = F(hxl + x2), h E F ,

Next, t interchanges xl and x2, SO t has cycles (Fxl, Fx2) and (F(hxl + x2), ~ ( h - l x l + x.~)), h E F', on C2. Then q5(t): z H l/z, so @(t) has cycles ( m , 0) and (h, h-') on r. Hence as (Fx1)a = m , (Fx2)a = 0, and F(hxl + x2)a =A, the proof is complete.

Chapter 5, Exercise 6. (1) Induct on m. By Jordan's Theorem, 15.17, G is 2-transitive on X, so the result holds when m = 1. Again by Jordan, G, is primitive on X - {x) = X' for x E Y. So as Y' = Y - {x) is of order m - 1 with G,,yf = G primitive on X' - Y' = X - Y, G, is m-transitive on X' by induction on m. Thus G is (m + 1)-transitive on X by 15.12.1.

(2) Let t be a transposition or cycle of length 3 in G and set Y =Fix(t). Then t E G and (t) is transitive on X - Y. Thus G is primitive on X - Y as IX - Y I is prime. Therefore G is (n - 2)-transitive on X by (I), so G contains the alternating group by 15.12.4.

(3) Let {x) = M(a) n M(b), y =xu, z =xb, and A = {x, y, z). Now

Fix(a) n Fix(b) = Fix((a, b)) 5 Fix([a, b]).

If v E M(a) - {x, y) then v, va-' E M(a) - {x) 5 Fix(b), so

Similarly if v E M(b) - {x, z) then v, vb-' E Fix(a), so

274 Appendix

Thus X - A C Fix([a, b]). Next xb-1 E Mov(b) - {x} C Fix(a), so

Y[a, b] =Ya-lb-lab =xb-lab =xb-lb =x.

Similarly xa-1 E Fix(b), so

x [a, b] =xa-lb-lab =xa~lab =xb = z.

Finally z =xb E Mov(b) - {x} c Fix(a) and y =xa E Fix(b), so

z[a, b] = za-lb-lab = zb-lab =xab = yb = y.

Thus (y, x, z) is a cycle of length 3 in [a, b], completing the proof of (3).(4) Assume G does not contain the alternating group on X and I Y I > n/2; we

must derive a contradiction. As I Y I > n /2, F = X - Y has order less than I Y I,so by minimality of IY1, Gr # 1. Pick a E G,, y E M(a), and set A = Y - {y}.Then I A I < IYI, so by minimality of IYI, Go # 1. Pick b E G. As Gy =1,y E M(b), while

M(a) fl M(b) c (X - F) n (X - A) = X - (I' U A) = {y},

so M(a) fl M(b) = (y). Thus [a, b] is a 3-cycle by (3), so (2) supplies a con-tradiction, establishing (4).

(5) Pick Y as in (4) and let H = Sym(X)y. Then I H I m!, where m = I X -Y Iand by (4), m > [(n + 1)/2]. Now H fl G = Gy = 1, so

ISym(X)1 > IHGI = IHI IGI = m! I GI ,

establishing (5).

Chapter 6, Exercise 2. (1) Let G be the semidirect product of H by A. Wemust show there exists an A-invariant Hall n-subgroup of H. Let A be theset of Hall n-subgroups of H. By Hall's Theorem, 18.5, there is K E A andH is transitive on A, so by a Frattini argument, G = HNG (K). By the Schur-Zassenhaus Theorem 18.1, there exists a complement B to NH(K) in NG(K)and B9 = A for some g E G. Hence J = K9 is an A-invariant Hall n-subgroupof H.

(2) We must show CH(A) is transitive on the set Fix(A) of fixed point ofA on A. But by Schur-Zaussenhaus, NH(K) is transitive on the set Ac flNG(K) of complements to NH(K) in NG (K), so by 5.21, NH(A) is transitiveon Fix(A).

(3) We must show each A-invariant n-subgroup X of H is contained in amember of Fix(A). The proof is by induction on I G I, so assume G is a minimalcounter example. Let M be a minimal normal subgroup of G contained in Hand G* = G/M. By minimality of G there exists an A-invariant Hall n-subgroup Y* of H* containing X*. Now IYI, = I HI,r, so Y contains a Halln-subgroup of H, and if H :h Y then X is contained in an A-invariant Hall

274 Appendix

Thus X - A G Pix([a, b]). ~ e x t xb-' E Mov(b) - {x} G Fix(a), so

Similarly xu-' E Fix(b), so

Finally z =xb E Mov(b) - {x} G Fix(a) and y =xa E Fix(b), so

Thus (y, x, z) is a cycle of length 3 in [a, b], completing the proof of (3). (4) Assume G does not contain the alternating group on X and IY I > n/2; we

must derive a contradiction. As I Y I > n/2, r = X - Y has order less than I Y I, so by minimality of IY I, Gr # 1. Pick a E G#,, y E M(a), and set A = Y - {y}. Then lAl < IYI, so by minimality of IYI, Ga # 1. pick b E G#,. As Gy = 1, y E M(b), while

so M(a) fl M(b) = {y}. Thus [a, b] is a 3-cycle by (3), so (2) supplies a contradiction, establishing (4).

(5)PickY asin(4)andletH =Sym(X)y.ThenIHI=rn!,wherern= IX-YI andby (4), rn > [(n + 1)/2]. Now H n G = G y = 1, so

establishing (5).

Chapter 6, Exercise 2. (1) Let G be the semidirect product of H by A. We must show there exists an A-invariant Hall n-subgroup of H. Let A be the set of Hall n-subgroups of H. By Hall's Theorem, 18.5, there is K E A and H is transitive on A, so by a Frattini argument, G = HNG(K). By the Schur- Zassenhaus Theorem 18.1, there exists a complement B to NH(K) in NG(K) and Bg = A for some g E G. Hence J = Kg is an A-invariant Hall n-subgroup of H .

(2) We must show CH(A) is transitive on the set Fix(A) of fixed point of A on A. But by Schur-Zaussenhaus, NH(K) is transitive on the set A' fl NG(K) of complements to NH(K) in NG(K), SO by 5.21, NH(A) is transitive on Fix(A).

(3) We must show each A-invariant n-subgroup X of H is contained in a member of Fix(A). The proof is by induction on I G 1, so assume G is a minimal counter example. Let M be a minimal normal subgroup of G contained in H and G* = G/M. By minimality of G there exists an A-invariant Hall n - subgroup Y* of H* containing X*. Now IYI, = IHI,, so Y contains a Hall n-subgroup of H, and if H # Y then X is contained in an A-invariant Hall

Appendix 275

7r-subgroup of Y by minimality of G. Thus H = Y, so HIM is a 7r-group andhence H = JM, so H* = J*.

By 9.4, M is a p-group for some prime p. If P E Tr then H is a 7r-group, soH is an A-invariant Hall 7r-subgroup of H containing X. Thus p 0 7r, so X is aHall 7r-subgroup of MX. Let Z = J fl XM. As H* = J*, Z* = X*, so Z and Xare A-invariant Hall 7r-subgroups of XM, and hence by (2) there is g E CH(A)with Z9 = X. Thus Jg is an A-invariant Hall 7r-subgroup of H containing X.

(4) Part (4) does not depend on p and hence remains valid.

Chapter 6, Exercise 3. As V is the permutation module for G = Alt(I) on Iand F = GF(2), we can identify V with the power set of I, and for u, v E V,u + v is the symmetric difference of u and v; that is,

u+v=uUv-(u fl v).

Define the weight of v to be the order I v I of the subset v of I.(1) Let 0:A W be an FG-submodule of V and W E W# of weight m. By

15.12.3, G is (n - 2)-transitive on I, so as n > 2, G is transitive on m-subsets ofI and hence on vectors of weight m. Thus for each v of weight m, v E w G C W.Assume w # I, the generator of Z. Then there is v of weight m such that w fl vis of order m - 1, so w + v is of weight 2. So as w + V E W, we may takem = 2. But as w is of weight 2, (wG) = U. Namely, if 0 # u E U, pick i E u,and observe u = Ei O jEU {i, j). Thus W = U or V.

(2) and (3): First observe that G = 02(G). If n > 4 this follows because Gis simple by 15.16. If n = 3 or 5 it is easy to check G = 02(G) directly.

Observe next that if n is odd then 10 U, so V = U ® Z and hence U = Uis of dimension n - 1. On the other hand, if n is even then 0 < Z < U < Vwith dim(U) = n - 2.

By Exercise 4.6, Z = Cv (G) and U = [V, G]. Further, if X < G is of oddorder, then by coprime action 18.7, CV(X) = CV(X), so as G = O2(G),CM(G) = CM(G) = 2 =0. By 8.5.3, [V, G] _ [V, G] = U.

As CU(G) = 0, by 17.11 there is a largest FG-module W such that [W, G] <U < W and CW(G) = 0. In particular as V is such a module, V < W. Also bydefinition, H1(G, U) = W/U, so it remains to show W = V, or equivalentlyH 1 (G, U) = 0 or F for n odd or even, respectively. We prove this by inductionon n.

If n = 3 then G = Z3 is of order prime to 2, so by 17.10, H '(G, U) = 0,as desired. Thus we may take n > 3. Let H = G 1 and J = I - { 1), so thatH =Alt(J) = A,i_1. For j E J, let vj = {1, j}. Then {vj: j E J} is a basis forU with vjh = Vjh for h E H, so U is the permutation module for H of degreen - 1. In particular, by Exercise 4.6.1, v = EjEJ vi generates Cu(H), and byinduction on n, (v) = CU(H). Notice if n is odd then v = J so (J) = CC, (H),while if n is even then v = I, so CU (H) = 0.

Appendix 275

n-subgroup of Y by minimality of G. Thus H = Y, so HIM is a n-group and hence H = JM, so H* = J*.

By 9.4, M is a p-group for some prime p. If p E n then H is a n-group, so H is an A-invariant Hall n-subgroup of H containing X. Thus p $ n , so X is a Hall n-subgroup of MX. Let Z = J fl XM. As H* = J*, Z* = X*, so Z and X are A-invariant Hall n-subgroups of XM, and hence by (2) there is g E CH(A) with Zg = X. Thus J g is an A-invariant Hall n-subgroup of H containing X.

(4) Part (4) does not depend on p and hence remains valid.

Chapter 6, Exercise 3. As V is the permutation module for G = Alt(I) on I and F = GF(2), we can identify V with the power set of I , and for u, v E V, u + v is the symmetric difference of u and v; that is,

Define the weight of v to be the order I v 1 of the subset v of I. (1) Let 0 # W be an FG-submodule of V and w E W# of weight m. By

15.12.3, G is (n - 2)-transitive on I , so as n > 2, G is transitive on m-subsets of I and hence on vectors of weight m. Thus for each v of weight m, v E w G G W. Assume w # I , the generator of Z. Then there is v of weight rn such that w f l v is of order rn - 1, so w + v is of weight 2. So as w + v E W, we may take rn = 2. But as w is of weight 2, (wG) = U . Namely, if 0 # u E U , pick i E u, and observe u = xi + j , ,{ i , j}. Thus W = U or V.

(2) and (3): First observe that G = 0 2 ( ~ ) . If n > 4 this follows because G is simple by 15.16. If n = 3 or 5 it is easy to check G = o ~ ( G ) directly.

Observe next that if n is odd then I $ U , so V = U @ Z and hence 0 E U is of dimension n - 1. On the other hand, if n is even then 0 < Z < U < V with dim(U) = n - 2.

By Exercise 4.6, Z = CV(G) and U = [V, GI. Further, if X 5 G is of odd order, then by coprime action 18.7, Cv(X) = CV(X), so as G = 0 2 ( ~ ) , Cv(G)=CV(G)=Z=O. By 8.5.3, [V, G]=[V, G]=U.

As Cc(G) = 0, by 17.1 1 there is a largest FG-module W such that [W, GI 5 u 5 W and Cw(G) = 0. In particular as V is such a module, V 5 W. Also by definition, H'(G, U) = w/U, so it remains to show W = V, or equivalently H' (G, U) = 0 or F for n odd or even, respectively. We prove this by induction on n.

If n = 3 then G E Z3 is of order prime to 2, so by 17.10, H1(G, 0 ) = 0, as desired. Thus we may take n > 3. Let H = G1 and J = I - {I}, so that H =Alt(J) Z A,-1. For j E J , let v , = {I, j}. Then { v j : j E J } is a basis for U with v, h = vjh for h E H , SO U is the permutation module for H of degree n - 1. In particular, by Exercise 4.6.1, v = C,,, vj generates CU(H), and by induction on n, (3) = Cu(H). Notice if n is odd then v = J so ( I ) = Cu(H), while if n is even then v = I , so Cu(H) = 0.

276 Appendix

Assume n is odd, so that U = U as an FG-module. Suppose W E Cw(H) - Uand let M= (wG). By Exercise 4.6, M is a homomorphic image of V as anFG-module. As CM(G) < Cw(G) = 0, 0 0 [M, G] < [W, G] = U, so a G isirreducible on U, U=[M, G] <M. Then as CM(G)=0 and by (1), U is theonly nontrivial image N of V with CN(G) = 0, we conclude M = U, contra-dicting w E W - U. Thus we have shown that Cw(H) = CU (H) = (J), so alsoCW1(j) (H) = 0 as H = 02 (H). Further U/(J) is the image of the permutationmodule U U for H modulo CO(H), so by induction on n, W/(J) = U/(J),so W = U, completing the proof in this case.

This leaves the case n even. Thus n - 1 is odd, so as U is the permuta-tion module for H, by induction on n, W = U ® Cw(H). Let K = G2. ThenCw(Hf1K)=Cw(H)®CU(H2)isofdimension d+l,where d= dim(Cw(H)).Thus Cw(H) and Cw(K) are hyperplanes of Cw(H fl K). Further

0 = Cw(G) = Cw((H, K)) = Cw(H) n CH(K),

so dim(Cw(H fl K)) < 2 and d < 1. Therefore

dim(W) = d + dim(U) < n - 1= dim(V ),

so again W = V, completing the proof.

Chapter 7, Exercise 5. Let u = ax2 + bxy + cy2 and v = rx2 + sxy + ty2 be inW. Then

Q(x+y)=(b+s)2-4(a+r)(c+t)= (b2 - 4ac) + (s2 - 4rt) + 2bs - 4(at + rc)

= TO + Q(v) + f (u, v).

Evidently f is bilinear and symmetric and Q(,ku) = A2Q(u) for E F. To prove(W, Q) is nondegenerate it suffices to observe that X = [x2, xy, y2} is a basisfor W with x2, y2 singular, Q(xy) = 1, f (x2, y2) = -4, and xy is orthogonal tox2 and y2.

(2) Let

g=lCa b

E dJG.

c

To prove (2), it suffices to check Q(wga) =,k(ga) Q(w) and f (wga, zga) _A(ga) f (w, z) for w, z E X, where k(ga) = det(g)2 = ad - bc. For example

Q(x2ga) = Q((ax + by)2) = Q(a2x2 + 2abxy + b2y2)

= 4a2b2 - 4a2b2 = 0 = X(ga)Q(x2),

Q(xy(ga)) = Q((ax + by)(cx + dy)) = Q(acx2 + (ad + bc)xy + bdy)

= (ad + bc)2 - 4abcd = (ad - bc)2 =gga) Q(xy),

276 Appendix

Assume n is odd, so that U Z 0 as an FG-module. Suppose w E Cw(H) - 0 and let M = (wG). By Exercise 4.6, M is a homomorphic image of V as an FG-module. As CM(G) ( Cw(G) = 0, 0 # [M, GI _( [W, GI = 0 , so a G is irreducible on 0 , 0 = [M, GI 5 M. Then as CM(G) = 0 and by (I), 0 is the only nontrivial image N of V with CN(G) = 0, we conclude M = 0 , contradicting w E W - 0. Thus we have shown that Cw(H) = Co(H) = ( J ) , so also Cw,,,-,(HI = 0 as H = 0 2 ( ~ ) . Further U/(J) is the image of the permutation module 0 Z U for H modulo Co(H), so by induction on n, W/(J) = o / ( J ) , so W = 0 , completing the proof in this case.

This leaves the case n even. Thus n - 1 is odd, so as U is the permutation module for H, by induction on n, W = 0 e3 Cw(H). Let K = G2. Then Cw(HnK) = Cw(H) @ C,7(H2) is ofdimensiond+l, whered = dim(Cw(H)). Thus Cw(H) and Cw(K) are hyperplanes of Cw(H n K). Further

o = Cw(G> = Cw((H, K)) = Cw(H) n CH(K),

so dim(Cw(H n K)) 5 2 and d ( 1. Therefore

so again W = V , completing the proof.

Chapter 7, Exercise 5. Let u = + bxy + cy2 and v = rx2 + SXJ + ty2 be in W. Then

= (b2 - 4ac) + (s2 - 4rt) + 2bs - 4(at + rc)

= Q(u> + Q<v> + f ( ~ 3 v).

Evidently f is bilinear and symmetric and Q(hu) = h2 ~ ( u ) for h t F. To prove (W, Q) is nondegenerate it suffices to observe that X = {x2, xy, y2} is a basis for W with x2, y2 singular, Q(q) = 1, f (x2, y2) = -4, and xy is orthogonal to x2 and y2.

(2) Let

To prove (2), it suffices to check Q(wga) = h(ga)Q(w) and f (wga, zga) = h(ga) f (w, z) for w, z t X, where h(ga) = det(,g)' =ad - bc. For example

~ ( x ~ g a ) = Q((ax + by)2) = ~ ( a ' x ' + 2abxy + b2y2)

Q(xy(ga)) = Q((ax + by)(cx + dy)) = ~ ( a c 2 + (ad + bc)xy + bdy2)

= (ad + hc)' - 4abcd = (ad - bc12 = h(ga)Q(xy),

Appendix 277

and

f (x2ga, y2ga) = f ((ax + by)2, (cx + dy)2)

= f(ax2 + 2abxy + b2y2,c2x2 + 2cdxy + d2y2)

= 8abcd - 4(a2d2 + c2b2)

_ -4(ad - bc)2 = )(ga)f (x2, y2).

(3) Observe that u is singular if b2 = 4ac, so the set S of singular points inW is So U {Fx2}, where

So = {F(e2x2 + 2exy + y2): e E F}.

The subgroup

T= {(i ?)F}of G is transitive on So, so G is 2-transitive on S and hence A (W) = Ga OF,Fy2.Now if h E OFx2 Fy2, then as Fxy = (x2, y2)', MX(h) is diagonal with x2h =µx2, y2h =.lµ-1 y2, and xyh = 5xy, for some µ E F#, where k _) (h). Onthe other hand ga E AFx2,Fy2, where

g(1 0-

o)EG,

andx2ga =x2, y2ga =µ-2y2, andxyga = soh E

Therefore (3) holds.(4) If U is a nondefinite 3-dimensional orthogonal space over F then U =

HID, where H is a hyperbolic line and D = Fd is definite. Multiplying thequadratic form Qu on U by a suitable scalar, we may assume Qu(d) = 1. Thus

(U, Qv) - (W, Q).(5) Part (4) implies (5), since if F is finite or algebraically closed, then no

3-dimensional orthogonal space over F is definite by lemmas 21.3 and 20.10.1.(6) Let A = O(W, Q). By Exercise 4.7.4,

ker(a)={aI:aEFand a2=1}=(-I).

Thus G(1)a - L2(F), so as 0 = SGa and S < Z(O), 0M = GO)a = L2(F).As 0/0(1) is abelian, rg E rO(1) for each r E R and g E 0, so rrg E 0M. Butby 43.12.1, L2(F) is simple unless I F I = 3, so except possibly in that case,0(1) = (rrh: h E 0(1)). Finally, if I Fl = 3 then

0

0)ERr= (-1

1

Appendix

and

(3) Observe that u is singular iff b2 = 4ac, so the set S of singular points in W is So U {FX~}, where

So= { ~ ( e ~ x ~ + 2exy + y2): e~ F}.

The subgroup

of G is transitive on So, so G is 2-transitive onS andhence A(W) = GaAF2,,,2. Now if h E AFX2,,,2, then as Fxy = (x2, y2) l , Mx(h) is diagonal with x2h = px2, y2h = A p - l y2, and xyh = A x y , for some p E F', where h = h(h). On the other hand g a E A,,2,,,2, where

andx2ga =x2, y2ga =hp-2y2,andxyga= Ap- 'xy,soh = g a . p I E Ga.S. Therefore (3) holds.

(4) If U is a nondefinite 3-dimensional orthogonal space over F then U = H I D , where H is a hyperbolic line and D = Fd is definite. Multiplying the quadratic form QU on U by a suitable scalar, we may assume QU(d) = 1. Thus

(U, Qu) r cw, Q). (5) Part (4) implies ( 3 , since if F is finite or algebraically closed, then no

3-dimensional orthogonal space over F is definite by lemmas 21.3 and 20.10.1. (6) Let A = A(W, Q). By Exercise 4.7.4,

ker(a) = { a t a E F and a2 = 1) = (-I).

Thus G(')CY Z L2(F), so as A = SGa and S ( Z(A), A(') = G(')a Z L 2(F). As A/A(') is abelian, rg E rA(') for each r E R and g E A, so rfl E A('). But by 43.12.1, L2(F) is simple unless I FI = 3, so except possibly in that case, A(') = ( r9 : h E A(')). Finally, if I F 1 = 3 then

278 Appendix

with center Fxy, and r inverts

g_ (1 1) EG1 1

of order 3, so again GO)a is generated by conjugates of rrg = g-1 of order 3.

Chapter 8, Exercise 11. Let V be a normal elementary abelian subgroup ofG of maximal rank, H = CG(V), and G* = G/H. By 23.16, V = Q1(H). Asm(G) > 2,m(V) > 2 by 23.17, so we may assume m (V) =3. Thus G* < GL(V ),so m(G*) < m(GL3(p)) = 2.

LetEp. A<Gwith n> 3.AsV=Q1(H),AnH=AnVisofrank atmost 2. Also m(A*) < m(G*) = 2, so

4 < m(A) = m(A*) + m(A n H) < 2 + 2 = 4

and hence all inequalities are equalities, so m(A)=4 and m(A*) =m(AnH)=2.

Let B = A n V. Then A centralizes the hyperplane B of V and m(A*) = 2, soA* is the full group of transvections with axis B. Therefore [A, VI = B = Cv (a )foreach a E A-B. Suppose D = Ep4 with D <AV. Then JAV: A(_ JAV: DI _p, so if A,- D then m(A n D) = 3, and hence there is a E A n D - B. ThenD < CAV(a) =ACv(a) =AB = A, a contradiction. Thus A is the unique E4-subgroup of AV, so A char AV.

Let K = NG(A); we may assume K# G, so there is g E NG (K) - K. As A* isof index pin G*, A* < G*, so (Ag)* = A* and hence A # A8 <AHn K = AX,where X = NH(A). Now

[A, AX]=[A, X] <A n H = B <Z(AH),

so AAg is of class at most 2. Hence by 23.11, AAg is of exponent 3. ThusAAg =AAg n X = A(AAg n x) with AAg n x < Q1(H) = V. Therefore AAg <AV, so as A is the unique Ep4-subgroup of AV, A = Ag, contradicting g K.

Chapter 9, Exercise 6. (1) Fix vi c VZ, i = 1, 2, and define

fV,,U2: V1 x V2 -* F,

(u1, u2) H f1(v1, u1)f2(v2, u2)

As f1 and f2 are bilinear, so is f,,,,,,2, so by the universal property of the tensorproduct, extends to a linear map f,2: V -* F with 11I,,,2(u1 (9 u2)

f,,,,,,2(u1, u2). Similarly

V1xV2-* V

(v1, v2) H fVhV2

Appendix

with center Fny, and r inverts

of order 3, so again G(')m is generated by conjugates of r+ = g-l of order 3.

Chapter 8, Exercise 11. Let V be a normal elementary abelian subgroup of G of maximal rank, H = CG(V), and G* = GIH. By 23.16, V = C21(H). As m(G) > 2,m(V) > 2by23.17, so wemay assumem(V) = 3.Thus G* ( GL(V), so m(G*) 5 m(GL3(p)) = 2.

LetE,. E A s G w i t h n > 3 . A ~ V = C 2 ~ ( H ) , A n H = A n V i s o f r a n k a t most 2. Also m(A*) ( m(G*) = 2, so

and hence all inequalities are equalities, so m(A) = 4 and m(A*) =

m(A n H) = 2. Let B = A n V. Then A centralizes the hyperplane B of V and m(A*) = 2, so

A* is the full group of transvections with axis B. Therefore [A, V] = B = Cv (a) foreacha E A- B. Suppose D Z Ep4 with D (AV.Then IAV: A1 = IAV: Dl = p, so if A # D thenm(A n D)=3, andhence t h e r e i s a ~ A n D - B. Then D 5 C ~ v ( a ) = ACv(a) = AB = A, a contradiction. Thus A is the unique Ep4- subgroup of AV, so A char AV.

Let K = NG(A); wemay assume K # G, so thereis g E NG(K)-K. As A* is ofindexp in G*, A* <I G*, so (Ag)* =A* andhence A# Ag (AHn K =AX, where X = NH(A). Now

so AAg is of class at most 2. Hence by 23.11, AAg is of exponent 3. Thus AAg =AAg n X = A(AAg n X) with AAg n X 5 C21(H) = V. Therefore AAg ( AV, so as A is the unique E$-subgroup of AV, A = Ag, contradicting g @ K.

Chapter 9, Exercise 6. (1) Fix vi E &, i = 1,2, and define

As f l and f2 are bilinear, so is f,, ,, , so by the universal property of the tensor product, f,,,,, extends to a linear map f ,,,,: V -+ F with f ,,,, ,(ul @I u2) =

fv, ,vz (u 1, u2). Similarly

v1 x vz -+ v*,

Appendix 279

is bilinear, so there is f E Hom(V, V*) with f (v1 (9 v2) = Thereforef : V x V -* F defined by f (x, y) = f (x)(y) is bilinear and

f (V1 ® v2, u1 (9 u2) = 1(V1 (9 v2)(ui (9 u2)

U U 2 ) -

A sAs the fundamental tensors generate V = V1 0 V2, f is unique subject to thisproperty. Finally

f (ul ® u2, v1 (9 v2) = f (u1, vi)f (u2, V2)= (-f (VI, ul))(-f (v2, u2))

= f(v1, ul)f(v2, u2) = f(V1 0 v2, u1 0 u2),

so as the fundamental tensors generate V1 ® V2, f is symmetric.(2) First f (v1 ® v2, V1 (9 v2) = fl (vi, vl)f2(v2, v2) = 0 as fi is symplectic.

But if char(F) : 2, recall from Chapter 7 that the unique quadratic form Qassociated to f satisfies Q(x) = f (x, x)/2. Thus Q(v10 v2) = f (v1(9 v2, v1®V2)/2 = 0, as desired. On the other hand:

Lemma. If char(F) = 2, X is a basis for an F-space U, f is a symplectic formon U, and ax E F for each x E X, then there is a unique quadratic form Q onU associated to f such that Q(x) = ax for each x E X.

Let Xi = {vi, ui) be a basis for V1; then X = X1 ®X2 is a basis for V, soby the lemma there is a unique quadratic form Q on V associated to f withQ (x 1 0 x2) = 0 for each x, E Xi. Now for w c V1, w = a v 1 + bu 1 and

Q(w (9 v2) = Q(a(v1 (9 v2) + b(u1 (9 v2))

= a2Q(vi (9 v2) + b2Q(ul (9 v2) + abf (v1 ® v2, u1 (9 V2) = 0.

Similarly Q(w (9 u2) = 0, so for z = cv2 + due E V2,

Q(w (9 z) = Q(c(w (9 v2) + d(w (9 u2))

= c2 Q(w (9 v2) + d2 Q(w (9 u2) + cdf (w ® v2, w (9 u2) = 0

completing the proof of (2).(3) Pick Xi to be a hyperbolic basis for Vi. Then x1 ® x2 is orthogonal to

x1 ® Y2 and y1 ®x2 for all xi, yi c Xi, while

f (V1 ® v2, u1 (9 u2) =1= f (V1 ® u2, u1 ® v2),

so X is a hyperbolic basis for V.(4) Let g = (g1, 92) E A. Then

f(V1 (9 v2)gn, (ul (9 u2)gn) = f(v1g1 ® v2g2, u1g1 (9 u2g2)

= fi(v1g1, u1g1)f2(v2g2, u2g2)

= A(g1)fl(vl,ul)?(g2)f2(v2,u2)

A(g1)X(g2)f (vl ® V2, U 1 (9 u2),

Appendix 279

is bilinear, so there is f E Hom(V, V * ) with f ( v l @ v2) = f v , , , , . Therefore f : V x V -+ F defined by f (x, y) = f ( ~ ) ( ~ ) is bilinear and

As the fundamental tensors generate V = Vl @ Vz, f is unique subject to this property. Finally

so as the fundamental tensors generate Vl @ V2, f is symmetric. (2) First f (vl @ v2, V I @ v2) = fl(v1, v l ) f2(v2, v2) = 0 as fi is symplectic.

But if char(F) # 2, recall from Chapter 7 that the unique quadratic form Q associated to f satisfies Q ( x ) = f ( x , x ) /2 . Thus Q(vl @ v2) = f (vl @ v2, vl @

v2)/2 = 0, as desired. On the other hand:

Lemma. If char(F) = 2, X is a basis for an F-space U , f is a symplectic form on U, and a, E F for each x E X , then there is a unique quadratic form Q on U associated to f such that Q ( x ) =a, for each x E X.

Let Xi = {vi , ui) be a basis for Vi; then X = X1 @ X2 is a basis for V , so by the lemma there is a unique quadratic form Q on V associated to f with Q(xl @ x2) = 0 for each xi E Xi. Now for w E V l , w =avl + bul and

2 = a Q(vl @ ~ 2 ) + b 2 e ( u l @ vz) + abf (vl @ vz , u l @ v2) =O.

Similarly Q(w @ uz) = 0, so for z = cv2 + duz E V2,

= c2 Q(W @ v2) + d 2 e ( w @ u2) + cdf (w @ vz, w @ u2) = 0

completing the proof of (2). (3) Pick Xi to be a hyperbolic basis for V,. Then xl @ x2 is orthogonal to

xl @ y2 and yl @ xz for all xi, yi E X i , while

f(v1 @ v 2 , u l @ ~ 2 ) = 1 = f(v1 @ U z , U 1 @ v 2 ) ,

so X is a hyperbolic basis for V . (4) Let g = (g i , g2) E A. Then

280 Appendix

so as the fundamental tensors generate V, gn E 0(V, f) with ) (grr) _.l(gi))l(g2). Similarly

Q((vl (9 v2)(g7)) = Q(v1g1 (9 v2g2) = 0 = Q(v1 (9 v2)

as Q(u1 (9 u2)=0 for all ui EVi, so by the lemma, gn also preserves Q.Of course gn c O(V, Q) if 1=,1(g) if ,l($2) =,l(gi)-1. Also g E ker(rr) ifx1 (&X2=(XI (&x2)g7r =xlg1 O X292 for all xi EX,, so as X is a basis for V,it follows that g c ker(rr) if gi = µi I for Ui E F# with A2 = µ i

(5) The map

Ta: V1 ® V2 - V,

(u, va) H v ®ua

is bilinear, so there exists a unique t = t,, c End(V) with ta(u (9 va) = v ® uafor all u, v c V1. Evidently t2 = 1. Further, as a is an isometry,

f ((vi (9 va)t, (ul (9 ua)t) = f (v ® via, u (9 ula)= fl(v, u) f2(vla, ula)

= fl (VI, ui)f2(va, ua)= f(vl (9 va, u1 ® ua),

sot is anisometry. Next, we may choose notation so that via = v2 and u 1 a = u2,so

(v1 (9 v2)t = (v1 (9 vla)t = (v1 (9 via) = v1 ® v2.

Similarly t fixes u 1®u2 and interchanges v1®u 2 and u1 ®v2, so t is a reflectionor transvection.

Let g = (g1, 1) E 01, so that gn E O1n. Then

(u (9 va)gt = (v (9 ua)gt = (vgi ® ua)t = u ® vgla = u ® (vagia*),

so as Ala* = 02, gt E O2n. Thus (Din)t = Len and similarly (Gin)` = Gen.(6) and (7): For Fv a point in V1 let Fv ® V2 = {v ® u : u E V2} and set

G1 = {Fv ® V2 : v E V141. For Fu a point in V2 define V1® Fu and G2 similarly.Observe G = G1 U G2 is a set of totally singular lines in V, as Q(vl (9 v2) = 0for all vi c Vi.

Claim the singular vectors are the fundmental tensors v1 ® v2 and G is theset of all totally singular lines. For if x E V# is totally singular then 0:0 x1 flFv® V2, so x E (v(gu)1 for some u E V20. Now as V is 4-dimensional hyperbolicspace and y = v ® u is totally singular, we have y1= Fy I H, where H is ahyperbolic line, and if H1 and H2 are the two totally singular points in H,then Fy + H1 and Fy + H2 are the two totally singular lines through Fy andcontain all totally singular points in y1. Thus x c Fy + Hi for some i, andas Fv ® V2 and V1 ® Fu are totally singular lines through y, we may takex E Fv ® V2. This shows each totally singular point is a fundamental tensor,and each totally singular line l is one of the two totally singular lines Fv ® V2

280 Appendix

so as the fundamental tensors generate V , gn E A(V, f ) with h(gn ) = h(gl)h(gz). Similarly

as Q(u1 @ uz) = 0 for all ui E Vi, SO by the lemma, gn also preserves Q. Of course gn E O(V, Q) iff 1 = h(g) iff h(gz) = A ( ~ ~ ) - ' . Also g E ker(n) iff xl @ xz = ( X I @ xz)gn =xlgl @ xzgz for all xi E Xi, so as X is a basis for V ,

1 it follows that g E ker(n) iff gi = p i I for ui E F' with p2 = p; . (5) The map

T,: Vl @ vz + v ,

( u , va ) H v @ ua

is bilinear, so there exists a unique t = t, E End(V) with t,(u @ va ) = v @ ua for all u, v E Vl . Evidently t2 = 1. Further, as a is an isometry,

= fl(V1, u1)fz(va, ua)= f (v l @ va, U 1 @ ua),

so t is an isometry. Next, we may choose notation so that vl a = vz and u 1 a = U Z ,

SO

Similarly t fixes u 1 @ uz and interchanges vl @uz and ul @ vz, so t is areflection or transvection.

Let g=(gl , 1 ) ~ A,, so that gn E A l n . Then

so as Ala* = A2, g' E Azn . Thus (A ln ) ' = A z n and similarly (Gin)' = G z n . (6 ) and (7): For Fv a point in Vl let Fv @ V2 = { v @ u : u E V2) and set

L1= { F v @ V2 : v E v:). For Fu apoint in V2 define Vl @ Fu and Lz similarly. Observe L = L1 U LZ is a set of totally singular lines in V , as Q(vl @ v2) = 0 for all vi E Vi.

Claim the singular vectors are the fundmental tensors vl @ vz and L is the set of all totally singular lines. For if x E V# is totally singular then 0 # x' n Fv@ V2, SO x E (v@u)' for some u E V$ Now as V is 6dimensional hyperbolic space and y = v @ u is totally singular, we have y' = Fy I H , where H is a hyperbolic line, and if H1 and Hz are the two totally singular points in H , then Fy + H1 and Fy + Hz are the two totally singular lines through Fy and contain all totally singular points in y'. Thus x E Fy + Hi for some i , and as Fv @ VZ and Vl @ Fu are totally singular lines through y, we may take x E Fv @ Vz. This shows each totally singular point is a fundamental tensor, and each totally singular line 1 is one of the two totally singular lines Fv @ Vz

Appendix 281

and V1 ® Fu through a point F(v 0 u) on 1, so L is the set of all totallysingular lines.

Observe G1 and G2 are the two classes of maximal totally singular subspacesof V described in 22.13, so by 22.13, the subgroup F of E = 0(V, Q) actingon G1 and G2 is of index two in E, with E = F(t).

Next by Exercise 7.1.1, Gl = SL(V, ). Further if X, _ (v ul) is our hyper-bolic basis, µ E F# and gµ,, E GL(V;) with v; gµ,; = v; and ui gµ,, = µu; , thengµ,? E Ai with )(gµ,i) = µ, so

Ai =SL(V,)(gµj: µ E F#) =GL(VV).

Now 01 n < ro, the stabilizer in F of V1®Fv2 and V1® Fu2. For x c Vl ® Fv2,define fx c (V1 ® Fu2)* by fx(y) = f (x, y). Then the map x i-* fx is a To-isomorphism of V1 ® Fv2 with (V1 (9 Fu2)*, so as Cr(Vi (9 Fv2) n Cr(Vi ®Fu2) = 1, ro is faithful on V1® Fv2, and thus, as 01 n acts faithfully as GL(V1(9Fv2) on V1 ® Fv2, it follows that F0 = Otn. Further 0271 is 2-transitive onthe points of A2 and hence also on G2, so F = F0L 27i = O7i. This completesthe proof of (6).

Finally c2 = c2(V, Q) is the derived group of O(V, Q), and O(V, Q) _ (O(V,Q) n r)(t) with

ro =O(V, Q) n F = ((gl, g2)7r:,1(g1)=A(g2)-'}

by (4) and (6). Thus Fo = (G1G2)2 Tom, where T = (gµ7r: µ E F#} and gN, =(gµ,1, 9,,-1,2) FurthergN,22r =h1h22r E (G1G2)n, where v1h1 = µ-1v1, u1h1 =

µu 1, v2h2 = µv2, and u2h2 = µ-1 u2, and g`µ = g.-I, so [Tn, t] < (G1 G2)jr andhence Q = Q(V, Q) < (G1G2)n. Conversely, if I FI > 3 then G;Tr = SL2(F)is perfect by 43.12.1, so Q = (G 1 G2)jr . If I F I = 3 it can be checked directlythat (G1G2) r <Q.

Chapter 9, Exercise 9. By Exercise 9.1, a = f 0 E HOMFG W, V0*) and asf # 0, a:0 0. Thus as G is irreducible on V, a is an isomorphism by Schur'sLemma 12.4. Let or = 0* regarded as an automorphism of GL,, (F), the groupof invertible n by n matrices over F, and regard G as a subgroup of GL,, (F).As V - VQ, g° = gB* for all g E G and some B E GL,, (F) from the discussionin section 13. Thus

gal = (gB*)° = g°(Bo )* = gB*(BQ)*= g(BB°)*

so V VQ2. However, 0 commutes with the transpose inverse map * onGL,,(F), so V°z =((V*)*)Bz = V° as (V*)* - V.

Similarly if V = Vo then V* - VB* = V, so by Exercise 9.1.4, G preservesa nondegenerate bilinear form on V. That is (1) holds in this case, so we mayassume V V0. Then as V -

V02,m is even, since otherwise 0 is a power of B2.

Appendix 28 1

and Vl @ Fu through a point F(v 8 u) on I , so L is the set of all totally singular lines.

Observe L1 and Lz are the two classes of maximal totally singular subspaces of V described in 22.13, so by 22.13, the subgroup r of X = A(V, Q) acting on L1 and L2 is of index two in X, with X = r ( t ) .

Next by Exercise 7.1.1, Gi = SL(&). Further if Xi = (vi, ui) is our hyperbolic basis, p E F' and g,,i E GL(&) with VigLL,i = Vi and UigCL,i = p i , then

g,,i E Ai with h(g,,i) = I I . 9 SO

Now Aln 5 ro , the stabilizer in r of Vl @ Fvz and Vl @ Fuz. For x E Vl @ Fvz, define f, E (Vl @ Fuz)* by f,(y) = f (x, y). Then the map x H f, is a ro- isomorphism of Vl @ Fvz with (Vl @ Fuz)*, so as Cr(Vl @ Fvz) n Cr(Vl @

Fuz) = 1, ro is faithful on Vl @ Fvz, and thus, as Aln acts faithfully as GL(Vl @

Fvz) on Vl @ Fvz, it follows that ro = Aln. Further A2n is Ztransitive on the points of Az and hence also on CZ, SO r = r0A2n = An. This completes the proof of (6).

Finally Q = Q(V, Q) is the derived group of O(V, Q), and O(V, Q) = (O(V, Q) n r ) ( t ) with

by (4) and (6). Thus ro = (G1G2)n . Tn, where T = {g,n: p E F') and g, = (g,,l, g,-~,~).Furtherg,m =hlh2n E ( G ~ G z ) ~ , wherevlhl = ~ l . - ~ v l , u ~ h l =

pul , v2h2 = ~ V Z , and U Z ~ Z = p-luz, and g; = g,-I, so [Tn, tl i (G1Gz)n and 1 hence Q = Q(V, Q) 5 (GIGz)n. Conversely, if IF1 > 3 then Gin 2 SLz(F)

is perfect by 43.12.1, so Q = (GIGz)n. If IF1 = 3 it can be checked directly that (G1Gz)n 5 Q.

Chapter 9, Exercise 9. By Exercise 9.1, a = f @ E H o m ~ ~ ( v , v'*) and as f # 0, a # 0. Thus as G is irreducible on V, a is an isomorphism by Schur's Lemma 12.4. Let CJ = O* regarded as an automorphism of GL,(F), the group of invertible n by n matrices over F , and regard G as a subgroup of GL,(F). As V 2 V", g" = gB* for all g E G and some B E GL,(F) from the discussion in section 13. Thus

so V 2 v"'. However, 6' commutes with the transpose inverse map * on GL,(F), so v"' = ((v*)*)" 2 V" as (V*)* 2 V.

Similarly if V 2 V' then V* 2 v'* 2 V, SO by Exercise 9.1.4, G preserves a nondegenerate bilinear form on V. That is (1) holds in this case, so we may assume V 7 v'. Then as V 2 v", rn is even, since otherwise 6' is apower of 6''.

282 Appendix

Let K be the fixed field of 02. Then (02) = Gal(F/K), so by 26.3, V = F ®KUfor some irreducible KG-submodule U of V. As 0 is an automorphism of K oforder 2, Exercise 9.1.4 says G preserves a nondegenerate hermitian symmetricform on U.

Chapter 9, Exercise 10. Let G1 = G2 = G. As or is 1-dimensional, or is irre-ducible and by hypothesis 7r is irreducible, so by 27.15, or ® n is an irreduciblerepresentation of G1 x G2. Recall that if we identify G with the diagonalsubgroup {(g, g): g E G} of G1 x G2 via the isomorphism g i-+ (g, g), then thetensor product representation or ® n of G is the restriction of the tensor productrepresentation or 0 r of G1 x G2 to the diagonal subgroup G, so it remains toshow the diagonal subgroup G is irreducible on U ® V, where U and V arethe representation modules for or and n, respectively.

But if u E U# then U = Cu and the map lo: au ® v H av is an isomorphismof U ® V with V such that x(g, g)(o ® n)cp =X(g)(x(p(g7r)) for g E G andx E U ® V, where u(ga) = ) (g)u and X(g) E C. Namely x = au ®v,

x(g, g)(a (9 7r)lp = (auger, vgn)1P = (a) (g)u, vgn)co = a) (g)vgn

and xcp(g7r) = avg7r. Thus co induces a bijection lp: W H Wlo of the CG-subspaces of U 0 V and V, so as G is irreducible on V, it is also irreducibleon U ® V.

Chapter 10, Exercise 6. (1) First, w =1 w with l(1) = 0, sow < w. Thus <is reflexive. Second, if u < w then w = xu with 1(w) = 1(x) +l (u), so as 1(x) > 0with equality iff x = 1, we have l(w) > 1(u) with equality iff x =1 and u = w.Then if also w < u, by symmetry, l(u) > l(w), so u = w. That is, < is anti-symmetric. Finally, assume u < w < v. Then w =xu and v = yw with 1(w) _I(x) + 1(u) and 1(v) = 1(y) + 1(w), so v = yw = yxu with

1(v) =1(y) + 1(w) =1(y) + 1(x) + 1(u).

Further l (yx) < 1(y) + 1(x), while as v = yxu,

l(yx) > 1(v) -1(u) = 1(y) + 1(w) - (1(w) - l(x)) =1(y) + 1(x),

so l(yx) = 1(y) + l(x) and hence u < v, proving < is transitive.(2) Let w be maximal with respect to < and suppose w wo. Then by

Exercise 10.3.2, 1(w) < IP1, so by 30.12, there is a E P with aw E P. Thenby 30.10, l(raw) > 1(w), so raw > w, contradicting the maximality of w.

(3) First, (b) and (c) are equivalent by 30.10.Next, if l(rw) < l(w), then by 30.10, l(w) = l(rw) + 1, so as 1(r) = 1, we

have 1(w) = l(rw) + 1(r) and w = r rw, so rw < w. That is, (b) implies (a).

282 Appendix

Let K be the fixed field of 0'. Then (0') = Gal(F/K), so by 26.3, V = F BK U for some irreducible KG-submodule U of V. As 0 is an automorphism of K of order 2, Exercise 9.1.4 says G preserves a nondegenerate hermitian symmetric form on U .

Chapter 9, Exercise 10. Let G1 = G2 = G. As a is 1-dimensional, a is irreducible and by hypothesis n is irreducible, so by 27.15, a €3 n is an irreducible representation of G1 x G2. Recall that if we identify G with the diagonal subgroup {(g, g): g E G) of G1 x G2 via the isomorphism g H ( g , g), then the tensor product representation a €3 n of G is the restriction of the tensor product representation a €3 n of G1 x G2 to the diagonal subgroup G, so it remains to show the diagonal subgroup G is irreducible on U €3 V, where U and V are the representation modules for a and n, respectively.

But if u E U' then U = Cu and the map (o: au €3 v H av is an isomorphism of U 8 V with V such that x(g , g)(a €3 n)(o = h(g)(x(o(gn)) for g E G and x E U €3 V, where u(ga) = h(g)u and h(g) E c'. Namely x = au €3 v,

and x(o(gn)=avgn. Thus (o induces a bijection (o: W H W(o of the CG- subspaces of U €3 V and V, so as G is irreducible on V, it is also irreducible on U €3 V.

Chapter 10, Exercise 6. (1) First, w = 1 . w with l(1) = 0, so w 5 w . Thus 5 is reflexive. Second, if u ( w then w =xu with l (w) = l ( x ) +l(u), so as l ( x ) 2 0 with equality iff x = 1, we have l (w ) 2 l (u ) with equality iff x = 1 and u = w. Then if also w ( u, by symmetry, l (u) 2 l(w), so u = w. That is, ( is anti- symmetric. Finally, assume u ( w ( v. Then w = xu and v = yw with 1 ( w ) = l ( x ) + l (u) and l ( v ) = l ( y ) + l (w) , so v = yw = yxu with

Further l(yx) 5 l ( y ) + l ( x ) , while as v = yxu,

so l(yx) = l ( y ) + l ( x ) and hence u ( v, proving ( is transitive. (2) Let w be maximal with respect to ( and suppose w # wo. Then by

Exercise 10.3.2, l (w ) < /PI , so by 30.12, there is a E P with aw E P. Then by 30.10,1(rffw) > l (w) , so raw > w , contradicting the maximality of w .

(3) First, (b) and (c) are equivalent by 30.10. Next, if 2(rw) 5 l(w), then by 30.10, l (w) = l (rw) + 1, so as l (r) = 1, we

have l (w) = l ( rw) + l ( r ) and w = r . rw , so rw ( w. That is, (b) implies (a).

Appendix 283

Assume rw <w. Then w = x rw with l(w)=l(rw)+l(x), so x = r andl(rw)=l(w) - 1=n - 1. Thus with r1 ER, so w=r(rw)=rr2 rn. That is, (a) implies (d). Finally, if (d) holds then visibly l(rw) _1(w) - 1, so (d) implies (b).

(4) As u<w, w=xu with l(w)=l(x)+l(u). As l(wr)<l(w), l(wr)=1(w) - 1 and similarly l(ur) =l(u) - 1. Now wr =xur with

1(wr)=1(w) - 1=l(x) + 1(u) - 1=l(x)+1(ur),

so ur < wr.

Chapter 11, Exercise 5. For 1 < u < v < n, let U,,,,, = U® ® Un+i ® ® U.For k = 2j or 2j - 1, let j (k) = j and 1(k) = j + 2 or j + 1, respectively. Let

U1 = Ul,j-1,U2 = Uj,l(k)-1, and U3 = Ul(k),n. Thus V = U1 ® U2 ® U3 and

and Qk(1+a®y)/,/2- or(a +,8)/,/2- for k even or odd, respectively. Let

as a aYP'a a , b- c= , d- e= .

Check that y,8 = -,8y and ya = -ay. Hence ya = -ay. Similarly, a,8 =- ,8a. Check also that a, ,8, y, and a are involutions, b2 = - 112=e2 , ac +ca=a, de+ed=e, c2 + 1/2 = c, and d2 + 1/2 = d.

(1)As if kisodd then Qk = a2 =1, while if k is even then

1®,8+a®y 21®,82+a®(.8Y+Y.8)+a2®Y2=1,

( 2 ) 2

as ,82 = a2 = y2 =1 and ,8y = -y,8. Therefore

Sk=(I ®ak(g k)2=I XQk

Next let V1=U1,j_1, V2=Uj,j+i, and V3=Uj+2,n. If k = 2j - 1 thenSk = I ® (a (9 y) where k = y ® ® y and Sk+1 = 1 ® ak+1 0 k+1Thus SkSk+1 = 10 (a ®y)ak+1 With SklSk+1 = y2 ®... ®Y2 =I asY2 = 1. Thus to show (SkSk+l)3 = -I we must show x3 = -I, where

x=(a®y)ak+l=(a®y)((l®,8+a®y)/,/2-)=a®b+c®1.Equivalently we must show x2 + 1= x. But

x2=a2®b2+(ac+ca)®b+c2®1=-1/2+a®b+c2®1,as a2 = 1, b2 = -1/2, and ac + ca= a, so as c2 + 1/2 = c, indeed

x2+1=a®b+c2®1+(1®1)/2==a®b+c®1=x,completing the proof that (S2j_1S2 j)3 = -I.

Appendix 283

Assume rw 5 w . Then w = x rw with l ( w ) = l ( r w ) + l ( x ) , so x = r and l ( r w ) = l ( w ) - l = n - 1. Thus r w = r z . . . r , with r i c R , so w = r ( r w ) = rrz . . . r,. That is, (a) implies (d). Finally, if (d) holds then visibly l ( r w ) = 1 (w ) - 1, so (d) implies (b).

(4) AS u ( w , w = x u with l (w)=l (x )+l (u ) . As l ( w r ) 5 l ( w ) , l (wr )= l ( w ) - 1 and similarly l (ur ) = l (u ) - 1. Now w r =xur with

so ur 5 wr.

Chapter 11, Exercise 5. For 1 5 u 5 v 5 n , let U,,, = U, @ U,+l €3.. . @ U,. For k = 2 j or 2 j - 1, let j (k) = j and l ( k ) = j + 2 or j + 1, respectively. Let

2 - u . U' = Ul , j - l , U - ,,l(k)-l, and u3 = U1(.),,. Thus V = U' €3 U' €3 u3 and sk = I €3 a k €3 6. where 6. = y €3 . . . €3 y and a . = (1 €3 /3 + a! €3 y) /& or (a! + /3)/& for k even or odd, respectively. Let

a!+B Y S aa! Ba Y a a=- , b = - c=- d = - e = - 1/Z &' 1/Z' &' &'

Check that y/3 = -/3 y and ya! = -my. Hence ya = -ay . Similarly, a/3 = - pa. Check also that a, /3, y , and a are involutions, b2 = - 112 = e2, ac +

2 ca=a ,de+ed=e ,c2+ 1/2=c,andd + 1 /2=d. (1 ) As y2= 1, at = ( y @ . . . @ Y ) 2 = Y 2 @ . + . @ Y 2 = 1. Similarly, if k is

odd then at = a2 = 1, while if k is even then

a;= ( ~ @ B ; @ Y ) ~ - ~ @ B ~ + . @ ( B Y + Y B ) + ~ ~ @ Y ~ - = 1, I 2

as = a!' = = 1 and /3 y = - y/3. Therefore

Next let V' = Ul , j - l , v2 = Uj, j+l , and v3 = Uj+2,,. If k = 2 j - 1 then sk=I @ ( a @ y ) @ 6 L 9 w h e r e 6 L = y @ . . . @ yandsk+l=I@ak+l@6k+l . Thus sksk+l = I @ (a @ y)ak+l 8 6Ltk+1 with 6;6k+l= y2 @ . . . @ y2 = I as

= 1. T ~ U S to show (S.S.+~)~ = -I we must show x3 = - I , where

Equivalently we must show x2 + 1 = x . But

completing the proof that (s2j-ls2j)3 = - I .

284 Appendix

Similarly let k = 2j. Then Sk = 10 ak 0 k and Sk+1 =1 ®(1 ® a) 0 k+1,so sksk+1=1® ak(1 0 a) 0 k k+1, and it remains to show y2 + 1= y, where

=(l 0 a)= (1®P + -Y)(1®a)y ark (l

But

y2=1®d2+a®(de+ed)+a2®e2=1®d2-1/2+a®e,as de + ed = e, a2 = 1, and e2 = - 1/2. Thus as d2 + 1/2 = d,

y2+ 1=a ®e+ 1 ®d2+(1 ®1/2)=a ®e+ 10 d = y,

completing the proof of (1).(2) Let k < i - 1; we claim either

(a) 1(k) < i(i), or(b) k = 2j and i = 2(j + 1).

For if j = j (k) then 2j - 1 < k < i - 1 < 2j(i) - 1, so j < j (i) - 1 and hencel (k) = j + e < j (i) unless e = 2 (so that k = 2j) and j (i) = j + 1. In the lattercase, as j (i) = j + 1 and k = 2j < i - 1, we have i = 2(j + 1), establishingthe claim.

In case (a) let Vl = Ui,j-1, V2 = Uj,1(k)-1, V3 = U1(k),j(i)-1, V4 = Uj(i),i(i)-1,and V5 = UI(i),n. Then V = V1 ® ® V5 and sr =sr,1 ® 0 Sr,5, where Srinduces sr,t on V. Further, Si,,, = Sk,1 = I for u < 3, sk,2 = ak, si,4 = ai, and si,5and Sk,,,, v > 2, are of the form y ® . ® y. In particular Sk,wSi,w = si,wsk,w forw 4, while sk,45i,4 = -si,45k,4, since ay = -ya, ,ay = -y,a, and ay = -yet.Therefore (SkSi)2 = J.

In case (b), let Wl = Ul,j_1, W2 = Uj,j+2, and W3 = Uj+3,n. Again V = W1®W2 ®W3 and Sr = sr,1 ®Sr,2 ®sr,3 with

Sr,1 =I, sr,3 of the form y ®... ®y, and

sk,2=(lOp ®Y+a®Y(&Y)1-12, si,2=(l®1®P+1(&a0Y)/J.

It remains to show (siSk)2 = -I, so we must show Sk,2Si,2 = -si,24,2. Thisfollows because Pa = -up, ay = -ya, and ,ay = -y,a.

(3) As (Sksi)2 = -I for i > k + 1, we have -I E G(1). Let G = G/(-I).Then G = (sk: 1 < k < 2n), and by (1) and (2), Isis j I = mi, j, where mi,i = 1,mi,i+1= mi+l,i = 3, and mi, j = 2 for Ji - j J > 1. Thus by 30.19, G = Sen. Letn > 2. As G - Stn with 2n > 4, G(1) = (1) = Aen is a nonabelian simplegroup by 15.16. Thus as -I E G(1), G(1) is quasisimple or G(1) = (-I) x G(2)with G(2) = A2n. In the latter case the projection of 5153 on G(2) is an involution,which is impossible, as (5153)2 = -I.

Chapter 11, Exercise 10. We first observe Z, is subnormal in Gy and hencealso in Gy,Z for each z E ry, since ZX < Gy,Z by definition of ZX . Let Qy = G y, ry .

284 Appendix

Similarly let k = 2j. Then sk = I €3 a k €3 Ck and sk+l= I €3 (1 €3 a) €3 Ck+l, so sksk+l= I C ~ J ak( l €3 a ) €3 (k(k+l, and it remains to show y2 + 1 = y, where

But

completing the proof of (1). (2) Let k i i - 1; we claim either

(a) l(k> i j(i), or (b) k = 2 j and i = 2(j + 1).

Forif j = j(k) then2j - 1 i k < i - 1 5 2 j ( i ) - 1, so j 5 j(i) - 1 and hence l(k)= j + c i j(i)unlessc=2(sothatk=2j)and j(i)= j+ 1.Inthelatter case, as j(i)= j + 1 and k = 2 j < i - 1, we have i =2 ( j + I), establishing the claim.

In case (a) let Vl = Ul,j-1, V2 = U;,l(k)-1, V3 = Ul(k),j(i)-1, V4 = Uj(i),l(i)-lr and Vg = Ul(i),n. Then V = Vl €3 . . . €3 V5 and s, = s , l €3 . . . €3 s,5, where sr induces s , , on V, . Further, s ~ , ~ = sk, 1 = I for u ( 3, Sk.2 = a k , s ~ , ~ = ai , and s i , ~ and ~ k , ~ , v > 2, are of the form y €3.. . €3 y. In particular Sk,WSi,w = Si,WSk,w for w # 4, whilesk,4si,4 = - ~ ~ , ~ s ~ , 4 , s i n c e a y = -ya, By = -yB,anday = -ya. Therefore (sksi)' = -I.

Incase (b), let W1= Ul, j-1, W2 = U;, j+2, and W3 = Uj+3,n. Again V = W1 €3 W2@W3 ands, =s , , 1€3~ , ,~€3~~ ,3 withs,,~ = I , s , ,~ oftheform y €3...€3y,and

~t remains to show (s~s~) ' = -I, SO we must show sk,zsi,z = -si,z~k,z. This follows because pa = -up, a y = -ya, and By = -yB.

(3) As (sksi)' = -I for i > k + 1, we have -I E G('). Let G = GI(-I) . Then G = (Sk: 1 i k < 2n), and by (1) and (2), lSiSjI =mi,j, where mi,i = 1, mi,i+1=mi+l, i=3,andmi9j=2forli- jl > l.Thusby30.19,G 2 Szn.Let n > 2. As G 2 Szn with 2n > 4, G(')= G(') 2 Ah is a nonabelian simple group by 15.16. Thus as -I E G('), G(') is quasisimple or G(') = (-I) x G ( ~ ) with G(') 2 Az,. In the latter case the projection of $ 1 ~ 3 on G ( ~ ) is an involution, which is impossible, as (~1~3) ' = -I.

Chapter 11, Exercise 10. We first observe Z, is subnormal in G, and hence also in G,,, foreachz E I',, since Z, i G,,, by definition of 2,. Let Q , = Gy,r,.

Appendix 285

Then Zx <Qy :!S G. <NG(Zx), so Zx a Qy a Gy, establishing the obser-vation.

Pick a prime q such that Zx is not a q-group, and for H < G let 0(H)Oq(H) = Oq(H)E(H). We next show 0(Gy) < Gx > 0(Gy,,). For as Zx is sub-normal in Gy and Gy,Z, we have E(Gy)E(Gy,,)<NGy(Zx) by 31.4, whileby hypothesis NG,(Zx) < Gx. Similarly, as Zx is not a q-group, 1 Oq(Zx),so NGy(Oq(Zx)) < Gx and by Exercise 11.9, Oq(Gy)Oq(Gy,z) < NG,,(Oq(Zx)).This completes the proof of the second claim.

Third we show 0(Gy)=0(Gx,y)=0(Qy). By claim 2, 0(Gy) <Gx,y, so as0(Gy) a Gy, 0(Gy) < 0(Gx,y) (cf. 31.3). By symmetry between x, y, z andz, x, y and claim 2, 0(Gx,y) < G, so 0(Gx,y) < Qy < NG(0(Gx,y)) and hence0(G) < 0 (Q y ). Finally, as 0(Q) a G, 0 (Q y) < 0 (G y ), establishing claim 3.

Now if Zy is not a q-group then we have symmetry between x and y, so byclaim 3, 0(Gy) = 0(Gx,y) = 0(Gx). Then if 0(Gx) 1,

Gy < NGy(0(Gy)) = NGy(0(Gx)) < Gx

and by symmetry, Gx < Gy, contradicting our hypothesis that Gx Gy. Thus0(Gx) = 1, so Oq(Gx) = E(Gx) = 1. But Zx 1, so F*(Gx) 1 and henceOp(GX) 1 for some prime p. However we have shown Oq(Gx) =1 unlessZx or Zy is a p-group, so interchanging the roles of x and y if necessary, wemay assume Zx is a p-group.

Next Qx a Gx,y, so 0(Qx) <0(Gx,y)=0(Qy). Hence

0(Qx) < n 0(Qy) < 0(Zx) =1,yErs

as Zx is a p-group. That is 0(Qx) =1. But Zy < Qx < Gx,y < NG(Zy), so0(Zy) < 0(Qx) = 1. Therefore F*(Zy) = Op(Zy). Thus as Zy 1, Op(Zy) 1,

so Zy is not a q-group for any prime q p, and hence by an earlier reduction,0(G,) = 9(Gx,y) = 0(G y) =1 for each q ¢ p. Therefore F*(Gx), F*(G y), andF*(Gx,y) are p-groups.

Chapter 11, Exercise 11. Let Y E Y, .F_ {Gx, Gy), and r = F(G, .T) be thecoset geometry defined in section 3. If Gx = Gy then Y = yGx = yGy = {y},contradicting the hypotheses that Y is nontrivial. Thus Gx G.

As G is primitive on X, Gx is maximal in G by 5.19. Thus if 10 H a Gx thenby maximality of G, NG(H) = Gx or G. But in the latter case, H < kerG. (G)by 5.7, while as G is faithful and transitive on X, kerGs (G) =1, contradictingH 1. Thus Gx = NG (H), so in particular NG, (H) < Gx . Thus, in the languageof Exercise 11.10, if Zx 1, then by Exercise 11.10, F*(Gx) is a p-group for

Appendix 285

Then Z, I Qy 5 G, 5 NG(Zx), so Z, I! Q, g G,, establishing the observation.

Pick a prime q such that Z, is not a q-group, and for H 5 G let B(H) = 8,(H) = Oq(H)E(H). We next show 8(Gy) ( G, > 8(GY,,). For as Z, is subnormal in G, and G,,,, we have E(G,)E(G,,,) 5 NGy (2,) by 3 1.4, while by hypothesis NGy(Zx) ( G,. Similarly, as Z, is not a q-group, 1 # Oq(Zx), so NG,(Oq(Zx)) 5 G, and by Exercise 11.9, Oq(G,)Oq(G,,,) 5 NG,(Oq(Zx)). This completes the proof of the second claim.

Third we show 8(Gy) = Q(G,,,) =8(Qy). By claim 2, 8(Gy) I G,,,, SO as 8(Gy) G,, 8(Gy)(8(G,,y) (cf. 31.3). By symmetry between x , y, z and z , x , y and claim 2, 8(GX,,) i Gz, so 8(G,,,) 5 Qy I NG(B(G,,,)) and hence B(G,,,) I 8(Qy). Finally, as 8(QY) 9 Gy,8(Qy) 5 8(Gy), establishingclaim 3.

Now if Z, is not a q-group then we have symmetry between x and y, so by claim 3, 8(Gy) = 8(G,,,) = 8(Gx). Then if Q(G,) # 1,

and by symmetry, G, ( G,, contradicting our hypothesis that G, # G,. Thus 8(G,) = 1, so Oq(G,) = E(G,) = 1. But Z, # 1, so F*(G,) # 1 and hence Op(G,) # 1 for some prime p. However we have shown Op(Gx) = 1 unless Z, or Z, is a p-group, so interchanging the roles of x and y if necessary, we may assume Z, is a p-group.

Next Qx 9 Gx,y, so g(Qx) 5 8(GX,,) = 8(Qy). Hence

as Z, is a p-group. That is 8(Q,) = 1. But Z, ( Q, 5 G,,, i NG(Z,), so 8(Zy) 5 8(Q,) = 1. Therefore F*(Z,) = Op(Zy). Thus as Z, # 1, Op(Zy) # 1, so Z, is not a q-group for any prime q # p, and hence by an earlier reduction, 8(G,) = 8(GX,,) =8(Gy) = 1 for each q # p. Therefore F*(G,), F*(G,), and F*(G,,,) are p-groups.

Chapter 11, Exercise 11. Let y E Y, F= {G,, G,), and r = r (G, F ) be the coset geometry defined in section 3. If G, = G, then Y = yGx = yG, = (y), contradicting the hypotheses that Y is nontrivial. Thus G, # G,.

As G is primitive on X, G, is maximal in G by 5.19. Thus if 1 # H g G, then by maximality of G,, NG(H) = G, or G. But in the latter case, H i kerG, (G) by 5.7, while as G is faithful and transitive on X, kerGx(G) = 1, contradicting H # 1. Thus G, = NG(H), soinparticular NGy(H) 5 G,. Thus, inthelanguage of Exercise 11.10, if Z, # 1, then by Exercise 11 .lo, F*(G,) is a p-group for

286 Appendix

some prime p. Therefore we may assume Z, = 1. Thus Gx is faithful on

0 = U c .yErx

Let m = I Y I. Then m = l Fx I =IF, I for each z E rx , so 101 <m2 and then, as

Gx is faithful on 0, 1 Gx I < m2 !. Thus we may take f (M) = M21 .

Chapter 12, Exercise 5. (1) The possible cycle structures for elements ofG are:

1, (1, 2), (1, 2, 3), (1, 2, 3, 4), (1, 2, 3, 4, 5), (1, 2)(3, 4, 5), (1, 2)(3, 4)

giving a set S = {gi : 1 < i < 71 of elements such that I gi I = i for i < 7 and1971= 2 by 15.2.4. Further S is a set of representatives for the conjugacy classesof G by 15.3.2.

(2) The representation of G on X is of degree 5 and 2-transitive by 15.12.2.Arguing as in Exercise 5.1, G has a transitive representation of degree 6 onthe coset space G/H, where H = NA(P) and P = (g5) E Sy15(A). Namely,counting the number of 5-cycles, G has 24 elements of order 5 and hence6 Sylow 5-subgroups, so by Sylow's Theorem, I G : H I = 6. Now the onlypossible cycle structure for the element g5 of order 5 on the 6-set G/H is onecycle of length 5 and one fixed point, so P is transitive on G/H - {H} andhence G is 2-transitive on G/H by 15.12.1.

Let 1/r be the permutation character of G on X. By Exercise 4.5.1, 1/r(g) isthe number of fixed points of g c G on X, so

1/r(gi) = 5, 3, 2, 1, 0, 0, 1

for i =1, ... , 7, respectively.Similarly let ip be the permutation character of G on G/ H. Let h = (1, 2, 4, 3).

By 15.3.1,g5=(2,4, 1,3,5)=g5,sohEHandH=P(h) as

I P(h)I = I P I I h I =20= IGI/IG : HI =1 H1.

Thus the elements in H are of conjugate to gi, g4, g5, and g7, so for all othergi, ip(gi) = IFixG/H(gi)I = 0. By Exercise 2.7, if Q E Sylq(H) then IFix(Q)I =I NG(Q) : NH(Q)I, so as H = NG(P), ip(g5) = 1 while as (h) E Sy12(H) withNG((h)) E Sy12(G) of order 8, ip(g4) = I c(h)I = 2. Thus g4 = (a, b, c, d)(e)(f)on G/H, so g7 = 924 = (a, c)(b, d) and hence ip(g7) = 2.

(3) By Exercise 12.1, G has I G : G(1) I linear characters. Let A = Alt(X);by 15.16, A is simple, and by 15.5, 1 G : A I = 2, so A = GM and G has twolinear characters, the principal character X1 and the sign character X2, where

286 Appendix

some prime p. Therefore we may assume 2, = 1. Thus G, is faithful on

Q = U r,. Y € ~ X

Let m = IYI. Then m = Ir,l= Ir,/ for each z E r,, so 131i m2 and then, as G, is faithful on 3 , IG, I m2!. Thus we may take f (m) = m2!.

Chapter 12, Exercise 5. (1) The possible cycle structures for elements of G are:

giving a set S = {gi: 1 I i ( 7) of elements such that Igi I = i for i < 7 and lg7 I = 2 by 15.2.4. Further S is a set of representatives for the conjugacy classes of G by 15.3.2.

(2) The representation of G on X is of degree 5 and Ztransitive by 15.12.2. Arguing as in Exercise 5.1, G has a transitive representation of degree 6 on the coset space GIH, where H = NA(P) and P = (gg) E Syl,(A). Namely, counting the number of 5-cycles, G has 24 elements of order 5 and hence 6 Sylow 5-subgroups, so by Sylow's Theorem, IG : HI = 6. Now the only possible cycle structure for the element gg of order 5 on the 6-set G I H is one cycle of length 5 and one fixed point, so P is transitive on GIH - {H) and hence G is 2-transitive on GI H by 15.12.1.

Let @ be the permutation character of G on X. By Exercise 4.5.1, @(g) is the number of fixed points of g E G on X, so

for i = 1, . . . ,7 , respectively. Similarly let rp be the permutation character of G on GI H. Let h = (1,2,4,3).

By 15.3.1,g,h=(2,4, 1 , 3 , 5 ) = g i , s o h ~ H a n d H = P ( h ) as

Thus the elements in H are of conjugate to gl, g4, gg, and g7, SO for all other gi, rp(gi) = IFixG/ff(gi)l = 0. By Exercise 2.7, if Q E Syl,(H) then IFix(Q)l = I NG(Q) : NH(Q)l, so as H = NG(P), rp(gs) = 1 while as (h) E Sy12(H) with N G ( ( ~ ) > E Sy12(G) of order 8, rp(g4) = Irp(h)l= 2. Thus g4 =(a, b, c, d)(e)(f) on GIH, so g7 = g: = (a, c)(b, d) and hence rp(g7) = 2.

(3) By Exercise 12.1, G has I G : G(')I linear characters. Let A = Alt(X); by 15.16, A is simple, and by 15.5, lG : A1 = 2, so A = G(') and G has two linear characters, the principal character and the sign character ~ 2 , where

Appendix 287

by Exercise 12.1, ker(X2) = A. That is X2(a) =1 for a E A and X2(9) _ -1 (theprimitive 2nd root of 1) for g c G - A.

(4) By Exercise 12.1, X1 and X2 are irreducible. By Exercise 12.6.3, X3 =1/r - X1 and X4 = p - X1 are irreducible characters. Then by Exercises 9.3 and9.10, Xi+2 = X2Xi, i = 3, 4, are also irreducible characters.

This gives 6 irreducible characters of G, exhibited in the first six rows of thetable below. As G has 7 conjugacy classes, G has 7 irreducible characters by34.3.1. Thus it remains to determine X7. Let ni = Xi(1); by 35.5.3,

7

120=IGI= n?i=1

so n7 = 6, giving the first column of the table. Then we use the orthogonalityrelation 35.5.2 to calculate X7(g0 for i > 1 to complete the table.

Character Table of S5

91 g2 g3 g4 g5 g6 g7

X1 1 1 1 1 1 1 1

X2 1 -1 1 -1 1 -1 1

X3 4 2 1 0 -1 -1 0

X4 5 -1 -1 1 0 -1 1

X5 4 -2 1 0 -1 1 0

X6 5 1 -1 -1 0 1 1

X7 6 0 0 0 1 0 -2

Chapter 13, Exercise 1. (1) Let M = CG(g). Then IHxMI = (M : MH I,where MHx is the stabilizer of Hx in M. Thus MHx = M fl Hx = CH=(g),so (IHxMI, p) = 1 if CHx(g) contains a Sylow p-subgroup of CG(g) ifCH(gx-') contains a Sylow p-subgroup of CG(gx-') iff gx-' is extremal in H.

(2) Let A = H/K, a: H A the natural map, and V the transfer of G intoA via a. Recall by 37.2 that V is a group homomorphism, so as A is abelianGf1> < ker(V) and hence it suffices to show gV 1.

Choose a set X of coset representatives for H in G as in 37.3. As IgI = p thelength ni of the ith cycle of g is 1 or p. If ni = p then 1 = gnh , so gnixr ' E gn' K;that is

gn'x`is

= (gn )a = (ga)n' = 1.

On the other hand ni = 1 if g fixes Hxi, in which case gxi ' E H. NowM-1

CG(g) permutes the fixed points of g on G/H and g(x`m)-' = gx1' for

Appendix 287

by Exercise 12.1, ker(x2) = A. That is x2(a) = 1 for a E A and x2(g) = - 1 (the primitive 2nd root of 1) for g E G - A.

( 4 ) By Exercise 12.1, ~1 and ~2 are irreducible. By Exercise 12.6.3, ~3 = @ - ~1 and x4 = rp - ~1 are irreducible characters. Then by Exercises 9.3 and 9.10, xi+2 = x2xi, i = 3 , 4 , are also irreducible characters.

This gives 6 irreducible characters of G , exhibited in the first six rows of the table below. As G has 7 conjugacy classes, G has 7 irreducible characters by 34.3.1. Thus it remains to determine ~ 7 . Let ni = ~ ~ ( 1 ) ; by 35.5.3,

so n7 = 6 , giving the first column of the table. Then we use the orthogonality relation 35.5.2 to calculate x7(gi) for i > 1 to complete the table.

Character Table of Ss

Chapter 13, Exercise 1. ( 1 ) Let M = CG(g). Then 1 H x M l = IM : M H x 1, where M H x is the stabilizer of H x in M . Thus MHx = M n H X = CHx(g) , so ( I H x M 1, p) = 1 iff CHI@) contains a Sylow p-subgroup of CG(g) iff c H ( g X - I ) contains a Sylow p-subgroup of cG(gX- I ) iff gX-I is extrernal in H .

( 2 ) Let A = H I K , a: H + A the natural map, and V the transfer of G into A via a. Recall by 37.2 that V is a group homomorphism, so as A is abelian G(' ) ( ker(V) and hence it suffices to show gV # 1.

Choose a set X of coset representatives for H in G as in 37.3. As IgI = p the -1

length ni of the ith cycle of g is 1 or p. If ni = p then 1 = ga , so gnlxl E gni K ; that is

On the other hand ni = 1 iff g fixes H x i , in which case gx;' E H . Now M = C G ( g ) permutes the fixed points of g on G I H and g("irn)-' = gx;l for

288 Appendix

each m E M, so if (HyjM: l < j < s} are the orbits of M on FixG/H(g)(Hx,: l < i < t} then

s

W a = r7 ((gy' )a)k,

=1 j1=11

where kJ = IHy,MJ is the length of the jth orbit. If p divides kj then(gyi')ki =

1 = (ga)ki. On the other hand, if p does not divide ki, then by (1), gyi ' is

extremal in H, so by hypothesis, gy" E g K and hence (gy ' )a = ga, so again(gyi 'a)ki _ (ga)ki. Thus using 37.3.3,

gV = fl ((gn`)x,')a = fl(ga)n' = (ga)ni=1 i=1

where n = E; _I n; _ IG : H1. Finally, by hypothesis, (IG : H1, p) = 1 andg E H - K, so ga 1, and hence IgaI = p. Therefore gV = (ga)n # 1,completing the proof.

Chapter 14, Exercise 9. (1) As u < w, w = xu for some x e W withn=1(w)=l(x)+l(u).Thenx=r1 rmandu=rm+1.. rnwith r, ES.By41.7, (C,: 0 < i < n) is a gallery from C to Cw in 8 where C; = Crn_,+i rnBy construction, Cl(") = Cn-m = Crm+1 "'rn = Cu.

(2) Let u = sw and n = 1(w). Thus w = su and by hypothesis, 1(u) < 1(w),so 1(u) = n - 1 and u < w. Hence by (1), Cu is adjacent to Cw, so E _Cu fl Cw is a wall of the chambers Cu and Cw.

Next BW fixes Cw and hence each subsimplex of Cw. Similarly B" fixeseach subsimplex of Cu, so B" BW fixes E. Thus B fl B"BW CNB(E) = A.

Let X be the set of galleries in 8 from C to E of length n - 1. By 41.7 and42.3, d(C, Cw) = n and d(C, Cu) = n - 1, so d(C, E) = n - 1. Thus X C Eby 42.3. As A acts on C and E, A acts on X. Further, Cu and Cw are the twochambers in E through E, so as d(C, Cw) = n, Cu is the terminal member ofeach member of X. Thus A < Gcu = B" = BSW

(3) By hypothesis,

l(s SW-1) = l(w-1) = (W)< l(ws) = l(sw-1),

so by (2), B fl B'B' c BW-'. Then conjugating by w, BW n BBS C B.(4) As u <w, w = xu with n = 1(w) =1(x) +1(u). Induct on l(x). If

l(x) = 0 then x = 1 and the lemma is trivial. Thus x = sy with s e S and1(y) = l(x) - 1, and u < yu, so by induction B fl By" < B". Also 1(sw) _l(yu) <1(w),soby(2),Bf1BW <By". Thus BflB' <Bf1By" <B"

Appendix 289

(5) Define

H = n BW.WEW

Then H is the pointwise stablilizer in G of E, so (a) and (b) are equivalent.By Exercise 10.6.2, w < wo for all w E W, so by (4), B fl BW0 < BW. ThusH = B fl BWO, so (b) and (c) are equivalent.

(6) Let T = (G, B, N, S). Recall H = B fl N a N with W = N/H and forw=nHEW,BW=Bn.Thus H<BWforallwEW,soH<H.

As T satisfies BN1, G = (B, N). Then as N < HN = N < G, G = (B, N).Also

BnN=BnRN =H(BfN)=HH=H.By construction, R is the pointwise stabilizer in G of E, so as N acts on E,N acts on H and hence H a HN = N. Thus T satisfies BN1. Also H fl Nis the kernel of the action of N on E, while by 43.5 and 41.8.2, W = N/H isregular on E, so H fl N = H. Let W = N/H. Then

W =N/H=NH/H=N/Hf1N=N/H=W,so BN2 is satisfied with S = (s:s e S), where w = nH for w = nH E S.Notice that as ft < B, we have wB = wB and BO = Bw, so also BO = Bw.Thus, as T satisfies BN3 and BN4, so does T.

Therefore T is a Tits system. By construction,

H=nBW=nBW,WEW WEW

so T is saturated.

Chapter 14, Exercise 10. (1) As aw > 0, l(sw) > 1(w) by Exercise 10.6.3,so l(w-Is) = l(sw) > 1(w) = l(w-1). Also by Exercise 10.6.2, w-Is < wo,and by Exercise 10.3.2, l (wos) < 1(wo). Therefore by Exercise 10.6.4, w-1 =(w-Is)s < wos. Then by Exercise 14.9.4,

Ba=BfBWOS<BW'.

(2) As aw < 0, 1(sw) < 1(w) by Exercise 10.6.3, so by Exercise 14.9.2,B n BSW BW c BSW. Conjugating this containment by w-1s, we concludeBW_is fl BBS C B, so BW_'S fl BS < B.

Next by hypotheses T is saturated, so by Exercise 14.9.5, B fl BWO = H.Thus

(BafBW )S=(B f1BWOSflBW-')S=B5nB'0nBW-'s<BnB'0=H

so BaflBW'=Hs =H.

Appendix 289

(5) Define

fl= n BW.WEW

Then H is the pointwise stablilizer in G of E, so (a) and (b) are equivalent.By Exercise 10.6.2, w < wo for all w E W, so by (4), B n BW ° < B'. ThusH = B n BWO, so (b) and (c) are equivalent.

(6) Let T = (G, B, N, S). Recall H = B n N a N with W = N/H and forw=nHEW,BW=B".Thus H<BWforallwEW,soH<H.

As T satisfies BN1, G = (B, N). Then as N < HN = N < G, G = (B, N).Also

BnN=BnRN =H(BnN)=HH=H.By construction, R is the pointwise stabilizer in G of E, so as N acts on E,N acts on H and hence H a RN = N. Thus T satisfies BN1. Also R n Nis the kernel of the action of N on E, while by 43.5 and 41.8.2, W = N/H isregular on E, so H n N = H. Let 4V = N/ H. Then

4V=N/H=NH/H=N/HnN=N/H=W,so BN2 is satisfied with S = (s": s E S), where w = nH for w = nH E S.Notice that as ft < B, we have wB = wB and BO = Bw, so also B"' = BI.Thus, as T satisfies BN3 and BN4, so does T.

Therefore T is a Tits system. By construction,

H=nBW=nBO,WEW WEW

so T is saturated.

Chapter 14, Exercise 10. (1) As aw > 0, l (sw) > I (w) by Exercise 10.6.3,so l(w-1s) = l(sw) > l(w) = l(w-1). Also by Exercise 10.6.2, w-1s < wo,and by Exercise 10.3.2, t(wos) < l(wo). Therefore by Exercise 10.6.4, w-1 =(w-1s)s < wos. Then by Exercise 14.9.4,

Ba=BnBWOS<BW'.

(2) As aw < 0, l(sw) < l(w) by Exercise 10.6.3, so by Exercise 14.9.2,B n B''WBW C B''W. Conjugating this containment by w-1s, we concludeBW-i8 n BBS C B, so BW-1S n BS < B.

Next by hypotheses T is saturated, so by Exercise 14.9.5, B n BWO = H.Thus

(BanB" )S= (BnBWOSnBW-')S=BSnBW°nBW-'S <BnBWO=H

so BanBW'=HI =H.

Appendix

(5) Define

H = n BW. W € W

Then H is the pointwise stablilizer in G of C, so (a) and (b) are equivalent. By Exercise 10.6.2, w 5 wo for all w E W, so by (4), B n BWO _( BW. Thus H = B i l BWO, so (b) and (c) are equivalent.

(6)Let F = (G, B, i?, S).Recall H = B n N r? N with W = N/H andfor w = n H ~ W , B ~ = B ~ . T h u s H _ ( B ~ f o r a l l w ~ W , s o ~ _ ( ~ .

AsTsatisfiesBN1,G = (B, N).ThenasN _( HN =i? _( G,G = (B,i?). Also

By construction, H is the pointwise stabilizer in G of C, so as N acts on C, N acts on H and hence H a AN = i?. Thus F satisfies BNl. Also H i l N is the kernel of the action of N on C , while by 43.5 and 41.8.2, W = N/H is regular on C , so H i l N = H. Let w = #/A. Then

so BN2 is satisfied with 3 = (S: s E S}, where B = n~ for w = n H E S. ~ot ice tha tas H 5 B,wehaveBB = wB andBB = Bw,soalso B' = Bw. Thus, as T satisfies BN3 and BN4, so does F .

Therefore F is a Tits system. By construction,

so ? is saturated.

Chapter 14, Exercise 10. ( 1 ) As uw > 0, l(sw) > l(w) by Exercise 10.6.3, so l(w-'s) = l(sw) > l(w) = l(w-'). Also by Exercise 10.6.2, w-'s 5 wo, and by Exercise 10.3.2, l(wos) 5 l(wo). Therefore by Exercise 10.6.4, w-' = (w-'s)s I wos. Then by Exercise 14.9.4,

(2) As aw < 0, l(sw) < l(w) by Exercise 10.6.3, so by Exercise 14.9.2, B i l BsWBW g BSW. Conjugating this containment by w-'s, we conclude Bw-Is n BBS E B, so Bw-Is n B" B.

Next by hypotheses T is saturated, so by Exercise 14.9.5, B i l BWO = H. Thus

290 Appendix

(3) As 1(w) < l(ws), l(w-1) < l(sw-1), so by 43.3.1, sBw-1 c Bsw-'B,and hence B c sBsw-1Bw = BSBW. Thus for b E B, there exist x c B' andy E Bw with b = xy. Then y = x-lb E Bw n BSB C B by Exercise 14.9.3,soy EBnBW andx=by-1 EBnBs.

(4) Notice as = -a < 0, so by (2), Ba n BS = H. Also, setting w = wos,1(w) < l(wo) = l(ws), so by (3), B = (B n BS)(B n Bw) = (B n BS)Ba.

(5) If Ba = H then by (4), B = H(B n Bs). Then as s is an involution,BS = HS(B n BS)S = H(B n BS) = B, contrary to BN4.

(6) If aw > 0 then Ba < Bw-' by (1). On the other hand, if aw < 0, thenby (2), H = Ba n Bw-', so Ba B'-' by (5).

(7) Define gyp: A (D by gyp: Ba i-H aw. Let P E r and w E W; we firstshow

(*) Ba = Bp if = aw.

First, if Ba = B then as B = B n Bw0'P, Ba = Bw-' n Bw0'pw-' . Thus by(6), aw > 0 < awrewo. Therefore awro < 0 by Exercise 10.3, so aw =by 30.7. Similarly if aw = P then awrpwo > 0 < caw, so by (6), Ba <Bw ' n Bw0'fw ' and hence Ba' < B n Bw0'P = B. As jw-1 = a, bysymmetry B' ' < B, so B = BC ,W Thus we have established (*).

By (*), Ba = B- if Ba U Bp if awu-1 = P if aw = 3u. Thus pis a well defined injection. By 30.9.3, p is a surjection. As Ba "cp = awu =(Ba )cpu, cp is W-equivariant.

(8) First notice

Bsw = (B n BwOw-'S)S = Bs n BwOw-'.

Next as aw < 0, aww0 > 0 by Exercise 10.3. Hence by (1), Ba < Bwow-', soBa <BnBw0w-' =Bw.

Again as awwo > 0, l(wwo) < l(swwo) by Exercise 10.6.3, so l(wow-'s) >l(w0w-').Then by Exercise 14.9.3, BS n Bw0w-' < B, so Bsw = BS n Bwow-' <B n BwOw Bw. Thus BaBsw S; Bw. On the other hand by (4), B =Ba(B n Bs), so as Ba < Bwow_I,

Bw = B n Bw0w_' = Ba(B n BwOw-' n Bs) = Ba(B n Bsw) C BaBsw

That is, BaBsw = Bw. Finally as = -a < 0, so by (2), Ba n BS = H. There-fore Ba n Bsw = Ba n BS n Bw°w-' = H.

(9) We apply (8) with w = wi_1w0, a = al, and s = ri for each 1 < i < n.By Exercise 10.6.3,aiwi_lwo < 0since l(riwi_1wo) = l(wiwo) < l(wi_lwo).Thus by (8),

(*) Bw;_iwo Bai Bw wo

290 Appendix

(3) As l(w) 5 l(ws), l(w-') 5 l(sw-'), so by 43.3.1, SBW-' 5 Bsw-' B, and hence B 5 s Bsw-' BW = BS BW. Thus for b E B, there exist x E BS and y E BW with b = xy. Then y = x-'b E BW i? BSB c B by Exercise 14.9.3, soy E B n BW andx = by-' E B n BS.

(4) Notice a s = -a < 0, so by (2), B, i? BS = H. Also, setting w = wos, Z(W) 5 z(wo) = ~(ws), SO by (3), B = (B n B ~ ) ( B n B ~ ) = (B n B~)B,.

(5) If B, = H then by (4), B = H(B i? BS). Then as s is an involution, BS = HS(B n Bs)S = H(B n BS) = B, contrary to BN4.

(6) If aw > 0 then B, 5 BW-' by (1). On the other hand, if aw < 0, then by (2), H = B, n BW-', SO B, 6 B ~ - ' by (5).

(7) Define p: A + @ by p: B," H aw. Let ,!? E n and w E W; we first show

(* 1 B," = Bp iff , !?=aw.

First, if B," = Bg then as Bg = B i? BWorB, B, = B"-' i? . Thus by (6), aw > 0 < awrpwo. Therefore awrp < 0 by Exercise 10.3, so aw = ,!? by 30.7. Similarly if aw = ,!? then awrpwo > 0 < aw, so by (6), B, 5 Bw-' n B W O ~ ~ W - ' and hence B," 5 B i? BWorB = Bp. As ,!?w-' = a , by symmetry B;' B,, so Bp = B,". Thus we have established (*).

By (*), B," = B; iff B,"u-' = Bp iff awu-' = ,!? iff aw = ,!?u. Thus p is a well defined injection. By 30.9.3, p is a surjection. As B,"'p = awu =

(B,")pu, p is W-equivariant. (8) First notice

Next as aw < 0, awwo > 0 by Exercise 10.3. Hence by (I), B, 5 BWow-I, so B, 5 B n BWOW-I = B,.

Againasawwo > O,l(wwo) 5 l(swwo)byExercise 10.6.3,sol(wow-'s) 2 l(wow-'). Thenby ~xercise 14.9.3, BS i? BWoW-' ( B, SO Biw = BS i? BWoW-' - < B i? Bwow-' = B,. Thus B,Biw 5 B,. On the other hand by (4), B = B,(B n BS), SO as B, 5 Bwow-I,

That is, B,Biw = B,. Finally a s = -a < 0, so by (2), B, i? BS = H. There- fore B, n BiW = B, n BS n BWow-' = H.

(9) We apply (8) with w = wi-lwo, a = ai, and s = ri for each 1 5 i < n. By Exercise 10.6.3, aiwi-lwo < Osince l(riwi-lwo) = l(wiwo) < l(wi-lwo). Thus by (8),

Appendix 291

and

(**) Bai fl B i,,. = H.

Claimk

(!) B"'i_1 Ba $WkWi-1W0 1wl-1 WkWO

j =i

for l < i < n and i < k < n. We induct on k. By (*),

Bw'-' - B 'i_I Br'wt_l

- Ba;w;_i BW1w;_Iwo a; w,wO - O

so (!) holds when k = i. Assume for k - 1. By (*),

BWk-1 - BWk-1 BrkWk-1 - B B WkWk_IWO ak WkWO - akWk-1 WkWO'

so by the induction assumption

k-1 kBWi-1 - B BWk-1 B Wk

wi-IwO ajwj_I Wk_1WO - ajwj_, * BWkWO,

i=i, 1=1

establishing (!). Notice that wn_lw0 = rn, so

BWn-1 = BW"-' = BW' ' = BW" 1WO rn a" anwn-1

Substituting this equality into (!) with k = n - 1, we get

n-1 n

(!!) BW' ' H B BW ' H BWi_1 WO = ajWj_1 W"-1 WO = ajwj_;

j=i j=i

Further B,,0 = B fl BwOwO = B, so applying (!!) at i = 1 we conclude

n

B = H Baj W j-1j=1

Next by (**) and (!!),

H = H" = (Bai fl BwiwO)w' 1 = Baiwi-1I I BWiWO

n

Baiwi-1 I I H Bajwj-ij=i+1

It remains to show 1Jrr: i H aiwi_1 is a bijection of I = {1, ..., n} with (D+.

We saw at the beginning of the proof that aiwi-1w0 < 0, so aiwi_1 > 0 andhence I1Jr c cp+, By Exercise 10.3, 1cI+J = n, so it remains to show iJr is aninjection. But if aiwi_1 = akwk_1 for some k > i, then

ai =CYkWk_1Wi 11 =akrk_1...ri,

Appendix

and

(**>

Claim

for 1 ( i < n and i ( k < n. We induct on k. By (*),

so (!) holds when k = i. Assume for k - 1. By (*),

so by the induction assumption

establishing (!). Notice that wn-lwo = rn, so

Substituting this equality into (!) with k = n - 1, we get

Further Bwo = B i? BWow0 = B, so applying (!!) at i = 1 we conclude

Next by (**) and (!!),

H = ~ w i - 1 - - (B,, n B ; ~ ~ ~ ) ~ ~ - ~ = B ~ ~ ~ ~ - ~ n B C : , ~ n

It remains to show +: i H aiwi-1 is a bijection of I = 11, . . . , n} with @+. We saw at the beginning of the proof that aiwi-lwo < 0, so aiwi-1 > 0 and hence I + C @+. By Exercise 10.3, ) @ + I = n, so it remains to show + is an injection. But if aiwi-1 = akwk-1 for some k > i, then

292 Appendix

so 0 > airi = akrk_1 ri+1, contrary to Exercise 10.6.3, since

l(rk ... ri+i) > l(rk-1 ... ri+i)

Chapter 15, Exercise 4. (1) To show H = CG(a) is balanced we must show

Or' (CH(X)) < Or'(H) for each X of order r in X.

But as H is solvable, this is immediate from 31.15.(2) We must show 0 is balanced. But for a, b E A#

0(b)nCG(a) = Or'(CG(b))nCG(a) = Or'(CCa(a)(X)) < Or'(CG(a)) = 0(a),

where X = (a, b), as CG(a) is balanced by the hypotheses of part (2).(3) Let 0 be the signalizer functor defined in (2); 0 is a signalizer functor by

(1) and (2). By the Solvable Signalizer Functor Theorem, 0 is complete; thatis, there exists an r'-subgroup Y of G such that 0(a) = Cy(a) for each a E A#.

For B a noncylic subgroup of A define

Y(B) = (0(b) : b E B#).

Now for g E G with a, ag E A#,

0(a)9 = Or,(CG(a))g = Or'(CG(ag)) = 0(ag),

so NG(B) < NG(Y(B)). Further, by construction Y(B) < Y, while by 44.8.1,Y < Y(B), so Y = Y(B). Thus NG(B) < NG(Y), so I'2,A(G) < NG(Y).But by hypothesis, G = 1'2,A(G), so Y < G, and hence as Y is an r'-group,Y _< Or'(G). Therefore Or (CG(a)) < Y < Or'(G).

In particular if Or (G) = 1 then Or (CG (a)) = 1. Hence, as CG (a) is solvable,

F*(CG(a)) = F(CG(a)) = Or(CG(a))

Chapter 16, Exercise 5. (1) By definition of strong embedding, H is properof even order, so there is an involution i E H. Indeed, 46.4.3 is one of theequivalent conditions for strong embedding, so

(*) (HnH91isoddforgEG-H.Let j E I; it suffices to show j E iG. Suppose jG _c H. Then for g E G,

jg E H n Hg, so IH n HgI is even and hence g E H by (*). As this holdsfor each g E G, G = H, contradicting H proper. This contradiction showsthere is a conjugate s of j in G - H, and it suffices to show s E iG. But ifnot, then by 45.2, lisI is even, so by 45.2.3 there is an involution z E (is)centralizing i and s. Now as 46.4.1 is one of the equivalent conditions definingstrong embedding, CG(t) < H for each t E I n H, so z E CG(i) < H, andthen also s E CG(z) < H, contrary to the choice of s.

(2)Leti,t E InH.By(1),ig = tforsomeg E G,sot E HnHgandhence, as above, g E H by (*). That is, H is transitive on I n H.

292 Appendix

so 0 > air; = akrk-1 . . ri+l, contrary to Exercise 10.6.3, since

Chapter 15, Exercise 4. (1) To show H = CG(a) is balanced we must show

Or! (CH ( X ) ) 5 Or/ ( H ) for each X of order r in X.

But as H is solvable, this is immediate from 3 1.15. (2) We must show O is balanced. But for a , b E A'

where X = (a , b ) , as CG(a) is balanced by the hypotheses of part (2). (3) Let 8 be the signalizer functor defined in (2); O i s a signalizer functor by

(1) and (2). By the Solvable Signalizer Functor Theorem, O is complete; that is, there exists an rl-subgroup Y of G such that O(a) = Cr(a) for each a E A'.

For B a noncylic subgroup of A define

Y ( B ) = (O(b) : b E B').

Now for g E G with a , ag E A',

O(alg = 0 , ! ( C ~ ( a ) ) ~ = Ort(Cc(ag)) = O(ag),

so NG(B) 5 NG(Y(B)). Further, by construction Y ( B ) ( Y , while by 44.8.1, Y _( Y ( B ) , SO Y = Y ( B ) . Thus NG(B) 5 NG(Y), so r z , ~ ( G ) 5 NG(Y). But by hypothesis, G = r2,A(G), SO Y <1 G , and hence as Y is an rl-group, Y < Orf(G). Therefore Or,(CG(a)) ( Y ( Orr(G).

In particular if Or! ( G ) = 1 then Or/ (CG (a)) = 1. Hence, as CG (a) is solvable,

F*(CG(a)) = F(CG(a>> = Or(CG(a)).

Chapter 16, Exercise 5. (1) By definition of strong embedding, H is proper of even order, so there is an involution i E H . Indeed, 46.4.3 is one of the equivalent conditions for strong embedding, so

Let j E I ; it suffices to show j E iG. Suppose jG E H . Then for g E G , jg E H n Hg, SO IH n Hgl is even and hence g E H by (*). As this holds for each g E G , G = H , contradicting H proper. This contradiction shows there is a conjugate s of j in G - H, and it suffices to show s E iG. But if not, then by 45.2, lisl is even, so by 45.2.3 there is an involution z E ( i s ) centralizing i and s. Now as 46.4.1 is one of the equivalent conditions defining strong embedding, CG(t) 5 H for each t E I n H , so z E CG(i) 5 H , and then also s E CG(z) ( H , contrary to the choice of s.

(2) Let i, t E 1 n H . By (I), i g = t for some g E G , so t E H fl Hg and hence, as above, g E H by (*). That is, H is transitive on I n H .

Appendix 293

(3) As u 0 H, D = H n H" is of odd order by (*).(4) Let j E J. Then j E D and j" = j-1 by definition of J. Thus (uj)2 =

ujuj = j"j = j-' j = 1, so Uj E I. As j E D, Uj E uD n I, so uJ cuD n I. On the other hand, suppose d E D with ud E I. Then 1 = (ud)2 =udud = dud, so d" = d-1 and hence d E J; that is, uD n I c uJ.

(5) Set Y = CG(j)(u). As J C D and I D I is odd by (3), 1 j I is odd, so asj 0 1, I j I > 2. Then as u inverts j, we have u 0 CG (j ). On the other handu E NG((j)), so Y < NG((j)) and hence CG(j) Y.

Suppose t E CG(j) and let K = H n Y. By construction, u E Y - H, soK is proper in Y and t E Y so K has even order. Further for Y E Y - K,y 0 H, so IK n KYI is odd by (*), and hence K is strongly embedded in Y.Thus applying (1) to Y, the involutions u and t are conjugate in Y. This isimpossible, as t E CG (j) < Y while u E Y - CG (j ).

(6) Let r, s be distinct involutions in uD. Then rD = uD and 10 rs E D,so

s= ErDnI =uDnI =Jby (4). But if rCG(t) = SCG(t), also rs E CG(t), contrary to (5). That is,rCG(t) 0 sCG(t).

(7) and (8): Define n = I G : H I and

0 = {(i, x, y): i c I and (x, y) is a cycle in i on G/H}.

As 46.4.4 is one of the equivalent conditions for strong embedding, i fixes aunique point of G/H, so i has (n -1)/2 cycles of length 2. Thus 101 =III (n -1)and III =n1I nHI =nm,so I0I =n(n - 1)m.

On the other hand, if A = X x X and 8(x, y) is the number of involutionswith cycle (x, y), then

101 = 8(x, y)(x,y) E A

Now, up to conjugation in G, we have x = H, and if 8(x, y) 0 0 then y = Hufor some u c I - H. Let 8 = 8(x, y), u1, ... , us be the involutions with cycle(x, y), and d1 = ulul for 1 < i < 8. Then dl c D < H, and if i 0 j thenby (6),

dl 1dj = utu1u1u3CG(t) = utu3CG(t) 0 CG(t),

so diCG(t) 0 dJCG(t). Thus

8<IH:Cx(t)1=ItG1=IInHI=mby (2). Hence

n(n-1)m=IQI= i 8(x,y)<IAIm=n(n-1)m,(x,y) E A

Appendix 293

(3) As u 4 H, D = H n Hu is of odd order by (*). (4) Let j E J . Then j E D and jU = j-' by definition of J. Thus ( ~ j ) ~ =

u j u j = j u j = j - ' j = 1 , s o u j E I . A s j E D , u j E u D n I , s o u J E u D n I. On the other hand, suppose d E D with ud E I. Then 1 = ( ~ d ) ~ = udud = dud , so du = d-' and hence d E J ; that is, u D n I 5 u J.

(5) Set Y = C c ( j ) ( u ) . AS J _C D and ID1 is odd by (3), Ijl is odd, so as j # 1, 1 jl > 2. Then as u inverts j , we have u 4 C c ( j ) . On the other hand

u E N c ( ( j ) ) , so Y 5 N G ( ( j ) ) and hence C G ( j ) Y . Suppose t E C G ( j ) and let K = H n Y. By construction, u E Y - H, so

K is proper in Y and t E Y so K has even order. Further for y E Y - K , y 4 H, so IK n Kyl is odd by (*), and hence K is strongly embedded in Y. Thus applying (1) to Y, the involutions u and t are conjugate in Y. This is impossible, as t E C G ( j ) <1 Y while u E Y - CG(j ) .

(6) Let r, s be distinct involutions in uD. Then r D = u D and 1 # rs E D, SO

by (4). But if r C ~ ( t ) = s C G ( ~ ) , also rs E CG(t), contrary to (5). That is, rCc(t> # sCc(t ) .

(7) and (8): Define n = I G : HI and

C2 = { ( i , x , y): i E I and ( x , y) is acycleini on GIH).

As 46.4.4 is one of the equivalent conditions for strong embedding, i fixes a uniquepointofG/H,soihas(n-1)/2cyclesoflength2.Thus~~~ = III(n-1) and 111 = nlI n HI = nm, so 1C21 = n(n - 1)m.

On the other hand, if A = X x X and 6(x , y) is the number of involutions with cycle ( x , y), then

Now, up to conjugation in G, we have x = H, and if 6(x , y) # 0 then y = Hu for some u E I - H. Let 6 = 6(x , y), u l , . . . , us be the involutions with cycle ( x , y), and di = ului for 1 5 i 5 6. Then di E D 5 H, and if i # j then by (61,

SO di CG ( t ) # d j CG(t). Thus

by (2). Hence

294 Appendix

so all inequalities are equalities and therefore S(x, y) = m for all x 0 y inG/H.

This establishes (7). Further, as diCG(t) 0 djCG(t) for i 0 j, we haveID: CD(t)I > S = m = I H : CH(t)I, so H = DCH(t) andhence (2) implies (8).

Chapter 16, Exercise 6. (1) Let z be an involution in Z(T). By hypothesis,CG(Z) is an elementary abelian 2-group, so as T < CG(Z), it follows thatCG(Z) = T. Thus (1) holds. Further as T is elementary abelian, each t E To isan involution in the center of T, so by symmetry between t and z, T = CG(t).Therefore T is a TI-set in G.

(2) As T is a TI-set in G, NG(S) < H for each 10 S < T, and as T ESy12(G), this is one of the equivalent conditions listed in 46.4 (with k = 1) forH to be strongly embedded in G.

(3) By (2), we may apply Exercise 16.5. In particular, by part (3) of thatexercise, D is of odd order. Thus ICD(u)I is odd, while as u is an involution,CD(U) is a 2-group, so CD(U) = 1. Hence u inverts D.

Adopt the notation of Exercise 16.5. Then I fl H = T #, so m = I I fl H I =q - 1, as T is elementary abelian of order q. By Exercise 16.5.2, H is transitiveonIflH,soq-1=IH:CH(t)I=IH:TI,andhence IHI=ITI(q-1)=q(q - 1). As u inverts D, J = uD is of order IDI, while J is the set ofinvolutions with cycle (H, Hu) on X, so by Exercise 16.5.7, IJI = m. ThusIDI = I J I = m = q - 1. Therefore, as D < H with T fl D = 1 as D is of oddorder, we have

ITDI = ITIIDI = q(q - 1) = IHI,

so H = TD and D is a complement to T in H.(4) First, as T is noncyclic, q > 2, so I D I = q - 1 > 1. Further, as u inverts

D, D is abelian, and T is abelian, while HIT - D, as D is a complementto T in H by (3). Thus H is solvable and D is a Hall 2'-subgroup of H, soDG fl H = DH by Phillip Hall's Theorem 18.5.

Let M = NG(D) and u E S E Sy12(M). As T is elementary abelian, so is S,so if S 0 (u) then S is noncyclic and hence D = (CD(s) : S E S#) by Exercise8.1. This is impossible, as D 0 1 while CD(S) = 1, since CG(S) is a 2-groupand I D I is odd.

Thus S = (u), so by 39.2, M = (u)E, where E = O(M). Let A = Fix(D)be the fixed points of D on X. As DG fl H = DH, M is transitive on A by5.21. Thus k = JAI = IM:MfHI.But

m fl H = NH(D) = NDT(D) = DNT(D) = D,

as H = TD and ND(T) = 1, as CD(t) = 1 for each t E To. Thus MID isregular on A.

294 Appendix

so all inequalities are equalities and therefore S(x, y) = m for all x # y in GIH.

This establishes (7). Further, as diCG(t) # djCG(t) for i # j, we have ID : CD(t)l 2 6 = m = IH : CH(t)l,so H = DCH(t)andhence(2)implies(8).

Chapter 16, Exercise 6. (1) Let z be an involution in Z(T). By hypothesis, CG(z) is an elementary abelian 2-group, so as T ( CG(Z), it follows that CG(z) = T. Thus (1) holds. Further as T is elementary abelian, each t E T# is an involution in the center of T, so by symmetry between t and z, T = CG(~) . Therefore T is a TI-set in G.

(2) As T is a TI-set in G, NG(S) 5 H for each 1 # S ( T, and as T E Sy12(G), this is one of the equivalent conditions listed in 46.4 (with k = 1) for H to be strongly embedded in G.

(3) By (2), we may apply Exercise 16.5. In particular, by part (3) of that exercise, D is of odd order. Thus ICD(u)l is odd, while as u is an involution, CD(u) is a 2-group, so CD(u) = 1. Hence u inverts D.

Adopt the notation of Exercise 16.5. Then I n H = T', so m = II n HI =

q - 1, as T is elementary abelian of order q. By Exercise 16.5.2, H is transitive on I n H, so q - 1 = IH: CH(t)l = IH: TI, and hence IHI = ITI(q - 1) = q(q - 1). As u inverts D, J = uD is of order IDI, while J is the set of involutions with cycle (H, Hu) on X, so by Exercise 16.5.7, I JI = m. Thus ID1 = IJI = m =q-l.Therefore,asD 5 H w i t h T n D = 1asDisofodd order, we have

so H = TD and D is a complement to T in H. (4) First, as T is noncyclic, q > 2, so I Dl = q - 1 > 1. Further, as u inverts

D, D is abelian, and T is abelian, while H I T S D, as D is a complement to T in H by (3). Thus H is solvable and D is a Hall 2'-subgroup of H , so DG n H = D~ by Phillip Hall's Theorem 18.5.

Let M = NG(D) and u E S E Sy12(M). As T is elementary abelian, so is S, so if S # (u) then S is noncyclic and hence D = (CD(s) : s E s') by Exercise 8.1. This is impossible, as D # 1 while CD(s) = 1, since CG(S) is a 2-group and I Dl is odd.

Thus S = (u), so by 39.2, M = (u) E, where E = O(M). Let A = Fix(D) be the fixed points of D on X. As D~ n H = D ~ , M is transitive on A by 5.21. T ~ U S k = la1 = I M : M n H I . B U ~

as H = TD and ND(T) = 1, as CD(t) = 1 for each t E T'. Thus M I D is regular on A.

Appendix 295

Let F be the set of triples (i, x, y) such that i E uE and (x, y) is a cycle of ion . As El is odd, CE(U) = 1,so uE=InMisoforderIEI,andasM/Dis regular on A, each i E uE is regular on A. Hence there are k members of Fwhose first entry is i, so

II'I = I EIk = IDIIE : Dik = IDIk2/2 = (q - 1)k2/2.

On the other hand, there are k(k - 1) choices for (x, y), and by Exercise 16.5.7there are m = q -1 involutions i with cycle (x, y), each in i Gx y = i D c uE,so IFI = k(k - 1)(q - 1). Thus k = 2, so D = E and (4) is established.

(5) Let x = H E X, y = xu, and z E X - {x} = X'. By Exercise 16.5.7,there is v E I with cycle (x, z), so by symmetry between u and v, Fix(Do) =(x, z) and Do = H fl H° is a Hall 2'-subgroup of H. Therefore there is h E Hwith Dh = Do by Phillip Hall's Theorem 18.5, as observed during the proofof (4). Now by (4)

(x, yh) = Fix(D)h = Fix(Dh) = Fix(Do) = {x, z},

so yh = z. Finally, as H = DT, we have h = dt for some d E D and t E T,so z = yh = ydt = yt, and therefore T is transitive on X. But by 46.4,{x} = Fix(t) for each t E TO, so T is regular on IX'I.

As T is regular on X', the action of D on X' is equivalent to its action on T byconjugation by 15.11. Thus as D is regular on T#, D is regular on X - (x, y),so G is 3-transitive on X and only the identity fixes 3 or more points of X.

(6) Let q = 2e, and regard T as an e-dimensional vector space over GF(2)and D as a subgroup of GL(T) as in 12.1. Let E = EndD (T ). As D is regular onT#, D is irreducible, so by Schur's Lemma 12.4, E is a finite division algebraover GF(2) of dimension at most e. Therefore by 26.1, E is a finite field andf = I E: GF(2)I < e. But D# < E#, so

q-1=IDI:5 IEI=2f-1<2e-1=q-1,

so E = F. Thus T can be regarded as the additive group of F, and D as themultiplicative group F#, and the action of D by conjugation is a: b H ab fora E D and b E T. Thus D and T are determined up to isomorphism andthe representation of D on T is determined up to quasiequivalence, so thesemidirect product H is determined up to isomorphism by 10.3. Hence thereexists an isomorphism 7r: H H* with D7r = D* _ (O(a, 0, 0, 1): a E F#),and the map

,B: HID -+ H*/D*,

Dt H D*(t7r)

Appendix 295

Let r be the set of triples (i, x, y) such that i E uE and (x, y) is a cycle of i on A. As [El is odd, CE(u) = 1, SO U E = I n M is of order /El, and as MID is regular on A, each i E uE is regular on A. Hence there are k members of r whose first entry is i, so

On the other hand, there are k(k - 1) choices for (x, y), and by Exercise 16.5.7 there are m = q - 1 involutions i with cycle (x, y), each in iG,,, = iD uE, so Irl = k(k - l)(q - 1). Thus k = 2, so D = E and (4) is established.

(5) Let x = H E X, y = xu, and z E X - {x) = XI. By Exercise 16.5.7, there is v E I with cycle (x, z), so by symmetry between u and v , Fix(Do) = (x, z} and Do = H n H U is a Hall 2'-subgroup of H. Therefore there is h E H with D~ = DO by Phillip Hall's Theorem 18.5, as observed during the proof of (4). Now by (4)

(x, yh} = Fix(D)h = F ~ x ( D ~ ) = Fix(Do) = {x, z),

so yh = z. Finally, as H = DT, we have h = dt for some d E D and t E T, so z = yh = ydt = yt, and therefore T is transitive on XI. But by 46.4, {x} = Fix(t) for each t E T', so T is regular on IX'I.

As T is regular on XI, the action of D on XI is equivalent to its action on T by conjugation by 15.1 1. Thus as D is regular on T', D is regular on X - (x, y}, so G is 3-transitive on X and only the identity fixes 3 or more points of X.

(6) Let q = 2", and regard T as an e-dimensional vector space over GF(2) and D as a subgroup of GL(T) as in 12.1. Let E = EndD(T). As D is regular on T', D is irreducible, so by Schur's Lemma 12.4, E is a finite division algebra over GF(2) of dimension at most e. Therefore by 26.1, E is a finite field and f = I E : GF(2)I 5 e. But D' ( E', so

so E = F . Thus T can be regarded as the additive group of F , and D as the multiplicative group F', and the action of D by conjugation is a : b H ab for a E D and b E T. Thus D and T are determined up to isomorphism and the representation of D on T is determined up to quasiequivalence, so the semidirect product H is determined up to isomorphism by 10.3. Hence there exists an isomorphism n : H 4 H* with D n = D* = (@(a, O,0, 1): a E F'}, and the map

296 Appendix

is a permutation isomorphism of the representations of H on HID and H* onH*/D*. Finally the map

y: H*/D* - F,

D*0(1, b, 0, 1) H b

is an equivalence of the representations of H* on H*/D* and F, so composingthese two isomorphisms, we get the isomorphism a of (6).

(7) As D is transitive on X - {H, Hu}, there is d E D with (la-1)ud =la-1. Let v = ud. Note that as u inverts D and D is abelian, v inverts D. Fora E F#, let *(a) = O(a, 0, 0, 1). Thus *(a)-1 = '(a-1) and *(a)n-1 E D,so 1'(a)n-1 is inverted by v. Therefore

(aa-1)v = ((lf'(a))a-1)v = (la-1)*(a)n-1v =(la-1)v(r(a)n-1)v

_ (la-1)(*(a)-1)n-1

= (lr(a-1))a-1 = a-1a-1

(8) As a: X - Y is an isomorphism of sets, a*: Sym(X) - Sym(Y) is anisomorphism of groups, where a*: g H a-1ga. Thus a*: L - L* is an iso-morphism, where L = (H, v), L* = (H*, v*), and v* = va*. However, byconstruction, v* fixes 1 and has cycles (0, oo), (a, a-1), for a E F - to, 11.That is, v* = 4(0, 1, 1, 0) E G*. Further a* extends n as xahn = (xh)a for allx E Xandh E H.ThusitremainstoshowthatG = (H, v) and G* = (H*, v*).But as G is 3-transitive on X, G is primitive on X by 15.14, so H is maximalin G by 5.19, and hence indeed G = (H, v). Similarly G* = (H*, v*).

296 Appendix

is a permutation isomorphism of the representations of H on H I D and H* on H *I D*. Finally the map

is an equivalence of the representations of H* on H*/D* and F , so composing these two isomorphisms, we get the isomorphism a of (6).

(7) As D is transitive on X - { H , H u ) , there is d E D with ( la- ')ud = la-'. Let v = ud. Note that as u inverts D and D is abelian, v inverts D. For a E F', let @(a) = $(a, O , 0 , 1). Thus @(a)-' = *(a-') and @(a)n-' E D , so @(a)nW' is inverted by v. Therefore

(8) As a : X + Y is an isomorphism of sets, a*: Sym(X) + Sym(Y) is an isomorphism of groups, where a*: g H a- lga . Thus a*: L + L* is an isomorphism, where L = ( H , v ) , L* = ( H * , v*), and v* = va*. However, by construction, v* fixes 1 and has cycles (0 , oo), ( a , a-'), for a E F - (0 , 1) . That is, v* = (b(0, 1, 1,O) E G*. Further a* extends n as x a h n = ( x h ) a for all x E X and h E H . Thusitremains to show that G = ( H , v ) and G* = ( H * , v*). But as G is 3-transitive on X, G is primitive on X by 15.14, so H is maximal in G by 5.19, and hence indeed G = ( H , v ) . Similarly G* = (H*, v*).

References

Al. Alperin, J., Sylow intersections and fusion, J. Alg. 6 (1967), 222-41.AL. Alperin, J. and Lyons, R., On conjugacy classes of p-elements, J. Alg. 19 (1971), 536-7.Ar. Artin, E., Geometric Algebra, Interscience, New York, 1957.As 1. Aschbacher, M., On finite groups of component type, Illinois J. Math. 19 (1975), 78-115.As 2 Aschbacher, M., Finite groups of rank 3, I, II, Invent. Math. 63 (1981), 357-402; 71 (1983),

51-162.Be 1. Bender, H., On groups with abelian Sylow 2-subgroups, Math. Z. 117 (1970), 164-76.Be 2. Bender, H., Goldschmidt's 2-signalizer functor theorem, Israel J. Math. 22 (1975), 208-13.BT. Borel, A. and Tits, J., Elements unipotents et sousgroupes paraboliques de groupes reductifs,

Invent. Math. 12 (1971), 97-104.Bo. Bourbaki, N., Groupes et algebras de Lie, 4, 5, 6, Hermann, Paris, 1968.BE Brauer, R. and Fowler, K., On groups of even order, Ann. Math. 62 (1965), 565-83.CPSS. Cameron, P., Praeger, S., Saxl, J. and Seitz, G., The Sims conjecture, to appear.Ca. Carter, R., Simple Groups of Lie Type, Wiley-Interscience, New York, 1972.Ch 1. Chevalley, C., The Algebraic Theory of Spinors, Columbia University Press, Momingside

Heights, 1954.Ch 2. Chevalley, C., Theorie des groups de Lie, Tome II, Hermann, Paris, 1951.Co. Collins, M., Some problems in the theory of finite insolvable groups, Thesis, Oxford

University, 1969.Di. Dieudonne, J., La geometrie des groupes classiques, Springer-Verlag, Berlin, 1971.FT Feit, W. and Thompson, J., Solvability of groups of odd order, Pacific J. Math. 13 (1963),

775-1029.FS. Fong, P. and Seitz, G., Groups with a (B, N)-pair of rank 2, I, II, Invent. Math. 21 (1973),

1-57; 24 (1974), 191-239.Gl 1. Glauberman, G., On solvable signalizer functors in finite groups, Proc. London Math. Soc.

23 (1976), 1-27.G12. Glauberman, G., Failure of factorization in p-solvable groups, Quart. J. Math. Oxford 24

(1973), 71-7.G13. Glauberman, G., Central elements in core-free groups, J. Alg. 4 (1966), 403-20.Gol 1. Goldschmidt, D., 2-signalizer functors on finite groups, J. Alg. 21 (1972), 321-40.Gol 2. Goldschmidt, D., Solvable signalizer functors on finite groups, J. Alg. 21 (1972), 137-48.Gor 1. Gorenstein, D., On the centralizers of involutions in finite groups, J. Alg. 11 (1969),

243-77.Gor 2. Gorenstein, D., The Classification of Finite Simple Groups, I, Plenum Press, New York,

1983.Gor 3. Gorenstein, D., Finite Simple Groups; An Introduction to their Classification, Plenum

Press, New York, 1982.Gor 4. Gorenstein, D., Finite Groups, Harper and Row, New York, 1968.GL. Gorenstein, D. and Lyons, R., The local structure of finite groups of characteristic 2 type,

Memoirs AMS 276 (1983), 1-731.GW. Gorenstein, D. and Walter, J., Balance and generation in finite groups, J. Alg. 33 (1975),

224-87.

References

Al. Alperin, J., Sylow intersections and fusion, J . Alg. 6 (1967), 22241. AL. Alperin, J. and Lyons, R., On conjugacy classes of p-elements, J. Alg. 19 (1971), 536-7. Ar. Artin, E., Geometric Algebra, Interscience, New York, 1957. As 1. Aschbacher, M., On finite groups of component type, Illinois J. Math. 19 (1975), 78-1 15. As 2 Aschbacher, M., Finite groups of rank 3, I, 11, Invent. Math. 63 (1981), 3 5 7 4 2 ; 71 (1983).

51-162. Be 1. Bender, H., On groups with abelian Sylow 2-subgroups, Math. Z. 117 (1970), 164-76. Be 2. Bender, H., Goldschmidt's Z-signalizer functor theorem, Israel J. Math. 22 (1975), 208-13. BT. Borel, A. and Tits, J., Elements unipotents et sousgroupes paraboliques de groupes reductifs,

Invent. Math. 12 (1971), 97-104. Bo. Bourbaki, N., Groupes et algebras de Lie, 4,5,6, Hermann, Paris, 1968. BF. Brauer, R. and Fowler, K., On groups of even order, Ann. Math. 62 (1965), 565-83. CPSS. Cameron, P., Praeger, S., Saxl, J. and Seitz, G., The Sims conjecture, to appear. Ca. Carter, R., Simple Groups of Lie Type, Wiley-Interscience, New York, 1972. Ch 1. Chevalley, C., The Algebraic Theory of Spinors, Columbia University Press, Morningside

Heights, 1954. Ch 2. Chevalley, C., Thkorie des groups de Lie, Tome 11, Hermann, Paris, 1951. Co. Collins, M., Some problems in the theory of finite insolvable groups, Thesis, Oxford

University, 1969. Di. Dieudonnt, J., La giometrie des groupes classiques, Springer-Verlag, Berlin, 1971. FT. Feit, W. and Thompson, J., Solvability of groups of odd order, Pacific J. Math. 13 (1963),

775-1029. FS. Fong, P. and Seitz, G., Groups with a (B, N)-pair of rank 2, I, 11, Invent. Math. 21 (1973),

1-57; 24 (1974), 191-239. G1 1. Glauberman, G., On solvable signalizer functors in finite groups, Proc. London Math. Soc.

23 (1976), 1-27. G12. Glauberman, G., Failure of factorization in p-solvable groups, Quart. J. Math. Oxford 24

(1973), 71-7. G13. Glauberman, G., Central elements in core-free groups, J . Alg. 4 (1966), 403-20. Go1 1. Goldschmidt, D., 2-signalizer functors on finite groups, J. Alg. 21 (1972), 32140. Go1 2. Goldschmidt, D., Solvable signalizer functors on finite groups, J . Alg. 21 (1972), 13748. Gor 1. Gorenstein, D., On the centralizers of involutions in finite groups, J. Alg. 11 (1969),

243-77. Gor 2. Gorenstein, D., The Classification of Finite Simple Groups, I, Plenum Press, New York,

1983. Gor 3. Gorenstein, D., Finite Simple Groups; An Introduction to their Classification, Plenum

Press, New York, 1982. Gor 4. Gorenstein, D., Finite Groups, Harper and Row, New York, 1968. GL. Gorenstein, D. and Lyons, R., The local structure of finite groups of characteristic 2 type,

Memoirs AMS 276 (1983), 1-731. GW. Gorenstein, D. and Walter, J., Balance and generation in finite groups, J . Alg. 33 (1975),

224-87.

298 References

HH. Hall, P. and Higman, G., The p-length of a p-solvable group and reduction theorems forBurnside's problem, Proc. London Math. Soc. 7 (1956), 1-41.

HKS. Hering, C., Kantor, W. and Seitz, G., Finite groups having a split (B, N)-pair of rank 1,J. Alg. 20 (1972), 435-75.

He. Herstein, I., Topics in Algebra, Zerox, Lexington, 1975.Hi. Higman, D., Finite permutation groups of rank 3, Math. Z. 86 (1964), 145-56.La. Lang, S., Algebra, Addison-Wesley, Reading, 1971.Mc. McBride, P., Nonsolvable signalizer functors on finite groups, J. Alg. 78 (1982), 215-38.Ri. Richen, F., Modular representations of split (B, N)-pairs, Trans. AMS 140 (1969), 435-60.Sh. Shult, E., On groups admitting fixed point free operator groups, Illinois J. Math. 9 (1965),

701-20.St. Steinberg, R., Lectures on Chevalley groups, Yale University, 1968.Su. Suzuki, M., Group Theory, Springer-Verlag, New York, 1982.Th 1. Thompson, J., Normal p-complements for finite groups, Math. Z. 72 (1960), 332-54.Th 2. Thompson, J., Normal p-complements for finite groups, J. Alg. 1 (1964), 43-6.Th 3. Thompson, J., Bounds for the orders of maximal subgroups, J. Alg. 14 (1970), 135-8.Ti. Tits, J., Buildings of Spherical Type and Finite (B, N)-Pairs, Springer-Verlag, Berlin, 1974.Wi 1. Wielandt, H., Eine Verallgemeinerung der invarianten Untergruppen, Math. Z. 45 (1939),

209-44.Wi 2. Wielandt, H., Finite Permutation Groups, Academic Press, New York.

References

HH. Hall, P. and Higman, G., The p-length of a p-solvable group and reduction theorems for Burnside's problem, Proc. London Math. Soc. 7 (1956), 1 4 1 .

HKS. Hering, C., Kantor, W. and Seitz, G., Finite groups having a split (B, N)-pair of rank 1, J . Alg. 20 (1972), 435-75.

He. Herstein, I., Topics in Algebra, Zerox, Lexington, 1975. Hi. Higman, D., Finite permutation groups of rank 3, Math. Z. 86 (1964), 145-56. La. Lang, S., Algebra, Addison-Wesley, Reading, 1971. Mc. McBride, P., Nonsolvable signalizer functors on finite groups, J. Alg. 78 (1982), 215-38. Ri. Richen, E, Modular representations of split ( B , N)-pairs, Trans. AMS 140 (1969), 435-60. Sh. Shult, E., On groups admitting fixed point free operator groups, Illinois J. Math. 9 (1965),

701-20. St. Steinberg, R., Lectures on Chevalley groups, Yale University, 1968. Su. Suzuki, M., Group Theory, Springer-Verlag, New York, 1982. Th 1. Thompson, J., Normal p-complements for finite groups, Math. Z. 72 (1960), 332-54. Th 2. Thompson, J., Normal p-complements for finite groups, J. Alg. 1 (1964), 43-6. Th 3. Thompson, J., Bounds for the orders of maximal subgroups, J. Alg. 14 (1970), 135-8. Ti. Tits, I., Buildings of Spherical Type and Finite ( B , N)-Pairs, Springer-Verlag, Berlin, 1974. Wi 1. Wielandt, H., Eine Verallgemeinerung der invarianten Untergruppen, Math. Z. 45 (1939),

2 0 9 4 . Wi 2. Wielandt, H., Finite Permutation Groups, Academic Press, New York.

List of symbols

Symbol Page SymbolH < G 2 (S)

IGI 2 H=GH<G 3 IG:HIHx, xH 3 ker(a)G/H 3 X8

XG 3 NG(X), CG(X)XY 4 II; G;GxH 4 Zn

n(G) 5p, n,

Ep. 5 mp(G)0, (G), 0- (G) 5 Z(G)

Q ,(G), S2n(G) 5 AutG (X)

aIR 6 C, 0,RG# 6 Aut(X)r(G, T) 8 Sym(X)GL(V) 9 Inn(G)Out(G) 11 Gy, G(Y), G1'xG 13 Fix(S)

np, IGIp 19 Sy1p(G)HG 23 H char G

[X, Y] 26, 45 [X, Y, Z]G(n) 27 Ln(G)

Zn(G) 28 S(A, G, r)G*H 33 LwrG, Lwr, GEndR(V) 38 HomR(U, V)Soc(V) 38, 158 Fnxn

Mx(g) 42 PdimTr, det 44 SL(V), PG(V)PSL(V), PGL(V) 44 SLn(F), Ln(F)PSLn(F), PGLn(F) 44 V*

BT 47 Mov(A)Cyca 54 Alt(X)U(G, V) 65 H'(G, V)nn, IGI 71 1Rad(V) 76 O(V, A O(V, Q)

List of symbols

Symbol H I G

1GI H g G Hx, XH

GIH xG XY G x H

n ( G )

EP. O,(G), O"(G) Qn(G), Qn(G) f f I R G #

U G , F) GL(V) Out(G) xG

np, lGlp H a a G

[ X , Yl ~ ( n )

Z"(G) G * H

E ~ ~ R ( V ) Soc(V)

Mx (g) Tr, det PSL(V), PGL(V)

PSLn(F), PGLn(F) B=

CYC, U(G, V ) 1 ~ x 7 lGln Rad(V)

Page 2 2 3 3 3 3 4 4 5 5 5 5 6 6 8 9 11 13 19 23 26,45

Symbol

(S) HLZG 1G:HI ker(a) X

NG(X), CG(X)

n, G, z n

P', n' mp(G) Z (G) Autc ( X ) C, Q, LQ Aut(X)

Sym(X) Inn(G)

G Y , G(Y) , GY Fix(S)

Sylp(G) H char G

[X , Y , ZI Ln(G) S(A, G , n ) Lwr G , Lwr,G

H ~ ~ R ( U , V ) Fnxn Pdim

SL(V), PG(V) SLn(F), Ln(F) v * Mov( A) Alt(X)

H ' (G , V ) I O W , f 1, O W , Q>

Page 2 2 3 3 3 3, 14 4 4 5 5 5 5 6 7 9 11 14 14 19 25 26 28 30 33 38 42 44 44 44 47 54 55 67 76 78

300 List of symbols

Symbol Page Symbol Page

A(V, ,f), A(V, Q) 78 J(X,f) 78

AB 78 NF 84

sgn(Q), sgn(V) 86, 87 Spn(F), Sp(V) 88

SO(V), O(V), Q(V) 88, 89 SPn(R), SUn(R) 89

OO(R), SOO(R) 89 cO(R) 89

PSpn (q), PGUn (q) 89 POO(R), POO(R) 89

G+ 97 (D(H) 105

Mode. , D2 107 SD2,,, Q2" 107Pi+2n

D8,- Qs 110,111 L(V1, ... , Vm; V) 117

V®U 118 UK ,K®FU 119

7rK 119 7r a, Va 121

Grp(Y : W) 140 l(h),IR(h) 143

W(E) 148 149

Comp(G) 157 E(G) 157

F(G) 158 Oc,(G) 158

F*(G) 159 On',R(G) 159

-W(G) 162 J(G) 162

J" (G, V) 163 aG(B) 175

char(G) 180 cl(G) 179

aG 188 (Fk(G) 246

-q(X) 245 rP,k(G), rp,k(G) 247

ek (G)o 248 e(G) 261

m2,p(G) 261

Symbol

A(V, f 1 3 A(V, Q ) A@

sgn(Q), sgn(V) so(v), O(V), Q(V) 0:(9) , so:(9) PSpn (9 1, PGUn (9 G +

Mod,. , Dzn

P ' + ~ , D:Qr V @ U Jr

Grp(Y : W )

W ( C ) Comp(G) F(G) F*(G)

P ( G , V ) char(G) aG

.@(XI &,P(G)O mz.,(G)

Page 78 78 86,87 88,89 89 89 97 107 110,111 118 119 140 148 157 158 159 162 163 180 188 245 248 26 1

List of symbols

Symbol

J ( X , f NF" Sp,(F), SP(V) Spn(q), SUn(9) QE(9) po:(9), pQ:(9) @(HI SD211, Q2n L(V1,. . ., v m ; V ) u K , K @ ~ U x u , vu Kh), l ~ ( h ) C+

E(G) O,(G) O,~,E(G> J (G) ac (B) cl(G) &;(GI r ~ , k ( G ) , r:,k(G)

e(G)

Page

78 84 88 89 89 89 105 107 117 119 121 143 149 157 158 159 162 175 179 246 247 26 1

Index

algebraic integer 184Alperin's Fusion Theorem 200alternating group 55apartment 215automizer 5automorphism 7

Baer-Suzuki Theorem 204bar convention 6basis

dual 47hyperbolic 81orthogonal 79orthonormal 79

BN-pair 218Borel subgroup 219Br-Property 263Brauer-Fowler Theorem 244building 215

Weyl group 219Burnside Normal p-Complement

Theorem 202Burnside pagb-Theorem 187

Cartan subgroup 219

category 6coproduct 7product 7

Cauchy's Theorem 20central product 32

with identified centers 33centralizer 3chamber 209character 49

degree 179generalized 180induced 189irreducible 179

character table 183characteristic value 127characteristic vector 127class function 179classical group 88Classification Theorem 260Clifford algebra 95Clifford group 96Clifford's Theorem 41cocycle 64

commutator 26complex 209

chamber 209chamber graph 209connected 209reflection 212thick 209thin 209

Component Theorem 263composition factors 24composition series 23conjugate 3Coprime Action Theorem 73covering 168Coxeter complex 211

diagram 141group 142matrix 141system 142, irreducible 146

critical subgroup 108cycle 54cycle structure 54

dihedral group 141direct product 4distance 8dual space 47Dynkin diagram 251

edge 8

exact sequence 47

short 47

split 47Exchange Condition 143extension, central 166

perfect 168universal 166

extension problem 10

FG-homomorphism 36FG-representation 35field, perfect 92Fitting subgroup 158

generalized 159fixed point 14flag 8folding 211

opposite 212

Index

algebraic integer 184 Alperin's Fusion Theorem 200 alternating group 55 apartment 215 automizer 5 automorphism 7

Baer-Suzuki Theorem 204 bar convention 6 basis

dual 47 hyperbolic 81 orthogonal 79 orthonomal 79

BN-pair 218 Bore1 subgroup 219 Bp-Property 263 Brauer-Fowler Theorem 244 building 215

Weyl group 219 Burnside Normal p-Complement

Theorem 202 Burnside paqb-~heorem 187

Cartan subgroup 219 category 6

coproduct 7 product 7

Cauchy's Theorem 20 central product 32

with identified centers 33 centralizer 3 chamber 209 character 49

degree 179 generalized 180 induced 189 irreducible 179

character table 183 characteristic value 127 characteristic vector 127 class function 179 classical group 88 Classification Theorem 260 Clifford algebra 95 Clifford group 96 Clifford's Theorem 41 cocycle 64

commutator 26 complex 209

chamber 209 chamber graph 209 connected 209 reflection 212 thick 209 thin 209

Component Theorem 263 composition factors 24 composition series 23 conjugate 3 Coprime Action Theorem 73 covering 168 Coxeter complex 2 1 1

diagram 141 group 142 matrix 141 system 142, irreducible 146

critical subgroup 108 cycle 54 cycle structure 54

dihedral group 141 direct product 4 distance 8 dual space 47 Dynkin diagram 25 1

edge 8 exact sequence 47

short 47 split 47

Exchange Condition 143 extension, central 166

perfect 168 universal 166

extension problem 10

FG-homomorphism 36 FG-representation 35 field, perfect 92 Fitting subgroup 158

generalized 159 fixed point 14 flag 8 folding 21 1

opposite 212

302

form, bilinear 75equivalent 78hermitian symmetric 76nondegenerate 76orthogonal 76quadratic 77radical 76sesquilinear 75similar 78skew hermitian 104skew symmetric 76symmetric 76symplectic 76Witt index 78

Frattini argument 20Frattini subgroup 105Frobenius complement 191Frobenius kernel 191Frobenius Normal p-Complement

Theorem 203Frobeinus Reciprocity Theorem 190Frobenius' Theorem 191fusion, control 199

G-isomorphism 9G-morphism 9gallery 209Gaschatz' Theorem 31general linear group 9general unitary group 89geometry 8

connected 8flag 8flag transitive 8morphism 8oriflamme 99polar 99rank 8

residually connected 8residue 8type 8

Glauberman Z*-Theorem 262graph 8

connected 8

connected component 8group algebra 36

center 5

characteristic p-type 261characteristically simple 25class 28complement 30covering 168

exponent 5

extension 30

free 138

Frobenius 191

Lie type 250k-connected 246

k-generated p-core 247metacyclic 203nilpotent 28perfect 27

representation by conjugation 11ring 36section 261solvable 27

Hall 7r -subgroup 71hyperbolic pair 80hyperbolic plane 80

induction map 189inner automorphism group 11integer

p-part 19nr-part 71

involution 5isometry 78isomorphism 7

Jordan's Theorem 58Jordan-Holder Theorem 24, 37

K-group 261

Levi factor 257Lie rank 250linear representation 9

absolutely irreducible 121dual 47enveloping algebra 42

indecomposable 37

induced 188

irreducible 37irreducible constituents 37principal 179regular 182tensor product 119written over a field 124

linear transformationdeterminant 44m-linear 118nilpotent 128semisimple 128semisimple part 131trace 44unipotent 128unipotent part 131

Maschke's Theorem 40maximal subgroup 5modular property of groups 6module

absolutely irreducible 121complement 37composition factor 37

IndexIndex

form, bilinear 75 equivalent 78 hermitian symmetric 76 nondegenerate 76 orthogonal 76 quadratic 77 radical 76 sesquilinear 75 similar 78 skew hermitian 104 skew symmetric 76 symmetric 76 symplectic 76 Witt index 78

Frattini argument 20 Frattini subgroup 105 Frobenius complement 191 Frobenius kernel 191 Frobenius Normal p-Complement

Theorem 203 Frobeinus Reciprocity Theorem 190 Frobenius' Theorem 191 fusion, control 199

G-isomorphism 9 G-morphism 9 gallery 209 Gaschiitz' Theorem 3 1 general linear group 9 general unitary group 89 geometry 8

connected 8 flag 8 flag transitive 8 morphism 8 oriflamme 99 polar 99 rank 8 residually connected 8 residue 8 type 8

Glauberman Z*-Theorem 262 graph 8

connected 8 connected component 8

group algebra 36 center 5 characteristic p-type 261 characteristically simple 25 class 28 complement 30 covering 168 exponent 5 extension 30 free 138 Frobenius 191 Lie type 250 k-connected 246

k-generated p-core 247 metacyclic 203 nilpotent 28 perfect 27 representation by conjugation 11 ring 36 section 261 solvable 27

Hall r-subgroup 71 hyperbolic pair 80 hyperbolic plane 80

induction map 189 inner automorphism group 11 integer

p-part 19 r-part 71

involution 5 isometry 78 isomorphism 7

Jordan's Theorem 58 Jordan-Holder Theorem 24,37

Levi factor 257 Lie rank 250 linear representation 9

absolutely irreducible 121 dual 47 enveloping algebra 42 indecomposable 37 induced 188 irreducible 37 irreducible constituents 37 principal 179 regular 182 tensor product 119 written over a field 124

linear transformation determinant 44 m-linear 118 nilpotent 128 semisimple 128 semisimple part 131 trace 44 unipotent 128 unipotent part 131

Maschke's Theorem 40 maximal subgroup 5 modular property of groups 6 module

absolutely irreducible 121 complement 37 composition factor 37

Index

condensed 125cyclic 38extension 37homogeneous 40homogeneous component 40indecomposable 37irreducible 37permutation 50semisimple 38simple 37socle 38split extension 37

morphism 6

normal p-complement 202normal series 22

A-invariant 22factors 23length 22

normalizer 3

Odd Order Theorem 260orbit 13orbital 55

diagonal 55paired 59self paired 59

ordinary Chevalley group 252orthogonal group 88orthogonality relations 183outer automorphism group 11

p-group 5elementary abelian 5extraspecial 108modular 107special 108symplectic type 109

p-rank 52-local 260

parabolic subgroup 221partition

G-invariant 18nontrivial 18

path 8length 8

permutationeven 55odd 55

permutation group 53permutation representation 9

by conjugation 15by right multiplication 15m-transitive 56primitive 18rank 55regular 56semiregular 56

303

transitive 14transitive constituents 16

Phillip Hall's Theorem 711r-group 5presentation 140projective geometry 43

dimension 44hyperplane 44lines 44points 44

projective general linear group 44projective special linear group 44

quaternion group 107

Rainy Day Lemma 216reflection 93

center 93representation 9

equivalence 9

faithful 9

quasiequivalence 9

residue 8root group 250root system 148

closed subset 255ideal 255ordering 149positive system 149simple system 149Weyl group 148

rootlong 251

short 251

scalar matrix 44scalar transformation 44Schreier Conjecture 160Schur's Lemma 38Schur multiplier 168Schur-Zassenhaus Theorem 70semidihedral group 107semidirect product 30signalizer functor 229

complete 229solvable 229solvably complete 229

similarity 78simplex 209Sims Conjecture 176Solvable 2-Signalizer Functor Theorem 229space

hyperbolic 81orthogonal 77symplectic 77unitary 77

special Clifford group 97special linear group 44

condensed 125 cyclic 38 extension 37 homogeneous 40 homogeneous component 40 indecomposable 37 irreducible 37 permutation 50 semisimple 38 simple 37 socle 38 split extension 37

morphism 6

normal p-complement 202 normal series 22

A-invariant 22 factors 23 length 22

normalizer 3

Odd Order Theorem 260 orbit 13 orbital 55

diagonal 55 paired 59 self paired 59

ordinary Chevalley group 252 orthogonal group 88 orthogonality relations 183 outer automorphism group 11

p-group 5 elementary abelian 5 extraspecial 108 modular 107 special 108 symplectic type 109

p-rank 5 2-local 260

parabolic subgroup 221 partition

G-invariant 18 nontrivial 18

path 8 length 8

permutation even 55 odd 55

permutation group 53 permutation representation 9

by conjugation 15 by right multiplication 15 m-transitive 56 primitive 18 rank 55 regular 56 semiregular 56

transitive 14 transitive constituents 16

Phillip Hall's Theorem 71 n-group 5 presentation 140 projective geometry 43

dimension 44 hyperplane 44 lines 44 points 44

projective general linear group 44 projective special linear group 44

quaternion group 107

Rainy Day Lemma 2 16 reflection 93

center 93 representation 9

equivalence 9 faithful 9 quasiequivalence 9

residue 8 root group 250 root system 148

closed subset 255 ideal 255 ordering 149 positive system 149 simple system 149 Weyl group 148

root long 251 short 251

scalar matrix 44 scalar transformation 44 Schreier Conjecture 160 Schur's Lemma 38 Schur multiplier 168 Schur-Zassenhaus Theorem 70 semidihedral group 107 semidirect product 30 signalizer functor 229

complete 229 solvable 229 solvably complete 229

similarity 78 simplex 209 Sims Conjecture 176 Solvable 2-Signalizer Functor Theorem 229 space

hyperbolic 81 orthogonal 77 symplectic 77 unitary 77

special Clifford group 97 special linear group 44

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