Simplicial Complexes in Finite Groups

(Cortona. July 2000)

Firenze, September 2000.

These notes contain the material presented at a Scuola Matematica Interuni-versitaria summer course held in Cortona in July 2000. For a large part they arebased on earlier unpublished seminar notes by Luis Ezquerro and myself on the samesubject. Thus, except for the errors, Luis should be regarded as a co-author.

I am graetly indebted to Jurgen Pulkus, whose sensible suggestions and com-ments led to many improvements both in presentaton and contents.

I also thank the students of the course who pointed out several misprints andinaccuracies.

Carlo CasoloDipartimento di Matematica ”U. Dini”Viale Morgagni 67/AI-50134 Firenze

e-mail : casolo@math.unifi.it

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Chapter 1

Simplicial Complexes

Definition. Let V be a finite non empty set. An (abstract) simplicial complexwith vertex set V is a collection Γ of non empty subsets of V satisfying the followingconditions:

1. v ∈ Γ for each v ∈ V ;

2. if ∅ 6= τ ⊆ σ ∈ Γ then τ ∈ Γ.

Let Γ be a simplicial complex with vertex set V . We then write V = vert(Γ),and call simplices the elements of Γ. More specifically, if σ ∈ Γ, and |σ| = n + 1,we say that σ is a n−simplex (or also that σ is a simplex of dimension n). We definethe dimension of Γ as the largest dimension of its simplices. For a n-simplex σ, weusually write σ = v0, v1, . . . , vn, where vi ∈ V . If σ is a simplex, any ∅ 6= τ ⊆ σis called a face of σ (by definition, τ ∈ Γ). Vertices (i.e. the elements of V ) areidentified with the 0-simplices.

A subcomplex of a complex Γ is a subset of Γ which is closed by taking all facesof its elements (clearly a subcomplex is in turn a complex). If n is the dimensionof Γ and 0 ≤ q ≤ n, then the set of all simplices of Γ of dimension at most q is acomplex, and it is called the q-skeleton of Γ.

Examples.

1) Let A be a finite graph with vertex set V . The clique complex K(A) associatedto A is the simplicial complex on Vert(K(A)) = V whose simplices are the non-emptysubsets of V that determine a complete subgraph (a ”clique”) of A.

2) Let V be a finite dimensional vector space over a finite field F . Then the setof all non-empty linearly independent subsets of V is a simplicial complex, whosedimension is dimF (V )− 1.

3) Let G be a finite group and F = H0, . . . ,Hn a family of subgroups of G.The coset complex ΓF associated to F is defined in the following way. The vertices

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of Γ are all the right cosets Hig with g ∈ G and i = 0, . . . , n. For 0 ≤ q ≤ n, aq-simplex is any set of q + 1 distinct cosets Hi0g0, . . . ,Hiqgq such that

Hi0g0 ∩ . . . ∩Hiqgq 6= ∅.

Since any element in the intersection can be taken as a representative for the samecosets, we have that a q-simplex is any family Hi0g, . . . ,Hiqg with g ∈ G and0 ≤ i0 < i1 < . . . < iq ≤ n.

Polyhedra.

An affine combination of points P0, . . . , Pm in an euclidean space Rn is

m∑i=0

λiPi withm∑i=0

λi = 1 .

An affine combination is convex if furthermore λi ≥ 0 for all i = 0, . . . ,m. Theconvex hull of a set of points is the set of all convex combinations of them.The points P0, . . . , Pm in Rn are affinely independent if none of them is an affinecombination of the others (or, in other words, if the vectors P1 − P0, . . . , Pm − P0

are linearly independent). If P0, . . . , Pm are affinely independent then their convexhull is denoted by σ = 〈P0, . . . , Pm〉 and it is called an affine m-simplex. The pointsP0, . . . , Pm are the vertices of σ, a face of σ is a simplex of the form 〈Pi0 , . . . , Piq〉with ∅ 6= Pi0 , . . . , Piq ⊆ P0, . . . , Pm. We leave it as an exercise to prove that asimplex determines its vertices.

A geometric simplicial complex (see [54]) is a collection K of affine simplices inan euclidean space such that if σ, τ ∈ K then any face of σ is in K and σ ∩ τ iseither empty or a face of both σ and τ . A topological space X is called a polyhedronif it is homeomorphic to a space |K| =

⋃σ∈K σ for some geometric complex K.

Now, to any geometric complex K there is a naturally associated abstract com-plex Ka whose simplices are the set of the vertices of the simplices of K. Conversely,it is easy to associate to any abstract complex Γ a geometric complex U(Γ) and itsunderlying topological space.In Rn the points E0 = 0, E1 = (1, 0, . . . , 0), E2 = (0, 1, . . . , 0), . . . , En = (0, 0, . . . , 1)are affinely independend; the simplex 〈E0, E1, . . . , En〉 is called the standard n-simplex and denoted by ∆n.

Let now Γ be an abstract complex with V = V ert(Γ) = v0, v1, . . . , vn. Thendefine a map u : Γ −→ ∆n which assignes to each simplex σ = vi0 , vi1 , . . . , viq ofΓ the face u(σ) = 〈Ei0 , Ei1 , . . . , Eiq〉 of ∆n. Finally put U(Γ) =

⋃σ∈Γ u(σ). Clearly

U(Γ) is a geometric complex.A topological space which is homeomorphic to |U(Γ)| is called a geometrical

realization of Γ. Up to homeomorphisms, we denote by |Γ| a geometric realizationof Γ.

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The geometrical realization allows to introduce concepts and language fromtopology in the study of abstract complexes. In these notes, I will refer from time totime to the geometric realization of a complex, as it might provide a more intuitivepoint of view, but I will generally try to avoid when possible topological subtlety,by keeping a more combinatorial approach. This will have the cost of losing someinformation on the topological features of the complexes we will study.

Notation. Given an n-dimensional complex Γ, for each 0 ≤ q ≤ n we denote bySq(Γ) the set of all q-simplices of Γ, and by αq the cardinality of Sq(Γ).

Definition. Let Γ be a complex of dimension n. The Euler-Poincare character-istic of Γ is the integer

χ(Γ) =n∑q=0

(−1)qαq .

The reduced Euler-Poincare characteristic is defined as: χ(Γ) = χ(Γ)− 1 .

Definition. Let Γ and Γ be simplicial complexes with vertex set V and V respec-tively. A simplicial map f : Γ −→ Γ is an application f : V −→ V which mapsall simplices of Γ into simplices of Γ (possibly of smaller dimension), i.e. such that

x0, x1, . . . , xq ∈ Γ ⇒ fx0, fx1, . . . , fxq ∈ Γ .

f : Γ −→ Γ is an isomorphism of simplicial complex if it is bijective and both fand f−1 are simplicial maps.

A simplicial map f : Γ −→ Γ of abstract simplicial complexes induces in astandard way a continuous map |f | : |U(Γ)| → |U(Γ)| of the geometric realizationsas we have described above. In fact, for every simplex σ of Γ the vertices of u(σ)are by construction affinely independent. Thus the restriction f |vert(σ) determines

a unique affine map fσ : |u(σ)| → |U(Γ)| which is continuous. This assignement isclearly consistent on faces, and allows, by piecing together the fσ, to define |f |.

It is then an immediate consequence that geometric realizations of isomorphicsimplicial complexes are homeomorphic. Also, it is clear that the converse is nottrue.

Barycentric subdivision

Let σ = 〈P0, . . . , Pm〉 be an affine simplex in Rn; the barycenter of σ is the point

Bσ =1

m+ 1(P0 + . . .+ Pm) .

Given a geometric complex K, one constructs a new complex Sd(K) whose vertexset is the set of all barycenters of the simplices of K, and for q ≥ 1 the q-simplices

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are the affine simplices 〈Bσ0 , Bσ1 , . . . , Bσq〉 where σ0, . . . σq are simplices of K suchthat σ0 ⊂ σ1 ⊂ . . . ⊂ σq. The complex Sd(K) just defined is called the barycentricsubdivision of K. It is easy to verify that the underlying space |Sd(K)| coincideswith |K|.

Now, let P be a finite partially ordered set. The complex of chains of P is thesimplicial complex ∆(P ) whose vertices are the elements of P , and simplices are allthe chains (i.e. linearly ordered subsets) of P . Thus, for q ≥ 0, the q-simplices of∆(P ) are all the chains of length q of P , and the dimension of ∆(P ) is the maximallength of a chain in P .

This construction for complexes is quite general. In fact, let Γ be a given abstractcomplex. Then we can view Γ as a partially ordered set, where the order relationamong simplices is just the set inclusion. Then the complex of chains ∆(Γ) is calledthe barycentric subdivision of Γ. It is not difficult to show that if |Γ| is a geometricrealization of Γ then its barycentric subdivision (in the geometric sense) Sd(|Γ|) isa geometric realization of ∆(Γ). Thus Γ and ∆(Γ) have homeomorphic geometricrealizations.

Let P, P ′ be partially ordedered sets. An application f : P −→ P ′ is an orderpreserving map if, for all x, y ∈ P

x ≤ y ⇒ f(x) ≤ f(y) .

An isomorphism of posets is a bijection f : P −→ P ′ such that f and its inverse f−1

are order preserving.

It is clear that an order preserving map on partially ordered sets induces in anatural way a simplicial map on the corresponding complexes of chains.

Examples. 1) Let V be a finite dimensional vector space over a finite field. Theset of all subspaces of V is finite, and ordered by inclusion. It then gives riseto a complex whose elements are all the chains (usually called flags) of subspacesV0 < V1 < . . . < Vq. The dimension of this complex is the dimension n of V as avector space. Observe that, in this complex, all maximal simplices have the samedimension n, and that the group of linear applications of V acts naturally as a groupof simplicial maps on the complex.

2) Let G be a finite group. Then the set L(G) of all subgroups of G ordered byinclusion, yields a complex of chains on which the group Aut(G) acts as a group ofsimplicial automorphisms. Observe that, apart from rather restricted cases, in theseexamples the maximal simplices have different dimensions.

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1.1 Homology

Definition. A Chain Complex is a family C = Cn, ∂n | n ∈ Z where, foreach n, Cn is an abelian group and ∂n : Cn −→ Cn−1 are group homomorphismssuch that ∂n∂n−1 = 0 (that is Im∂n ⊆ ker ∂n−1).

If C is a chain complex, then for all n ∈ Z the n-th homology group of C isdefined as Hn(C) = ker ∂n/Im∂n+1. We refer to the homology of C to mean thefamily of groups H∗(C) = Hn(C) | n ∈ Z.

Let Γ be a simplicial complex. An orientation of Γ is a partial ordering onVert(Γ) which induces a total ordering on each simplex of Γ. A complex with afixed orientation is said to be oriented. Observe that a total ordering on Vert(Γ) isalways an orientation of Γ; thus every complex can be oriented.

Given an oriented complex Γ, we recall the fundamental construction of theassociate integral chain complex Cq(Γ), ∂q | q ∈ Z . We fix n to denote thedimension of the complex Γ.

For 0 ≤ q ≤ n, and for each element σ of Sq(Γ), we now introduce the symbol[σ] = [v0, v1, . . . , vq] where v0, v1, . . . , vq are the vertices of σ written in the ordergiven by the orientation.

Then, for 0 ≤ q ≤ n, we denote by Cq(Γ) the free abelian group with freegenerating set whose elements are all symbols [σ] with σ ∈ Sq(Γ). For q ≥ n+ 1 weset Cq(Γ) = 0.

We now define the maps ∂q, called the differentiations (of degree q)

- For q = 0 we define ∂0 : C0(Γ) −→ 0 to be the zero map.

- For 1 ≤ q ≤ n, ∂q : Cq(Γ) −→ Cq−1 is the homomorphism obtained by extendingby Z-linearity the mapping

∂q([x0, x1, . . . , xq]) =

q∑i=0

(−1)i[x0, . . . , xi, . . . , xq]

where the notation xi means that the element xi has been removed.

Theorem 1.1 Let Γ be a simplicial complex of dimension n. Then, with the defi-nitions given, the sequence

C∗(Γ), ∂∗ : 0→ Cn(Γ)→ Cn−1(Γ)→ · · · → C1(Γ)→ C0(Γ)→ 0

is a chain complex.

Proof. We have only to show that, for each q, Im∂q+1 ⊆ ker ∂q, that is ∂q∂q+1 = 0.We have

∂q∂q+1([x0, x1, . . . , xq+1]) = ∂q

(q+1∑i=0

(−1)i[x0, . . . , xi, . . . , xq+1]

)=

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=

q+1∑i=0

(−1)i∂q([x0, . . . , xi, . . . , xq+1]) =

=

q+1∑i=0

(−1)i

i−1∑j=0

(−1)j [x0, .., xj , .., xi, .., xq+1] +

q+1∑j=i+1

(−1)j−1[x0, .., xi, .., xj , .., xq+1]

= 0

Now, For each q we put

Zq(Γ) = ker ∂q the group of q-cycles of Γ

Bq(Γ) = Im∂q+1 the group of q-boundaries of Γ

The q-th homology group of Γ is the factor group

Hq(Γ) =Zq(Γ)

Bq(Γ)

We denote by H∗(Γ) the sequence of groups Hq(Γ).

It is a fundamental fact in algebraic topology that the homology of a simplicialcomplex coincides with the homology of its geometrical realization (see [54], [51]).In particular it does not depend on the chosen orientation. In the following we willalways assume that a complex is already oriented.

Let Γ be a complex of dimension n. Then the differentiation ∂n+1 is the zeromap, whence Hn(Γ) = Zn(Γ) ≤ Cn(Γ) is a subgroup of a free abelian group, and soit is free (possibly trivial). We make this observation a proposition.

Proposition 1.2 Let Γ be a complex of dimension n, then Hn(Γ) is a free abeliangroup.

In several arguments, it is more convenient to work with the reduced homologyH∗(Γ). It is defined via the augmented chain complex

C∗(Γ), ∂∗ : 0→ Cn(Γ)→ Cn−1(Γ)→ · · · → C0(Γ)→ Z→ 0

where ∂q = ∂q for q ≥ 1, ∂−1 = 0, and ∂0 is defined by

∂0

∑v∈S0(Γ)

λvv

=∑

v∈S0(Γ)

λv.

It is easy to verify that Hq(Γ) = Hq(Γ) for q ≥ 1, and H0(Γ) = H0(Γ)⊕ Z.

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Definition. Let A be a finitely generated abelian group. Then A is the directproduct of cyclic subgroups. The number of infinite cyclic group that appear insuch a decomposition is an invariant of A, that we call the rank of A, and denote byr0(A).

The following two facts are not difficult to prove. In the first statement, thetensor product is as Z-modules, and the dimension is that of a Q-vector space.

Proposition 1.3 Let A be a finitely generated abelian group. Then

r0(A) = dim(A⊗Q)

Lemma 1.4 Let A be a finitely generated abelian group, let A1 ≤ A, and writeA2 = A/A1. Then

r0(A) = r0(A1) + r0(A2) .

This fact can be restated by saying that given an exact sequence

0 −→ A1 −→ A −→ A2 −→ 0

of finitely generated abelian groups, then r0(A) = r0(A1) + r0(A2).

Theorem 1.5 Let Γ be an oriented simplicial complex of dimension n. Then

χ(Γ) =

n∑q=0

(−1)qr0(Hq(Γ)) .

Proof. Consider the chain complex associated to Γ:

0 −→ Cn(Γ) −→ Cn−1(Γ) −→ . . . −→ C1(Γ) −→ C0(Γ) −→ 0

with differentials ∂i. Then, for each 0 ≤ q ≤ n, we have αq = r0(Cq(Γ)) (by definitionof Cq(Γ)) and an exact sequence:

0 −→ Zq(Γ) −→ Cq(Γ) −→ Bq−1(Γ) −→ 0

where we put B−1 = 0. Thus, by Lemma 1.4

r0(Cq(Γ)) = r0(Zq(Γ)) + r0(Bq−1(Γ)) .

On the other hand, by the very definition of homology groups, for each 0 ≤ q ≤ n,we have the exact sequence

0 −→ Bq(Γ) −→ Zq(Γ) −→ Hq(Γ) −→ 0

and so, again by Lemma 1.4

r0(Zq(Γ)) = r0(Bq(Γ)) + r0(Hq(Γ))

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where Bn(Γ) = 0. Therefore

χ(Γ) =∑n

q=0(−1)qr0(Cq(Γ)) =∑n

q=0(−1)qr0(Zq(Γ)) +∑n

q=0(−1)qr0(Bq−1(Γ))

=∑n

q=0(−1)qr0(Zq(Γ)) +∑n

q=0(−1)q+1r0(Bq(Γ))

=∑n

q=0(−1)q(r0(Zq(Γ))− r0(Bq(Γ))) =∑n

q=0(−1)qr0(Hq(Γ))

Definition. A complex Γ is acyclic if H0(Γ) = Z and Hq(Γ) = 0 for all q ≥ 1.

Equivalently, if its reduced homology is zero (i.e. Hq(Γ) = 0 for all q ≥ 0).

By Theorem 1.5, if Γ is an acyclic complex then χ(Γ) = 1.

Let X, Y be oriented simplicial complexes, and let f : X −→ Y be a sim-plicial map. Then, for each simplex σ = x0, . . . , xq ∈ Sq(X), its image fσ =fx0, . . . , fxq is a simplex of Y . Quite possibly, the dimension of fσ could be lessthan that of σ. Also, if fσ has the same dimension q, the ordering fx0, . . . , fxqmight not be the one compatible with the given orientation in Y ; thus we denote byπσ the permutation on the indices 0, . . . , q such that fxπσ0 < . . . < fxπσq in theordering of Y .

Then, for q ≥ 0, the map f induces a group homomorphism fq : Cq(X) −→ Cq(Y )by setting, for all σ ∈ Sq(X),

fq(σ) =

sign(πσ)[fσ] if fσ ∈ Sq(Y )0 if fσ 6∈ Sq(Y )

and extending by Z-linearity.

One also defines f−1 to be the zero map in the ordinary case, and the identity 1Z inreduced homology.

With the given definition, it is not difficult to check the following fundamentalproperty.

Proposition 1.6 Let X, Y be oriented complexes. For all q ≥ 0 let ∂q, ∂q denotethe q-differentiation on X and on Y respectively. Let f : X −→ Y be a simplicialmap. Then for all q ≥ 0:

∂q fq = fq−1 ∂q .

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This propositions shows that, for each q ≥ 0, one has

∂q fq(Zq(X)) = fq−1 ∂q(Zq(Y )) = 0

that is fq(Zq(X)) ≤ Zq(X). Also

fq(Bq(X)) = fq ∂q+1(Cq+1(X)) = ∂q+1 fq+1(Cq+1(X)) ≤ Bq(Y ).

These observations allow to define, for each q ≥ 0 an application

f∗q : Hq(X) −→ Hq(Y )

as the homomorphism induced by fq , that is, for all a ∈ Zq(X),

f∗q(a+Bq(X)) = fq(a) +Bq(Y ).

A morphism h : C −→ D, of two chain complexes of abelian groups C = Cn, ∂nand D = Dn, δn is a family hn | n ∈ Z of group homomorphisms hn : Cn −→ Dn,such that hn−1∂n = δnhn for all n ∈ Z.A sequence 0→ A→ B→ C→ 0 of morphisms of chain complexes is called exactif it arises from short exact sequences of abelian groups 0 → An → Bn → Cn → 0for all n ≥ 0.

Thus, Proposition 1.6 states that any simplicial map induces a morphism of chaincomplexes. Observe that if X and Y are simplicial complexes, and f : C∗(X) −→C∗(Y ) is a morphism of chain complexes, then the images f0(x) of the verticesx ∈ V ert(X) need not be vertices of Y . This is the case if f is induced by asimplicial map. Then, we say that a morphism f : C∗(X) −→ C∗(Y ) is a chain andvertex morphism if f0 maps V ert(X) (as a distinguished set of free generators ofC0(X)) in V ert(Y ).

The following result is useful in many cases. We omit the proof which is obtainedby chasing along map diagrams, and it is not difficult.

Proposition 1.7 Given a short exact sequence 0 → A → B → C → 0 of chaincomplexes, there is a long exact sequence

. . .→ Hq+1(C)→ Hq(A)→ Hq(B)→ Hq(C)→ Hq−1(A)→ . . .

Proof. See Rotman [51], pages 93,94.

A particular case occurs in the direct sum of chain complexes:

Proposition 1.8 Let C and D be chain complexes. Then

H∗(C⊕D) = H∗(C)⊕H∗(D).

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We now consider the case in which a complex X is covered by two (or more)subcomplexes X1 and X2. This simply means that X = X1 ∪ X2. In order tomake the statements more complete, it is convenient to deal also with the empty set(which might occur as X1 ∩ X2). So we adopt the convention that, for all q ≥ 0,Cq(∅) = Hq(∅) = 0 (and that χ(∅) = 0).

Lemma 1.9 Let X1 and X2 be subcomplexes of the complex X such that X1∪X2 =X. Then for each q ≥ 0 there is an exact sequence

0→ Cq(X1 ∩X2)→ Cq(X1)⊕ Cq(X2)→ Cq(X)→ 0.

Proof. Consider the four simplicial inclusions

i1 : X1 ∩X2 → X1, i2 : X1 ∩X2 → X2, j1 : X1 → X, j2 : X2 → X.

Observe that the definition of the groups Cq implies that Cq(X1 ∩X2) = Cq(X1) ∩Cq(X2) and Cq(X) = Cq(X1) + Cq(X2). Then we have natural definitions for ho-momorphisms

i : Cq(X1 ∩X2)→ Cq(X1)⊕ Cq(X2) j : Cq(X1)⊕ Cq(X2)→ Cq(X)

given by, for any s ∈ Sq(X1 ∩X2) and any s1 ∈ Sq(X1), s2 ∈ Sq(X2)

i(s) = (i1(s),−i2(s)) j(s1, s2) = j1(s1) + j2(s2).

By the remark made above, it is immediate that j is surjective, and i is injective.Also, if s ∈ Sq(X1∩X2), then j(i(s)) = j(s,−s) = s− s = 0, that is Im(i) ≤ ker(j).Conversely, given a =

∑λss ∈ Cq(X1) and b =

∑µrr ∈ Cq(X2), write a = a′ + ca,

b = b′+cb, where ca, cb are the contribution given by generators in Sq(X1)∩Sq(X2).If j(a, b) = 0, we then have 0 = a + b = a′ + b′ + (ca + cb) and so, by linearindependence, a′ = b′ = 0 = ca + cb. Thus a = −b and a, b belong to the groupgenerated by Sq(X1) ∩ Sq(X2). In other words (a, b) = (a,−a) = i(a), proving thatker(j) ≤ Im(i). In conclusion the sequence

0→ Cq(X1 ∩X2)→ Cq(X1)⊕ Cq(X2)→ Cq(X)→ 0

is exact.

Theorem 1.10 (Mayer-Vietoris) Let X1, X2 be subcomplexes of the complex Xsuch that X = X1 ∪X2. Then there is the long exact sequence:

. . .→ Hq+1(X)→ Hq(X1 ∩X2)→ Hq(X1)⊕Hq(X2)→ Hq(X)→ . . .

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Proof. By Proposition 1.8, the homology of C∗(X1) ⊕ C∗(X2) is H∗(X1) ⊕H∗(X2) Consider then the short exact sequence of Lemma 1.9, and apply Proposition1.7 to get the long sequence.

There is a similar statement for reduced homology, which is in fact more suitablefor calculations. Here, the convention for the empty set is Hq(∅) = 0 for q ≥ 0, and

H−1(∅) = Z (while, if X 6= ∅, H−1(X) = 0).

Theorem 1.11 Let X1, X2 be subcomplexes of the complex X such that X = X1 ∪X2. Then there is the long exact sequence:

. . .→ Hq+1(X)→ Hq(X1 ∩X2)→ Hq(X1)⊕ Hq(X2)→ Hq(X)→ . . .

where the final maps are

→ H1(X)→ H0(X1 ∩X2)→ H0(X1)⊕ H0(X2)→ H0(X)→ H−1(X1 ∩X2)→ 0 .

Definition. Let X be a complex. A path from a vertex v to a vertex w of X isjust a path in the 1-skeleton of X viewed as a (undirected) graph; i.e. a sequence of1-simplices σ0, . . . , σk such that

s0 = v = v0, v1, s1 = v1, v2, . . . , sk−1 = vk−1, vk, sk = vk, vk+1 = w.

X is connected if every two vertices of X are joined by a path. A subcomplex Yof X is a connected component if Y is connected and there is no path joining anyvertex v ∈ V ert(Y ) to any w ∈ V ert(X) \ V ert(Y ).

Observe that in our definition of path we make no reference to an orientation.It is not hard to verify that a complex is connected if and only if its geometricalrealization is path-connected in the topological sense. Also it is easy to see that theconnected components of a complex are uniquely determined.

Exercises.

1) A circuit in a complex is a non empty path joining a vertex v to itself. Provethat if s0, . . . , sk is a path in X from the vertex v to w, then (given a suitableorientation) ∂1(s0 + . . . + sk) = v − w. Thus prove that every circuit is associatedto a 1-cycle.

2) Let F = H0, . . . ,Hn be a family of proper subgroups of the group G, andlet ΓF be the associated coset complex. Prove that ΓF is connected if and only if〈H0, . . . ,Hn〉 = G.

Proposition 1.12 Assume the complex X has d connected components X1, . . . , Xd.Then, for each q ≥ 0; Hq(X) = Hq(X1)⊕Hq(X2)⊕ · · · ⊕Hq(Xd).

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Proof. This is obtained by an easy application of the Mayer-Vietoris sequenceand induction on d.

Proposition 1.13 Let Γ be a connected complex, and let V =Vert(Γ). Then

i) B0(Γ) = ∑

v∈V λvv |∑

v∈V λv = 0 .

ii) H0(Γ) = Z.

Proof. i) Clearly, B = ∑

v∈V λvv |∑

v∈V λv = 0 is a subgroup of C0(Γ),and it contains B0(Γ), since for all [v, w] ∈ S1(Γ), ∂([v, w]) = v − w ∈ B.Fix a vertex a in Γ. Then for all v ∈ V ert(Γ) there is a sequence so, s1, . . . , skof 1-simplices of Γ connecting v to a. Then, by choosing a suitable orientation,∂(so + s1 + · · ·+ sk) = v − a ∈ B0(Γ). Now, if

∑v∈V λv = 0 then∑

v∈Vλvv =

∑v∈V

λvv −∑v∈V

λva =∑v∈V

λv(v − a) ∈ B0(Γ)

proving that B0(Γ) = B.ii) By i) the map au : C0(Γ) −→ Z defined by au

(∑v∈V λvv

)=∑

v∈V λv is asurjective homomorphism with kernel B0(Γ). Then H0(Γ) = C0(Γ)/B0(Γ) = Z.

Corollary 1.14 For any complex Γ, H0(Γ) = Zd where d is the number of connectedcomponents of Γ.

If (X,x0) and (Y, y0) are pointed topological spaces, then their wedge X ∨ Yis the quotient space of their disjoint union, where the basepoints are identified.A similar construction can be given for simplicial complexes: a wedge X ∨ Y ofconnected complexes X and Y is a complex whose vertex set is the disjoint union ofV ert(X) and V ert(Y ) with a single vertex identified, and whose simplices are justthose of X and those of Y .

Exercise. Let Y,Z be connected complexes. Prove that, for n ≥ 1: Hn(Y ∨Z) ∼= Hn(Y ) ⊕Hn(Z), and H0(Y ∨ Z) = Z. More generally, let Y, Z be connectedsubcomplexes of the complex X = Y ∪Z, and assume that Y ∩Z is the set of facesof a single simplex. Prove that H∗(X) ∼= H∗(Y )⊕ H∗(Z).

What we have described in this section is the integral homology of a complex,which is the most common in topology. Fixed an abelian group A it is possibleto similarly define homology with coefficients in A, by considering for each q ≥ 0,instead of a free abelian group Cq(Γ) the Z-tensor product Cq(Γ;A) = Cq(Γ) ⊗ A,and differential maps ∂q ⊗ 1. The homology groups of the resulting chain complex

13

are denoted by Hq(Γ;A). In topology it is often useful the case in which one takes asA some (integral) homology group of the complex. In group theory one is interestedin the case in which A is a field; we have seen that if a group G acts as a groupof simplicial maps on a complex Γ, then there are natural induced actions of G onthe homology groups. If coefficients are taken in a field this gives raise to linearrepresentations of G.

For those who are familiar with the theory of Z-modules, we recall that the rela-tion between integral and arbitrary homology is described by the so-called UniversalCoefficients Theorem:

Hq(Γ;A) ∼= (Hq(Γ)⊗A)⊕ Tor(Hq−1(Γ), A)

In particular if A is torsion-free, then Hq(Γ;A) ∼= Hq(Γ) ⊗ A. Thus Hq(Γ;Q) = 0if and only if Hq(Γ) is a finite group; we then say that Γ is Q-acyclic if, H0(Γ) = Zand Hq(Γ) is finite for all q ≥ 1.Another case that we like to mention is when A = Zp is the additive group of thefield of p elements. We say that Γ is Zp-acyclic if H0(Γ;Zp) = Zp and Hq(Γ;Zp) = 0for q ≥ 1. By the UCT Theorem this happens if and only if H0(Γ) = Z and, for allq ≥ 1, Hq(Γ) is a finite group of order not divided by p. In fact, if G is an abeliangroup, then G⊗Zp ∼= G/pG and Tor(G,Zp) ∼= x ∈ G | px = 0. Thus, since Hq(Γ)is a finitely generated abelian group, Hq(Γ;Zp) ∼= (Hq(Γ)⊗ Zp)⊕ Tor(Hq−1(Γ),Zp)is trivial if and only if Hq(Γ) is a finite group with no elements of order p.

1.2 Homotopy equivalence

In this section, following [51], we describe a way of establishing that two complexeshave the same homology. Thus, let X and Y be simplicial complexes, and denoteby ∂q and δq the differentiation in X and in Y respectively.

Definiton. Two simplicial maps f, g : X −→ Y are (homotopy) equivalent(written f ' g) if for each q ≥ 0 there exists a homomorphism πq : Cq(X) −→Cq+1(Y ) such that

δq+1πq + πq−1∂q = fq − gqwhere π−1 = 0.

Homotopy equivalence of simplicial maps is readily seen to be an equivalencerelation (exercise).

Theorem 1.15 If f, g : X −→ Y are equivalent simplicial maps then f∗q = g∗q forall q ≥ 0.

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Proof. Let f ' g and, for each q ≥ 0 let πq : Cq(X) → Cq+1(Y ) be ahomomorphism such that δq+1πq + πq−1∂q = fq − gq, and π−1 = 0. Let z ∈ Zq(X),then

(fq − gq)(z) = (δq+1πq + πq−1∂q)(z) = δq+1πq(z) ∈ Imδq+1 = Bq(Y ).

Hence f∗q(z +Bq(X)) = fq(z) +Bq(Y ) = gq(z) +Bq(Y ) = g∗q(z +Bq(X)), and sof∗q = g∗q.

Definition. Two simplicial complexes X and Y are equivalent (written X ' Y ) ifthere exist simplicial maps f : X −→ Y and g : Y −→ X such that g f ' 1X andf g ' 1Y .

(Of course it is possible to define similarly the more general notion of equivalence ofchain complexes - see [51] - but we will not insist on that.) Complexes having home-omorphic geometrical realizations are easily seen to be equivalent. In particular, anycomplex is equivalent to its barycentric subdivision.

Theorem 1.16 If X and Y are equivalent complexes then for all q ≥ 0

Hq(X) ∼= Hq(Y ) .

Proof. Let f : X → Y and g : Y → X be morphisms such that g f ' 1Xand f g ' 1Y . Then, for q ≥ 0, we have from Theorem 1.15,

g∗q f∗q = (g f)∗q = (1X)∗q = 1Hq(X).

Similarly, f∗q g∗q = 1Hq(Y ). Hence Hq(X) ∼= Hq(Y ).

A simplicial map is called a constant if all vertices have the same image. We saythat a complex is (chain) contractible if it is equivalent to the complex of a singlepoint.

Proposition 1.17 Every contractible complex is acyclic; in particular its Eulercharacteristic is 1.For a complex X the following conditions are equivalent

(i) X is contractible;

(ii) the identity map 1X is equivalent to a constant;

(iii) for each q ≥ 0 there exists a homomorphism cq : Cq(X) −→ Cq+1(X) and avertex a of X such that ∂q+1cq+cq−1∂q = 1Cq(X) for q ≥ 1, and ∂1c0(v) = v−afor all v ∈ Vert(X).

15

Proof. The first claim follows immediately from Theorem 1.16.Suppose that X is contractible. Then it is equivalent to a point v, and there

exist simplicial maps f : X → v and g : v → X such that g f ' 1X andf g ' 1v. But then g f is just the constant application which maps everyx ∈ V ert(X) in g(v), and this proves (i)⇒ (ii).

Suppose now that the identity map 1X is equivalent to a constant a : X → X,where for all x ∈ V ert(X), a(x) = a for a fixed vertex a. Then by definition,for each q ≥ 0 there exists a homomorphism πq : Cq(X) → Cq+1(X) such that∂q+1πq + πq−1∂q = (1X)q − aq. Now, (1X)q = 1Cq(X), aq = 0 for q ≥ 1, anda0(x) = a for all x ∈ V ert(X). We get (iii) by simply renaming πq by cq. Hence(ii)⇒ (iii).

Finally, assume that (iii) holds. Put πq = cq for all q ≥ 0, π−1 = 0, and let Y bethe complex with the single vertex a. Define f : X → Y and g : Y → X by settingf(v) = a for all v ∈ V ert(X) and g(a) = a. Then f g = 1Y and g f = a. Forq ≥ 1 we have ∂q+1πq + πq−1∂q = 1Cq(X) = (1X)q − aq. Also, for all v ∈ V ert(X)we have (∂1π0 + π−1∂0)(v) = (∂1c0)(v) = v − a, whence ∂1π0 + π−1∂0 = (1X)0 − a0.Thus a = g f ' 1X , proving (iii)⇒ (i).

Remark. Our definition of contractible simplicial complex is indeed equivalent toacyclicity. This is because the chain groups Cq(X) are finitely generated free abeliangroups. In the literature on complexes and finite groups, a contractible complex isusually one whose geometric realisation is homotopy equivalent to a point in thetopological sense. We have, however, given a homological definition because it seemscloser to the combinatorial point of view of these notes. Moreover, all the results thatwe will present, proving the contractibility of a certain complex, unless explicitelystated, hold in the topological sense also.

1.3 Examples

Let n ≥ 0 and In = v0, . . . , vn be a set of cardinality n + 1. We denote by ∆n

the complex of all non-empty subsets of In (a geometric realization of it is just thestandard n-simplex with all its faces). As an exercise show that ∆n is acyclic. Thisis in fact a particular case of what is called a cone.

Definiton. A complex X is a cone if there exists a vertex x of X such thatσ ∪ x ∈ X for all σ ∈ X.

Observe that the complexes ∆n defined above are cones.

Proposition 1.18 Let the complex X be a cone. Then X is contractible. In par-ticular χ(X) = 1.

Proof. Let v be a vertex of the cone X such that σ ∪ v ∈ X for all σ ∈ X.Fix an orientation on X such that v is the least element.

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For each q ≥ 0 define a homomorphism πq : Cq(X) −→ Cq+1(X) by setting, forall q-simplex σ = [x0, . . . , xq] ∈ Sq(X):

πq(σ) =

[v, x0, . . . , xq] if x0 6= v0 otherwise

and extending by Z-linearity. Let q ≥ 1. Then, if x0 6= v:

(∂q+1πq + πq−1∂q)(σ) = ∂q+1πq(σ) + πq−1∂q(σ) =

= ∂q+1([v, x0, . . . , xq]) + πq−1

q∑i=0

(−1)i[x0, . . . , xi, . . . , xq] =

= [x0, . . . , xq] +

q∑i=0

(−1)i+1[v, x0, . . . , xi, . . . , xq] +

q∑i=0

(−1)i[v, x0, . . . , xi, . . . , xq] =

= [x0, . . . , xq] = σ .

If x0 = v:

(∂q+1πq + πq−1∂q(σ) = πq−1∂q)(σ) = πq−1

q∑i=0

(−1)i[x0, . . . , xi, . . . , xq] = [x0, . . . , xq].

Thus ∂q+1πq + πq−1∂q = 1 if q ≥ 1. For q = 0 we have ∂1π0(x) = x − v for allx ∈Vert(x)\v, and ∂1π0(v) = 0 = v − v. By Proposition 1.17, X is contractible.

Suppose, for example, that P is a partially ordered set with a greatest (or a least)element. Then the complex of chains ∆(P ) is a cone, and so by Propositon 1.18 itis contractible. We will obatin significant generalizations of this fact in chapter 3.

Let n ≥ 0. The complex of all proper faces of ∆n+1 is called the n-dimensionalsphere, and is denoted by Sn. As an abstract complex it is the set of all propernon empty subsets of a set of cardinality n+ 2.

Proposition 1.19 Let Sn be the n-dimensional sphere. Then

1) if n ≥ 1, then H0(Sn) = Hn(Sn) = Z and Hq(Sn) = 0 for 0 6= q 6= n.

2) H0(S0) = Z× Z and Hq(S0) = 0 for q ≥ 1.

Proof. S0 is the set of two disjoint vertices, while S1 is the set of verticesand edges of a triangle. In these cases the statement is verified by a straightforwardcalculation.

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Thus let n ≥ 2, and observe that Sn is connected, whence H0(Sn) = Z. We thenproceed by induction on n.Fix a vertex v0 of Sn, and let X = Sn \ v1, . . . , vn+1 be the complex obtainedby removing from Sn the face ”opposite” to v0 (this subcomplex is called the starof v0 and is the set of all simplices that are faces of some simplex that has v0 asa vertex). By its definiton, X is a cone, and thus it is contractible. Let Y be thecomplex of all simplices of Sn that do not contain v0. Y is just the set of all non-empty subsets of v1, . . . , vn+1, so it is of type ∆n and it is also contractible. ThusHq(X) = Hq(Y ) = 0 for q ≥ 1, and H0(X) = H0(Y ) = Z.Clearly Sn = X ∪ Y , and Z = X ∩ Y is just Sn−1. By inductive hypothesis we thenhave H0(Z) = Hn−1(Z) = Z and Hq(Z) = 0 for 0 6= q 6= n− 1.

Mayer-Vietoris theorem now gives the exact sequence:

0→ Hn(Sn)→ Hn−1(Z)→ Hn−1(X)⊕Hn−1(Y )→ Hn−1(Sn)→ . . .. . .→ H1(Sn)→ H0(Z)→ H0(X)⊕H0(Y )→ H0(Sn)→ 0

which reduces to

0→ Hn(Sn)→ Z→ 0→ · · · → 0→ H1(Sn)→ Z→ Z⊕ Z→ Z→ 0

yielding easily the desired conclusion.

Corollary 1.20 If n is even then χ(Sn) = 2; if n is odd χ(Sn) = 0.

The statement of Proposition 1.19 can be more concisely expressed by usingreduced homology: for n ≥ 1

Hn(Sn) = Z and Hq(Sn) = 0 for q ≤ n− 1

Definiton. Given n ≥ 0, a complex Γ is n-spherical if Hq(Γ) = 0 for q 6= n. Acomplex is spherical if it is n-spherical for some n ≥ 0.

Exercise. Let 0 ≤ n ≤ m. Study the homology of the set X of all non-emptysubsets of the set 0, 1, . . . ,m that have cardinality at most n.

Our final example is considering complexes of dimension 1. These are just (undi-rected and loop-free) graphs. We leave as an exercise to prove the following.

Proposition 1.21 Let X be a complex of dimension 1. Then the following condi-tions are equivalent.

i) X is connected and χ(X) = 1;

ii) X is acyclic;

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iii) X is contractible;

iv) X is connected and does not have any circuit;

v) for each pair of vertices x, y of X there exists one and only one path connectingx to y.

A 1-dimensional complex satisfying the properties of Proposition 1.21 is calleda tree.

1.4 Carriers

We now describe briefly an important tool, that has wide applications and whichwe will use in Chapter 3. In what follows, X and Y are given abstract simplicialcomplexes.

Definitions. 1) A carrier of X in Y is a map C that assigns to each simplex σ ofX a subcomplex C(σ) of Y such that if τ is a face of σ then C(τ) ⊆ C(σ).A carrier C is contractible (acyclic) if C(σ) is contractible (acyclic) for all σ ∈ X.A morphism of chain complexes g : C∗(X)→ C∗(Y ) is carried by C if, for all q ≥ 0,gq(σ) ∈ Cq(C(σ)) for any σ ∈ Sq(X).

Lemma 1.22 Let C be an acyclic carrier from X to Y . Then

(i) There exists a morphism g : C∗(X)→ C∗(Y ), carried by C.

(ii) If C is contractible, then any two morphism g, f : C∗(X)→ C∗(Y ) carried by Care equivalent.

Proof. (sketch). (i) For each x ∈ V ert(X) choose a vertex y ∈ C(x) (whichis not empty); let g0(x) = y and extend by linearity to obtain a homomorphismg0 : C0(X) → C0(Y ). Then, proceed by induction to construct homomorphismsgn : Cn(X) → Cn(Y ), such that gn−1∂n = ∂ngn. Let gn−1 be defined. For allσ ∈ Sn(X), the complex C(σ) is acyclic, thus ∂n−1gn−1∂n(σ) = gn−2∂n−1∂n(σ) = 0,whence gn−1∂n(σ) ∈ ker ∂n−1 = Im∂n, as Hn−1(C(σ)) = 0. (the differentials hereand in many following occurences are the restrictions to C(σ)). Then choose s ∈Cn(C(σ)) such that ∂n(s) = gn−1∂n(σ), and put gn(σ) = s. Do the same for allσ ∈ Sn(X) and then extend by linearity to define gn. The homomorphisms gn thusconstructed define a morphism g : C∗(X)→ C∗(Y ), which by construction satisfyies∂ g = g∂, and is carried by C. (Observe that, for the proof, it is enough to assumethat for all σ ∈ X, Hi(C(σ)) = 0 for −1 ≤ i < dim(σ).)

(ii) One constructs homomorphisms πn : Cn(X)→ Cn+1(Y ) satisfying ∂n+1πn+πn−1∂n = fn− gn in a similar way, proceeding by induction. For each x ∈ V ert(X),as C(x) is contractible, there exists a homomorphisms πx0 : C0(C(x)) → C1(C(x))

19

such that for all y ∈ V ert(C(x)), ∂1πx0 (y) = y − a for a fixed vertex a of C(x).

Since f0(x)− g0(x) ∈ C0(C(x)), we define π0(x) = πx0 (f0(x)− g0(x)), and extend bylinearity.

If πk is defined for all k < n, let us define πn. Let σ ∈ Sn(X). Then∂n(πn−1∂n(σ) − (fn(σ) − gn(σ))) = 0. Now, the simplices that appear in ∂n(σ)are faces of σ, and so πn−1∂n(σ) ∈ Cn(C(σ)). Also, fn(σ) − gn(σ) ∈ Cn(C(σ)).Therefore πn−1∂n(σ) − (fn(σ) − gn(σ)) belongs to Cn(C(σ)) ∩ ker ∂n. Again, sinceC(σ) is acyclic, there exists s ∈ Cn+1(C(σ)) ⊆ Cn+1(Y ) such that ∂n+1(s) =πn−1∂n(σ) − (fn(σ) − gn(σ)). We then set πn(σ) = s and extend by linearity todefine a homomorphism πn : Cn(X) → Cn+1(Y ). The πn thus constructed satisfythe required conditions to establish that f and g are equivalent.

20

Chapter 2

Group Actions

Let a group G act by permutations on a set Ω (not necessarily faithfully). Foreach x ∈ Ω and each g ∈ G we denote by xg the image of x under the permutationinduced by g. We also fix the following notations:

- xG = xg | g ∈ G (the orbit of x under G);

- Gx = g ∈ G | xg = x (the stabilizer of x in G);

- Ωg = x ∈ Ω | xg = x (the set of fixed points of g);

- ΩG = x ∈ Ω | xg = x for all g ∈ G (the set of fixed points of G).

Recall that the distinct orbits of G on Ω form a partition of Ω, and that for eachx ∈ Ω, |xG| = |G : Gx|. We remind now a well known formula that we will use lateron.

Lemma 2.1 Let the group G act by permutations on the (finite) set Ω. Then∑g∈G|Ωg| = t|G|

where t is the number of distinct orbits of G on Ω.

Proof. We count the order of the set S = (x, g) ∈ Ω × G | xg = x . Letx1, x2, . . . , xt be a set of representatives of the distinct orbits. We have

∑g∈G|Ωg| = |S| =

∑x∈Ω

|Gx| =t∑i=1

|xGi ||Gxi | =t∑i=1

|G : Gxi ||Gxi | = t|G| .

21

Recall that an automorphism of a simplicial complex Γ is a bijective simplicialmap Γ → Γ (in this case it is easy to prove that the inverse map is again simpli-cial). Clearly, the set Aut(Γ) of all automorphisms of a complex Γ is a group undercomposition. A group G acts simplicially on Γ if there is a group homomorphismG → Aut(Γ). In this case we say that Γ is a G-complex. Observe that if G actson Γ, then, for all q ≥ 0 and all g ∈ G, the map g∗q is an automorphism of Hq(Γ),and the correspondence g 7→ g∗q defines a linear representation (over the integers inordinary homology) G→ Aut(Hq(Γ)).

If f is a simplicial map on a complex Γ of dimension n, then for each 0 ≤ q ≤ nwe denote by Sq(Γ)f the set of all q-simplices of Γ fixed by f . (Observe that itis not always the case that f actually permutes the elements of Sq(Γ), as it mightmap some of those to simplices of smaller dimension; however it certainly inducespermutations if it is an isomorphism). Now, denote by Γf the union of all Sq(Γ)f

(i.e. the set of all simplices of Γ fixed by f). In general, Γf might well not be acomplex (think for instance to a rotation of a triangle, it is clearly simplicial and fixesthe triangle itself but none of its proper faces). This remark leads to the followingdefinition.

Definition. Let Γ be a simplicial complex. A simplicial map f : Γ −→ Γ isadmissible if for every simplex σ = x0, x1, . . . , xn in Γ

fx0, fx1, . . . , fxn = σ ⇒ fxi = xi, i = 0, 1, . . . , n .

The following facts are easily verified

Proposition 2.2 Let f : Γ −→ Γ be an admissible simplicial map on the complexΓ. Then Γf is a subcomplex of Γ and, for any positive q, Sq(Γ)f = Sq(Γ

f ).

Proposition 2.3 Let P be a poset and let f : P −→ P be an order preservingmap. Then f induces an admissible simplicial map on ∆(P ). In particular, everysimplicial map f : Γ −→ Γ on the complex Γ induces an admissible map on thebarycentric subdivision of Γ.

It is also clear that the inverse of an admissible automorphism is admissible, andthat the composition of admissible automorphisms is admissible. Thus, the set ofadmissible automorphisms of a complex Γ is a subgroup of Aut(Γ) (prove that it isa normal subgroup).

Proposition 2.4 Let G be a group of admissible automorphisms of the complex Γ.Let n be the dimension of Γ, and for 0 ≤ q ≤ n, let tq be the number of orbits of Gon Sq(Γ). Then ∑

g∈Gχ(Γg) = |G|

n∑q=0

(−1)qtq =∑σ∈Γ

(−1)dim(σ)|Gσ|

22

Proof. By Lemma 2.1 we have∑

g∈G |Sq(Γ)g| = |G|tq for all 0 ≤ q ≤ n. Hence

∑g∈G

χ(Γg) =∑g∈G

n∑q=0

(−1)q|Sq(Γg)| =n∑q=0

(−1)q

∑g∈G|Sq(Γ)g|

=

=n∑q=0

(−1)q|G|tq = |G|n∑q=0

(−1)qtq .

On the other hand, if for a 0 ≤ q ≤ n, σ1, . . . , σtq are representatives of the orbits

of G on Sq(Γ), we have tq|G| =∑tq

i=1 |G : Gσi ||Gσi | =∑

σ∈Sq(Γ) |Gσ|. Thus

∑g∈G

χ(Γg) = |G|n∑q=0

(−1)qtq =∑σ∈Γ

(−1)dim(σ)|Gσ| .

Corollary 2.5 Let G be a group of admissible automorphisms of the complex Γ.Assume that χ(Γg) = 1 for all 1 6= g ∈ G. Then

χ(Γ) ≡ 1 (mod |G|) .

Proof. By the previous Proposition (and with the same notations)

|G|n∑q=0

(−1)qtq = χ(Γ) +∑g 6=1

χ(Γg) = χ(Γ) + |G| − 1 .

2.1 Lefschetz number

Let A be a free abelian group of finite rank, and x1, . . . , xn be a free system ofgenerators of A. For any f ∈ End(A) are defined coefficients αij ∈ Z such that forevery i = 1, . . . , n

f(xi) =

n∑j=1

αijxj .

The trace of f is then the integer

Tr(f) =n∑i=1

αii .

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As in the familiar case of vector spaces over a field, one verifies that the trace of fdoes not depend on the free system of generators.

In general, the trace of an endomorphism f of a finitely generated abelian groupG is defined as the trace of the endomorphism f induced by f on the factor groupG/T (G), where T (G) denotes the torsion subgroup of G (i.e. the set of all theelements of finite order of G; it is a well known and easy fact that G/T (G) is afinitely generated free abelian group).

Lemma 2.6 Let G be a finitely generated abelian group, f ∈ End(G), G1 a sub-group of G such that f(G1) = G1, and G2 = G/G1. Then f induces by re-striction a homomorphism f1 of G1, and a homomorphism f2 of G2 by the rulef2(g +G1) = f(g) +G1, and we have

Tr(f) = Tr(f1) + Tr(f2) .

Proof. Exercise.

Now, let Γ be a simplicial complex of dimension n, and f : Γ −→ Γ a simplicialmap. For each 0 ≤ q ≤ n, f induces a homomorphism fq on Cq(Γ). We consider thefree base Sq(Γ) of Cq(Γ). Since fq permutes the elements of Sq(Γ), or sends someof them to 0, the rows of the matrix representing fq with respect to Sq(Γ) all haveαq − 1 zero entries and the remaining entry is either 1 or 0. Thus, the trace of fq isthe number of 1-entries in the diagonal, i.e. the number of q-simplices fixed by fq.Hence, for all 0 ≤ q ≤ n, we have

Tr(fq) = |Sq(Γ)f | .

The Lefschetz number Λ(f) of f is now defined as follows

Λ(f) =n∑q=0

(−1)qTr(fq) .

Similarly if, for 0 ≤ q ≤ n, f∗q is the endomorphism induced by f on Hq(Γ), andf∗ = (f∗0, . . . , f∗n), we set

Λ(f∗) =n∑q=0

(−1)qTr(f∗q) .

Theorem 2.7 Let f : X −→ X be a simplicial map on the complex X. Then

Λ(f) = Λ(f∗) .

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Proof. Let n be the dimension of X, and consider the chain complex withdifferentiations ∂q

0→ Cn(X)→ Cn−1(X)→ . . .→ C1(X)→ C0(X)→ 0.

Since, for all q, ∂qfq = fq−1∂q, we have that fq(Zq(X)) ≤ Zq(X) and fq(Bq(X)) ≤Bq(X); so the restriction fZq of fq to Zq(X) is a homomorphism of Zq(X), and the

restriction fBq to Bq(X) is a homomorphism of Bq(X).Thus, by applying Lemma 2.6 to the short exact sequence

0 −→ Zq(X) −→ Cq(X) −→ Bq−1(X) −→ 0

we get (setting fB−1 = 0)

Tr(fq) = Tr(fZq ) + Tr(fBq−1).

On the other hand, the definition of homology groups is the exact sequence

0 −→ Bq(X) −→ Zq(X) −→ Hq(X) −→ 0

and, again applying Lemma 2.6 (here: fBn = 0),

Tr(fZq ) = Tr(fBq ) + Tr(f∗q).

Putting things together

Λ(f) =

n∑q=0

(−1)qTr(fq) =

n∑q=0

(−1)q(Tr(fZq ) + Tr(fBq−1)) =

=

n∑q=0

(−1)qTr(fZq ) +n∑q=0

(−1)qTr(fBq−1) =

=n∑q=0

(−1)qTr(fZq ) +n∑q=0

(−1)q+1Tr(fBq ) =

=

n∑q=0

(−1)q(Tr(fZq )− Tr(fBq )) =

n∑q=0

(−1)qTr(f∗q) = Λ(f∗).

In the statement of the next Theorem, and in what follows, we adopt the con-vention: χ(∅) = 0.

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Theorem 2.8 Let f : Γ −→ Γ be an admissible simplicial map on the complex Γ.Then

Λ(f) = χ(Γf ) .

Proof.

Λ(f) =n∑q=0

(−1)qTr(fq) =n∑q=0

(−1)q|Sq(Γ)f | =n∑q=0

(−1)q|Sq(Γf )| = χ(Γf ) .

Corollary 2.9 Let f : Γ −→ Γ be a simplicial map such that Λ(f) 6= 0. Then

1. If f is admissible, it admits a fixed point (on V ert(Γ))

2. f fixes at least one simplex of Γ.

Proof. 1) Follows immediately from the previous Theorem.2) Consider the simplicial map f ′ induced by f on the barycentric subdivision

∆(Γ) of Γ. Then f ′ is admissible. Also, for 0 ≤ q ≤ n = dim(Γ), we have Hq(Γ) =Hq(∆(Γ)), and f∗q = f ′∗q. Hence

Λ(f ′) = Λ(f ′∗) = Λ(f∗) = Λ(f) 6= 0 .

By point 1), f ′ has a fixed point on ∆(Γ), i.e. f fixes some simplex of Γ.

Corollary 2.10 Let f : Γ −→ Γ be an admissible simplicial map. If Γ is acyclicthen χ(Γf ) = 1 (in particular, f fixes some vertices of Γ).

Proof. By hypothesis we have Hq(Γ) = 0 for all q ≥ 1, and H0(Γ) = Z.Therefore, Tr(f∗q) = 0 for all q ≥ 1, and Tr(f∗0) = 1 (in fact, if v ∈ V ert(Γ) thenv+B0(Γ) is a generator of H0(γ) = Z, also f(v) is a vertex and so ∂0(f(v)− v) = 0,and f∗0(v + B0(Γ)) = f(v) + B0(Γ) = v + B0(Γ)). By Theorem 2.7 and Theorem2.8 we get χ(Γf ) = Λ(f) = Λ(f∗) = Tr(f∗0) = 1.

Observe that Corollary 2.10 remains true if one replaces the hypothesis of acyclic-ity of Γ by just assuming that Γ is connected and Hq(Γ) is a finite group for all q ≥ 1(or, in other words, Γ is Q-acyclic).

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2.2 Orbits

Let G be a group of admissible automorphisms of the complex X. Then each Sq(X)partitions into its G-orbits. This suggests the possibility of thinking of a complexwhose elements are the orbits of G on X. However, what appears to be a naturaldefinition for it does not always work. Let us look at a simple example. Supposex, y is an edge of X; then we would like xG, yG to be an edge in a complex oforbits. The simplest way is to ask whether xG, yG = x, yG. To see that thisis not always the case, think of a 1-sphere X = x, y, z, [x, y], [y, z], [z, x] and tothe automorphism g of X that permutes cyclically x, y, z. Then both S0(X) andS1(X) have a unique orbit under G = 〈g〉, whence the set of orbits cannot be givena structure of a complex (if we want that orbits of 1-simplices be again 1-simplices).We are then lead to consider the following properties for a group of automorphismsG of a complex X.

(A) For all x ∈ V ert(X) and all g ∈ G, if x, xg ∈ X then x = xg.

(B) for every simplex x0, x1, . . . , xq of X (vertices not necessarily distinct) andg0, g1, . . . , gq ∈ G, if xg00 , x

g11 , . . . , x

gqq ∈ X then there exists g ∈ G such that

xg00 , xg11 , . . . , x

gqq = xg0, x

g1, . . . , x

gq.

Lemma 2.11 Property (B) implies property (A).

Proof. Suppose the group G satisfies (B) in its action on the complex X, andlet x ∈ V ert(X) and g ∈ G be such that x, xg ∈ X. Then there exists h ∈ G suchthat x, xg = xh, xh, and this forces x = xg.

Note that property (A) is stronger than admissibility, but it is still verified byautomorphisms of posets. (Observe that the 1-sphere example considered beforeshows that admissibility does not imply property (A)).

Lemma 2.12 Let G be a finite group of automorphisms of a poset P . Then Gsatisfies (A) on the complex of chains ∆(P ).

Proof. Let x ∈ P and g ∈ G be such that x, xg is a chain in P , say x ≤ xg.Since g has finite order m and is order preserving we get x ≤ xg ≤ . . . ≤ xg

m= x

whence xg = x.

Exercise. Assume that G is a group of automorphisms of the complex X. Provethat G satisfies property (A) if and only if for all σ ∈ X and all g ∈ G, every vertexin σ ∩ σg is fixed by g.

Definition. A group of automorphisms G of a complex X is regular if all subgroupsof G satisfy property (B) in the induced action.

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Theorem 2.13 Let G be a group of automorphisms of the complex X satisfying(A). Then G is regular on the barycentric subdivision ∆(X).

Proof. It will be enough to prove that G satisfies property (B), since (A) is aproperty which is obviously hinerited by all subgroups of G.

Let Σ = σ0, σ1, . . . , σq be a simplex of ∆(X). Then σ0, . . . , σq are simplicesof X and we have chosen indices so that σ0 ⊂ σ1 ⊂ . . . ⊂ σq. Let g0, g1, . . . , gq beelements of G such that

Σ′ = σg00 , σg11 , . . . , σ

gqq ∈ ∆(X) .

Then, since the elements of G are automorphisms, σg00 ⊂ σg11 ⊂ . . . ⊂ σ

gqq . We prove

by induction on q that there exists g ∈ G such that Σ′ = Σg.If q = 0 we have nothing to prove. Thus, let q ≥ 1. By inductive hypothesis we havea g ∈ G such that

σg11 , . . . , σgqq = σg1 , . . . , σ

gq.

Now σg11 is the least element of the chain, so σg00 ⊂ σg11 = σg1 . Let h = g0g−1. Then

σh0 ⊂ σ1. If x is a vertex of σ0 then x, xh ⊆ σ1 is a simplex in X. By property (A)this forces x = xh. Hence σ0 = σh0 , that is σg00 = σg0 , thus completing the proof.

Corollary 2.14 Let G be a finite group of automorphisms of a poset P . Then Ginduces a regular group of automorphisms on the barycentric subdivision of ∆(P ).

Corollary 2.15 Let G be a group of automorphisms of the complex X. Then Ginduces a regular group of automorphisms on the second barycentric subdivision ofX.

For groups satifying (B) we may now define the orbit complex.

Proposition 2.16 Let G be a regular group of automorphisms of the complex X,and let X/G be the set of all the G-orbits on X (we mean the set of the orbits onall subsets Sq(X)). Then X/G has a natural structure of simplicial complex.

Proof. Exercise.

Definition. The complex X/G defined in Proposition 2.16 is called the orbitcomplex of G over X.

Observe that, if G is regular on the complex X and σ ∈ X, then the G-orbitσG is a simplex of X/G whose faces are the orbits of the faces of σ. Thus, writingτG ⊆ σG it means that there exist g, h ∈ G such that τ g ⊆ σh.

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Assume G is a regular group of automorphisms of a complex X. We are nowinterested in describing relations among the homology of X and of the orbit complexX/G.Observe first that the projection π : X −→ X/G defined by mapping each simplexs of X in its G-orbit, π(s) = sG, is a surjective simplicial map. Extending bylinearity we have surjective homomorphisms π : C∗(X) −→ C∗(X/G) and in turnhomomorphisms on homology

π∗ : H∗(X) −→ H∗(X/G)

such that π∗(s+B∗(X)) = π(s) +B∗(X/G), for all s ∈ Z∗(X).Now consider the application σ defined by setting σ(s) =

∑g∈G s

g for all s ∈Cq(X); it is clearly an endomorphism for each q ≥ 0.

Lemma 2.17 With the above notations, kerσ = kerπ.

Proof. Observe that the Lemma will follow if for each simplex s ∈ X, anintegral combination c =

∑g∈G ngs

g belongs to kerσ if and only if it belongs tokerπ.

Now, for all simplices s, σ(s) = σ(sg). Thus σ(c) = (∑

g∈G ng)σ(s), and c ∈ kerσif and only if

∑g∈G ng = 0.

On the other hand π(c) = (∑

g∈G ng)sG and again c ∈ kerπ if and only if∑

g∈G ng = 0.

Theorem 2.18 There exists a homomorphism µ∗ : H∗(X/G) −→ H∗(X) such that,for all s ∈ Z∗(X)

µ∗π∗(s+B∗(X)) = σ(s) +B∗(X)

π∗µ∗(sG +B∗(X/G)) = |G|(sG +B∗(X/G))

Proof. For every q ≥ 0 we have

Cq(X/G) ∼= Cq(X)/ kerπ = Cq(X)/ kerσ ∼= Imσ ≤ Cq(X)

explicitely, we have the monomorphism µ : Cq(X/G) −→ Cq(X) given by

µ(sG) = µ(π(s)) = σ(s),

which is well-defined. The induced homomorphism µ∗ : Hq(X/G) −→ Hq(X) givenby, for all s ∈ Zq(X),

µ∗(sG +Bq(X/G)) = σ(s) +Bq(X)

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has the required properties. In fact

µ∗π∗(s+Bq(X)) = µ∗(sG +Bq(X/G)) = σ(s) +Bq(X)

π∗µ∗(sG +Bq(X/G)) = π∗(σ(s) +Bq(X)) =

∑g∈G

π(sg) +Bq(X/G) =

= |G|(sG +B∗(X/G)).

Definition. The homomorphism µ∗ defined by Theorem 2.18 is called transfer.

Proposition 2.19 Let G be a regular group of automorphisms of the complex X,and let X/G be the orbit complex. Then

i) if, for some q ≥ 0, Hq(X) = 0 then Hq(X/G) is a finite group whose exponentdivides |G|. In particular r0(Hq(X/G)) = 0.

ii) If X is connected, then X/G is connected.

iii) If X is acyclic, then χ(X/G) = 1.

Proof. i) If Hq(X) = 0 for some q, then 0 = πqµq(Hq(X/G)). This meansthat for every h ∈ Hq(X/G) one has |G|h = 0. Thus Hq(X/G) is an abelian groupof exponent dividing |G|.ii) If X is connected then any two vertices x, y of X are joined by a path ρ in the1-skeleton of X. Clearly the orbits of the edges of ρ form a path from xG to yG inX/G, and so X/G is connected.

iii) If X is acyclic, then it follows from i) and ii) that χ(X/G) = r0(H0(X/G)) = 1.

A complex X endowed with an action of a group of automorphisms G is calleda G-complex.Let X and Y be two G-complexes, we say that X is G-isomorphic to Y if there existsan isomorphism f : X −→ Y such that for all verticies v ∈ V ert(X), and all g ∈ G,f(vg) = (f(v))g. If X and Y are isomorphic G-complexes, then for each q ≥ 0 thepermutation actions of G on Sq(X) and Sq(Y ) are similar, and the groups Hq(X),Hq(Y ) are isomorphic ZG-modules.More generally, it is possible to define equivalence among G-complexes. Let X andY be two G-complexes. A morphism of X in Y as G-complexes is a simplicial mapthat commutes with the action of G (as in the isomorphism case). Two morphisms

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of G-complexes f, g : X → Y are equivalent (written f 'G g) if for each q ≥ 0 thereexists a homomorphism of ZG-modules πq : Cq(X) −→ Cq+1(Y ) such that

δq+1πq + πq−1∂q = fq − gq

where π−1 = 0. Then we say that X and Y are equivalent as G-complexes if thereexists morphisms f : X → Y and g : Y → X such that g f 'G 1X and f g 'G 1Y .We leave it as an exercise to show that if X, Y are equivalent G-complexes and f isa G-equivalence, then, for each q ≥ 0, f∗q : Hq(X) −→ Hq(Y ) is an isomorphisms ofZG-modules. Also, if X, Y are G-equivalent then X/G, Y/G are equivalent (withtrivial G-action).

A even more general, and maybe more natural, approach is to define equivalenceof G-chain complexes. As an exercise, give appropriate definitions, and compare thechain with the simplicial point of view.

Exercises.

1) Let F = H0, . . . ,Hn be a family of proper subgroups of the group G, andlet ΓF be the associated coset complex.Define an action of G on V ert(ΓF ) = Hx |H ∈ F , x ∈ G by setting (Hx)g = Hxg.Prove that this induces an action of G as a regular group of automorphisms of ΓF ,and that the kernel of this action is

⋂ni=0(Hi)G (where for a subgroup H of G, HG

denotes the largest normal subgroup of G contained in H). Prove then that theorbit complex ΓF/G is contractible, and that the set of G-fixed points is empty.

2) Let X be a complex and G a regular group of automorphisms of X. Assumefurther that G permutes transitively the maximal simplices of X. Prove that thereexists a family F of subgroups of G such that X is G-equivalent to ΓF .

Proposition 2.20 Let G be a regular group of automorphisms of the complex Γ.Then ∑

g∈Gχ(Γg) = |G|χ(Γ/G).

Proof. by Proposition 2.4 we have∑

g∈G χ(Γg) = |G|∑n

q=0(−1)qtq, where tqis the number of distinct orbits of G on Sq(Γ), i.e. tq = αq(Γ/G). Therefore

|G|χ(Γ/G) = |G|n∑q=0

(−1)qtq =∑g∈G

χ(Γg).

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Corollary 2.21 Let p be a prime, and G = Cp be a cyclic group of order p actingregularly as a group of automorphisms of the complex Γ. Then

χ(Γ) + (p− 1)χ(ΓG) = pχ(Γ/G).

If, in particular, Γ is acyclic, then χ(ΓG) = 1.

Proof. Apply Proposition 2.20, observing that χ(Γ1) = χ(Γ), and for eachi = 1, . . . , p− 1, χ(Γg

i) = χ(Γg) = χ(ΓG).

If Γ is acyclic, then χ(Γ/G) = 1 by Proposition 2.19 (iii). The previous formulanow gives χ(Γg) = 1.

2.3 Fixed Points

In this section, starting from what we already learned by using the Lefschetz num-ber, we discuss the problem of the existence of fixed points for group actions oncomplexes. Most of the existence theorems apply to acyclic complexes, and we willuse some of them later on, to show that certain interesting complexes arising fromthe inner structure of a group are not acyclic. The bulk of this section is a somewhatsimplified version of part of chapter III in Bredon text [19].

Theorem 2.22 (P. Smith) Let p be a prime and P a p-group of admissible auto-morphisms of the complex Γ. Then

1. χ(ΓP ) ≡ χ(Γ) (mod p)

2. if χ(Γ) = 1 then P fixes some vertex of Γ.

Proof. 1) Let |P | = pm. We argue by induction on m. If m = 1 then P isgenerated by an element g of order p. Clearly, for each i = 1, . . . , p − 1, we haveΓg

i= ΓP . Let n be the dimension of Γ, and, for 0 ≤ q ≤ n, let tq be the number of

P -orbits on Sq(Γ). Then by Proposition 2.4

pn∑q=0

(−1)qtq =

p−1∑i=0

χ(Γgi) = χ(Γ) + (p− 1)χ(ΓP )

proving that χ(ΓP ) ≡ χ(Γ) (mod p).Let now m ≥ 2. Let N be a maximal subgroup of P ; then |N | = pm−1 and the

factor group P = P/N has order p. Clearly ΓP ⊆ ΓN . If ΓN is empty, such is ΓP

and, by inductive assumption, χ(Γ) ≡ 0 (mod p). Let ΓN 6= ∅. Since N is normal inP it is easy to observe that ΓN is a P -invariant subcomplex of Γ, and ΓP = (ΓN )P .Also we have an action of P on ΓN defined by xgN = xg for x ∈ ΓN and gN ∈ P .

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We then have ΓP = (ΓN )P = (ΓN )P . By case m = 1 and the inductive assumptionwe get

χ(ΓP ) = χ((ΓN )P ) ≡ χ(ΓN ) ≡ χ(Γ) (mod p) .

2) is an immediate consequence of 1).

Definition. Let p be a prime number. A group G is said to be p-hyperelementaryif it admits a normal cyclic subgroup C such that G/C is a p-group.

Proposition 2.23 A p-hyperelementary group of admissible automorphisms of anacyclic complex fixes at least one vertex.

Proof. Let G be a p-hyperelementary group of automorphisms of the acycliccomplex Γ, and let C be a cyclic normal subgroup of G with G/C a p-group. Thenby Corollary 2.10 χ(ΓC) = 1. Now ΓC is a G/C-invariant subcomplex of Γ, andΓG = (ΓC)G. By Smith’s Theorem G fixes a vertex of ΓC .

By passing to the baricentric subdivision, we then have

Corollary 2.24 A p-hyperelementary group of automorphisms of an acyclic com-plex leaves at least one simplex invariant.

Exercises. 1) Let G be a hyperlementary group and F = H0, . . . ,Hn a familyof proper subgroups of the group G. Show that the coset complex ΓF cannot beacyclic.

2) Let G be an elementary abelian p-group. Find a connected graph (i.e. a1-dimensional spherical complex) X such that Xg = ∅ for all 1 6= g ∈ G.

In fact something more than Proposition 2.23 can be said. Let D denote theclass of all finite groups G that admit normal subgroups N ≤ M such that N isa p-group, M/N is cyclic, and G/M is a q-group (where p, q are not necessarilydistinct primes). We then have:

Theorem 2.25 (Oliver [43]) Let G be a group of admissible automorphisms of anacyclic complex. If G belongs to the class D then G fixes at least one vertex.

For the proof we need the following strenghtening of Theorem 2.22 (see Chapter IIIof Bredon book [19] for a far more complete treatment).

Theorem 2.26 (P. Smith) Let p be a prime, and G be a p-group of admissibleautomorphisms of the complex X. If X is Zp-acyclic, then XG is Zp-acyclic.

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Proof. Let X be a Zp-acyclic complex. Remember that this means that thatHq(X;Zp) = 0 for q ≥ 1, and H0(X;Zp) = Zp; and is equivalent to require that Xis connected and the integral homology groups in dimension q ≥ 1 are finite groupsof order coprime to p.Just for this proof, we write Hq(X) = Hq(X;Zp); thus Hq(X) is a finite dimensionalvector space over Zp.

Let then G be a p-group of admissible automorphisms of the Zp-acyclic complexX (of dimension n). By applying an easy inductive argument as in the proof ofTheorem 2.22, we reduce to the case in which G = 〈g〉 is a cyclic group of order p.

Fixed a 0 ≤ q ≤ n, we look at g as an automorphims of Cq(X) (which for thistime we will write on the left), and consider the homomorphisms

σ = 1 + g + g2 + . . .+ gp−1 τ = 1− g.

Then στ = τσ = 1− gp = 0; that is σCq(X) ≤ ker τ and τCq(X) ≤ kerσ.Let x ∈ Sq(XG), then τ(x) = x − g(x) = 0 and σ(x) = x + g(x) + . . . + gp−1(x) =px = 0. Thus, Cq(X

G) ≤ kerσ ∩ ker τ .Let Cq be the subspace of Cq(X) generated by Sq(X) \ Sq(XG). Clearly, Cq is

G-invariant and Cq(X) = Cq ⊕ Cq(XG). In particular, σCq(X) = σCq ⊆ Cq andτCq(X) = τCq ⊆ Cq. Thus

σCq(X)⊕ Cq(XG) ⊆ ker τ and τCq(X)⊕ Cq(XG) ⊆ kerσ.

Now, let S be a set of representatives of the G-orbits on Sq(X) \ Sq(XG); then

σCq(X) =∑s∈S

λsσ(s) | λs ∈ Zp

= ker τ ∩ Cq

This yields ker τ = σCq(X)⊕Cq(XG) and, by counting dimensions (as σ is a homo-morphism of a vector space) kerσ = τCq(X)⊕ Cq(XG)

Thus, for all q ≥ 0, we have the following exact sequences

0→ σCq(X)⊕ Cq(XG)→ Cq(X)→ τ(Cq(X))→ 0

0→ τCq(X)⊕ Cq(XG)→ Cq(X)→ σ(Cq(X))→ 0

which provide, as it easy to verify, short exact sequences of chain complexes. Theseyield, by Proposition 1.7 long exact sequences

. . .→ Hq+1(τX)→ Hq(σX)⊕Hq(XG)→ Hq(X)→ Hq(τX)→ . . .

. . .→ Hq+1(σX)→ Hq(τX)⊕Hq(XG)→ Hq(X)→ Hq(σX)→ . . .

If X is Zp-acyclic, starting with 0 = Hn(X) = Hn(σX) = Hn(τX) = Hn(XG), oneeasily obtains Hq(X

G) = 0 for q ≥ 1. For q = 0 we have

. . .→ H1(τX)→ H0(σX)⊕H0(XG)→ H0(X)→ H0(τX)→ 0

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which reduces to

0→ H0(σX)⊕H0(XG)→ Zp → H0(τX)→ 0

Now, we know by Corollary 2.10 that XG 6= ∅, and H0(XG) 6= 0, thus the lastsequence gives H0(XG) = Zp.

Proof. (of Theorem 2.25) Let G be a group in the class D, and let N ≤ Mbe normal subgroups of G such that N is a p-group, M/N is cyclic, and G/M is aq-group. Suppose that G acts as a group of admissible automorphisms on the acycliccomplex X. Then, by Theorem 2.26, XN is Zp-acyclic. This means, in particular,that H0(XN ) = Z and, for q ≥ 1, Hq(X

N ) is a finite group. As M/N is cyclic, wehave by Corollary 2.10 (and the remark following it) that χ(XM ) = χ((XN )M ) = 1.Finally, since G/M is a q-group, by Theorem 2.22 we conclude that G fixes somevertex in XM .

Oliver’s Theorem is in a sense best possible. In fact Oliver proves [43] that everyfinite group not in the class D acts without fixing any vertex as a group of admissibleautomorphism of a suitable acyclic complex.

Thus, to obtain fixed points results for large classes of finite groups, one has toimpose different or stronger conditions than acyclicity on the complexes involved.The case of a sphere is an interesting one, and one can consult Bredon’s book (oralso, from our abstract point of view, an exercise in the next chapter). In imposingstronger conditions than acyclicity, one natural idea is to consider complexes of smalldimension. The 1-dimensional case is described in the next two statements, whoseproofs we leave as an exercise. Recall that a tree is a 1-dimesional acyclic complex.

Lemma 2.27 Let f be an automorphism of the tree X. Then f sends extremalvertices (i.e. vertices that belong to only one edge) to extremal vertices. Also f fixesthe central vertex or the central edge of every maximal path of X.

Proposition 2.28 Let G be a group of admissible automorphisms of the tree X.then XG is a non-empty tree.

For two dimensional complexes, the most general result has beeen obtained byY. Segev [52].

Theorem 2.29 (Segev) Let G be a finite group of admissible automorphisms of anacyclic 2-dimensional complex X. Then

1. XG is either empty or acyclic.

2. If G is solvable then XG is acyclic.

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We isolate part of the proof in the following Lemma.

Lemma 2.30 Let G be a group of admissible automorphisms of a 2-dimensionalcomplex X. Assume that H2(X) = H1(X) = 0. Then H2(XG) = H1(XG) = 0 (weallow XG = ∅)

Proof. Suppose first that |G| = p for some prime p. Let Y be a connectedcomponent of X. Then Y is acyclic and so, by Theorem 2.26, Y G is Zp-acyclic; inparticular H1(Y G) is a finite group (clearly H2(Y G) ≤ H2(Y ) = 0) . If s ∈ Z1(Y G)then s ∈ B1(Y ), and there exists u ∈ C2(Y ) with ∂2(u) = s. Now, for some n ≥ 1,ns ∈ B1(Y G), so ∂2(nu) = n∂2(u) = ns ∈ B1(Y G). Since ∂2 is injective, we havenu ∈ C2(Y G) which forces u ∈ C2(Y G) (because C2(Y G) is a direct summand ofC2(Y )) and thus s ∈ B1(Y G), proving that H1(Y G) = 0.Since the homology of XG is the direct some of the homology of its connectedcomponents, we have proved the claim for |G| = p. Proceding by induction on |G|in the usual way, we get the result for G cyclic.

Now suppose G be any group. Then H2(XG) = 0 because, since X is 2-dimensional, H2(X) = 0 implies that the first differentiation ∂2 is injective. Letz ∈ Z1(XG). Then, z ∈ Z1(Xg) for all g ∈ G. Hence, by the cyclic case, there existsug ∈ C2(Xg) such that ∂2(ug) = z. Again, by injectivity of ∂2, u = ug is the samefor all g ∈ G. Thus u ∈ C2(XG) and z ∈ B1(XG), proving that H1(XG) = 0.

Proof. (of Theorem 2.29). We first deal with 2. Thus assume G is soluble, andproceed by induction on |G|. Let N be a maximal normal subgroup of G. Then G/Nis a group of order p for some prime p. By inductive hypothesis, XN is acyclic. Now,the same argument in the proof of the previous Lemma gives that XG = (XN )G/N

is acyclic.

1. Let G be a finite group of admissible automorphisms of the acyclic 2-dimensional complex X, and assume that XG is not empty. We proceed by in-duction on |G|. Suppose that G has a proper non-trivial normal subgroup N . Thenby inductive hypotesis XN is acyclic. Moreover, G/N acts as a group of admissi-ble automorphisms on it, and (XN )G/N = (XN )G = XG is not empty. Again byinductive hypotesis, we have that XG is acyclic.Then we may assume that G is a non abelian simple group. Then, by Feit andThompson Theorem, G has even order; in particular, G is generated by its invo-lutions. Remind that, in any group, the subgroup generated by two involutions isdihedral (and soluble). So if x, y are involutions of G the fixed point complex X〈x,y〉

is acyclic by point 2.Let x1, . . . , xn be a minimal set of involutions that generates G, and, for all i =1, . . . , n let Xi = Xxi . Let K = 〈x3, . . . xn〉; K is a proper subgroup of G, so byinductive hypothesis XK is acyclic. Then, Maier-Vietoris sequence gives (keeping

36

in mind that H2 is trivial for all subcomplexes of X):

0→ H1(XK ∪X1)→ H0(XK ∩X1)→ H0(XK)⊕H0(X1)→ H0(XK ∪X1)→ 0.

Now, XK ∩X1 = X〈x1,x3,...,xn〉 is acyclic by inductive hypothesis, and XK ∪X1 isconnected since XK , X1 are connected and their intersection is not empty. Thusthe sequence above reduces to

0→ H1(XK ∪X1)→ Z→ Z2 → Z→ 0

forcing H1(XK ∪X1) = 0. Similarly one proves H1(X1 ∪X2) = 0.Now, XG = X1 ∩ X2 ∩ . . . ∩ Xn = X1 ∩ X2 ∩ XK . Let Z = X2 ∩ XK ; then Z =X〈x2,...,xn〉 is acyclic by inductive hypothesis. Also X1∪Z = (X1∪X2)∩ (X1∪XK).Then Maier-Vietoris sequence gives

0→ H1(X1 ∪ Z)→ H1(X1 ∪X2)⊕H1(X1 ∪XK)→ . . .

so, by what we have proved before we get the exact sequence

0→ H1(X1 ∪ Z)→ 0→ . . .

yielding H1(X1 ∪ Z) = 0.Finally, as XG = X1 ∩ Z, by another application of Maier-Vietoris sequence, wehave

0→ H1(X1 ∪ Z)→ H0(XG)→ H0(X1)⊕H0(Z)→ H0(X1 ∪ Z)→ 0 ,

and in turn, by what we have already observed

0→ H0(XG)→ Z2 → Z→ 0

which forces H0(XG) = Z. By Lemma 2.30, we then conclude that XG is acyclic.

Segev then gives an example showing that case XG = ∅ might in fact occur. LetG = A5 be the alternating group on 1, 2, 3, 4, 5. Let A be the stabilizer of thepoint 1, B the stabilizer of the set 1, 2, and C the normalizer in G of the subgroupgenerated by (1 2 3 4 5). Then A ∼= A4 has index 5 in G, B ∼= S3 has index 10, andC ∼= D10 has index 6. Also, |A ∩ C| = 2 = |B ∩ C| and |A ∩ B| = 3. Let X bethe coset complex of the family A,B,C. Clearly, G acts regularly on X by rightmultiplication with no fixed vertices. Some work it is then needed (see [52]) to checkthat X is acyclic.

Segevs result on soluble groups has been extended to most simple groups byAschbacher and Segev in [7]. Recently, Oliver and Segev ([44]) have finally givena complete description of the finite groups which admit an action on an acyclic2-dimensional complex without fixed points.

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It is to be mentioned that the complex X in the example above, as well as thoseemployed by Oliver and Segev in their recent paper, is acyclic but its geometric real-ization is not contractible (in the topological sense). Indeed, the following conjectureappears to be still open.

Conjecture (W. Dicks). Every group of admissible automorphisms of a contractiblecomplex of dimension two fixes some vertex.

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Chapter 3

Partially ordered sets

We use the word poset to mean a partially ordered set. Although many of ourdefinitions make sense in larger classes of partially ordered sets, we will alwaysassume that a poset is finite, and usually denote by ”≤” the order relation.

Let P be a poset. A chain in P is a non-empty totally ordered subset of P ,that is a σ = x0, . . . , xn such that, for a suitable choice of the indices, x0 ≤ x1 ≤. . . ≤ xn. If σ is a chain then the integer |σ| − 1 is called the lenght of σ. Wehave already recalled the definitions of order preserving maps and of isomorphismsof posets. All the other familiar terminology associated to posets (maximal andminimal elements, inf, sup, etc.) is assumed. If a poset P has a least element inf(P )or a greatest element sup(P ) we denote them, respectively, by 0 and 1. We alsowrite P = P \ 0, 1.

Let P be a poset; remind that the complex of chains of P is the simplicial com-plex ∆(P ) whose q-simplices are all the chains of length q of P (thus the dimensionof ∆(P ) is the maximal length of a chain in P ).

In order to simplify notations, if P is a poset, we write C∗(P ), H∗(P ), and soon, instead of respectively C∗(∆(P )) and H∗(∆(P )). Similarly, we refer to P theproperties of ∆(P ); thus we say that P is contractible, spherical, etc.

3.1 Equivalence

An order preserving map of posets f : P −→ P ′ induces in a natural way a simplicialmap, which we still denote by f , from ∆(P ) to ∆(P ′). If σ = (x0 < x1 < . . . < xq) isa q-simplex of ∆(P ) then f(σ) = (f(x0) ≤ f(x1) ≤ . . . ≤ f(xq)) ∈ ∆(P ′) (possibly,f(σ) has smaller dimension than σ). It is plain that if P = P ′ and f is a bijectionthen the induced simplicial map on ∆(P ) is admissible. In particular we then havethe following observation.

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Proposition 3.1 Let f : P −→ P ′ be an order preserving map of posets. Then finduces a homomorphism of homology

f∗ : H∗(P ) −→ H∗(P′) .

If P and Q are posets, the product poset P × Q is the set of all ordered pairs(x, y), x ∈ P, y ∈ Q with the order relation (x, y) ≤ (x1, y1) if x ≤ x1 and y ≤ y1.

Theorem 3.2 (Quillen) Let X, Y be posets and f, g : X −→ Y two order preserv-ing maps, such that f(x) ≤ g(x) for all x ∈ X. Then f is equivalent to g.

Proof. Given y ∈ Y we set Y≤y = z ∈ Y | z ≤ y . For each simplexσ = x0 < x1 . . . < xq in ∆(X) we then define C(σ) = ∆(Y≤g(xq)). Since g is an orderpreserving map, C is a carrier from ∆(X) to ∆(Y ). Also, since Y≤g(x) has a greatestelement g(x), ∆(Y≤g(x)) is contractible and so C is a contractible carrier. Now, foreach simplex σ = x0 < . . . < xq in ∆(X), gq(σ) is a simplex of C(σ) by the verydefiniton of C, while fq(σ) is a simplex of C(σ) by our hypothesis that f(x) ≤ g(x)for all x ∈ X. Thus f and g induce morphisms on the chain complexes C∗(∆(X))and C∗(∆(Y )) that are both carried by C. By Lemma 1.22, f and g are equivalent.

Definiton. An element a of a poset X is conjunctive (respect. subjunctive)if for each x ∈ X there exists a ∨ x = supa, x in X (respect. if there existsa ∧ x = infa, x for all x ∈ X).

The following proposition is a generalization of the contractibility of a cone, andis quite useful in applications.

Proposition 3.3 Let X be a poset. Suppose that there exists an element a ∈ Xand an order preserving map f : X −→ X such that x ≤ f(x) ≥ a for all x ∈ X.Then ∆(X) is contractible.In particular, if X admits a conjunctive element then ∆(X) is contractible.

Proof. Consider the following poset endomorphisms of X,i) the identity map 1X ;ii) the constant map c defined by c(x) = a for all x ∈ X;iii) the map f .Then, for all x ∈ X we get 1X(x) ≤ f(x) ≥ c(x). By Quillen’s Theorem, we thenhave 1X ' f ' c. Therefore the identity is equivalent to a constant and thus X iscontractible.

If, in particular, a ∈ X is a conjunctive element, one applies this fact by consid-ering the map f : X −→ X defined by f(x) = a ∨ x for all x ∈ X.

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Corollary 3.4 Let f : P −→ P be an order preserving map of the poset P . If Padmits a conjuntive (subjunctive) element then f has a fixed point.

Exercise. Let X, Y be posets and f, g : X −→ Y two order preserving maps suchthat f(x) is comparable to g(x) for all x ∈ X. Prove that f is equivalent to g.

Definition. Let X, Y be posets, and f : X −→ Y an order preserving map. Foreach y ∈ Y we define

i) f/y = x ∈ X | f(x) ≤ y (the lower fiber of y); and

ii) y/f = x ∈ X | y ≤ f(x) (the upper fiber of y).

Exercise. Let f : X −→ Y be an order preserving map of posets. Suppose that foreach y ∈ Y , f/y is (non-empty) connected. Prove that X is connected if and onlyif Y is connected.

The following Theorem, due to Quillen, is one of the most effective tools whenworking with complexes of chains of posets.

Theorem 3.5 (Quillen’s criterion) Let X, Y be posets, and f : X −→ Y be anorder preserving map.

1. If f/y (or resp. y/f) is contractible for all y ∈ Y then f induces a homotopyequivalence between the complexes ∆(X) and ∆(Y ).

2. If f/y (or resp. y/f) is acyclic for all y ∈ Y then H∗(∆(X)) = H∗(∆(Y )).

Proof. We prove only point 1). Let X, Y be posets, and f : X −→ Y be anorder preserving map such that f/y is contractible for all y ∈ Y .

For a simplex σ ∈ ∆(Y ), we denote by m(σ) its greatest element, and denoteby C(σ) the subcomplex of ∆(Y ) whose elements are the chains τ of Y such thatm(τ) ≤ m(σ). Clearly, C(σ) is a cone, and so it is contractible. and for all faces τof σ, C(τ) ⊆ C(σ). Thus C defines a contractible carrier from ∆(Y ) to itself, and theidentity 1Y is carried by C.

Also, to each σ ∈ ∆(Y ) we associate the complex D(σ) = ∆(f/m(σ)), whichis by assumption a contractible subcomplex of ∆(X). This defines a contractiblecarrier D from ∆(Y ) to ∆(X).

Finally, we define a carrierDf from ∆(X) to ∆(X) by setting, for each σ ∈ ∆(X),Df (σ) = D(f(σ)). Df is contractible, and σ ∈ Df (σ) for all σ ∈ ∆(X). The identity1X is carried by Df .

Now, by Lemma 1.22, there exists a morphism g : C∗(Y ) → C∗(X) which iscarried by D. Consider the morphism f g : C∗(Y ) → C∗(Y ). Observe that ifσ ∈ ∆(Y ) and τ ∈ D(σ), then f(m(τ)) ≤ m(σ), whence f(τ) ∈ C(σ). Suppose

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σ ∈ ∆(Y ) is a q-simplex. Then gq(σ) ∈ Cq(D(σ)), and so gq(σ) =∑

i λiτi where theτi are q-simplices in Sq(D(σ)). Therefore, fq gq(σ) =

∑i λifq(τi) ∈ Cq(C(σ)), and

this shows that f g is carried by C. By point (ii) of Lemma 1.22, g f ' 1Y .Conversely, for any q-simplex σ ∈ ∆(X), gq fq(σ) ∈ Df (σ). Thus, g f : C∗(X)→C∗(X) is a morphism carried by Df , and again g f ' 1X .

A more combinatorial proof appears in a paper of Walker [61]. Also, it is ofcourse of interest to have a similar criterion which establishes G-equivalence forcomplexes associated to G-posets. This is done in an article by Webb and Thevenaz[60].

We now illustrate the power of Quillen’s criterion by proving a sort of Mayer-Vietoris type theorem.

A covering F of a complex X is a set of subcomplexes X1, X2, . . . , Xd of X suchthat X = X1 ∪ X2 ∪ . . . ∪ Xd. Then nerve of the covering is the complex N(F)whose simplices are the non-empty subsets i0, i1, . . . , iq of 1, 2, . . . , d such thatXi0 ∩ . . . ∩Xid 6= ∅.

Theorem 3.6 (Folkman-Leray). Let F = X1, X2, . . . , Xd be a covering of thecomplex X. Suppose that for any ∅ 6= I ⊆ 1, . . . , d the intersection

⋂i∈I Xi is

either empty or contractible. Then X is equivalent to N(F).

Proof. Let F = X1, X2, . . . , Xd be a covering of the complex X satisfyingthe property in the statement, and let I = 1, . . . , d. For each σ ∈ X let f(σ) be theset of all indices i ∈ I such that σ ∈ Xi. f(σ) is not empty because F is a coveringof X, and also σ ∈

⋂i∈f(σ)Xi. Thus f defines an application X −→ N(F). f is

clearly an order preserving map from (X,⊆) to Fop = (N(F),⊇). By construction,for each D = i0, i1, . . . , iq ∈ F , the fiber is f/D = Xi0 ∩ . . .∩Xid . Thus, fibers arecontractible by assumption, and so by Quillen’s criterion, the baricentric subdivisionof X is equivalent to the baricentric subdivision of N(F) (clearly, reversing the orderdoes not have any effect on the complex of chains), whence X is equivalent to N(F).

Definition. A subset Z of a poset X is closed (or also an ideal) if for all z ∈ Z andx ∈ X, if x ≤ z then x ∈ Z.

Proposition 3.7 (Quillen [48]). Let X, Y be posets, and let Z be a closed subsetof the product poset X × Y . For each x ∈ X let Zx = y ∈ Y | (x, y) ∈ Z. If Zx iscontractible for all x ∈ X then Z is equivalent to X.

Proof. Let π : Z −→ X be the restriction to Z of the projetion of X ×Y ontoX. Then, for each a ∈ X we have the fiber a/π = (x, y) ∈ Z | a ≤ x. Considerthe following order preserving maps.

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i) u : Za −→ a/π defined by u(y) = (a, y) for each y ∈ Za.ii) v : a/π −→ Za defined by v(x, y) = y for each (x, y) ∈ a/π.

We leave as an exercise to show that u and v are order preserving.Now, for each y ∈ Za we have (v u)(y) = y, i.e. v u = 1Za . On the other

hand, for each (x, y) ∈ a/π we have (u v)(x, y) = u(y) = (a, y) ≤ (x, y), whence byTheorem 3.2, u v ∼ 1a/π. Therefore, for all a ∈ X, a/π is equivalent to Za whichis contractible by assumption. By Quillen’s criterion, Z is equivalent to X

Corollary 3.8 (Quillen [48]) Let X, Y be posets, and let Z be a closed subset ofthe product poset X × Y . For each x ∈ X, let Zx = y ∈ Y | (x, y) ∈ Z and foreach y ∈ Y , let Zy = x ∈ X | (x, y) ∈ Z. If Zx is contractible for all x ∈ X andZy is contractible for all y ∈ Y then X is equivalent to Y .

Given posets P , Q, their join poset P ∗Q is the disjoint union P tQ with theorder relation defined by x ≤ y if x, y ∈ P and x ≤P y, or x, y ∈ Q and y ≤Q x, orx ∈ P , y ∈ Q.If P is a poset, we define the cone CP on P as the poset obtained from P byattaching to it a least element 0. Thus ∆(CP ) is a cone as a complex. In particularit is contractible.

Lemma 3.9 (Quillen [48]) Let P , Q be posets. Then ∆(P ∗ Q) is equivalent to∆(CP × CQ \ (0, 0)).

Proof. Let D = CP ×CQ \ (0, 0), and define f : D −→ P ∗Q by setting, forall (x, y) ∈ D:

f(x, y) =

x if x 6= 0y if x = 0

Then f is an order preserving map. Now, let a ∈ P ∗ Q. If a ∈ P then a/f =(x, y) ∈ D | f(x, y) ≥ a = (x, y) ∈ D | a ≤ x ∈ P, y ∈ CQ has a least element(a, 0) and so it is contractible.If a ∈ Q then a/f = (0, y) ∈ D | a ≤ y ∈ Q has a least element (0, a) and iscontractible. Thus, by Theorem 3.5, f induces an equivalence between ∆(P ∗ Q)and ∆(D).

Exercises. 1) Let P and Q be posets. A relation R ⊆ P ×Q is an ideal relation ifR is a closed subset of the product poset P × Q. Let R ⊆ P × Q. In the disjointunion of P and Q define a relation ′ ≤′ by x ≤ y if and only if x, y ∈ P and x ≤P y,or x, y ∈ Q and y ≤Q x, or x ∈ P , y ∈ Q and (x, y) ∈ R. Prove that this defines apartial ordering if and only if R is an ideal relation.

2) (Walker) Let R be an ideal relation among posets P and Q. Suppose that for allx ∈ P , y ∈ Q, R(x) and R−1(y) are contractible. Prove that P and Q are equivalent.

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3.2 Lattices

Recall that a lattice is a poset P in which for each pair of elements a, b ∈ P thereexist a ∨ b = supa, b and a ∧ b = infa, b. A finite lattice admits, by definition,a maximum element 1 and a minimum element 0.

Let L be a lattice with maximum 1 and minimum 0, and let x ∈ L. An elementy ∈ L is a complement of x if x∨y = 1 and x∧y = 0. A lattice L is complementedif every element of L has a complement.

Proposition 3.10 Let L be a non-complemented lattice, and let P be the posetL \ 0, 1. Then ∆(P ) is contractible. In particular every order preserving map ofP has a fixed point.

Proof. Let L and P be as in the hypotheses, and let a ∈ P be a non-complemented element in L. Now, V = b ∈ P | a ∨ b 6= 1 is a closed subset ofP . Also, a ∈ V and for all b ∈ V we have a ∨ (a ∨ b) = a ∨ b 6= 1. Hence a is aconjunctive element of V , and ∆(V ) is then contractible by Proposition 3.3.

Now consider the inclusion map ι : V −→ P and let c ∈ P . If c ∈ V thenι/c = x ∈ V | x ≤ c has a maximum, which is c, and so it is contractible. If c 6∈ Vthen a ∨ c = 1 and thus, by chioce of a, a ∧ c 6= 0. Since a ∧ c ∈ V , a ∧ c ∈ ι/c, andif b ∈ ι/c we have b∨ (a∧ c) ≤ c and a∨ (b∨ (a∧ c)) = a∨ b 6= 1, i.e. b∨ (a∧ c) ∈ V .Hence a ∧ c is a conjunctive element of ι/c and so ι/c is contractible. By Theorem3.5, ∆(V ) is equivalent to ∆(P ). Since ∆(V ) is contractible it follows that ∆(P ) iscontractible.

In fact, in [10] , Baclawski and Bjorner prove a stronger result

Proposition 3.11 Let L be a lattice, P = L \ 0, 1. For x ∈ P , let [x] denote theset of all complements of x in L. Then P \ [x] is contractible for all x ∈ P .

Exercises. 1) Let P be a poset, and suppose that there exists s ∈ P such thati) for all x ∈ P there exists either s ∨ x or s ∧ x;ii) if x < y, s ∧ x does not exists and s ∧ y exists, then there exists (s ∧ y) ∨ x.Prove that P is contractible.

2) Let f, g : P −→ L be two order preserving maps of posets P and L. Suppose thatL is a lattice and that for all x ≤ y in P , g(y) is not a complement of f(x). Provethat f and g are equivalent.

A lattice L is modular if for all a, b, c ∈ L with a ≤ c one has (a∨b)∧c = a∨(b∧c);For instance, the set of all normal subgroups of a group is modular. An analysis ofthe homology of (semi)modular lattices is given by Folkman in [29].

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An important particular case of a complemented modular lattice is the set of allsubspaces of a vector space ordered by inclusion. We will apply the next result lateron.

Theorem 3.12 Let V be a vector space of dimension n ≥ 2 over a finite field F ,and let T (V ) denote the partially ordered set of all proper non-trivial subspaces ofV . Then ∆(T (V )) is spherical, and Hn−2(T (V )) 6= 0.

Proof. Clearly, the dimension of ∆(T (V )) is n − 2, so Hq(T (V )) = 0 forq ≥ n− 1.

Denote by Γ the complex whose q-simplices are the sets U0, U1, . . . , Uq wherethe Ui are 1-dimensional subspaces of V and U0 + U1 + . . .+ Uq 6= V . Define mapsf : Γ −→ T (V ) and g : T (V ) −→ Γ by setting

f(U0, U1, . . . , Uq) = U0 + U1 + . . .+ Uq and g(W ) = U ≤W | dim(U) = 1

Then, for all A ∈ Γ, g f(A) ⊇ A, whence by Quillen’s Theorem 3.2, g f ' 1Γ. Onthe other hand f g = 1T (V ). Hence f induces an equivalence (which we denote againby f) from ∆(Γ) to ∆(T (V )). In particular, for all q ≥ 0, it induces isomorphismsf∗q : Hq(∆(Γ)) −→ Hq(∆(T (V ))),

Now, let Γ0 be the poset of all non-empty subsets of 1-dimensional subspaces ofV . Clearly Γ0 is contractible. Now, for q ≤ n−2, the q-skeleton of Γ0 coincides withthe q-skeleton of Γ. This implies that for 0 ≤ q < n− 2, Hq(Γ) = Hq(Γ0). Thus for1 ≤ q < n− 2 we have

Hq(∆(T (V ))) = Hq(∆(Γ)) = Hq(Γ) = Hq(Γ0) = 0

and H0(∆(T (V ))) = H0(Γ0) = Z.To finish the proof we have to show that the kernel of the differention ∂n−2 :

Cn−2(Γ) −→ Cn−3(Γ) is not trivial. To do that, consider a linearly independentfamily of n 1-dimensional subspaces Σ = U0, . . . , Un−1, viewed as a (n−1)-simplexin Γ0. Then

0 6= σ = ∂n−1(Σ) ∈ Cn−2(Γ0) = Cn−2(Γ)

and now, ∂n−2(σ) = ∂n−2∂n−1(Σ) = 0, whence 0 6= σ ∈ Zn−2(Γ).

Exercise. Let L be a finite boolean lattice. Then it is well known that L isisomorphic to the lattice of all subsets of a finite set 0, 1, . . . , n. Set P = L\0, 1.Then P is just the barycentric subdivision of the (n−1)-dimensional sphere Sn−1 andso ∆(P ) is homeomorphic to Sn−1. By 1.19, P has two non-zero homology groupsH0(P ) = Z, and Hn−1(P ) = Z. Let f : P −→ P be a poset isomorphisms; then f isnaturally induced by a permutation π on the set of minimal elements (which is justthe set 0, . . . , n). Prove that Tr(f∗0) = 1, and Tr(f∗n−1) = sign(π). Completethe study of fixed points of f . Prove that if P f 6= ∅ then P f ∪ 0, 1 is a booleanlattice.

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3.3 Shellability

Shellability is a property of posets introduced by Stanley at al. (see [12]) which is inmany cases useful in studying posets of subgroups of a finite group. Before givingthe definiton, let us fix some notation and terminology.

We say that a poset P is pure if its maximal chains have all the same lenght.Let P be a poset and x, y ∈ P , with x ≤ y. We write [x, y] to denote the closedinterval

[x, y] = a ∈ P | x ≤ a ≤ y ,

and (x, y) to denote the open interval (i.e. (x, y) = [x, y] \ x, y). We also writexl y (and say that y covers x) if x ≤ y, x 6= y and [x, y] = x, y.If P has a least element 0, we say that a x ∈ P is a atom if x covers 0. Dually, ifP has a greatest element 1, x ∈ P is a co-atom if 1 covers x.

Definition. A finite poset P is shellable if P is pure and there exists a totalordering of the maximal chains C1, . . . , Ck of P , such that the following condition issatisfied:

for all i < j there exists t < j such that Ci ∩Cj ⊆ Ct ∩Cj and |Ct ∩Cj | = |Cj | − 1.

Let P be a shellable poset, and let ∆(P ) be the chain complex associated to P .Then all maximal simplices of ∆(P ) have the same dimension n, and there existsa total ordering on them σ1, . . . σk with the property that for all i < j there existst < j such that σi ∩ σj is a face of σt ∩ σj and σt ∩ σj is a (n-1)-simplex.

Abstract complexes which satisfy these conditions are said to be shellable, andan ordering of the maximal simplices which satisfies the definiton is called a shellingof the complex. Thus, the complex of chains of a shellable poset is shellable, andconversely the barycentric subdivision of a shellable complex is a shellable poset.We now prove that a shellable complex is spherical. We begin with a simple butuseful Lemma.

Lemma 3.13 Let X be a shellable complex, and let σ1, . . . , σk be a shelling of X.Then for each 2 ≤ m ≤ k there is a unique face σm of σm which is minimal suchthat it is not a face of any σi for i ≤ m− 1.

Proof. Let Y be the complex of all faces of σ1, . . . , σm−1, and let σ be a face ofσm minimal such that σ 6∈ Y (observe that σm 6∈ Y and so such a minimal elementexists). Let x be a vertex of σ; then there is 1 ≤ j ≤ (m−1) such that σ\x is a faceof σj . By shellability there is then a j ≤ k < m such that σ \ x ⊆ σk, and σk ∩σmis a maximal face of σm. Since σ is not a face of σk, it must be σk ∩ σm = σm \ x.Now, if τ is another face of σm with τ 6∈ Y , then τ is not in σk, and so x ∈ τ . Thisholds for every vertex x of σ and thus σ ⊆ τ .

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Exercise. Suppose that the maximal simplices of the complex X all have the samedimension and their set can be linearly ordered in such a way that the conclusion ofLemma 3.13 is verified. Prove that X is shellable.

From our point of view the relevance of shellable complexes is that they arespherical.

Theorem 3.14 Let X be a shellable complex of dimension n. Then X is spherical.More precisely, Hq(X) = 0 for all 0 ≤ q ≤ n − 1. If σ1, . . . , σk is a shelling of X,

then the rank of Hn(X) equals the number of indices 2 ≤ i ≤ k such that σm = σm.

Proof. Let σ1, . . . , σk be a shelling of X, and for 1 ≤ i ≤ k, let Xi denote thesubcomplex of X whose simplices are all the faces of σi.We argue by induction on k. If k = 1, then X is the set of faces of a simplex and istherefore 0-spherical.Let then k ≥ 2, and let Y = X1 ∪ X2 ∪ . . . ∪ Xk−1. Observe that σ1, . . . , σk−1 isthen a shelling of Y , and dim(Y ) = n. We then have the inductive hypothesis on Y .

We denote, as in Lemma 3.13, by σk the smallest face of σk that does not belongto Y .

Assume first that σk 6= σk. Then there exists a vertex x ∈ σk \σk. If τ ∈ Y ∩Xk

then τ ∪ x 6⊇ σk, and so τ ∪ x ∈ Y ∩Xk. Thus Y ∩Xk is a cone and thereforecontractible. As Xk is also contractible, and Y ∪Xk = X, a straightforward appli-cation of Mayer-Vietoris sequence gives Hq(X) = Hq(Y ) for all q. Thus inductionhypothesis gives the result for X.

Suppose then σk = σk. In this case Y ∩Xk is the complex of proper faces of σk;thus Hq(Y ∩Xk) = 0 for 0 ≤ q ≤ n− 2 and Hn−1(Y ∩Xk) = Z. Since Xk has zerohomology, Mayer-Vietoris sequence gives

0→ Hn(Y )→ Hn(X)→ Hn−1(Y ∩Xn)→ 0

which, by the inductive assumption, gives the desired result in dimension n. Forsmaller dimension, Mayer-Vietoris and induction give zero homology for X.

Definition. Let Γ be a complex and σ ∈ Γ. The link of σ is the subcomplex

Link(σ) = τ ∈ Γ \ σ | σ ∪ τ ∈ Γ .A complex Γ is Cohen-Macaulay if it is spherical and Link(σ) is spherical for everyσ ∈ Γ.

Exercise. Prove that in a shellable complex the link of every simplex is shellable.Conclude that a shellable complex is Cohen-Macaulay.

The following definitions could be given in a way that makes sense in a boundedposet. However, for semplicity, we restrict to the case of a lattice, which will beenough for the applications that will follow in Chapter 4.

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Definition. Let L be a lattice. A CA-ordering on L is a total ordering ≺ on theset of all co-atoms of L such that the following condition is satisfied:

For all co-atoms a, b, if a ≺ b there exists a co-atom c ≺ b, such thata ∧ b ≤ c ∧ bl b.

Definition. Let L be a lattice. We say that L has a recursive CA-ordering if eitherL = 0, 1, or L has a CA-ordering ≺1 and for each co-atom a of L the interval [0, a]has a recursive CA-ordering such that the co-atoms of [0, a] which are covered bysome co-atom b ≺1 a of L come first.

Proposition 3.15 Let L be a lattice which is pure and admits a recursive CA-ordering. Then P is shellable. In particular ∆(P ) is spherical.

Proof. Since L is assumed to be pure, we have only to define a shelling on it.Observe that by definition of recursive CA-ordering, for each 0 6= a ∈ L we are givena CA-ordering ≺a on the interval [0, a]. Thus let

σ : 0 = x0 l x1 l . . .l xn = 1σ′ : 0 = y0 l y1 l . . .l yn = 1

be two distinct maximal chains of L. Let 1 ≤ s ≤ n− 1 be the maximal index suchthat xs 6= ys, and write a = xs+1 = ys+1. Now, xs and ys are distinct co-atoms of[0, a]; we put σ < σ′ if xs ≺a ys.The fact that this defines a total ordering on the set of maximal chains of L followseasily as all CA-orderings ≺a are total.

We now prove that the relation thus defined is a shelling. Let σ and σ′ be twodistinct maximal chains of L as fixed above, with σ < σ′. Argunig by induction on s,we find a maximal chain σ′′ such that σ′′ < σ′, σ∩σ′′ ⊆ σ′∩σ′′ and |σ′∩σ′′| = |σ′|−1.

If s = 1 then σ and σ′ differ only at x1 6= y1 and we take σ′′ = σ.Thus, let s ≥ 2 and let 0 ≤ t < s be maximal such that xt = yt.

Suppose first t = s− 1. Consider the maximal chain

σ′′ : 0 = y0 l . . .l yt l xs l ys+1 l . . .l yn = 1 .

Then σ′′ < σ′ by definition of the order relation, and σ′′ satisfies the requiredconditions.Now, suppose t < s − 1. If there exists a co-atom z of [0, a] such that z ≺a ys andz ∧ ys = ys−1 then the chain

σ′′ : 0 = y0 l . . .l ys−1 l z l a = ys+1 l . . .l yn = 1

has the desired properties. Otherwise, for all co-atoms z of [0, a] with z ≺a ys wehave z ∧ ys 6= ys−1 and so, by recursive CA-ordering, if w is a co-atom of [0, ys] with

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w ≤ z then w ≺ys ys−1. Now, the fact that ≺a is a CA-ordering of [0, a] implies theexistence of a co-atom b of [0, a] such that

xt ≤ xs ∧ ys ≤ b ∧ ys l ys

and, by what we have observed above, b∧ ys ≺ys ys−1. Let σ be a maximal chain ofL such that

σ : 0 = y0 l y1 l . . .l yt l . . .l b ∧ ys l ys l ys+1 = al . . .l yn = 1 .

Then σ ∩ σ′ ⊆ σ ∩ σ′′ and, by inductive hypothesis, there exists a maximal chain σ′′

such that σ′′ < σ′, σ ∩ σ′′ ⊆ σ′ ∩ σ′′ and |σ′ ∩ σ′′| = |σ′| − 1.

Definition. A lattice L is lower semimodular if for all x, y ∈ L,

xl x ∨ y ⇒ x ∧ y l y .

For example, the lattice of all subgroups of a finite nilpotent group is lowersemimodular. We shall use this fact in the next Chapter. We leave the proof of itas an exercise, and the same we do for the following lemma.

Lemma 3.16 Let L be a lower semimodular lattice. Then L is pure, and everyinterval of L is a lower semimodular lattice.

Definition. A lattice L is strongly CA-orderable if every total ordering of the setof its co-atoms is a CA-ordering.

Proposition 3.17 Let L be a lattice. The following properties are equivalent

(a) L is lower semimodular;

(b) for every 0 6= x ∈ L the interval [0, x] is strongly CA-oredrable.

Proof. (a) ⇒ (b). Let L be lower semimodular. Then every interval in L is lowersemimodular, so it will be enough to prove that L is strongly CA-orderable. This isimmediate as if x, y are distinct co-atoms of L then x∨ y = 1 so by semimodularityx ∧ y l y.

(b) ⇒ (a). Let L satisfy (b), and let x, y ∈ L such that xlx∨ y. We prove thatx ∧ y l y by induction on the length of the interval [y, x ∨ y]. Let b be a co-atom[0, x ∨ y such that y ≤ b. By assumption, by ordering the co-atoms of [0, x ∨ y]in such a way that x and b are consecutive, we get a CA-ordering, and this forcesx ∧ b l b. But b = (b ∧ x) ∨ y, and the length of [y, b] is strictly less than that of[y, x ∨ y]. By inductive hypothsis x ∧ y = (b ∧ x) ∧ y l y.

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Proposition 3.18 Let L be a lower semimodular lattice. Then every total order-ing of the set of co-atoms of L is compatible with a recursive CA-ordering of L.Consequently, L is shellable.

Proof. Let L be a lower semimodular lattice. We prove the claim by proceeding byinduction on the length (of a maximal chain) of L. If L has length 1, then L = 0, 1and the assertion is trivial. Thus suppose that the length of L is greater than 1,and let x1 ≺ x2 ≺ . . . ≺ xm be a given total ordering of the co-atoms of L. For eachco-atom a = xk of L, by semimodularity, the elements xi∧a with i 6= k are co-atomsof [0, a], and a total ordering ≺a of the co-atoms of [0, a] can be given such thatelements that are co-atoms of xj for j < k come first. By inductive hypothesis ≺ais compatible with a recursive CA-ordering of [0, a]. This shows that the orderingon co-atomes of L is compatible with a recursive CA-ordering.

From this proposition it is immediately deduced the following useful (see Chapter4) remark.

Proposition 3.19 Let L be a lattice such that [0,M ] is lower semimodular for eachco-atom M . Then L has a recursive CA-ordering if and only if L admits a CA-ordering.

Exercise. Let T be the poset of all proper non-trivial subspaces of the vector spaceV = GF (q)n, n ≥ 2. Prove that T is shellable, and compute the rank of its tophomology group.

3.4 Mobius function

If P is a poset and x, y ∈ P , we denote by (x, y) the open interval with extremes xand y, i.e.

(x, y) = z ∈ P | x < z < y with the ordering induced by P . Clearly, an interval might also be empty.

Definition. Let P be a poset. The Mobius Function on P is the application

µP : P × P −→ Z

defined by, for all a, b ∈ P :

1) µP (a, a) = 1;

2) µP (a, b) = 0 if a 6≤ b;

3) if a < b then µP (a, b) is the number of chains of even length in (a, b) minus thenumber of chains of odd length (including the empty chain, which we view ashaving length −1).

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From the definition we get immediately the following observation, which seems tohave been first noticed by P. Hall.

Proposition 3.20 Let P be a poset. Then, for all a, b ∈ P with a < b:

µP (a, b) = χ((a, b))− 1 .

If P is a poset, and a, b ∈ P with a < b. Let us denote by Cb the set of all chainsσ such that a < inf σ and b = supσ. Then the chains of even (odd) length in (a, b)are in 1-1 correspondence to the chains of odd (even) length of Cb. Thus

µP (a, b) =∑σ∈Cb

(−1)|σ| .

Having observed that, we now prove a useful property of Mobius function.

Proposition 3.21 Let P be a poset, and a, b ∈ P with a < b. Then

3′ )∑a≤x≤b

µP (a, x) = 0 =∑a≤x≤b

µP (x, b) .

Proof. Let a, b ∈ P with a < b. We denote by C the set of all non-emptychains of (a, b), and for σ ∈ C we write m(σ) = supσ. Then, by definition of Mobiusfunction:

µ(a, b) = −1−∑σ∈C

(−1)|σ| = −µ(a, a)−∑a<x<b

∑m(σ)=x

(−1)|σ|

whence, by the observation made above,

µ(a, b) = −µ(a, a)−∑a<x<b

µ(a, x) = −∑a≤x<b

µ(a, x)

proving the first equality. The second one is proved in a similar way.

Indeed, it is important and not difficult to observe (see Rota [50]) that properties1), 2) and 3′) completely determine the function µP .

Exercises. 1) if N is the set of all positive integers ordered by divisibility, thenµ(n) = µN(1, n) is just the classical Mobius function. It is defined by µ(1) = 1, andfor n ≥ 1, µ(n) = (−1)s if n is the product of s distinct primes and µ(n) = 0 if nhas a square factor.

2) if P(X) is the set of all subsets of the set X ordered by inclusion, then, forall A,B ∈ P(X) with A ⊂ B

µP(X)(A,B) = (−1)|B|−|A| .

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The fundamental property of Mobius function is the so-called Mobius Inver-sion Formula.

Theorem 3.22 Let P be a poset, and G an abelian (additive) group.

1) Given an application f : P −→ G, define a map F : P −→ G by setting, for allx ∈ P

F (x) =∑x≤y

f(y) .

Then, for all x ∈ Pf(x) =

∑x≤y

µP (x, y)F (y).

2) Conversely, given a map H : P −→ G, if h : P −→ G is defined by

h(x) =∑

x≤y µP (x, y)H(y), then

H(x) =∑x≤y

h(y) .

Proof. 1) For all x ∈ P , by applying property 3′) we get∑x≤y

µP (x, y)F (y) =∑x≤y

(µP (x, y)

∑y≤z

f(z))

=

=∑x≤z

(f(z)

∑x≤y≤z

µP (x, y))

= f(x)µP (x, x) = f(x).

2) For all x ∈ P , by applying again property 3′) we get∑x≤y

h(y) =∑x≤y

(∑y≤z

µP (x, z)H(z))

=

=∑x≤z

(H(z)

∑x≤y≤z

µP (y, z))

= H(x)µP (x, x) = H(x).

Remark. It is clear that Mobius inversion formula holds, exchanging ≤ with ≥,also for the function F (x) =

∑y/leqx f(y). More generally, if X is the opposite

poset of X, then for all a, b ∈ X, µX(b, a) = µX(a, b).

One of the first appearances of the (generalized) Mobius function of a poset wasindeed in group theory. P. Hall [33] introduced it as a tool to study the number ofgenerating sets of a finite group.

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Proposition 3.23 (P. Hall) Let G be a finite group, and s ≥ 1. Denote by φs(G)the number of ordered s-tuples (x1, . . . , xs) such that G = 〈x1, . . . , xs〉. Then

φs(G) =∑H≤G

µL(H,G)|H|s

where L = L(G) is the poset of all subgroups of G ordered by inclusion.

Proof. Clearly, for all s ≥ 1 and H ≤ G, we have |H|s =∑

K≤H φs(K). Byapplying Mobius inversion formula, we get the result.

P. Hall then determined the Mobius function for L(G) when G is a group of typePSL(2, p). We will return to this and other facts about Mobius functions of groupsin Chapter 6.

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Chapter 4

The Brown Complex

Let G be a finite group and p a prime divisor of |G|. We denote by Sp(G) the setof all non-trivial p-subgroups of G ordered by inclusion. The Brown complex atthe prime p of G is the complex associated to the poset Sp(G), which we againdenote by Sp(G), instead of ∆(Sp(G)). This was introduced by K.Brown in [20] as atool for describing group cohomology over a finite field; its study as an independentmathematical object started with D. Quillen’s fundamental paper [48].

Examples. 1) Let G = A5 be the alternating group on 5 points. Then the elementsof the Brown complex of G at p = 2 are:

- five Sylow 2-subgroups; these are isomorphic to the Klein group Z2 × Z2 oforder 4.

- fifteen subgroups of order 2, each of those is generated by the product of twodisjoint transpositions.Each subgroup of order 2 is contained in a single Sylow 2-subgroup, and in turn eachSylow 2-subgroup contains three subgroups of order 2. Thus the complex Γ = S2(G)is a non-connected graph, with five connected components, each of those consistsin three edges with a common vertex. We have C1(Γ) = Z15, C0(Γ) = Z20, whileH1(Γ) = 0 and H0(Γ) = Z5.

2) Let G = Q8 be the quaternion group of order 8, and p = 2. Then G hasfive non-trivial subgroups: G, the centre Z(G) which has order 2, and three cyclicsubgroups of order 4, each containing Z(G). As S2(G) in this case has a maximum,the complex associated to S2(G) is contractible; indeed it is formed by three triangles(i.e. 2-simplices) glued toghether along the same edge.

We start with a numerical property of the Euler characteristic of the Browncomplex that reminds one of the Sylow theorems. If G is a group and p a prime, wedenote by |G|p the p-part of |G|. If |G|p = pn, then the dimension of Sp(G) is n− 1.

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Theorem 4.1 (K.Brown) Let G be a finite group, and p a prime divisor of |G|.Then

χ(Sp(G)) ≡ 1 (mod |G|p) .

Proof. Let P be a Sylow p-subgroup of G, and consider the action of P onSp(G) by conjugation. Let 1 6= g ∈ P . Then 〈g〉 ∈ Sp(G) and if Y is a non trivial p-subgroup of G such that Y g = Y , then 〈Y, g〉 is a p-subgroup of G. This shows that〈g〉 is a conjunctive element in the poset of g-fixed points Sp(G)g. By Proposition3.3, Sp(G)g is contractible and in particular χ(Sp(G)g) = 1, for all 1 6= g ∈ P . ByCorollary 2.5 we then have χ(Sp(G)) ≡ 1 (mod |P |).

Exercise. Let G be a finite group.

a) Let S be the poset of all non-trivial soluble subgroups of G, ordered by inclusion.Prove that χ(S) ≡ 1 (mod |G|).b) (S.Lucido) Let N be the poset of all non-trivial nilpotent subgroups of G orderedby inclusion, and let F = Fit(G) be the Fitting subgroup of G (i.e. the maximalnormal nilpotent subgroup of G). Prove that χ(N ) ≡ 1 (mod |F |).

If G is a group, we denote by Ap(G) the ordered set of all non-trivial elementaryabelian subgroups of G (the complex of chains of Ap(G) is called the Quillen complexof G at the prime p).The following replacement theorem is most useful. Remind that, if P is a p-group,then Ω1(P ) = 〈x ∈ P | xp = 1〉.

Proposition 4.2 (Quillen) Let G be a finite group, and p a prime divisor of |G|.Then Sp(G) is homotopy equivalent to Ap(G).

Proof. Consider the inclusion map i : Ap(G) −→ Sp(G) (as a map on posets).Then, for every B ∈ Sp(G), Ω1(Z(B)) 6= 1. So

Ω1(Z(B)) ∈ i/B = A ∈ Ap(G) | A ≤ B .

Now, let A0 ∈ i/B. Then Ω1(Z(B))A0 is elementary abelian, whence Ω1(Z(B))A0 ∈i/B. This shows that Ω1(Z(B)) is a conjunctive element of i/B, for each B ∈Sp(G). By Quillen’s criterion, i induces a homotopy equivalence on the associatedcomplexes.

In fact, it can be proved (see [60]) that Sp(G) and Ap(G) are equivalent asG-complexes, under the action of G induced by conjugation.

Definition. If G is a group and p is a prime the p-rank rp(G) of G is the maximalrank of an elementary abelian p-subgroup of G. (i.e. if A is an elementary abelianp-subgroup of G of maximal order then |A| = prp(G)).

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Thus, the dimension of Ap is rp(G)− 1. The equivalence established in 4.2 says,in particular, that for q ≥ rp(G), Hq(Sp(G)) = 0.

Exercises. 1) Let G be a finite group, and p a prime divisor of |G|. Let Ip(G) be theposet of all non-trivial p-subgroups of G that are intersections of Sylow p-subgroupsof G. Prove that Ip(G) is equivalent to Sp(G).

3) (commuting complex). Let G be a finite group, and p a prime divisor of |G|.Let Λp(G) be the set of all elements of G that have order p. Define Cp(G) to be thesimplicial complex whose simplices are the non-empty subsets x0, . . . , xq of Λp(G)such that xixj = xjxi for i, j = 0, . . . , q. Prove that Cp(G) is equivalent to Ap(G).

Proposition 4.3 (Quillen) Let G1, G2 be groups, and p a prime. Then Ap(G1×G2)is equivalent to the join Ap(G1) ∗ Ap(G2).

Proof. Let T be the subposet of Ap(G1 ×G2) consisting of all p-subgroups ofthe form A1×A2 where Ai is an elementary abelian p-subgroup of Gi, and not bothof them are trivial. Then T is just CAp(G1) × CAp(G2) \ (0, 0) (remember thedefinition of a cone of a poset on page 41, here (0, 0) is the pair of trivial subgroups).By Lemma 3.9, T is equivalent to Ap(G1) ∗ Ap(G2), so it is enough to show thatAp(G1 ×G2) is equivalent to T .We consider the inclusion map i : T → Ap(G1×G2), and the product of projectionsr : Ap(G1 × G2) → T defined by, for all A ∈ Ap(G1 × G2), r(A) = (π1(A), π2(A)).Then r i = 1T , and i r ' 1Ap(G1×G2) by Theorem 3.2. Thus T is equivalent toAp(G1 ×G2) and we are done.

4.1 Quillen’s Conjecture

Let G be a finite group, and p a prime number. Then Op(G) denotes the maximalnormal p-subgroup of G. More generally, if π is a non-empty set of primes, Oπ(G)denotes the largest normal subgroup of G whose order is a product of primes in π.

Lemma 4.4 Let G be a finite group and p a prime divisor of |G|. If Op(G) 6= 1then Sp(G) is contractible.

Proof. If Op(G) 6= 1, then Op(G) is a conjunctive element in Sp(G). ByProposition 3.3, Sp(G) is contractible.

In his paper [48], Quillen provided much evidence that the converse of this state-ment should also be true. A lot of the subsequent work on Brown complexes hasthen been motivated by this conjecture of Quillen’s

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Conjecture. Let G be a finite group and p a prime divisor of |G|. Then Op(G) 6= 1if and only if Sp(G) is contractible.

Of course one direction (the easy one) is Lemma 4.4. The converse, in its gener-ality, is still an open question. Quillen proved it when G is soluble or a group of Lietype in characteristic p. Later, it has been confirmed for p-soluble groups (see nextsection) and for all almost simple groups by Alperin, Aschbacher, Kleidman, Smithand others. In the sequel we try to give an idea of these proofs. The most generalresult is due to Aschbacher and Smith [8] and is the following one.

Theorem 4.5 Let G be a finite group and p a prime divisor of |G| with p ≥ 7.Suppose Op(G) = 1 and G has no composition factor isomorphic to a unitary group

U(n, q) with q ≡ −1 (mod p). Then H∗(Sp(G)) 6= 0 (i.e. Sp(G) is not acyclic and,a fortiori, not contractible).

Exercises. 1) (Bouc complex) Let G be a finite group, and p a prime divisorof |G|. Let Bp(G) be the poset of all non-trivial p-subgroups H of G such thatH = Op(NG(H)), ordered by inclusion. Prove that Bp(G) is equivalent to Sp(G).

2) (Benson complex) Let G be a finite group, and p a prime divisor of |G|.Let Zp(G) be the poset of all non-trivial elementary abelian p-subgroups A of Gsuch that A = Ω1(Op(Z(CG(A))), ordered by inclusion. Prove that Zp(G) is equiv-alent to Ap(G) (hint: consider the map defined by, for any A ∈ Ap(G), A 7→Ω1(Op(Z(CG(A))). Prove that such a map is order preserving and that its image isZp(G). Then conclude by applying Quillen’s criterion).

4.2 Simple groups

Let us begin with a significative example; here we follow the original discussion byQuillen.

Proposition 4.6 Let F be a finite field of characteristic p, and G = SL(n, F ) (withn ≥ 2). Then the complex Sp(G) is (n− 2)-spherical and Hn−2(Sp(G)) 6= 0.

Proof. Let Γ be the set of all non-empty chains of proper non-trivial sub-spaces of the vector space V = Fn, ordered by inclusion (thus Γ is the barycentricsubdivision of the poset T (V ) of all proper non-trivial subspaces of V ).

Consider the following subset of the partially ordered set Γ×Sp(G) (here Sp(G)is the poset of p-subgroups of G, not the associated complex):

Z = (σ,A) ∈ Γ× Sp(G) | σx = σ for all x ∈ A .

Thus (σ,A) ∈ Z if and only if σ ∈ ΓA (i.e. A ≤ Gσ). Observe that if σ ⊇ τ ∈ Γthen Gσ ≤ Gτ , whence Z is a closed subset of Γ× Sp(G).

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Now, for all σ ∈ Γ and all A ∈ Sp(G), we prove the contractibility of thecomplexes associated to the posets

Zσ = A ∈ Sp(G) | (σ,A) ∈ Z and ZA = σ ∈ Γ | (σ,A) ∈ Z .

Note first of all that Zσ = Sp(Gσ) and ZA = ΓA.a) Zσ is contractible. Let σ be the chain of subspaces 0 6= V1 ≤ . . . ≤ Vd 6= V .

Choosing a base of V obtained by progressive extensions of a base of Vi to a base ofVi+1 (i = 1, 2, . . . , d− 1) and finally extending the base of Vd to V , we have that Gσis represented by the set of matrices with determinant 1 that have diagonal blockscorresponding to the factors V1, V2/V1, . . . , V/Vd, and zeros below the diagonalblocks. Hence Op(Gσ) 6= 1; in fact Op(Gσ) is just the set of such matrices whosediagonal blocks are identity matrices. By Lemma 4.4, Zσ = Sp(Gσ) is contractible.

b) ZA is contractible. Observe that ZA is the complex associated to the posetY = T (V )A of all proper non-trivial subspaces of V that are A-invariant. Since Ais a p-group and char(F ) = p, for each W ∈ Y we have

CW (A) = w ∈W | wA = w 6= 0 .

and CW (A) = M ∩ W , where M = CV (A) 6= V . Denote by K the poset of allnon-trivial subspaces of M . As K has a greatest element M , the complex ∆(K) iscontractible. Consider the map f : Y −→ K defined by f(W ) = CW (A) for eachW ∈ Y . By Quillen’s criterion this yields the equivalence of ZA = ∆(Y ) and ∆(K).Thus ZA is contractible.

Finally, Sp(G) is equivalent to Γ by Corollary 3.8. But Γ is just the barycentricsubdivision of the complex associated to T (V ) which is, by what we have observedbefore, spherical with non-zero homology in dimension n− 2.

Remark. Let G = SL(n, F ) as in the previous Proposition. We proved that Sp(G)is (n− 2)-spherical. Then

V = Hn−2(Sp(G))⊗ C = Hn−2(Sp(G);C)

has a natural structure of CG-module. Moreover, it can be proved that it is irre-ducible, and coincides with the so-called Steinberg module of SL(n, F ). Its dimen-sion is the rank of Hn−2(Sp(G)) and equals the order of the Sylow subgroup of G,i.e. qn(n−1)/2 (in fact, the field C can here be replaced by an algebraically closed fieldK of characteristic p. As Sp(G) is spherical Hn−2(Sp(G);K) = Hn−2(Sp(G)) ⊗K,this has the same dimension of V and is a simple projective KG-module).

This is an instance of a more general fenomenon, which is perhaps the reasonof the importance of the Brown complexes. To each group of Lie type G (likefor instance, unitary, symplectic or orthogonal groups), J. Tits has associated in a

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uniform way a geometry, called the Tits Building of G. The building is in fact asimplicial complex with special properties. In the case of the general linear groupin its natural action on V , the Tits building is just the complex of all chains (calledflags) of non-trivial proper subspaces of V . The relevant fact is that for all groupsof Lie type in characteristic p, the Tits building is homotopy equivalent to Sp(G)(Quillen [48]); more precisely, its barycentric subdivision is isomorphic to the Bouccomplex Bp(G) defined in an exercise above.

Now, the Theorem of Solomon-Tits asserts that the building of a Lie type groupis (n − 1)-spherical, where n is the Lie rank of the group. It follows that for anyLie group G in characteristic p, Sp(G) is spherical. Moreover its top homologygroup (coefficients in C or in an algebraically closed field of characteristic p) is theSteinberg module for G (see Carter [23] for a precise definition).

The proof that Quillen’s conjecture holds for all finite simple groups consistsin eventually checking through the list of them. This is the origin of the followingLemma, which is then a consequence of the classification of finite simple groups.

Lemma 4.7 Let G be a non abelian simple group. Then for any prime divisor p of|G| there exists a prime q 6= p such that a Sylow q-subgroup of G does not normalizeany non trivial p-subgroup of G.

Proof. See [24].

Theorem 4.8 Let G be a non abelian simple group, and p a prime divisor of |G|.Then χ(Sp(G)) 6= 1 (in particular, Sp(G) is not acyclic).

Proof. Assume, by contradiction, χ(Sp(G)) = 1 and, by Lemma 4.7, choose aprime q 6= p such that, ifQ is a Sylow q-subgroup ofG, thenQ does not normalize anynon-trivial p-subgroup of G. Now, Q acts by conjugation as a group of simplicialautomorphisms of Sp(G). If χ(Sp(G)) = 1 then, by Theorem 2.22, Q fixes somevertex of Sp(G), that is Q normalizes some non-trivial p-subgroup, a contradiction.

Indeed, a similar statement holds for all almost simple groups, that is groups Gwith a unique minimal normal subgroup S which is a non abelian simple group. Ofcourse Theorem 4.8 does not give any information on the homology of the Browncomplex of a simple group. Having more information is often needed, for examplewhen dealing with extensions. If G is a group of Lie type and p is its defining char-acteristic, then Sp(G) is spherical by Solomon-Tits Theorem; the homology of Sp(G)for some classes of groups of Lie type when p is different from the defining charac-teristic has been studied by Das [26] [27] and Pulkus [47], as well as by Aschbacherand Smith.

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Example. Following Alperin ([1]), we show that if G is the symmetric group Snand 5 ≤ p ≤ n, then the Quillen complex Ap(G) has non-zero homology in its topdimension. Let p ≤ n; it is an easy fact that r = rp(Sn) = [n/p] (the integral partof n/p). If for each i = 1, . . . , r we denote by Gi the subgroup of Sn that fixes allelements outside p(i− 1) + 1, . . . , pi, then Gi ∼= Sp. For each index i, we take Ai acyclic subgroup of order p of Gi. Then A1 ×A2 × . . .×Ar is a maximal elementaryabelian p-subgroup of Sn. Now, if p ≥ 5, then in Gi ∼= Sp there exits a subgroup Biof order p, disjoint from Ai. Consider the chains in Ap(Sn) of the following type

X1 < X1X2 < . . . < X1X2 · · ·Xr

where Xi is either Ai or Bi. Conseder, given a suitable orientation, the element s ofCr−1(Ap(Sn)) which is the sum of all the chains of the above type that have an evennumber of subgroups of type B minus the sum of those that have an odd number ofsubgroups of type B. Then prove that s is a non-zero (r− 1)-cycle in Ap(G). Sincethe dimension of the complex is r − 1, this shows that Hr−1(Ap(G)) 6= 0.

Remark. I do not know of any group G whose order is divided by p, such thatOp(G) = 1 and χ(Sp(G)) = 1. It would be interesting to know if, at least for solublegroups, the condition χ(Sp(G)) = 1 is sufficient for Op(G) 6= 1. This is true, byProposition 4.15 in the next section, if G is an extension of a soluble p′-group by an(elementary) abelian p-group. Isaacs and Hawkes [35] have proved that it holds alsofor any extension of a p′-group by an abelian p-group.

Exercise. Let G be a finite group, and p a prime such that rp(G) ≤ 2. Prove thatOp(G) 6= 1 if and only if χ(Sp(G)) = 1. (hint: replace Sp(G) by Ap(G) and, forrp(G) = 2 observe that if χ(Ap(G)) = 1 then Ap(G) is a tree).

4.3 p-soluble groups

We now discuss the soluble and p-soluble cases. In the course of proving Quillen’sconjecture for such groups, we will obtain more detailed informations on the homol-ogy of the Brown complex in some special configurations, where the concept of ashellable complex will provide a good insight.

We start by listing a few elementary but important facts that can be found in anyintroductory text on finite groups. I recommend those of Aschbacher [3], Kurzweiland Stellmacher [40], or Suzuki [56].

Lemma 4.9 (Frattini Argument) Let N be a normal subgroup of G, and P a Sylowp-subgroup of N , for some prime p. Then G = NNG(P ).

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Lemma 4.10 Let A be an abelian group acting faithfully on an elementary abelianp-group N . If N is irreducible for A (i.e. no proper non-trivial subgroup of N isnormalized by A), then A is cyclic.

Lemma 4.11 (Three Subgroup Lemma) Let A,B,C be subgroups of the group G.If [A,B,C] = 1 and [B,C,A] = 1, then [C,A,B] = 1.

Lemma 4.12 Let A be a p-group acting on the p′-group N . If [N,A,A] = 1, then[N,A] = 1.

We now consider soluble groups. The fact that Quillen’s conjecture holds in thiscase is orignally due to Quillen himself.

Let G be a group and p a prime divisor of its order. In what follows we denoteby P the poset obtained by adding to Sp(G) a least element 0 and a largest element1 (if G is not a p-group, you may think of 0 as being the trivial subgroup 1 and of1 as being the whole G). Then P is a lattice, it is pure, and the co-atoms of P arejust the Sylow p-subgroups of G. Now, for each such subgroup A the lattice [0, A] islower semimodular, and so by Proposition 3.19, P admits a recursive CA-orderingif and only if it admits a CA-ordering (in this case we say that Sp(G) admits aCA-ordering). By Proposition 3.15, we have immediately

Proposition 4.13 Let p be a prime divisor of order of the group G. If Sp(G) admitsa CA-ordering, then Sp(G) is shellable and therefore it is spherical.

Lemma 4.14 Let N be a normal subgroup of G, and let P be a Sylow p-subgroupof G. If both Sp(G/N) and Sp(NP ) admit a CA-ordering, then Sp(G) admits aCA-ordering.

Proof. By assumption, we are given a total ordering P 1, . . . , Pn of the Sylowp subgroups of G/N which is a CA-ordering of Sp(G/N). For each i = 1, . . . , n fixa Sylow p-subgroup Pi of G such that P i = NPi/N . Now, for any i = 1, . . . , n,NPi is isomorphic to NP , so it admits a CA-ordering ≺i of its Sylow subgroups.Now each Sylow p-subgroup of G is contained (as a Sylow p-subgroup) in one andjust one of the NPi. Let Q, S be Sylow p-subgroups of G. We set Q ≺ S if eitherQ ≤ NPi, S ≤ NPj and i < j, or Q,S ≤ NPi and Q ≺i S. Then, clearly ≺ is atotal ordering of the Sylow p-subgroups of G. We show that it is a CA-ordering.Suppose Q ≺ S. If Q and S are contained in the same NPi then, since ≺i is aCA-ordering, there exists a Sylow p-subgroup T of NPi (and of G) such that T ≺i S(and so T ≺ S), Q ∩ S ≤ T ∩ S, and T ∩ S is maximal in S. Otherwise Q ≤ NPi,S ≤ NPj with i < j. By the CA-ordering on G/N there is an index k < j such thatNPi ∩ NPj ≤ NPk ∩ NPj and |NPj : NPk ∩ NPj | = p. Now, let H = NPk ∩ S;then Q ∩ S ≤ NPi ∩NPj ∩ S ≤ NPk ∩ S = H. Also, since NS = NPj ,

|S : H| = |S||H|

=|S ·NPk||NPk|

=|NPj ·NPk||NPk|

=|NPj |

|NPj ∩NPk|= p .

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It follows that if K is a Sylow p-subgroup NPk containing H, then K ∩ S = H ≥Q∩S. Finally, K ≺ S by definition of ≺, and K ∩S is maximal in S. We have thenproved that ≺ is a CA-ordering of Sp(G).

Proposition 4.15 (J. Pulkus) Let G = NA be a soluble group, where N is a normalp′-subgroup and A an elementary abelian p-group of rank n+1. Then Sp(G) = Ap(G)

admits a CA-ordering and so it is shellable. If Op(G) = 1 then Hn(Sp(G)) 6= 0.

Proof. Observe that Op(G) = CA(N), and proceed by induction on |G| toshow that Ap(G) admits a CA-ordering.Let N/M be a chief factor of G. Then N/M is an elementary abelian t-group, forsome prime t 6= p. By Lemma 4.10, A/CA(N/M) is either trivial or cyclic of orderp.

By inductive hypothesis, Ap(MA) admits a CA-ordering. If M 6= 1, then byinductive hypohesis again, Ap(G/M) admits a CA-ordering. Then by Lemma 4.14Ap(G) admits a CA-ordering.Thus suppose M = 1. Then G is the extension of the elementary abelian t-groupN by A. If A = CA(N) then A is normal and so it is the only Sylow p-subgroupof G and we have nothing to prove. Otherwise, CA(N) E G is a maximal subgroupof A, and all Sylow p-subgroups of G (that are the conjugates of A) intersect A inCA(N). It then readily follows that Ap(G) admits a CA-ordering. Observe that inthis case if Op(G) = CA(N) = 1, then n = 0 and Ap(G) is just a set of at lest two

disconnected points, whence H0(Ap(G)) 6= 0.

We have now to prove that, assuming Op(G) = CA(G) = 1, then Hn(Ap(G)) 6= 0.We proceed by induction on |G|. Let N/M be a chief factor as above, and let C =CA(N/M). Now, if CA(M) = 1, then by inductive hypothesis 0 6= Hn(Ap(MC)) ⊆Hn(Ap(G)). Thus, let CA(M) 6= 1; then C 6= A (for, if C = A, then CA(M) =CA(G)), and C acts faithfully on M . By Lemma 4.10, A/C is cyclic and so thep-rank of MC is n. By inductive hypothesis, Hn−1(Ap(MC)) 6= 0. Now, the caseM = 1 we have already discussed above; so let M 6= 1, end let P1/MC, P2/MCtwo distinct Sylow p-subgroups of G/MC. For i = 1, 2 let Xi = Ap(Pi). Then Xi

is isomorphic to Ap(MA), and so it is contractible because Op(MA) = CA(M) 6= 1.Finally, observe that X1 ∩X2 = Ap(MC). Mayer-Vietoris sequence now gives

0→ Hn(X1 ∪X2)→ Hn−1(X1 ∩X2)→ Hn−1(X1)⊕ Hn−1(X2) = 0

whence Hn(X1 ∪X2) is isomorphic to Hn−1(X1 ∩X2) 6= 0. As n is the dimension ofAp(G) this yields Hn(Ap(G)) 6= 0.

In general, the p-Quillen complex of a soluble group is not spherical. The fol-lowing example is due to Alperin, and you may try to work it out by exercise. Let

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p, q be primes with p|q − 1, and let H be the non-abelian group of order qp (thus,H is the extension of a normal subgroup of order q by a subgroup of order p). LetCp be a cyclic group of order p. Finally, let G be the wreath product HwrCp. ThenAp(G) has dimension p − 1, it is connected, Hp−1(Ap(G)) 6= 0 6= H1(Ap(G)), andHk(Ap(G)) = 0 for k 6= 0, 1, p− 1.

We now move to the larger class of p-soluble groups, aiming at studying a similarsituation of an extension of a p′-group N by an elementary abelian p-group A. Inthe case in which N is not soluble we will just be content to prove that the tophomology group of the p-Quillen complex is not trivial. We will do that by usingthe idea of generalized Fitting subgroup, so I shall briefly introduce this concept.For a more detailed treatment see one of the mentioned texts [3][40][56].

Definition. Let p be a fixed prime. A finite group G is said to be p-soluble if itadmits a (finite) normal series

1 = G0 E G1 E . . . E Gn = G

all of whose factors Gi+1/Gi is either a p-group or a p′-group (i.e. a group whoseorder is not diveded by p).

It is easy to prove that a soluble group is p-soluble for every prime p. Also, ifG is p-soluble and Op(G) = 1 then Op′(G) 6= 1, where Op′(G) denotes the largestnormal subgroup of G whose order is not divided by p.

Lemma 4.16 Let G be a p-soluble group with Op(G) = 1. Then CG(Op′(G)) ≤Op′(G).

Proof. Let G be a p-soluble group with Op(G) = 1. Write C = CG(Op′(G)),K = Op′(G)C/Op′(G) and suppose, by contradiction, C 6≤ Op′(G) (i.e. K 6= 1).Then K E G/Op′(G), and Op′(K) = 1. In fact, if T/Op′(G) = Op′(K), then T E Gand |T | = |T/Op′(G)||Op′(G)| is a p′-number, whence by definiton T = Op′(G).Since G (and therefore G/Op′(G)) is p-soluble, this forces X/Op′(G) = Op(K) 6= 1.Observe that X = Op′(G)(C ∩X) and C ∩X/C ∩Op′(G) ' X/Op′(G). Let P be aSylow p-subgroup of C∩X; then P is be a Sylow p-subgroup of X and X = Op′(G)P .But P centralizes Op′(G) and P ∩Op′(G) = 1, whence X = Op′(G)×P . In particularP is a characteristic p-subgroup of X. Since X is normal in G, P is also normal,and therefore P ≤ Op(G) = 1, a contradiction.

The generalized Fitting subgroup.

A finite group H is quasisimple if H = H ′ and H/Z(H) is simple. The compo-nents of a finite group G are its subnormal quasisimple subgroups (of course a groupneed not have components).

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Lemma 4.17 Let H be a quasismple group, and K EE H. Then either K ≤ Z(H)or K = H.

Proof. Let Z = Z(H). Then ZK/Z is subnormal in H/Z. Since H/Z issimple we have either ZK = Z or ZK = H. The first case implies K ≤ Z. Thesecond case gives K E H and H/K ' Z/(Z ∩K) abelian; since H = H ′ this forcesK = H.

Lemma 4.18 Let H be a component of G and K EE G. Then either H ≤ K or[H,K] = 1. In particular, distinct components of G commute.

Proof. Let G be a minimal counterexample. Then K 6= G so, as K issubnormal there exists a proper L E G with K ≤ L and, by minimalty of G, H 6≤ L.So, by Lemma 4.17, L∩H ≤ Z(H). If H = G then L ≤ Z(H) and [H,K] ≤ [H,L] =1. Then H 6= G and there exists a proper T E G with H ≤ T . By minimality of G,[H,L∩T ] = 1. Now, [H,K] ≤ L∩T , so [H,K,H] ≤ [L∩T,H] = 1 and, by the threesubgroups Lemma, [H,H,K] = 1. Since H = H ′ we get [H,K] = [H,H,K] = 1.

We now denote by E(G) the subgroup of G generated by all its components. Asif H is a component and α is an automorphism of G, then Hα is also a component,E(G) is a characteristic subgroup of G. Now, let F (G) be the Fitting subgroup ofG. Then the generalized Fitting subgroup of G is defined by

F ∗(G) = E(G)F (G) .

Proposition 4.19 Let C the set of all distinct components of the group G. LetZ = Z(E(G)). Then

1. Z = 〈Z(H) | H ∈ C〉, and HZ/Z ' H/Z(H) for all H ∈ C;

2. E(G)/Z is the direct product of the groups HZ/Z where H ∈ C.

Proof. Part 1) is clear from Lemma 4.18; for part 2) we make induction onthe number n of distinct components. If n = 1 we have noting to prove. Let n ≥ 2,H a component, and T the subgroup generated by all the components differentfrom H. Then by Lemma 4.18 H centralizes T , so T 6= G and H ∩ T ≤ Z(T ).Hence T has n − 1 components and, by inductive hypothesis T/Z(T ) is the directproduct of them modulo Z(T ). Also, as [H,T ] = 1 and E(G) = TH, we have thatE(G)/Z ' HZ/Z × TZ/Z and the proof is complete.

Lemma 4.20 For any group G, CG(F ∗(G)) = Z(F (G)).

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Proof. Exercise (try to arrange the proof of Lemma 4.16).

At this point let us also remember that a non-abelian chief factor of a finite groupis (isomorphic to) a direct product S1×S2×. . .×Sr of isomorphic non-abelian simplegroups Si, and G acts transitively by conjugation on the set S1, S2, . . . , Sr.

We finally need the following fact, which is a consequence of the classification ofthe finite simple groups.

Proposition 4.21 Let G be a finite simple group whose order is not divided by theprime p. Then any p-group of automorphisms of G is cyclic.

Lemma 4.22 Let p be a prime, and A a p-group of acting on the p′-group N . Thenfor every prime divisor q of the order of N , there is a A-invariant Sylow q-subgroupof N .

Proof. Consider the semidirect product G = NA, defined by the action of A onN . For a prime divisor q of the order of N , let Q be a Sylow q-subgroup of N .By the Frattini argument, G = NNG(Q). It follows that NG(Q) contains a Sylowp-subgroup of G, that is, for some g ∈ N

Ag ≤ NG(Q) .

By conjugating by g−1, we get that A normalizes Qg−1

.

Lemma 4.23 Let the elementary abelian p-group A act faithfully on the p′-groupN . Assume that for every proper A-invariant subgroup H of N we have CA(H) 6= 1.Then N is nilpotent.

Proof. We work in the semidirect product G = NA, and take G to be a minimalcounterexample. Now, faithfullness of A on N just means CA(N) = 1. Let F ∗ be thegeneralized Fitting subgroup of G, and let X = CA(F ∗). Then X ≤ CG(F ∗) E G.By Lemma 4.20, we have therefore

[N,X,X] ≤ [F ∗, X] = 1

whence, by Lemma 4.12, X = 1. This in particular shows that F ∗ is contained inN , and by our assumption on the action of A on N , we get N = F ∗. If N is solubleF ∗ = F (N) is nilpotent, and we are done.Thus suppose that N is not soluble. Then, by the structure of N = F ∗ thereexists a non-abelian chief factor N/M of G. Let Y = CA(M). Now, N/M is thedirect product of non-abelian simple groups, the components of G/M . Since N/Mis a minimal normal subgroup of G/M , such components are all isomorphic and

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transitively permuted by A. Let B be the stabilizer in A of a fixed component S(i.e. the normalizer in A of S). Because A is abelian and acts transitively on theset of components, B is the kernel of such action. Also, if C = CB(S) then, forall a ∈ A, C = Ca = CB(Sa), whence C = CA(N/M). By Lemma 4.21, B/C iscyclic of order p. Now, let q be a prime divisor of |N/M |; then, by Lemma 4.22,there exists a Sylow q-subgroup Q/M of N/M which is invariant for A. By ourassumption, CA(Q) 6= 1. Now, the centralizer in A of Q centralizes M and clearlyfixes every component in N/M , that is CA(Q) ≤ B ∩ Y . Since B/C is of order p,and C ∩ Y = 1, it must be 1 6= CA(Q) = B ∩ Y . But this happens for any primedivisor q of the order of N/M , and so B centralizes N/M , that is B = C, and (for anA-invariant Sylow subgroup Q of N) CA(Q) = 1 which contradicts our assumptions.

Theorem 4.24 Let G be a p-soluble group with Op(G) = 1, and p||G|. Then

Hn(Sp(G)) 6= 0, where n+ 1 = rp(G).

Proof. Let G be a p-soluble group with Op(G) = 1, with n = rp(G) − 1 ≥ 0.Since Sp(G) is equivalent to Ap(G) , we consider the latter one, which is a complexof dimension rp(G)− 1 = n.

If n = 0 then Ap(G) is a set of more than 2 isolated points (because Op(G) = 1),

whence H0(Ap(G)) 6= 0.We then assume n ≥ 1, and proceed by induction on |G|. Let N = Op′(G), and

fix an elementary abelian p-subgroup A of G of maximal rank (thus |A| = pn+1).Observe that 0 is the only n-boundary of Ap(G). Thus, in order to prove that

Hn(Sp(G)) is not trivial it is enough to show that there exists a non-zero n-cycle inAp(G). For that, it is enough to find a non-zero n-cycle in some Ap(H) where H isa subgroup of G containing A.

Consider the subgroup NA of G. Let X = Op(NA); then X E NA and so[X,N ] ≤ X ∩ N = 1, whence X ≤ CG(N), which by Lemma 4.16 is contained inN . This forces X = 1. Now, since the p-rank of NA is equal to the p-rank of G, ifNA is a proper subgroup of G then, by inductive hypothesis, there exists a non-zerocycle in Cn(Ap(NA)) and so there exists a non-zero cycle in Cn(Ap(G)), and we aredone. So NA = G.Now, let H be a proper A-invariant subgroup of N . Then HA is a subgroup ofG, and Op(HA) = CA(H). If Op(HA) = 1 then by inductive hypothesis we find anon-zero n-cycle in Ap(HA) and we are done. Thus, we may assume that, for everyproper A-invariant subgroup H of N , CA(H) 6= 1. But then we are in a position toapply Lemma 4.23 and obtain that N is nilpotent. By the soluble case we are done.

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Corollary 4.25 Let p be a prime, and G a p-soluble group. If the Brown complexSp(G) is acyclic then Op(G) 6= 1.

By analogy with the soluble case, one is led to ask whether the p-Brown complexof any extension of a p′-group by an elementary abelian p-group is always spherical,but this still open question appears to be a rather difficult one.

Exercise. Let G be a finite group and p a prime. We denote by Op(G) the smallestnormal subgroup N of G such that the factor G/N is a p-group, and by Sp(G) theposet (and also the associated complex) of all proper subgroups of G whose index isa power of p. Prove that Op(G) 6= G if and only if χ(Sp(G)) = 1 (the combinatoricsand the homology of Sp(G) is studied by Weidner and Welker in [65]. They show,in particular, that Sp(G) is spherical).

4.4 Connectedness

Acyclicity is just one among many topological properties that can be assigned andstudied in the Brown or other complexes. Given one such property of the Browncomplex, it might be intersting to ask (as in the Quillen’s conjecture) for a descriptionof equivalent group theoretical conditions.

Maybe the easiest complex property to investigate is connectedness. It is clearthat the group G acts by conjugation on the set of connected components of Sp(G).The following Theorem collects rather strightforward and classical results. In thisform it appears in Quillen’s paper [48].

Theorem 4.26 Let p be a prime divisor of the order of the group G, and let M asubgroup of G. The following conditions are equivalent:

i) M contains the stabilizer in G of some connected component of Sp(G);

ii) there exists a Sylow p-subgroup P of G such that NG(H) ≤ M for every non-trivial subgroup H of P ;

iii) there exists a Sylow p-subgroup P of G such that NG(P ) ≤ M , and every p-subgroup of G that has non-trivial intersection with M is contained in M ;

iv) p divides |M | and p does not divide |M ∩Mx| for all x ∈ G \M .

Proof. i) ⇒ ii). Let X be a connected component of Sp(G) whose stabilizeris contained in M . Clearly each connected component of Sp(G) contains a Sylowp-subgroup of G. Thus, let P be a Sylow p-subgroup belonging to X. Let H be anon-trivial subgroup of P , and let g ∈ NG(H). Then X g is a connected componentof Sp(G) that has a common vertex H with X . Thus X g = X , and so g belongs tothe stabilizer of X g, which by assumption is contained in M .

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ii) ⇒ iii). Assume M satisfies ii) with respect to the Sylow p-subgroup P .Then NG(P ) ≤ M directly by assumption. Let then H be p-subgroup of G suchH∩M 6= 1. Then, by Sylow Theorem, there exists a x ∈M such that (H∩M)x ≤ P .By assumption NG(H ∩M)x = NG((H ∩M)x) ≤ M , and so NG(H ∩M) ≤ M . Inparticular we have NH(H ∩M) = H ∩NG(H ∩M) ≤ H ∩M . Since H is a p-group,this forces H = H ∩M , whence H ≤M .

iii)⇒ iv). Assume M satisfies iii) with respect to the Sylow p-subgroup P . Thenp divides |M |. Let x ∈ G, and suppose that p divides |M ∩Mx|. Then M ∩Mx

contains a non-trivial p-subgroup H. Let Q be a Sylow p-subgroup of G containingH. Then, by assumption, Q ≤ M . Observe also that Hx−1 ≤ Qx

−1 ∩M whenceQx−1 ≤M . Thus there exist elements y, z of M such that Q = P y and Qx

−1= P z.

It follows that zxy−1 ∈ NG(P ) ≤M , and so x ∈M , and M satisfies iv).iv) ⇒ i). Assume that M satisfies iv). For all x ∈ G we have

|MxM | = |MxM | = |Mx||M ||M ∩Mx|

=|M |2

|M ∩Mx|.

Since G is the disjoint union of the distinct double cosets MxM , we get

|G : M | = |G||M |

=1

|M |∑|MxM | =

∑ |M ||M ∩Mx|

.

If x 6∈ M then p divides the index |M : M ∩Mx| by assumption, while the doblecoset M = MgM for g ∈ M contributes as 1 to the above sum. Thus |G : M | ≡ 1(mod p). Consequently, M contains a Sylow p-subgroup P of G. Let N be thestabilizer in G of the connected component of Sp(G) that contains P , and let x ∈ N .Then P and P x belong to the same connected component. It is now easy to seethat this implies the existence of Sylow p-subgroups P = P0, P1, . . . , Ps = P x suchthat 1 6= Pi−1 ∩ Pi. For 0 ≤ i ≤ s − 1 let xi ∈ G such that P xii = Pi+1. Wemay choose xs−1 in a way that x0x1 · · ·xs−1 = x. Suppose that Pi ≤ M . Then1 6= Pi ∩ Pi+1 ≤ M ∩Mxi . By our assumption on M it follows that xi ∈ M andthus Pi+1 ∈ M . Since P0 = P ∈ M by repeating this argument we get xi ∈ M forall indices i and, consequently, x ∈M . This shows that M ≤ N .

Corollary 4.27 Let p be a prime divisor of the order of the group G, and let P bea Sylow p-subgroup of G. Then the stabilizer in G of the connected component ofwhich contain P is 〈NG(H) | H ≤ P 〉.

Further results and remarks about the connection between the Brown and otherG-complexes and the Sylow structure of the group (like Alperin’s fusion theorem)can be found in Quillen’s article [48], and in a more recent paper by L. Barker [11].

A proper subgroup M of the group G that satisfies any (and therefore all) of theproperties in the statement of Theorem 4.26 is called a strongly p-embedded subgroup.

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It is a concept that played a relevant role in the classification of finite simple groups.In particular one has the fundamental classification, due to H. Bender, of groupswhich admit a strongly 2-embedded subgroup (i.e. groups in which S2(G) is notconnected), which is an important step in the classification.According to Bender’s Theorem, if G is a group admitting a strongly 2-embeddedsubgroup, then either

i) the Sylow 2-subgroups of G are cyclic or generalized quaternion, orii) the section O2′(G)/O2′(G) is a simple group of type PSL(2, q), PSU(3, q) or

a Suzuki group Sz(q), where q is a power of 2 and q > 4.

For p an odd prime, groups which admit a strongly p-embedded subgroup canalso be described, but only by using the classification of finite simple groups (see[32]).

Recall that the finite p-groups that admit just one subgroup of order p are thecyclic and (for p = 2) the generalized quaternion groups. Thus rp(G) = 1 if andonly if the Sylow p-subgroups of G are of these types. If rp(G) = 1 then Sp(G) isequivalent to Ap(G) which is a set of non-connected points; if further Op(G) = 1then Sp(G) is not connected. On the other hand in the group A5, we have thatS2(G) is not connected and r2(A5) = 2.

Exercise. Let p a prime divisor of the order of the group G, and suppose thatOp(G/Op′(G)) 6= 1. Prove that if Sp(G) is not connected then rp(G) = 1. Deducethat for a soluble group G, Sp(G) is not connected if and only if Op(G) = 1 andrp(G) = 1.

Simply connected Brown complexes.

A much more difficult and relevant question is to determine those pairs (G, p)such that the Brown complex at the prime p of the group G is simply connected.By that, we mean that the geometric realization of Sp(G) is simply connected asa topological space (i.e. its fundamental group π1(Sp(G)) is trivial). As in thiscourse I have not recalled the notion of the fundamental group of a complex (it ispossible to give a combinatorial definition in terms of chain complexes, as it is donefor instance in the book of Rotman [51]), I will just refer to Aschabacher’s paper [4]and survey [5], where this question is studied and discussed.If Op(G) 6= 1 then Sp(G) is simply connected by rather easy remarks (indeed itfollows trivially from the fact that, in this case, Sp(G) is contractible). So, thequestion really arises when Op(G) = 1. Then, a necessary condition for Sp(G) beingsimply connected is rp(G) ≥ 3. Aschbacher’s point is that, under these conditions,Sp(G) is almost always simply connected. In fact he shows that that would be thecase if it is true for simple groups of p-rank at most 3, and the following conjectureis provedConjecture. If G = F ∗(G)A, where A is an elementary abelian p-group of rank at

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most 3 and F ∗(G) is the product of conjugates of a single component K of p′-order,then Sp(G) is simply connected.

In [4], Aschbacher essentially reduces the proof of his conjecture to the casewhere K is a simple group of Lie type of Lie rank 1 or a sporadic group. He andSegev [53] have then proved the conjecture in a number of cases, while Das [26],[27]has studied the simple connectivity of the Brown complex in some types of simplegroups.

4.5 Orbits of the Brown complex

The conjugation action of a group G on a Brown complex Sp(G) is clearly simplicialand admissible, but in general it is not regular. Consequently, we cannot form theorbit simplicial complex without passing to the barycentric subdivision. However,it is possible to look at the orbit space Sp(G)/G as the topological quotient spaceobtained from a geometric realization of Sp(G) factored modulo G-action. It is nota simplicial complex, but still has a natural structure of a CW-complex (see Spanier[54] or any other text in algebraic topology for the definitions). As such it is treatedin the literature. Recently, P. Symonds [57] has confirmed a conjecture of Webb [63]by proving the following remarkable result.

Theorem 4.28 Let G be a group, and p a prime divisor of its order. Then the orbitspace Sp(G)/G is contractible.

The dual case of the orbits of subgroups of p-power (or of p′) index is studied byWeidner and Welker [66].In these notes, we will content to compute the Euler characteristic of the orbit spaceof the Brow complex, and prove that it is equal to 1. To be consistent with oursimplicial point of view, let us define χ(Sp(G)/G) to be the Euler characteristic ofthe orbit complex of the barycentric subdivision of Sp(G) (the action of G on thebarycentric subdivision of Sp(G) is regular by Corollary 2.14, so its G-orbit spaceis a simplicial complex. Its Euler characteristic can be shown to coincide with theEuler characteristic of the quotient space Sp(G)/G).

We begin with an observation which is essentially due to Quillen.

Lemma 4.29 Let p be a prime divisor of the order of the group G, and H a p-subgroup of G. Let V = Sp(NG(H)/H) and S>H = Y ∈ Sp(G) | H < Y . Thenfor any g ∈ NG(H) the complex VgH is equivalent to Sg>H .

Proof. If H is a Sylow p-subgroup of G, then V and S>H are both empty.Thus, suppose that H is not a Sylow p-subgroup of G. Then, as it is well known,p divides |NG(H)/H|. Better, if R is a p-subgroup properly containing H, thenH < NR(H) = R ∩ NG(H). Also, if g ∈ NG(H) normalizes R, then gH fixes the

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section NR(H)/H. Thus we can define an order preserving map f : Sg>H → VgH , by

setting f(R) = NR(H)/H for each R ∈ Sg>H . Let T = T/H ∈ VgH , then f(T ) = T ,

and T is the smallest element of the fiber T/f = R ∈ Sg>H | f(R) ≥ T. Hence

T/f is contractible. By Quillen’s criterion, f induces an equivalence between thecomplexes associate to Sg>H and to VgH .

Lemma 4.30 Let p be a prime divisor of the order of the group G. Then∑g∈G

χ(Sp(G)g) = |G| .

Proof. We set, for convenience, S = Sp(G) (the poset) and ∆ = ∆(S) (theassociated complex). For each chain σ ∈ ∆, as usual, we denote by Gσ the stabilizerin G of σ (if σ = H then GH = NG(H)).

Let H be a p-subgroup of G which is not a Sylow subgroup (allowing the caseH = 1), and set V = VH = Sp(NG(H)/H) and S>H = Y ∈ S | H < Y . ThenGH acts admissibly by conjugation on both these posets; observe that ∆(S>H)g =∆(Sg>H) for all g ∈ GH . By applying Proposition 2.4 to the complex ∆(S>H), wehave ∑

g∈GH

χ(∆(S>H)g) =∑

σ∈∆(S>H)

(−1)dim(σ)|Gσ ∩GH |.

By the previous Lemma∑σ∈∆(S>H)

(−1)dim(σ)|Gσ ∩GH | =∑g∈GH

χ(VgH) = |H|∑

gH∈GH/H

χ(VgH).

We now prove the Proposition by arguing by induction on |G|. If σ ∈ ∆, we writeH|σ if H is the least element of σ. Then∑

H|σ

(−1)dim(σ)|Gσ| = |NG(H)| −∑

σ∈∆(S>H)

(−1)dim(σ)|Gσ ∩GH |.

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Denote by S] the set of all non-trivial and non-Sylow p-subgroups of G. By applyingProposition 2.4 once more we get∑

g∈Gχ(∆g) =

∑σ∈∆

(−1)dim(σ)|Gσ| =∑H∈S

∑H|σ

(−1)dim(σ)|Gσ| =

=∑H∈S

|NG(H)| −∑

σ∈∆(S>H)

(−1)dim(σ)|Gσ ∩GH |

=

=∑

P∈Sylp(G)

|NG(P )|+∑H∈S]

|NG(H)| − |H|∑

gH∈GH/H

χ(VgH)

=

= |G|+∑H∈S]

|NG(H)| − |H|∑

gH∈GH/H

χ(VgH)

.

Now, by inductive hypothesis,∑

gH∈GH/H

χ(VgH) = |NG(H)/H|. Hence

∑g∈G

χ(Sp(G)g) = |G|+∑H∈S]

(|NG(H)| − |NG(H)|) = |G|

thus completing the proof.

Theorem 4.31 Let p be a prime divisor of the order of G. Then χ(Sp(G)/G) = 1.

Proof. Since the action of G on the barycentric subdivision S ′ of Sp(G) is regular,by applying Proposition 2.20 and Lemma 4.30 we get

χ(Sp(G)/G) = χ(S ′/G) = |G|−1∑g∈G

χ(S ′g) = |G|−1∑g∈G

χ((Sp(G)g)′) =

= |G|−1∑g∈G

χ(Sp(G)g) = |G|−1|G| = 1.

4.6 Some applications.

Let Zp denote the p-adic integers, and let X be the Brown or the Quillen complexat the prime p of the group G. P. Webb [62] [63] has proved that the chain complexwith coefficients in Zp

C∗(X) : Cn(X;Zp)→ Cn−1(X;Zp)→ . . .→ C0(X;Zp)

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has an acyclic split augmented subcomplex D∗ such that all the modules of thequotient complex are projective ZpG-modules. This means that for each q ≥ 0,Cq(X) decomposes as Cq(X) = Dq ⊕ Pq, the augmented chain complex Dn →. . . → D0 → Zp → 0 has zero homology, and all Pq are projective (see also G.Robinson [49]). Webb uses this fact and its generalizations [64] to obtain formulasin cohomology. He also defines, for any group G, the Steinberg module of G over Zpas the alternating sum in the Green ring

Stp(G) =

n∑q=0

(−1)qCq(Sp(G);Zp)

then, he uses the above result to prove that Stp(G) is the difference of two projectivemodules (if G is a group of Lie type then this is, up to the sign, the usual Steinbergmodule). Similar objects are described by Bouc in [16].

The Brown complex at a prime p has also been instrumental in reformulationsof some important conjectures in modular representation theory, like the so-calledAlperin weight conjecture [2].

Let p be a prime divisor of the order of the group G, and let K be an algebraicallyclosed field of characteristic p. We denote by lp(G) the number of irreducible rep-resentation of G over K (which is the number of conjugacy classes of p′-elements ofG), and by bp(G) the number of those that are projective KG-modules. Thus bp(G)is just the number of p-blocks of defect zero, and by basic modular representationtheory, it is the number of irreducible complex representation of G whose degree isa multiple of |G|p. Finally, let P be a set of representatives of conjugacy classes ofthe p-subgroups of G. Alperin’s conjecture can be stated as follows

Alperin’s conjecture. For every finite group G, lp(G) =∑

P∈P bp(NG(P )/P ).

The equality in Alperin’s conjecture has been proved for soluble groups, groupsof Lie type in characteristic p, symmetric groups, general linear groups, and a fewother classes of groups. In [37], Knorr and Robinson were the first to find equivalentformulations of Alperin’s conjecture, that involve the p-Brown complex of G. Here,k(G) denotes the number of irrreducible complex representations of G, that is thenumber of conjugacy classes of G.

Proposition 4.32 (Knorr and Robinson) The following two statements are equiv-alent to Alperin’s conjecture.

1. For every group G, lp(G)− bp(G) =∑

σ∈Sp(G)/G(−1)dim(σ)lp(Gσ)

2. For every group G, k(G)− bp(G) =∑

σ∈Sp(G)/G(−1)dim(σ)k(Gσ)

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Note that we are not claiming that, for a given a group G, the formulas inAlperin’s conjecture and those in the reformulation are equivalent.

J. Thevenaz [58] has also proposed an equivalent formulation of Alperin’s con-jecture, namely

Conjecture. For every group G, k(G)− bp(G) =∑

[g] χ(Sp(G)g/CG(g)), where [g]runs over all conjugacy classes of G.

By applying Proposition 2.20 we may reformulate this conjecture once more: forevery group G,

k(G)− bp(G) =1

|G|∑gh=hg

χ(Sp(G)〈g,h〉)

where the sum runs over all non-ordered pairs of commuting elements of G.

It might be interesting to report that sums which are very similar to the rightterm of the above statement of Alperin’s conjecture are of interest in string theory.In fact, people in that area are interested in what it is called the ’orbifold Eulercharacteristic’

1

|G|∑gh=hg

χ(X〈g,h〉)

where G is a finite group acting on a compact differentiable manifold X. I am notcompetent enough to give further informations on this, but the interested readermight start with the article [34] by Hirzerbruch and Hofer.

Exercise. Prove the equivalence of Thevenaz conjecture and Knorr and Robinsonformulation 2).

Problem. Show that if Thevenaz (and thus Alperin’s) conjecture is true thenQuillen’s conjecture is true.

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Chapter 5

Other complexes in finite groups

While it is the Brown complex that has attracted the more attentions by grouptheorists, other complexes associated in a natural way to a finite group have beenor could be considered. Indeed (although this is not the only way, as we shall see inexamples) any set of subgroups of a group G is naturally a poset ad thus becomes acomplex. It is clear that most interesting istances occur when we get a complex onwhich G acts in some natural way; this is the case if the set of subgroups is chosento be closed by conjugation in G.

5.1 The poset of all subgroups

If G is a group, we denote by L(G) the set of all subgroups of G ordered by inclu-sion. It is a lattice with minimum 1 and maximum G, so the complex of chainsassociated to it is contractible, and thus not very interesting from the point of viewof homology. Thus we are led to rather consider the poset L(G) = L(G) \ 1, G.

A subgroup H of G is complemented in L(G) if there exists a K ≤ G such thatH ∩ K = 1 and 〈H,K〉 = G. Thus L(G) is a complemented lattice if and onlyif each subgroup of G has a complement (for instance, the alternating group A5 iscomplemented). From Proposition 3.10 we get immediately,

Proposition 5.1 If L(G) is not complemented then L(G) is contractible.

I am not aware whether the converse holds, although the finite complementedgroups are known (in a sense, ”most” groups are not complemented). In [38] Kratzerand Thevenaz remark that if G = PSL(2, 7), then L(G) is complemented andH1(L(G)) 6= 0 6= H2(L(G)), but χ(L(G)) = 1.

For the following result see also Bouc [15]. It is an application of a more generalresult of Folkman [29] on the homology of modular lattices. The formula for theEuler characteristic is also implicit in a paper of Gaschutz of 1959. We give here adirect proof.

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Theorem 5.2 (Kratzer and Thevenaz). Let G be a soluble group, not of primeorder, with L(G) complemented. Let 1 = N0 E N1 E . . . E Ns = G be a chief seriesof G, and for i = 1, . . . , s let mi be the number of distinct complements of Ni/Ni−1

in G/Ni−1. Then L(G) is (s-2)-spherical and r0(Hs−2(L(G)) = m1m2 · · ·ms (andso χ(L(G)) = 1 + (−1)sm1m2 · · ·ms).

Proof. We proceed by induction on s ≥ 2 (if s = 1 then G is cyclic of prime orderand L(G) is empty).If s = 2 then G is the split extension of an elementary abelian normal subgroup N1

by a cyclic group of prime order (remember that L(G) is complemented). Then L(G)is made by m1 isolated vertices, corresponding to the complemets of N1, and by aconnected component that is the complex associated to L(N1) \ 1. The latter hasa greatest element and so it is contractible. Thus L(G) is equivalent to a complexof m1 + 1 isolated points. It is 0-spherical and H0(L(G)) has rank m1.Let s ≥ 3, and set N = N1, H = G/N and k = m2m3 . . .ms. By inductivehypothesis, L(H) is (s − 3)-spherical, and H0(L(G)) has rank k. Let ∆ be thecomplex associated to the poset of all non-trivial subgroups R of G such that NR 6=G. In this poset, N is a conjunctive element, and so ∆ is contractible. Let m = m1,and let H1, . . . ,Hm be the distinct complements of N in G. For each i = 1, . . . ,mlet Yi be the complex associated to L(Hi)\1, and let Xi = ∆∪Yi. Now, for everyi = 1, . . . ,m, Yi is contractible, and for each q ≥ 0, Mayer-Vietoris sequence yields

Hq+1(∆)⊕ Hq+1(Yi) = 0→ Hq+1(Xi)→ Hq(∆ ∩ Yi)→ Hq(∆)⊕ Hq(Yi) = 0

whence Hq+1(Xi) = Hq(∆ ∩ Yi).If R is a proper subgroup of Hi, then, since N is normal, NR 6= G. From thisobservation it easily follows that, for all i = 1, . . . ,m, ∆ ∩ Yi coincides with L(Hi)which in turn is clearly isomorphic to L(H). Hence, for q ≥ 0

Hq+1(Xi) ∼= Hq(∆ ∩ Yi) ∼= Hq(L(H)) .

Observe now that L(G) = X1∪X2∪. . .∪Xm. In fact, let σ : Q0 < Q1 < . . . < Qt = Qbe a chain in L(G). If NQ 6= G then σ is a chain in ∆; otherwise NQ = G and,since N 6= G and N is a minimal normal subgroup of G, N ∩Q = 1 and Q is one ofthe complements Hi. Also, if σ belongs to Xi∩Xj and not to ∆, then Hi = Q = Hj .Hence, for all i 6= j, Xi ∩ Xj = ∆. Now, recalling that ∆ is contractible, Mayer-Vietoris sequence and an easy induction on m yield, for all q ≥ 0,

Hq(L(G)) = Hq(X1)⊕ Hq(X2)⊕ · · · ⊕ Hq(Xm) .

In particular, Hq(L(G)) = 0 for 0 ≤ q 6= s − 2, while Hs−2(L(G)) is free of rankmk = m1m2 · · ·ms, which is what we wanted to prove.

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Corollary 5.3 Let G be a soluble group. The following conditions are equivalent.

i) L(G) is not complemented;

ii) L(G) is acyclic;

iii) χ(L(G)) = 1;

iv) G admits a non-complemented chief factor.

Proof. i) ⇒ ii) follows from Proposition 5.1. ii) ⇒ iii) is standard, while iii) ⇒iv) follows from Theorem 5.2. Finally, if N/M is a non-complemented chief factorof the soluble group G, then the normal subgroup N is not complemented in L(G),and so iv) ⇒ i).

S.Bouc ([15]) has generalized similar results to the case of a a section of type(H : G) = K | H < K < G in a group G with operators.

Remark. A theorem of Iwasawa says that a finite group G is supersoluble if andonly if all the maximal chains in L(G) have the same length. Thus, if for a group G,L(G) is shellable, then G is supersoluble. Bjorner ([12]) has proved that the converseis also true. If, in the definition of a shellable poset we remove the requirement thatthe poset is pure, then we have what is called a non-pure shellable poset (see [14]).Recently, Shareshian has shown that a finite group G is soluble if and olnly if L(G)is non-pure shellable.

5.2 Nilpotent subgroups

Given a finite group G, we denote by N (G) and A(G) the posets of, respectively, allnon-trivial nilpotent subgroups of G, and of all non-trivial abelian subgroups of G,and by abuse of notation, also the complexes of chains associated to them. Then,for each prime divisor p of |G|, the Brown complex at p is a subcomplex of N (G).

Proposition 5.4 The complexes N (G) and A(G) are equivalent.

Proof. Consider the inclusion map ι : A(G) −→ N (G). It is clearly an orderpreserving map of posets. Let H be a non-trivial nilpotent subgroup of G, and Z(H)its centre (which is not trivial). Then, if A is an abelian subgroup of H, Z(H)A isalso an abelian subgroup of H. This shows that Z(H) is a conjunctive element ofthe fiber ι/H = K ∈ A(G) | K ≤ H. Thus, ι/H is contractible. By Quillen’scriterion, A(G) is equivalent to N (G).

Exercise. Prove that N (G) is equivalent to the complex associated to the posetof all non-trivial abelian subgroups of G that are generated by elements of primeorder.

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The connectedness of N (G) is studied by S. Lucido in [42]. To state her resultwe recall the notion of the prime graph Π(G) of a group G. The vertices of Π(G)are all prime divisors of the order of G, and two vertices p, q are joined by an edgeif and only if G admits an element of order pq.

Proposition 5.5 The complex N (G) is connected if and only if the prime graphΠ(G) is connected.

Proof. Suppose first that Π(G) is not connected, and let Π1, Π2 be two distinctconnected components of Π(G). Assume by contradiction that N (G) is connected.If (H,K) is a 1-simplex in N (G), then either H ≤ K or K ≤ H. In any case forany prime divisors r of |H| and s of |K| (with r 6= s) r, s is an edge in Π(G). Letthen H be a non-trivial nilpotent Π1-subgroup of G, and K a non-trivial nilpotentΠ2-subgroup of G, such that the distance d = d(H,K) in the 1-skeleton of N (G)is minimal. Then there is a sequence H = H0, . . . ,Hd = K of non-trivial nilpotentsubgroups such that, for each 0 ≤ i ≤ d − 1 either Hi ≤ Hi+1 or Hi+1 ≤ Hi.But then, by the prevoius remark, H1 is a non-trivial nilpotent Π1-subgroup of G,contradicting the minimality of d.

Suppose now that N (G) is not connected. Observe that if, for a fixed prime ptwo connected components of N (G) contain a p-subgroup, then they are conjugateby some element of G, since they both must contain a Sylow p-subgroup. If x, y ∈G] = G \ 1, write x ≈ y if 〈x〉 and 〈y〉 belong to the same connected componentof N (G). Then ≈ is an equivalence relation, and conjugation induces an actionof G on G]/ ≈. If this action is transitive then |G| − 1 = |[x]||G : G[x]| for someequivalence class [x]. This implies that G = G[x], and therefore there is just oneconnected component in N (G), which is against our assumption. Thus there existat least two equivalence classes (and so two connected components of N (G)) thatare not conjugate in G. By the remark made above, this means that there existsa prime p and a connected component K of N (G) such that p does not divide theorder of any vertex of K. Let q be a be a prime divisor of some non-trivial nilpotentsubgroup belonging to K. Then there exists a Sylow q subgroup Q of G belonging toK. Assume by contradiction that there is a path p = p0, p1, . . . , pn = q connecting pto q in Π(G). Then G admits an element of order pn−1q and so there exists a Sylowpn−1-subgroup Pn−1 of G belonging to K. By repeating this argument we get thatthere exists a Sylow p-subgroup P of G belonging to K, which is a contradiction.Thus p and q are not connected in Π(G), and therefore the prime graph Π(G) is notconnected.

We leave as an exercise to show by an example that, in general, the number of con-nected components of Π(G) might differ from the number of connected componentsof N (G).

The general structure of a finite group whose prime graph is not connected hasbeen given by Gruenberg and Kegel (umpublished). In particular, thay describe

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all soluble groups with non-connected prime graph. From the point of view of thecomplexN (G) this is summarized in the following result. In its statement we denote,as usual, by F1(G) = Fit(G) the Fitting subgroup of the finite group G (i.e. thelargest normal nilpotent subgroup of G) and by F2(G) the inverse image in G ofFit(G/F1(G)).

Theorem 5.6 ([42]). Let G be a finite group with F2(G) 6= F1(G), and suppose thatN (G) is not connected. Then

1) Either G is a Frobenius group, or both F2(G) and G/F1(G) are Frobenius groupsand G/F2(G) is cyclic.

2) N (G) is equivalent to a set of |F1|+ 1 isolated points. G acts on it by fixing onepoint and transitively permuting the others.

Simple groups with non-connected prime graph have been classified by Williamsand by Kondratev, and the quasi simple ones by S.Lucido. Observe that the prop-erties of point (ii) in the previous Theorem do not hold in general. In fact, ifG = PSL(2, 2n) then the connected components N (G) are all contractible and splitinto three orbits under the action of G, all containing more than one element. On theother hand, if G = A7 then N (G) is not connected and has a G-invariant connectedcomponent which is not contractible.I do not know anything relevant about the topology of N (G) in the connected case.For instance, it would be interesting to know whether contractibility of N (G) forcesany significative group theoretical property of G.

Let now denote by D(G) the poset and the associated complex of all non-trivialsubgroups of the group G, ordered by subnormality (i.e. H ≤D(G) K if and only ifH is a subnormal subgroup of K). We leave the proof of the following observationas an exercise.

Proposition 5.7 If the group G is soluble, then D(G) is equivalent to N (G).

If G is not soluble, D(G) and N (G) need not be equivalent. For example, if G isthe alternating group A5, then N (G) is equivalent to a set of 21 isolated points (allthe Sylow subgroups of G), while D(G) is equivalent to a set of 22 isolated points(the 21 components already in N (G), and a further component G). It mightbe interesting to know more about the connections between N (G) and D(G). Forinstance: is it true that if D(G) is connected then it is equivalent to N (G) ?Finally, denote by PS(G) the poset and the associated complex of all subgroups ofG which are product of non-trivial simple groups, ordered by subnormality; thenprove the following fact

Proposition 5.8 For any group G the complexes D(G) and PS(G) are equivalent.

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5.3 Commuting complexes

Let U be a non-empty normal subset of a group G (by normal we mean that xg ∈ Ufor all x ∈ U and g ∈ G). The commuting complex K(U) associated to U is thecomplex that has U as set of vertices and whose simplices are all finite non-emptysubsets of U whose elements commute pairwise (i.e. the finite subsets of U thatgenerate a non-trivial abelian subgroup).

Suppose that p is a prime divisor of |G|, and let P be the set of all non-trivialp-elements of G. Then the map defined by H 7→ Z(H) \ 1 for all non-trivialp-subgroups H of G, establishes the equivalence of the Brown complex Sp(G) andthe (barycentric subdivision) of K(P ). Similarly, if G] = G \ 1, then one provesthat K(G]) is equivalent to N (G).

An interesting case of commuting complex has been studied by Bouc in [17].Following his notation, for n ≥ 2, we denote by d2(n) the commuting complexK(Tn), where Tn is the set of all transpositions of the symmetric group Sn. Aremarkable feature of these complexes is that in some cases their fundamental andhomology groups have torsion. We collect in the following statement part of Bouc’sresults.

Theorem 5.9 (S. Bouc [17]). Let d2(n) = K(Tn), where Tn is the set of all trans-positions of the symmetric group Sn. Then

1) If n ≥ 5, d2(n) is connected.

2) If n ≥ 8, then d2(n) is simply connected. The fundamental groups of d2(5) andd2(6) are free abelian of rank 6 and 16 respectively; while π1(d2(7)) ∼= Z/3Z.

3) Hq(d2(n)) = 0 if 3q + 4 < n; and Hq(d2(3q + 4)) ∼= Z/3Z if q ≥ 2.

Alperin and Glauberman have later confirmed a conjecture of Bouc, by showingthat the fact that π1(d2(7)) has order 3 is related to the existence of a non splittingcentral extension of the cyclic group of order three by the alternating group A7.

5.4 Wielandt complexes

Let F be a normal family of subgroups of the group G (again, normal means thatF is closed under conjugation). The Wielandt complex W(F) associated to F isthe complex whose vertices are the elements of F and simplices are all the subsetsH0, . . . ,Hq of F such that Hi is subnormal in 〈H0, . . . ,Hq〉 for each i = 0, . . . , q.

If F is the family of all non-trivial, respectively subgroups, nilpotent subgroups,or p-subgroups of G, then W(F) is equivalent to D(G), to N (G) or to the Browncomplex Sp(G).

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The minimal cases occur when F is the conjugacy class of a single subgroup H ofG, that is F = [H] = Hg | g ∈ G. The case in which the corresponding Wielandtcomplex turns out to be acyclic is described by the following result.

Theorem 5.10 ([24]) Let H ≤ G; then the following are equivalent.

i) H is subnormal in G;

ii) W([H]) is contractible;

iii) χ(W([H])) = 1.

The proof is a corollary of the following subnormality criterion, which is provedusing the classification of finite soluble groups.

Theorem 5.11 Let G be a finite group and H ≤ G. Suppose that for every primedivisor p of G there exists a Sylow p-subgroup P of G such that H is subnormal in〈P,H〉. Then H is subnormal in G.

Proof. (of Theorem 5.10). i) ⇒ ii). If H is subnormal in G, then W([H]) isthe complex of faces of a simplex and is therefore contractible.

ii) ⇒ iii). This needs no proof.iii) ⇒ i). Assume χ(W([H])) = 1, and consider the action of G on W([H])

by conjugation, which is clearly simplicial. Let p be a prime divisor of |G|, andlet P be a Sylow p-subgroup of G. Then by Theorem 2.22, P fixes some simplexof W([H]) (observe that here the action is not by admissible automorphisms, so wehave to pass to the barycentric subdivision in order to apply Smith’s Theorem), thatis there exists a simplex Hg0 , . . . ,Hgn of W([H]) which is fixed by conjugation byP , and this means that P normalizes the subgroup 〈Hg0 , . . . ,Hgn〉 = T . It followsthat

Hg0 EE T E 〈T, P 〉

and, by conjugating by g−10 , we get that H is subnormal in 〈H,P g

−10 〉. By Theorem

5.11 we then conclude that H is subnormal in G.

Exercises. 1) Let H ≤ G. Let m be the G.C.D. of the indices |G : T | where Truns over all subgroups of G that contain H as a subnormal subgroup. Prove thatm divides χ(W([H])).

2) Let p be an odd prime, Cp a cyclic group of order p, and D8 the dihedral groupof order 8 ancting irreducibly on Cp × Cp. Let G be the semidirect product

G = (Cp × Cp) >C D8 ,

u a non-central involution of D8, and H = 〈u〉. Show that W([H]) is a completebipartite graph on p+ p vertices, and χ(W([H])) = p(2− p).

Problem. Study the homology of the complexes W([H]).

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5.5 Coset Poset

Recently, K. Brown ([22]) has considered the poset C(G) of all proper right cosetsof the group G, ordered by set inclusion, and its associated complex (observe thatHx ⊆ Ky if and only if H ≤ K and Ky = Kx).

Let G be a group, and s a positive integer. We remember from section 3.4 thatthen φs(G) denotes the number of ordered s-tuples of elements of G that generateG. Let P (G, s) be the probability that an ordered s-tuple of elements of G generatesG. Thus

P (G, s) =φs(G)

|G|s.

By P.Hall’s Theorem 3.23 we then have

P (G, s) =∑H≤G

µ(H,G)|H|s

|G|s=∑H≤G

µ(H,G)

|G : H|s=∑n≥1

αnn−s

where µ is the Mobius function on the poset of all subgroups of G, and

αn =∑

H≤G, |G:H|=n

µ(H,G) .

Thus we can extend P (G, s) to a function of a complex variable s, as a Dirichletseries. In particular, we may consider its value at s = −1:

P (G,−1) =∑H≤G

µ(H,G)|G : H| .

Exercise. Let G and H be groups of coprime order. Prove that P (G ×H,−1) =P (G,−1)P (H,−1).

Now, K. Brown observes the following fact.

Proposition 5.12 For any group G, P (G,−1) = −χ(C(G)).

Proof. Given a group G, let C be the poset obtained by adding to C(G) a greatestelement (which we may well take as G) and a least element (that we denote by 0).Let µ denote the Mobius function on C. Then (see Proposition 3.20)

µ(0, G) = χ(C(G)) .

Now, let Hx be a proper coset of G. Then the interval [H,G] in the lattice L(G) ofall subgroups of G is isomorphic to the interval [Hx,G] in C via the map K 7→ Kx.Thus

µ(Hx,G) = µ(H,G) .

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We then have, applying Proposition 3.21,

P (G,−1) =∑H≤G

µ(H,G)|G : H| = µ(G,G) +∑

Hx∈C(G)

µ(Hx,G) =

= µ(G,G) +∑

Hx∈C(G)

µ(Hx,G) = −µ(0, G) =

= −χ(C(G))

which is what we wanted to show.

To state the main Theorem in [22] we need to introduce the following notation.Let p be a prime, and Q a p-subgroup of the group G; we denote by N∗(Q) thesubgroup of G generated by all x ∈ G such that 〈Q,Qx〉 is a p-group.

Theorem 5.13 (K. Brown). Let G be a group, and p a prime. Denote by P a Sylowp-subgroup of G, and by Q a subgroup of P of maximal order such that N∗(Q) = G.Then |P : Q| divides P (G,−1).

Finally, let me report a couple of questions raised by Brown in his paper.

Problems. 1) Does there exist any group G such that C(G) is contractible ?2) Is it possible to characterize soluble groups G in terms of the combinatorial

topology of C(G) ?

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Chapter 6

The Mobius function

The Mobius function on a poset provides a somehow coarser but still interestingapproach to some of the topics we have treated so far. Although it leads to aless complete theory than a more topological approach, it often allows to keep thearguments at a purely combinatorial and more elementary level. Moreover, theMobius function on (a subposet of) the lattice of all subgroups of a group is aninteresting subject in itself. For the fundamental technics, and also a historicalintruduction related to it, we refer to the paper of Hawkes, Isaacs and Ozaydin [36].The following proposition, for instance, corresponds to Proposition 3.3.

Proposition 6.1 Let X be a poset in which every non-empty subset has a leastelement (a poset with this property is called a meet-semilattice). Let a, b ∈ X witha < b. Suppose that a is not the meet of a set of co-atoms of [0, b]. Then µX(a, b) = 0.

Proof. Let a, b ∈ X, with a < b and µX(a, b) 6= 0, and suppose by contradictionthat a is not the intersection of co-atoms of [0, b]. Assume also that a is chosen to bemaximal for such property. Then a < d where d is the intersection of all co-atomsof [0, b] that are above a. Then∑

a≤z≤bµX(z, b) = 0 and

∑d≤t≤b

µX(t, b) = 0 .

By subtracting the first equality from the second one, we get∑d≤t≤b,d 6≤t

µX(t, b) = 0 .

Now, it is clear that none of the elements t in this sum is an intersection of co-atomsof [0, b]. Thus, by the maximality of a, we have µX(t, b) = 0 for a < t. This in turnyields µX(a, b) = 0, against our assumption.

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We are interested in particular in the lattice L(G) of all subgroups of a groupG. Following a by now standard notation, we set

µ(G) = µL(G)(1G, G) = −∑H<G

µL(G)(1G, H).

In other words, µ(G) = χ(L(G)) − 1, where L(G) is the complex defined in sec-tion 5.1. Observe that if H ≤ G, µL(G)(1G, H) does not depend on the way H isembedded in G, but just on H itself (and, clearly, µL(G)(1G, H) = µ(H)).

Examples. 1) One has µ(S3) = 3, µ(A4) = 4, µ(D2p) = p (where p is an odd prime,and D2p is the dihedral group of order 2p), µ(A5) = −60, µ(PSL(2, 7)) = 0.

2) If Cn is a cyclic group of order n, then µ(Cn) coincides with the value at nof the classical Mobius function (i.e. µ(Cn) = 0 if n is divided by the square of aprime, and µ(Cn) = (−1)s if n is the product of s distinct primes).

3) If p is a prime, then µ(Cp × Cp) = p.4) Let Φ(G) be the Frattini subgroup of the group G (i.e. the intersection of all

maximal subgroups of G). Then by Proposition 6.1, µ(G) = 0 if Φ(G) 6= 1G.

By refering to Theorem 5.2, one can give a formula for µ(G) when G is soluble.In this form, the result is due to Gaschutz [30]. A direct proof of it can be found inthe original paper of Gaschutz, or also in [36].

Theorem 6.2 Let G be a soluble group, and 1 = N0 E N1 E . . . E Ns = G a chiefseries of G. For i = 1, . . . , s denote by mi the number of distinct complements ofNi/Ni−1 in G/Ni−1. Then µ(G) = (−1)sm1m2 · · ·ms.

From this Theorem it follows easily that if G is soluble and µ(G) 6= 0, then everyprime divisor of µ(G) divides the order of G. That this fact is not true in generalhas been observed by Hawkes, Isaacs and Ozaydin [36], who prove that if G is thedirect product of n copies of A5 then µ(G) = (−60)n(1 · 3 · 5 . . . (2n− 1)). Also, asa particular case of it, we have the following Corollary of Theorem 6.2.

Corollary 6.3 Let p be a prime and G an elementary abelian p-group of order pn.

Then µ(G) = −1 if n = 1, and µ(G) = (−1)np(n2) if n ≥ 2.

More generally, we may prove the following

Corollary 6.4 Let p be a prime, G a p-group, and H ≤ G. Then

1) µL(G)(H,G) = 0 if Φ(G) 6≤ H;

2) µL(G)(H,G) = −1 if H is a maximal subgroup;

3) µL(G)(H,G) = (−1)np(n2) if Φ(G) ≤ H and |G/H| = pn with n ≥ 2.

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Proof. If Φ(G) ≤ H then H is normal in G. Hence µL(G)(H,G) = µ(G/H), andpoints 2) and 3) follow from Corollary 6.3. If Φ(G) 6≤ H, then the intersection of allmaximal subgroups of G containing H is strictly larger than H, and point 1) followsfrom Proposition 6.1

Exercise. Let p be a prime divisor of the order of the group G. For d ≥ 1 let md

denote the number of distinct elementary abelian p-subgroups of G of rank d. Provethat

χ(Sp(G)) = m1 +∑d≥2

(−1)d−1p(d2)md.

In the literature, the Mobius function has been effectively used to derive somegeneral divisibility properties of various kinds. Here we prove one of such results,that we will use in the next section, and suggest a few more as exercises. For manyresults of this kind, we refer to the mentioned papers of Bouc [15], Brown andThevenaz [21], Hawkes, Isaacs and Ozaydin [36].

A group G is perfect if G coincides with its derived subgroup G′, i.e. if G has nonon-trivial abelian homomorphic images.

Lemma 6.5 Let S be a normal perfect subgroup of the group G. Then, for anysubgroup H of G,

|NG(H) : H| divides∑

K≥SHµ(H,K) .

Proof. If H = G we have nothing to prove. If S ≤ H < G then, by Proposition

3.21,∑

K≥SHµ(H,K) =

∑K≥H

µ(H,K) = 0, and we are done. Thus, assume S 6≤ H,

set V = K ≤ G | H < K, S 6≤ K , ordered by inclusion, and V = V∪H,G. ByProposition 3.20, we have χ(V) = µV(H,G) + 1, and by applying twice Propositon3.21,

χ(V) = −∑K∈V

µV(H,K) = −∑K∈V

µ(H,K) =

=∑

K≥SHµ(H,K) + µ(H,H) =

∑K≥SH

µ(H,K) + 1 .

Observe that if g ∈ NG(H) and K ∈ V, then H = Hg < Kg 6≥ Sg = S. ThusNG(H)/H acts by conjugation on V. Now, let g ∈ NG(H) \ H, and let K ∈ VHg(i.e. Kg = K). We prove that T = 〈K, g〉 belongs to VHg. To do that, we have toshow that T does not contain S. Suppose to the contrary that T ≥ S. Then, asK E T , K ∩ S E S and

S

K ∩ S∼=SK

K≤ T

K

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is cyclic. Since S is perfect, this forces S = S ∩ K and the contradiction S ≤ K.Hence 〈H, g〉 is a conjunctive element of VHg, which is therefore contractible. Inparticular, χ(VHg) = 1 for all 1 6= Hg ∈ NG(H)/H. By Corollary 2.5 we then havethat |NG(H)/H| divides χ(V)− 1 =

∑K≥SH µ(H,K).

Exercises. 1) ([36]) Let H be a subgroup of the group G, and let m be the prod-uct of the distinct prime divisors of |G : G′H|. Prove that |NG(H) : H| dividesmµL(G)(H,G).

2) ([36]) Let n be a divisor of |G|, m the product of the distinct prime divisors of|G : G′|, and Q the set of all subgroups of G which are either trivial or of order that

does not divide n. Prove that |G| divides mn∑X∈Q

µL(G)(X,G).

3) ([36]) Let H be a subgroup of the group G, and n a divisor of |G|. Denote by Xthe set of all subgroups of G that contain H and whose order divides n. Prove thatn divides

|G||NG(H) : H|

∑X∈X

µL(G)(H,X).

Now, in the final two sections, we give some applications of the Mobius function.

6.1 The Burnside Ring

Let G be a (finite) group. A G-set A is a finite set on which G acts (on the right)as a group of permutations. Two G-sets A and B are isomorphic if there exists abijection ϕ : A→ B such that ϕ(ag) = (ϕ(a)g for all a ∈ A and g ∈ G. If A and Bare G-sets, then their disjoint union A+B and the cartesian product A×B are alsoG-sets in a natural way. This allows to consider disjoint union and cartesian productas operations on the set of all isomorphism classes of G-sets, that respect the axiomsof a commutative halfring (being, respectively, the sum and the multiplication). Theassociated ring Ω(G) is called the Burnside ring of the group G.

Let us look better at what this definition means. Before, we fix the followingnotations. Given a group G and a subgroup H, we denote with the boldface letterH the conjugacy class of H in G, and we write C = C(G) for the set of all conjugacyclasses of subgroups of G. A G-set A is transitive if for all x, y ∈ A there existsg ∈ G such that y = xg. Any G-set decomposes in a unique way as the sum (withrespect to the above definition) of its G-orbits, each of those is a transitive G-set.Addictively, Ω(G) is then the free abelian group with free generating set the set ofall isomorphism classes of transitive G-sets. Now, if A is a transitive G-set and His the stabilizer of a point, then A is isomorphic to the set G/H of all right cosetsmodulo H, with G acting on it by right multiplication. Also, two coset sets G/H

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and G/K are isomorphic G-sets if and only if H is conjugate to K in G. For H ≤ Gwe denote by αH the isomorphism class of the G-set G/H. Thus, a free generatingset for the additive group of Ω(G) is αH | H ∈ C. Then the multiplication onΩ(G) may be defined on the generators (i.e. αHαK is the isomorphism class of theG-set G/H ×G/K decomposed as the sum of its transitive constituents) and thenexteded by distributivity (viewed in this way, the ring Ω(G) is what is called theGrothendieck ring of the category of G-sets). In this ring the identity element 1 isαG.

Use of the Mobius function in the study of Ω(G) has been introduced by Gluck[31] and Yoshida [67]. In this section, we borrow ideas from Gluck to give a proof ofa striking theorem of Dress [28], which says that a group G is soluble if and only ifthe only idempotents of Ω(G) are 0 and 1 (see also [59] for another different proof).The Mobius funtion µ we will use is the Mobius function on the lattice L(G) of allsubgroups of G.

Let A be a G-set and H ≤ G. We denote by φH(A) the number of points ofA fixed by H. φH(A) is then a positive integer which, given A, depends only onthe conjugacy class of H. Moreover, it is immediate that, if A,B are G-sets, thenφH(A + B) = φH(A) + φH(B) and φH(A × B) = φH(A)φH(B). In other words,φH : Ω(G) → Z is a ring homomorphism. We first need an observation about thevalues of φH on (isomorphism classes of) transitive G-sets. If H,K are subgroupsof G, we denote by βHK the number of distinct G-conjugates of H contained in K,and by λHK the number of distinct G-conjugates of K containing H. Finally, giventwo conjugacy classes H,K of subgroups of G, we write H ≤ K if there exists aconjugate of H which is contained in K.

Lemma 6.6 Let H,K be subgroups of the group G. Then

φH(αK) = βHK|NG(H)||K|

= λHK|NG(K)||K|

.

In particular, φH(αK) 6= 0 if and only if H ≤ K.

Proof. If H and K are subgroups of G, then φH(αK) is the number of distinctcosets Kg such that KgH = Kg, i.e.

φH(αK) = |Kg | gHg−1 ≤ K| = |Kg | H ≤ Kg|.

Now, the set Γ = g ∈ G | gHg−1 ≤ K is the disjooint union of the cosets Kg suchthat gHg−1 ≤ K. Therefore, |Γ| = |K|φH(αK). On the other hand, Γ is the disjointunion of the cosets gNG(H) such that gHg−1 ≤ K, Whence |Γ| = βHK |NG(H)|.Comparison of the two numbers yields φH(αK)|K| = βHK |NG(H)|, proving the firstequality of the Lemma. We leave as an exercise the proof of the second equality.

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Lemma 6.7 Let G be a group, and X, Y ∈ Ω(G). Then X = Y if and only ifφH(X) = φH(Y) for all H ≤ G.

Proof. In one direction the implication is obvious.

Thus, let X =∑K∈C

xKαK and Y =∑K∈C

yKαK (with integers xK and yK) be

distinct elements of Ω(G). Then there exists a subgroup H of G such that xH 6= yH,but xK = yK for all H < K. Therefore

φH(X) =∑K∈C

xKφH(αK) = xHφH(αH)∑K>H

xKφH(αK) 6=

yHφH(αH)∑K>H

xKφH(αK) = yHφH(αH)∑K>H

yKφH(αK) = φH(Y).

Lemma 6.8 Let G be a group and X =∑K∈C

xKαK ∈ Ω(G). Then, for all H ≤ G

xH|NG(H)||H|

=∑K≥H

µ(H,K)φK(X).

Proof. By Lemma 6.6, for each H ≤ G we have

φH(X) =∑K≥H

xKφH(αK) =∑K≥H

xKλHK|NG(K)||K|

=∑K≥H

xK|NG(K)||K|

.

By the Mobius inversion formula, we then get the desired equality.

Let now X ∈ Ω(G). Since for all H ≤ G, the application φH is a homomorphism,by Lemma 6.7 it follows that X is idempotent (i.e. X = X2) if and only if φH(X) =φH(X)2 for all H ≤ G. Since φH(X) is an integer, we have that X is idempotent ifand only if φH(X) ∈ 0, 1 for all H ≤ G.

Theorem 6.9 (Dress). A finite group G is soluble if and only if 0 and 1 are theonly idempotents of Ω(G).

Proof. Let G be soluble, and let X =∑K∈C

xKαK be an idempotent of Ω(G).

We prove by induction on |G| that φH(X) = φ1(X) for all subgroups H of G. IfH < G, then X can be seen as an element of Ω(H) in a natural way, and as such it

89

is idempotent. Hence, by inductive assumption, φH(X) = φ1(X). Now, since G issoluble, there exists a prime p and a normal subgroup N of G such that |G : N | = p.By Lemma 6.8,

xN · p = µ(N,N)φN (X) + µ(N,G)φG(X) = φN (X)− φG(X) = φ1(X)− φG(X)

and so p divides φ1(X) − φG(X). Since X is idempotent in Ω(G), we must haveφH(X) ∈ 0, 1 for all H ≤ G, and this forces φ1(X) = φG(X).In conclusion we have that either φH(X) = φ1(X) = 0 for all H ≤ G (whenceX = 0), or φH(X) = φ1(X) = 1 for all H ≤ G (whence X = αG = 1).

We now prove the reverse implication. If G is a group, we denote by G(n) then-th term of the derived series of G. Then there exists a n such that G(n) = G(n+1).If n is such an integer, we write S = G(n). Then S is a normal perfect subgroup ofG. We then define

XS =1

|G|∑H≤G

|H| ∑K≥SH

µ(H,K)

αH .

The coefficient of αH in XS is

1

|G||G : NG(H)||H|

∑K≥SH

µ(H,K) =|H|

|NG(H)|∑

K≥SHµ(H,K)

which is an integer by Lemma 6.5. Thus XS ∈ Ω(G). If G is soluble, then S = 1and XS = αG = 1. If G is not soluble, S 6= 1 and XS is not equal to 1 (nor to 0) inΩ(G). We prove that XS is an idempotent of Ω(G). As we have observed above, itwill be sufficient to show that φU (XS) ∈ 0, 1 for all subgroups U of G. Thus, letU ≤ G, then

φU (XS) =1

|G|∑H≤G

|H| ∑K≥SH

µ(H,K)

βUH|NG(U)||H|

=

=|NG(U)||G|

∑H≤G

∑K≥SH

µ(H,K)

βUH =

=|NG(U)||G|

∑T≥S

∑H≤T

µ(H,T )βUH

.Now, βUH is the number of distinct conjugates of U contained in H, which we may

write as∑A≤H

δUA, where δ is the Kronecker symbol. Applying Mobius inversion

90

formula, we then get∑H≤T

µ(H,T )βUH = δUT. Therefore

φU (XS) =|NG(U)||G|

∑T≥S

δUT .

Finally, observing that∑T≥S

δUT is just the number of conjugates of U containing S,

we get

φU (XS) =

1 if S ≤ U0 if S 6≤ U

This shows that XS is an idempotent of Ω(G) and concludes the proof of the The-orem.

Indeed, Dress proves that there is a 1-1 correspondence between the non zeroidempotents of Ω(G) and the conjugacy classes of perfect subgroups of G. Use ofthe Mobius function allows to make this correspondence explicit. We outline thatin the following exercise.

Exercise. Any group A has a maximal normal soluble subgroup, that we denoteby Sol(A). Let G be a given group, S a perfect subgroup of G, and let N ≤ G suchthat N/S = Sol(NG(S)/S).

1) Prove that NG(S) = NG(N).

2) Let S ≤ H ≤ N , and g ∈ G. Prove that if Hg ≤ N then g ∈ NG(S). Deducethat, for S ≤ H ≤ N , βHN = |NG(N) : NG(H)|.3) Prove that XS defined below is a non zero idempotent of Ω(G).

XS =1

|NG(S)|∑H≤N

|H| ∑SH≤K≤N

µ(H,K)

αH .

4) Prove that if S, T are perfect subgroups of G then XS = XT if and only if S andT are conjugate in G.

6.2 Subnormalizers

Our second application of the Mobius function and of some other facts that we haveproved earlier, concerns a result which is by no means as important as the DressTheorem on the Burnside ring. Consequently, I will not give fully detailed proofs.However, I hope it will serve as an example of how combinatorial techniques maybe used to prove some genuine group theoretical results.

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A subgroup H of the group G is subnormal (write H EE G) if there exists afinite series H = H0 E H1 E . . . E Nd = G. Given a subgroup H of G we define thesubnormalizer of H in G to be the set

SG(H) = g ∈ G | H EE 〈H, g〉 .

In general this subset is not a subgroup (see the book [41] for more details on thistopic). Our aim is to provide a formula for its order in the case in which H is ap-subgroup for some prime p.

Let g be an automorphism of the poset X. Then, as usual, Xg is the subposetof all fixed points of g. We denote by gX the subposet of all elements x ∈ X suchthat there exists y ∈ Xg with x ≤ y (i.e. gX is the closure of Xg in X). Let a ∈ X,and fix the following further notations.

- Xg>a = y ∈ Xg | a < y;

- gX>a = y ∈ gX | a < y;- if G is a group of automorphisms of X, sG(a) = g ∈ G | a ∈ gX.

Finally, we recall that a meet-semilattice is a poset X in which for any pair ofelements a, b ∈ X there exists a ∧ b = infa, b.

Lemma 6.10 Let g be an automorphism of the meet-semilattice X. Then for alla ∈ X the complexes associated to Xg

>a and to gX>a are equivalent.

Proof. If Xg>a is empty, then also gX>a is empty. If Xg

>a is not empty, considerthe inclusion map ι : Xg

>a → gX>a. Then use the fact that X is a meet-semilatticeand Quillen’s criterion to show that ι induces an equivalence.

Exercise. Prove that the previous Lemma does not hold in general if X is not asemilattice.

Lemma 6.11 Let G be a group of automorphisms of the meet-semilattice X. Thenfor all x ∈ X, ∑

y>x

µX(x, y)|sG(y)| = −∑g∈G

χ(Xg>x).

Proof. Exercise.

We recall that if H,K are subgroups of the group G, then with λHK we denotethe number of distict conjugates of K that contain H. Fixed a group G and a primedivisor p of |G|, we denote by Sp the set of all p-subgroups of G (trivial subgroupincluded). Then Sp = Sp(G)∪1G is a meet-semilattice. Also, if H is a p-subgroupof G, and we consider the conjugation action of G on Sp, then sG(H) is just the

92

subnormalizer SG(H) as defined at the beginning of the section. In fact, if g ∈ sG(H)then g normalizes some p-subgroup T containing H; whence H EE T E 〈T, g〉 and,consequently, H EE 〈H, g〉, i.e. g ∈ SG(H). Conversely, if g ∈ SG(H), then H isa subnormal p-subgroup of 〈H, g〉, and this implies H ≤ Op(〈H, g〉) = T E 〈H, g〉.Thus, g normalizes the p-subgroup T which contain H; so g ∈ sG(H).

Theorem 6.12 ([25]). Let p be a prime divisor of the order of the group G, and Pa Sylow p-subgroup of G. Then, for every p-subgroup H of G,

|SG(H)| = λHP |NG(P )|.

Proof. We firts prove the following equality, for any p-subgroup H of G.∑H≤K∈Sp

µ(H,K)|SG(K)| =|NG(H)| if H ∈ Sylp(G)0 if H 6∈ Sylp(G)

where Sylp(G) denotes the set of all Sylow p-subgroups of G, and µ is the Mobiusfunction on Sp. To prove that, we put, for any H ∈ Sp,

φ(H) =∑

H≤K∈Sp

µ(H,K)|SG(K)|.

If H ∈ Sylp(G), then clearly φ(H) = |SG(H)| = |NG(H)|. Suppose now that H ∈ Spis not a Sylow p-subgroup. By the remark we made above and Lemma 6.11, we have

φ(H) = |SG(H)|+∑H<K

µ(H,K)|SG(K)| = |SG(H)|+∑g∈G

χ(Sg>H).

If g 6∈ SG(H), then Sg>H = ∅ and χ(Sg>H) = 0. If g ∈ SG(H) \ NG(H), then gnormalizes a p-subgroup containing H, but does not normalize H itself; this meansthat if T the intersection of all p-subgroups in Sg>H , then H < T and so T ∈Sg>H . Thus, Sg>H has a least element and it is therefore contractible; in particular,χ(Sg>H) = 1. Hence, the total contribution in the above sum of the elements g 6∈NG(H) is equal to |SG(H)| − |NG(H)|. Writing N = NG(H) we then have

φ(H) = |NG(H)| −∑g∈N

χ(Sg>H).

Now, if V is the complex Sp(N/H), then by Lemma 4.29, the complexes VgH andSg>H are equivalent for all g ∈ N . Therefore

φ(H) = |NG(H)| − |H|∑

gH∈N/H

χ(VgH).

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Now we apply Lemma 4.30, which ensures that∑

gH∈N/H

χ(VgH) = |N/H|, and get

φ(H) = |NG(H)| − |H|∑

gH∈N/H

χ(VgH) = |NG(H)| − |H||NG(H)/H| = 0

which proves our assertion.Finally, we apply Mobius inversion formula to the functions Φ(K) = |SG(K)|

and φ(H) =∑

H≤K∈Spµ(H,K)|SG(K)| to deduce

|SG(H)| =∑

H≤K∈Sp

φ(K) = λHP |NG(P )|

thus completing the proof.

Corollary 6.13 Let p be a prime divisor of the order of the group G, and H ap-subgroup of G. Then H EE G if and only if H EE 〈H, g〉 for all g ∈ G.

Proof. In one sense the implication is clear. Conversely, if H EE 〈H, g〉 for allg ∈ G, then by Theorem 6.12, λHP |NG(P )| = |G|, which gives λHP = |G : NG(P )|.Thus H is contained in the intersection of all Sylow p-subgroups of G, that is H ≤Op(G), and this implies H EE G.

Of course, this Corollary is just a particular case of a famous Theorem ofWielandt, which ensures that the same conclusion holds for any subgroup H ofG, with no assumptions on its order. It is not clear to me what might be a possiblegeneralization, if it exists, of Theorem 6.12 to arbitrary subgroups.

Exercise. Prove that, with the assumptions and notations of Theorem 6.12, then|SG(H)| = βHP |NG(H)|.

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