Topics in Infinite Groups

Lectured by J. Button

Lent Term 2014

Course schedule

1 Introduction 1

2 A brief guide to abelian groups 8

3 Free groups 9

4 Presentations of groups 16

5 Soluble, polycyclic, nilpotent groups 21

6 Finite index subgroups and virtual properties 24

7 Maximal (normal) subgroups 30

8 Residual finiteness 31

9 The Generalised Burnside Problem 35

Last updated: Sun 16th Aug, 2015These haven’t been proof-read, so please let me know of any corrections: glt1000@cam.ac.uk

Course description

This course is a general introduction to infinite groups, with particular emphasis on finitelygenerated and finitely presented groups.

The following is a summary of the lectures:

Review of basic definitions and results; brief mention of Abelian groups.

Free (non-abelian) groups and free products, Nielsen-Schreier theorem and index formula;presentations of groups, free products with amalgamation and HNN extensions; nilpotent,polycyclic and soluble groups.

Subgroups of finite index and virtual properties; maximal and maximal normal subgroups;infinite simple groups; residual finiteness and Hopficity; Baumslag-Solitar groups.

The generalised Burnside Problem.

Desirable Previous Knowledge

Any introductory undergraduate group theory course as well as some basic algebraic topology,up to covering spaces and the fundamental group.

Introductory Reading

Any introductory text in group theory, of which there are plenty. To list but two:

1. J. F. Humphreys, A course in group theory, Oxford Science Publications

2. W. Ledermann, Introduction to group theory, Longman

The necessary algebraic topology can certainly be found in either of:

1. J. M. Lee, Introduction to topological manifolds, (GTM 202), Chapters 7, 10, 11, 12

2. A. Hatcher, Algebraic topology, CUP, Chapter 0 and Sections 1.1, 1.2, 1.3, 1.A

Reading to complement course material

Some of the results from the course can be found in:

1. R. C. Lyndon and P. E. Schupp, Combinatorial group theory, Springer, Sections I.1,II.1, II.2, IV.1, IV.2, IV.4

1. IntroductionLecture 1

The first chapter was given out in printed form in the first lecture, and annotations weremade to fill in some examples and sketch proofs. This is included here, which is why Lecture1 looks a bit long.

Example 1.1.

1. X a set, S(X) the group of permutations of X .

2. F a field, Mn(F ) = {n× n matrices with entries in F}.

Subgroups

Proposition 1.2. H ⊆ G is a subgroup of G ⇔ H 6= ∅ and for all a, b ∈ H , ab−1 ∈ H . Wewrite H 6 G and H < G if H 6= G.

Notation. Subgroup: H 6 G. Proper subgroup: H < G. Trivial subgroup I = {e}.

Proposition 1.3.

(i) L 6 H and H 6 G ⇒ L 6 G.

(ii) If Hi 6 G for all i ∈ I then⋂

i∈I Hi 6 G.

For H1, H2 6 G, H1 ∪H2 is not generally a subgroup. But. . .

Proposition 1.4 (Ascending sequence of subgroups). If H1 6 H2 6 . . . 6 G (whichmeans Hn 6 G and Hn 6 Hn+1 for all n) then

⋃∞

n=1Hn 6 G.

If we have groups Gn (n ∈ N) for which Gn 6 Gn+1 and we form the set X =⋃∞

n=1Gn thenX is a group.

Example 1.5.

1. S1 ⊂ {C \ {0},×}. Take Hn = {e2πik/2n

: 1 6 k 6 2n}.

X =⋃∞

n=1Hn. (Do for prime p: quasicyclic p-group.)

2. G = S(Z), Hn = {permutations of {−n, . . ., n}, fixes rest} ∼= S2n+1 6 Hn+1.⋃Hn = S0(Z). Note that f(z) = z + 1 ∈ S(Z) \ S0(Z).

Generators

Let X ⊆ G.

Definition 1.6. The subgroup 〈X〉 generated by X is⋂H , where this intersection is over

all H with X ⊆ H 6 G. It is the smallest subgroup of G containing X . We write〈x1, . . . , xk〉, 〈X,Y 〉, 〈Xi : i ∈ I〉, etc.

Definition 1.7. G is finitely generated (fin.gen./f.g./fg) if G = 〈g1, . . . , gk〉. OtherwiseG is infinitely generated (inf.gen./i.g.).

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Example 1.8. G = 〈g〉 means that G is cyclic. Either G = Z or, if G has order n, we writeG = Cn.

Cyclic results.

1. A subgroup of a cyclic group is cyclic: Cn,Z

2. Suppose there is no H with I < H < G. Then G = Cp (p prime) or I.

Proposition 1.9. If X ⊆ G then the elements of 〈X〉 are (e and)

xn1

1 xn2

2 . . . xnk

k

for n1, . . . , nk ∈ Z, and x1, . . . , xk ∈ X (but not necessarily distinct).

So G finite ⇒ G finitely generated ⇒ G countable.

If we have an ascending sequence of subgroups H1 6 H2 6 . . . 6 G then either:

• there is some N such that HN = HN+n for all n ∈ N (the chain terminates); or

• (on throwing away duplicates) H1 < H2 < . . . (the chain is strictly ascending)

Definition 1.10. G has max (satisfies the maximal condition) if every ascending sequenceof subgroups terminates.

Theorem 1.11. G has max ⇔ H is finitely generated for all H 6 G.

Warning! We can have G finitely generated but H < G with H infinitely generated. Wewill see examples later; indeed it can be argued that many finitely generated groups donot have max.

Cosets

If H 6 G then the left cosets of H in G are the sets gH = {gh : h ∈ H} for each g ∈ G.

Proposition 1.12 (Lagrange for infinite groups). The left cosets of H in G form a par-tition of G and any left coset is in bijection with H .

Note. We have right cosets Hg in bijection with H too. Also there is a bijection from theset of left cosets to the set of right cosets given by gH goes to Hg−1 (not to Hg).

Definition 1.13.

(i) The index of H in G is the cardinality of the set of left (or right) cosets, written[G : H ] if finite.

(ii) A left (or right) transversal is a choice of left (or right) coset representatives,with exactly one for each coset.

Normal subgroups

For g, x ∈ G and H 6 G the conjugate of x by g is gxg−1 ∈ G and the conjugate of H byg is gHg−1 6 G. Conjugacy is an equivalence relation.

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Definition 1.14. The subgroupN is normal inG (writtenN�G) if the following equivalentconditions hold:

(i) gN = Ng for all g ∈ G

(ii) gNg−1 = N for all g ∈ G

(iii) gNg−1 6 N for all g ∈ G

(iv) N is a union of conjugacy classes of G.

Examples.

(i) I �G and G�G

(ii) If G is abelian then H �G for all subgroups H (but not the converse)

(iii) If [G : H ] = 2 then H �G.

Proposition 1.15. (cf. Proposition 1.3)

(i) If N �G and H 6 G then N⋂H �H . But M �N , N �G 6⇒ M �G.

(ii) Ni �G ∀i ∈ I ⇒⋂

i∈I Ni �G.

Proposition 1.16. (cf. Proposition 1.4)

If N1 6 N2 6 . . . 6 G with Nn �G ∀n ∈ N then⋃∞

n=1Nn �G.

Let X ⊆ G.

Definition 1.17. (cf. Definition 1.6)

The normal closure 〈〈X〉〉, or 〈〈X〉〉G to be clear, of X in G is⋂N over all N with

X ⊆ N �G. It is the smallest normal subgroup of G containing X .

Note. We have X ⊆ 〈X〉 6 〈〈X〉〉 but 〈〈X〉〉 can be much bigger than 〈X〉.

Proposition 1.18. (cf. Proposition 1.9)

If X ⊆ G then the elements of 〈〈X〉〉 are (e and)

g1xn1

1 g−11 g2x

n2

2 g−12 . . . gkx

nk

k g−1k

for n1, . . . , nk ∈ Z, x1, . . . , xk ∈ X and g1, . . . , gk ∈ G (but not necessarily distinct).

Set products

If A,B 6 G then the set product is AB = {ab : a ∈ A, b ∈ B}. It is not in general asubgroup of G.

Proposition 1.19.

(i) AB 6 G⇔ AB = BA.

(ii) If so then AB = 〈A,B〉.

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(iii) For N �G and H 6 G we have NH = HN .

Homomorphisms

A function θ : G → H for groups G,H is a homomorphism if θ(xy) = θ(x)θ(y) for allx, y ∈ G.

It is an isomorphism if θ is bijective (both surjective (‘onto’) and injective (‘1-1’)), which isequivalent to the existence of an inverse θ−1 : H → G, (where ‘inverse’ here means ‘two-sidedinverse’). In this case, θ−1 is unique and is also a homomorphism. We write G ∼= H , andthey are then the same as abstract groups.

Sets and Functions

If X,Y are sets and f : X → Y is a function then for U ⊆ X the image (or pushforward)of U is f(U) = {f(x) : x ∈ U} ⊆ Y , and for V ⊆ Y the inverse image (or pullback) of V isf−1(V ) = {x ∈ X : f(x) ∈ V } ⊆ X .

Lemma 1.20.

(i) f−1f(U) ⊇ U and is equal if f is injective.

(ii) ff−1(V ) ⊆ V and is equal if f is surjective.

Theorem 1.21.

If θ : G→ H is a homomorphism and A 6 G, B 6 H , we have θ(A) 6 H , θ−1(B) 6 G.

If B �H then θ−1(B) �G. If θ is surjective then A�G⇒ θ(A) �H .

Consequently the image θ(G) of θ is a subgroup of H and the kernel ker θ = θ−1(I) is anormal subgroup of G. In fact we have

Corollary 1.22. For θ : G→ H with A 6 G, B 6 H and K = ker θ we have θ−1θ(A) = AKand θθ−1(B) = B ∩ θ(G).

Quotients and the Isomorphism Theorems

If N � G then the set of (left) cosets forms a group under the well-defined multiplicationxN · yN = (xy)N . We say the group G/N is a quotient of G.

Theorem 1.23 (Homomorphism Theorem).

If θ : G→ H is a homomorphism then G/ ker θ ∼= θ(G).

What are the subgroups of G/N? If N 6 H 6 G then N �H and H/N 6 G/N .

Theorem 1.24 (Correspondence Theorem).

If N �G then the subgroups of G/N are exactly H/N for N 6 H 6 G and the normalsubgroups are exactly L/N for N 6 L�G.

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If π : G→ G/N is the natural homomorphism and H 6 G then π(H) = HN/N .

Theorem 1.25 (Product Isomorphism Theorem).

If H 6 G and N �G then H/(H ∩N) ∼= HN/N .

Theorem 1.26 (Quotient Isomorphism Theorem).

Let N,L�G with N 6 L. Then (G/N)/(L/N) ∼= G/L.

Definition 1.27. Given θ : G → H and N � G with π : G → G/N the natural homo-morphism, we say that θ factors through G/N if there exists θ : G/N → H withθ = θ ◦ π.

We must have N 6 ker θ and this is sufficient by setting θ(gN) = θ(g).

Extensions

If G/N ∼= Q we say that G is an extension of N by Q. We write G is N -by-Q but thisdoes not necessarily determine G uniquely!

Lemma 1.28. If G is A-by-(B-by-C) then G is (A-by-B)-by-C.

Proof. G/A ∼= Q where Q/B ∼= C. Thus B is N/A for some N � G (correspondencetheorem). So C ∼= (G/A)/(N/A) ∼= G/N . 2

Note, ‘⇐’ is false. E.g., A4 is (C2-by-C2)-by-C3, but not C2-by-(C2-by-C3) as A4 does nothave a normal subgroup of order 2.

Group properties

These only depend on the abstract structure of the group. It is always worth asking whethera group property is preserved by (i) Subgroups, (ii) Quotients, (iii) Extensions (namely ifG/N ∼= Q and N and Q both have this property then G does too).

For instance, what about finite, countable, cyclic, abelian, finitely generated, max?

Finite Ctble Cyclic Abelian FG MaxSubgroups y y y y n yQuotients y y y y y1 yExtensions y y n n y2 y3

To see entry 1, that fg is preserved by quotients, note that θ(〈g1, . . ., gn〉) = 〈θ(g1), . . ., θ(gn)〉.

Theorem 1.29. The properties finitely generated and max (entries 2 and 3) are preservedby extensions.

Proof.

2. If G/N = Q, N = 〈n1, . . ., nr〉 and Q = 〈g1N, . . ., gsN〉, then for all g ∈ G, we havegN = g0N for some g0 ∈ 〈g1, . . ., gs〉. So g = g0n. So G = 〈n1, . . ., nr, g1, . . ., gs〉.

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3. H < G. Then H/H ∩N ∼= HN/N 6 G/N . G/N has max, so HN/N is fg. H hasmax, so H ∩N is fg. Use 2 just proved. So H is fg, so G has max.

Actions

We say the group G acts on the set X (on the left) if there is a function ψ : G ×X → Xsuch that

ψ(e, x) = x and ψ(g1, ψ(g2, x)) = ψ(g1g2, x) for all g1, g2 ∈ G, ∀x ∈ X.

Note for each g ∈ G the function x 7→ ψ(g, x) is a permutation of X (put in g−1, g and theng, g−1 to get an inverse).

Equivalently there is a homomorphism ρ : G→ S(X) given by ρ(g)(x) = ψ(g, x).

We say G acts faithfully (effectively) if ρ is injective. We can then unambiguously writeg(x) for ρ(g)(x); we sometimes do this anyway. We say the action is (fixed point) free ifρ(g)(x) = x⇒ g = e (which implies faithful).

For x ∈ X the orbit Orb(x) = {ρ(g)(x) : g ∈ G} ⊆ X and the stabiliser Gx = {g ∈ G :ρ(g)(x) = x} 6 G. The orbits form a partition of X and the action is transitive if there’sone orbit. If y = ρ(g)(x) then Gy = gGxg

−1 so stabilisers are rarely normal.

Theorem 1.30 (Orbit-Stabiliser). If G acts on X then for x ∈ X the sets Orb(x) and{left cosets of Gx} are in bijection.

Example 1.31.

(i) G acts on itself via ρ(g)(x) = gx. This is transitive, and also free so G 6 S(G).

(ii) G acts on itself by conjugation: ρ(g)(x) = gxg−1. Here Orb(x) is the conjugacyclass of x while the stabiliser is the centraliser {g ∈ G : gx = xg} of x in G,written CG(x) or sometimes just C(x) when clear. Note 〈x〉 6 C(x) but it’s notgenerally abelian.

We also have for H 6 G the centraliser CG(H) = {g ∈ G : gh = hg for all h ∈ H} =⋂h∈H CG(h). We set CG(G) = Z(G), the centre of G, which is abelian and normal.

Moreover G acts on the set of its subgroups by conjugation: ρ(g)(H) = gHg−1 6 G. ThenOrb(H) is the set of subgroups conjugate to H and the stabiliser is the normaliser N(H) ={g ∈ G : gHg−1 = H} 6 G. It is the largest subgroup of G in which H is normal. AlsoC(H) 6 N(H).

Automorphisms

An isomorphism (homomorphism) from G to G is an automorphism (endomorphism).

Example 1.32. For any g ∈ G, αg(x) = gxg−1 is an automorphism so H ∼= gHg−1 forall H 6 G. These are the inner automorphisms and form a group Inn(G) undercomposition.

We have αg = e ⇔ g ∈ Z(G) so G/Z(G) ∼= Inn(G). Moreover all automorphisms form agroup Aut(G), with Inn(G) � Aut(G) and the quotient is defined to be the outer automor-phism group Out(G) of G.

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Definition 1.33. The subgroup C of G is characteristic in G if α(C) = C ∀α ∈ Aut(G)(but α(C) 6 C ∀α is enough), so C �G.

Proposition 1.34.

(i) A characteristic in B, B characteristic in C ⇒ A characteristic in C.

(ii) A characteristic in B and B � C ⇒ A� C.

Direct products

We can form the direct product G1 × G2 from groups G1, G2 via (x1, x2) · (y1, y2) =(x1y1, x2y2). This is external, we can also do this internally.

Proposition 1.35. If M,N �G with MN = G and M⋂N = I then θ :M ×N → G given

by θ(m,n) = mn is an isomorphism.

We can extend this drastically:

Definition 1.36. Given groups Gn for n ∈ N, the Cartesian (unrestricted) product is theset of all sequences with nth component an element of Gn and pointwise multiplication;we write

∏n∈N

Gn. The direct product ×n∈NGn is the subgroup of sequences whichare eventually e.

Note: If Gn 6= I for infinitely many n then ×Gn is infinitely generated and∏Gn is uncount-

able.

Semidirect products

Definition 1.37. Given groups G1, G2 and a homomorphism ϕ : G2 → Aut(G1), thesemidirect product G1 ⋊ϕ G2 is the set of ordered pairs with multiplication

(x1, x2) · (y1, y2) = (x1(ϕ(x2)(y1)), x2y2).

Example 1.38.

(i) ϕ the trivial homomorphism gives the direct product.

(ii) Take Z = 〈z〉 (written additively) and C2 = {e, c}. Then Z ⋊ϕ C2 with ϕ(c)(z) =−z is the infinite dihedral group.

Proposition 1.39. (cf. Proposition 1.35)

If H 6 G and N � G with NH = G and N⋂H = I (so G/N ∼= H by 1.25) then

θ : N ⋊ϕ H → G given by θ(n, h) = nh and ϕ(h)(n) = hnh−1 ∈ N is an isomorphism.

So again the internal and external versions are equivalent. There is another point of view:

If G/N = Q with π : G → G/N the natural projection, we say that the extension splits if∃H 6 G such that π : H → Q is an isomorphism.

Now forG = NH a semidirect product, π restricted toH is an isomorphism asH⋂ker π = I.

But a split extension implies H⋂N = I and NH = G as π(H) = π(G), so they’re the same.

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Example 1.40. SL(2,C) = {A ∈ GL(2,C) : detA = 1}� GL(2,C) with quotient C \ {0}.If I2 is the 2 × 2 identity matrix then PSL(2,C) = SL(2,C)/{±I2}. This extensiondoes not split because we have elements g (of order four) in SL(2,C) with π(g) of ordertwo in PSL(2,C), but the only element of order 2 in SL(2,C) is −I2 with π(−I2) = e.

Example 1.41. Let H = S0(Z) 6 S(Z). Recall H is not fg. But H is generated by{(n, n + 1) : n ∈ Z}, as any h ∈ H is a permutation of −N, . . ., N and so h is someproduct of (−N,−N + 1), . . ., (N − 1, N).

Now consider G = 〈H, f〉 where f(z) = z+1. Then (n, n+1) = fn(01)f−n (composedright-to-left), so G = 〈(01), f〉. So G is a 2-generator group, but H is a subgroup that’snot finitely-generated.

2. A brief guide to abelian groups

Theorem 2.1. If G is a finitely-generated abelian group, then

1. (Rational) G ∼= Zr × Cd1× . . .× Cds

with 1 < d1 | d2 | . . . | ds (uniquely).

2. (Primary) G ∼= Zr × P1 × . . .× Pt, where for each Pi we have a different prime psuch that Pi = Cpe1 × . . .× Cpej with 1 6 e1 6 . . . 6 ej (uniquely).

Corollary 2.2. If G is fg abelian, then G has max.

Proof. Z has max, so Zk has max (extensions). Now Zr+s ։ G, which has max.(Note: ‘։’ means a surjection.)

Definition. G is torsion if all elements have finite order, and is torsion-free if only e hasLecture 2finite order. (So the group I is both torsion and torsion-free!)

If G is fg, let d(G) be the minimum size of a generating set.

Proposition 2.3. For a fg group G ∼= Zr×Cd1× . . .×Cds

(as in 2.1), we have d(G) = r+s.

Proof. We have d(G) 6 r + s, so take prime p | d1, and then there exists θ : G ։ Cr+sp .

The image is a vector space over the field Fp, with generating set a spanning set, butθ(generating set for G) is a generating set for image, of dimension r + s.

Examples of non-fg abelian groups: Q, Q/Z, R, R/Z ∼= S1 (see Example 1.5(i)).

If P is a property preserved by subgroups, say that G is locally P if H has P for all fgH 6 G.

Examples.

1. If H =⟨

p1

q1, . . ., pk

qk

⟩6 Q, then H 6

⟨1

q1...qk

⟩, so Q is locally cyclic.

2. Q/Z is locally cyclic (similarly) and pq +Z has finite order, so Q/Z is locally finite

(and torsion).

Imre Leader moment 1. If A × Z ∼= B × Z then we can have A 6∼= B, but if they areabelian then A ∼= B. (We prove the abelian case.)

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Proof. Suppose Γ = A× Y = B × Z where Y, Z ∼= Z.

1. Say A 6 B then Γ/A ∼= Z ։ Γ/B ∼= Z, so have an isomorphism with A = B. (Ifb ∈ B \A then b is in the kernel.)

2. So suppose we have A 6< B and B 6< A. Then A/A ∩B ∼= AB/B 6 Γ/B ∼= Z andB < AB, so A/A ∩B ∼= Z.

Since abelian, we have A ∼= (A ∩B)× Z, and swapping A and B gives the same.

Abelianisation

For G any group with x, y ∈ G, the commutator [x, y] = xyx−1y−1 ∈ G.

Definition 2.4. The commutator/derived subgroup G′ of G is 〈[x, y] : x, y ∈ G〉.

Proposition 2.5. G′�G with G/G′ abelian.

Proof. For α an automorphism of G, we have α([x, y]) = [α(x), α(y)] ∈ G′, so G′ is in factcharacteristic.

In G/G′, xyG′ = yx([x−1, y−1]G′). 2

Corollary 2.6. G/G′ is the largest abelian quotient of G. If G/N is abelian then G′ 6 N ,so G։ G/N factors through G/G′.

Definition 2.7. For any G, the abelianisation G of G is this abelian quotient G/G′.

Note: if G is fg, then (2.1) apply to G.

3. Free groups

Definition 3.1. Let X (say {a, b, . . .}) be a set of symbols and let X−1 ( = {a−1, b−1, . . .})be a set disjoint from X .

We write X±1 = X ∪X−1.

A word in X±1 is a finite sequence of elements of X±1 (letters). Note: ∅ is a word.

Let W be the set of all words, and let W0 ⊂ W be the set of reduced words – thesecontain no subword xx−1 or x−1x.

Aside. We want to define the free group on X as the set of reduced words.

Multiply w1 and w2 by concatenation, cancelling any disallowed subwords that occur.

But we won’t define it this way (some books do), as the proof of associativity is fiddly,because of the cancelling.

Definition 3.2. The free group F (X) on set X = {xi} is the subgroup of S(W0) generatedby elements

χi(w) =

{xiw if w does not start with x−1

i

w′ if w = x−1i w′

This does have image in W0 and inverse χ−1i (w).

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Proposition 3.3. The map M :W0 → F (X) given by replacing x±1i by the respective χ±1

i ,and multiplying in F (X) is an injection.

Proof. If M(w1) =M(w2) for w1, w2 ∈ W0 then note M(wi)(∅) = wi by (3.2). 2

Corollary 3.4. M : W0 → F (X) is surjective, and given a (not necessarily reduced) wordw ∈ W , if we delete all subwords xx−1 and x−1x in any order, we reach a uniquew0 ∈ W0.

Proof. Extend M to a map W → F (X). This is surjective by (1.9), and each deletionreduces the length of w, so we reach some w0 ∈W0. Deletions do not change the groupelement, so M(w) =M(w0), so w0 is unique by (3.3). 2

Theorem 3.5 (Universal property of free groups). Any map f : X → G (G any group)extends uniquely to a homomorphism f∗ : F (X)→ G such that f∗(χi) = f(xi).

Proof. First define f(x−1i ) = f(xi)

−1 ∈ G. Then let f∗(t1. . .tk) = f(t1). . .f(tk), wheretj ∈ {χ

±1i } is represented by letter tj ∈ X±1.

Then f∗ is well-defined by (3.3), (3.4), and is a homomorphism by definition. Any suchhomomorphism must satisfy the property in the first paragraph. 2

Proposition 3.6. If F (X) and F (Y ) are free groups on X and Y then F (X) ∼= F (Y ) ⇐⇒Lecture 3|X | = |Y |.

Proof. (⇐) If f : X → Y is a bijection, then extend to a homomorphism f∗ : F (X) →F (Y ) and (f−1)∗ : F (X) → F (Y ). But (f−1)∗f∗ fixes X and so is the identityhomomorphism (by uniqueness). And the same for f∗(f−1)∗, so f∗ is bijective.

(⇒) For any group G, let SG � G be 〈g2 : g ∈ G〉. Then G/SG is abelian with allnon-identity elements of order 2, and thus it is a vector space over the field F2.

Now, the image of X in F (X)/SF (X) is linearly independent and spans and so |X | isthe dimension. 2

We can define Fn, the free group of rank n when |X | = n. Then F0 = I, F1 = Z, but ifa 6= b ∈ X then ab 6= ba in F (X), so F (X) is non-abelian, and in fact 〈a, b〉 = F2 6 F (X).

Corollary 3.7. Every (fg) group is a quotient of a (fg) free group.

Proof. If G = 〈gi : i ∈ I〉, then take X = {xi : i ∈ I} and use 3.5. 2

Corollary 3.8. If Gθ։ Fn then Fn 6 G and G = ker θ ⋊ Fn.

Proof. Take g1, . . ., gn ∈ G with θ(gi) = xi, then extend. Have f(xi) = gi. We have θ ◦ fis the identity homomorphism (by uniqueness), so f : Fn → G is an isomorphism fromFn to f(Fn).

For K = ker θ, Kf(Fn) = G and K ∩ f(Fn) = I. Use 1.39. 2

Definition 3.9. We say the set S = {si} ⊂ F (X) is a free basis for F (X) if 〈S〉 = F (X)(‘generates’), and for any reduced word w0 6= ∅ on S±1 we have w0 6= id when evaluatedin F (X) (‘free’).

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Proposition 3.10. Free bases for F (X) have cardinality |X |.

Proof. Given free basis S for F (X), define homomorphism f∗ : F (S)→ F (X) by extending

s ∈ Sf7→ s ∈ F (X). Then f∗ is surjective and injective, so F (S) ∼= F (X). Use 3.6. 2

Proposition 3.11. The automorphisms of F (X) are exactly (extensions of) bijective func-tions f : X → S, with S a free basis.

Proof. An automorphism α must send a free basis bijectively to a free basis as α(w(xi)) =w(α(xi)), and it is determined by this. Moreover, given f , extend to f∗ : F (X)→ F (X)uniquely. Then f(X) = S means f is surjective, and S free means f is injective. 2

Definition 3.12. A word w in X±1 is cyclically reduced if it is reduced and the first andlast letters of w are not mutual inverses.

We can write any reduced word w0 = u|w1|u−1 where w1 is cyclically reduced and | means

no cancellation between two words.

Proposition 3.13. If w,w′ are cyclically reduced then they are conjugate iff w′ is a cyclicpermutation of w = ℓ1|ℓ2|. . .|ℓn for ℓi ∈ X±1. I.e., w′ = ℓk|ℓk+1|. . .|ℓn|ℓ1|. . .|ℓk−1 forsome k.

Proof. If w′ = cwc−1 (with c reduced and c 6= ∅), then as w′ is cyclically reduced there existscancellation in either cw or wc−1 but not both. Wlog, say c = d|ℓ−1

1 and w = ℓ1|v, thenw′ = dvℓ1d

−1, but vℓ1 is a cyclic permutation of w. So either d = ∅, or continue. 2

Corollary 3.14. A free group is torsion free.

Proof. Any reduced word w0 6= ∅ can be written as u|w|u−1 for w cyclically reduced. Butfor n > 0, we have wn

0 = u|wn|u−1, so there is no cancellation and so wn0 6= ∅. 2

Example 3.15. Take F2 free on a, b. Let Hn = 〈a, bab−1, . . ., bnab−n〉 6 F2.

By 1.9, h ∈ Hn implies h = bi1ak1bi2−i1ak2 . . .b−im for 0 6 ij 6 n.

Hence bn+1ab−(n+1) /∈ Hn. So H =⋃Hn is not fg, by 1.11.

Corollary 3.16. A fg group G containing a non-abelian free group (⇔ F2 6 G) does nothave max.

Free Products

Definition 3.17. Let {Gλ : λ ∈ Λ} be an indexed family of groups. A reduced sequencein {Gλ} is a finite sequence g1. . .gr where gi ∈

⊔λ∈Λ

Gλ, such that each gj 6= e and no

successive gj, gj+1 are in the same Gλ.

Let A be the set of all sequences in⊔Gλ, and let R be the set of all reduced sequences

(including ∅).

Definition 3.18. The free product ∗λ∈ΛGλ is the subgroup of S(R) generated by elements

11

γ(g,λ) for λ ∈ Λ and (g, λ) ∈ Gλ \ I, where

γ(g,λ)(g1. . .gn) =

{(g, λ)g1. . .gn if g1 /∈ Gλ((g, λ)g1

)g2. . .gn if g1 ∈ Gλ (remove if e)

with γ−1(g,λ) = γ(g−1,λ).

Proposition 3.19. The function f : A → ∗Gλ given by f(g1. . .gn) = γg1 ◦ . . . ◦ γgn restrictsto a bijection from R.

Proof. If g1, g2 ∈ Gλ then γg1 ◦ γg2 = γg1g2 , so gather γs from the same group to get asurjection (remove e).

Then f(g1. . .gn)(∅) = γg1 . . .γgn(∅) = g1. . .gn 2

Note. ιλ : Gλ → ∗λ∈ΛGλ given by ιλ = (g, λ) is a homomorphism, injective.Lecture 4

Theorem 3.20 (Universal property). For any group H and any collection of homomor-phisms θλ : Gλ → H , there exists a unique homomorphism θ : ∗λ∈ΛGλ → H such thatθλ = θ ◦ ιλ for all λ.

Proof. Define θ(g1. . .gn) = θλ1(g1). . .θλn

(gn), where gj ∈ Gλ.

Similarly, this is a homomorphism, unique. 2

If F (X) is free on X = {xi : i ∈ I} then it is ∗i∈IGi, where Gi = {xni : n ∈ Z}, infinite cyclic.

Note G1 ∗ G2 is infinite if G1, G2 6= I, and for non-identity g1 ∈ G1 and g2 ∈ G2 we have(g1g2)

n 6= e for n > 0, and g1g2 6= g2g1.

Suppose X is a topological space and G a group of homeomorphisms of X . We say thatS ⊂ X is a G-packing if g(S) ∩ S = ∅ for all g ∈ G \ I.

Theorem 3.21 (Klein, 1883). If G1, G2 6 Homeo(X) with Gi-packing Si such thatS1 ∪ S2 = X and S1 ∩ S2 6= ∅, then G = 〈G1, G2〉 = G1 ∗G2.

Proof. Note for x ∈ S1, g(x) /∈ S1 for any g ∈ G1 \ I so g(x) ∈ S2 (swap 1 and 2). Takes ∈ S1 ∩ S2 and a reduced sequence g1, . . ., gn with (wlog) gn ∈ G1.

Then gn(s) ∈ S2 \ S1, gn−1gn(s) ∈ S1 \ S2, etc, so g1. . .gn(s) 6= s.

Thus the homomorphism in (3.20), θ : G1 ∗G2 ։ 〈G1, G2〉, is an isomorphism. 2

Example 3.22.

(i) Let X = R2 with reflections a, b in lines x = 0 and x = 12 .

a

0

b

1

2

✛

✛

✛

✲

✲

✲

(detail later!)

Then 〈a, b〉 = 〈a, ba〉 = C2 × C2. This is the infinite dihedral group (1.38) asba(x) = x+ 1 and a(ba)a−1 = (ba)−1.

12

(ii) Mobius Transformations. Let f(z) =az + b

cz + dwith a, b, c, d ∈ C, ad− bc 6= 0.

This is a bijection of C ∪ {∞}, with inverse f−1(z) =dz − b

−cz + a.

Let f(z) = z + 2 and g(z) =z

2z + 1.

.

.........

..........

.........

..................

........................................................

..................

.........

..........

......

.............

..........

.........

..................

........................................................

..................

.........

..........

......

...

−1 0 1

Proposition 3.23. Let F =

(1 20 1

), G =

(1 02 1

)∈ SL(2,Z).

Then F,G freely generate F2.

Proof. f, g ∈ PSL(2,Z)π← SL(2,Z)/±I2, and 〈f〉, 〈g〉 infinite so 〈f, g〉 = F2 by (3.21).

Now, send f, g 7→ F,G and extend using (3.5) to θ with π ◦ θ = id, so θ(F2) = F2. 2

At this point, there was a handout.

Topics in Infinite Groups – Topological Background

We follow the books Lee and Hatcher as in the course summary.

X is a topological space which is always assumed to be path connected and locally pathconnected:∀x ∈ X and ∀ open U ⊆ X with x ∈ U , we have an open path connected set P withx ∈ P ⊆ U .

Let f, g : X → Y be continuous maps (and let A ⊆ X).A homotopy between f and g (relative to A, written rel A) is a continuous map H :X × [0, 1]→ Y with H(., 0) = f and H(., 1) = g (with H(a, t) = f(a) ∀a ∈ A and ∀t ∈ [0, 1],so f and g must agree on A). This is an equivalence relation, written f ≃ g (or f ≃ g rel A).A (strong) deformation retraction of X onto A ⊆ X is a homotopy rel A from IdX tor : X 7→ X with r(X) ⊆ A and r|A = IdA.

For x0 ∈ X the fundamental group π1(X, x0) is the group of homotopy classes of closedpaths γ with start and end at x0 (in other words γ : [0, 1] → X is continuous with γ(0) =γ(1) = x0). Changing the basepoint x0 “doesn’t matter” as we obtain an isomorphic group,written π1(X).

Any continuous map f : X → Y induces a homomorphism f∗ : π1(X) → π1(Y ). If X ishomeomorphic to Y (which is also written X ∼= Y ) then π1(X) ∼= π1(Y ).

X is homotopy equivalent to Y (written X ≃ Y ) if there exist continuous maps f : X → Yand g : Y → X such that gf ≃ IdX and fg ≃ IdY . If so then f∗ : π1(X) → π1(Y ) is anisomorphism.

X is contractible if X ≃ {x}, which implies that X is simply connected (meaning thatπ1(X) is the trivial group).

13

Seifert - Van Kampen Theorem (Hatcher Theorem 1.20 in the case where the intersec-tions are simply connected): If X = ∪αAα where all of the Aα are path connected and openin X , with x0 ∈ ∩αAα and each pairwise and triple intersection Aα1

∩Aα2, Aα1

∩Aα2∩Aα3

is simply connected thenπ1(X) ∼= ∗α π1(Aα).

If A ⊆ X and ι : A → X is the inclusion map then sadly ι∗ : π1(A) → π1(X) might not beinjective or surjective (see picture). But in a deformation retraction X ≃ A (by taking f = rand g as inclusion of A = Y in X) so X and A have isomorphic fundamental groups.

A continuous map p : X → X is a covering map (with X a covering space for X) if∀x ∈ X there exists an open neighbourhood V of x with p−1(V ) a disjoint union of open

sets in X, each of which is mapped homeomorphically by p onto V . As X is connected, thecardinality of p−1(x) (which is non-zero as we assume X,X 6= ∅) is constant: the degree ornumber of sheets.

Given a path γ : [0, 1]→ X and a point x above γ(0), there exists a unique path γ : [0, 1]→ X

lifting γ (namely pγ = γ). Moreover p∗ : π1(X)→ π1(X) is injective.

Now assume X is locally contractible.

Classification of Coverings Theorem: For each H 6 π1(X) there exists a cover p : X →

X with p∗(π1(X)) = H (this cover is “unique”). The degree is the index of H in π1(X).

A deck transformation of the cover p : X → X is a homeomorphism ϕ of X such thatpϕ = p. They form a group D and for any x0 ∈ X the action of D on p−1({x0}) ⊆ X

is transitive if and only if p∗π1(X, x0) is normal in π1(X, x0), in which case we say p is aregular cover.

End of handout

Definition 3.24. A graph Γ (a 1-dim CW complex) is made up of a set V of vertices withdiscrete topology and a set E of edges Iα (with intIα = eα), each edge being a copy of[0, 1]. The edge endpoints are attached to points in V , giving f0, f1 : E → V so that

Γ = V⊔

α

Vα

/Iα(0) = f0(Iα), Iα(1) = f1(Iα)

Notes.

1. Subset S of Γ is open (closed) ⇐⇒ S ∩ eα open (closed) in eα for all α.

2. Γ is locally path-connected and locally contractible.

(Here, connected ⇐⇒ path-connected.)

3. Basic open sets for Γ are these and open intervals in eα.

14

A subgraph of Γ is a union ∆ of edges and vertices such that eα ⊂ ∆ =⇒ eα ⊂ ∆.Lecture 5

Proposition 3.25. If X is a covering space of the (connected) graph Γ then X is a graphwith vertices and edges the lifts of those in Γ.

Proof. Let V (X) = p−1(V ) for p : X → Γ, and for edges take map from Iα into Γ and v

above f0(Iα) to get a unique lift. Thus X is a graph, and the topologies agree on basicopen sets, so are the same. 2

If S is a simple graph, an edge path in S is a finite sequence of vertices such that any twoconsecutive vertices span an edge in S. A cycle v0, . . ., vn is a closed (v0 = vn) edge path.A tree in S is a connected simple graph with no reduced cycles.

Proposition 3.26. Given any connected graph Γ and any vertex v0 in Γ, there exists asubgraph ∆ ≃ {v0} such that ∆ contains all of V (Γ).

Proof. Let Γ0 = {v0} ⊂ Γ1 ⊂ Γ2 ⊂ . . . be a sequence of subgraphs where Γi+1 is Γi togetherwith edges eα for all eα ∈ Γ \ Γi with an endpoint in Γi.

Then⋃Γi is open (neighbourhood of point in Γi is in Γi+1) and closed in Γ(union of

closed edges), so is Γ.

Next, ∆0 = Γ0 ⊂ ∆1 ⊂ ∆2 ⊂ . . ., where ∆i has same vertices as Γi but Γi+1 is Γi withone edge from each v ∈ Γi+1 \ Γi to ∆i.

Now, ∆i+1 deformation retracts onto ∆i, so set ∆ =⋃∆i (contains V (Γ)), and ∆

deformation retracts to ∆0 by performing homotopies in the interval [1/2i+1, 1/2i],with homotopy ∆→ ∆ being continuous. 2

Corollary 3.27. A tree T is contractible.

Proof. Apply (3.26) to get ∆ ⊂ T , contractible. If edge e is spanned by v 6= w not in ∆,then take reduced edge paths v0. . .uv and wx. . .v0 in ∆. Note u 6= w and v 6= x ase /∈ ∆. So v0. . .vw. . .v0 is a reduced cycle in T . //\\ 2

Theorem 3.28. The fundamental group of a (connected) graph Γ is free.

Proof. Let T ⊂ Γ (wlog simple) be ∆ in (3.26), so T is a tree, with {eα : α ∈ A} the edgesin Γ \ T , and wα, w

′α endpoints of eα. For each α, take a reduced cycle cα from v0

including wαw′α so that

cα = v0u1u2. . .wα w′α. . .v1v0

First assume one edge. If c has ui = vj in it, then replace with ui. . .vj which stillcontains wαw

′α, else it is a reduced cycle in T .

So now have (renamed) v0. . .v0 with no repeats, giving a loop L with L ∼= S1, soπ1(L) ∼= Z.

But Γ ≃ L as the components K of Γ \ L and in T and so are trees. If a 6= b ∈ K ∩ L,then go from a to b in K and then back via L (not via wαw

′α) to get a reduced cycle

in T .

Thus by (3.27) we can deformation retract each K (simultaneously) onto the point inK ∩ L, giving π(Γ) = Z.

15

For general Γ, let mα be the ‘midpoint’ of eα and set Aα =(Γ \

⋃a∈Ama

)∪mα.

Then Aα is open in Γ and path-connected with Aα ∩ Aβ(∩Aγ) = Γ \⋃

ama, whichdeformation retracts onto T , so (3.27) implies simply connected. Also, Aα retracts ontoT ∪ eα, so π1(Aα) ∼= Z, giving (by Seifert-Van Kampen) that π1(Γ, v) ∼= ∗αZ. 2

Corollary 3.29. For every free group F (X), there exists a simple graph Γ with π1(Γ) =F (X).

Proof. Take a loop ℓα for each xα ∈ X , all joined at v, turn it into a triangle and then use(3.28). 2

Theorem 3.30 (Nielsen-Schreier). A subgroup of a free group is free.

Proof. For H 6 F (X), use (3.29), subgroups ↔ coverings (X → X , π(X) = H), so (3.25)and (3.28) give H is free. 2

Theorem 3.31 (Nielsen-Schreier index formula). If H has index i in the free group Fn

then H has rank i(n− 1) + 1.

Proof. For finite graph Γ, define χ(Γ) = |V | − |E|. If p : Γ → Γ has degree i, thenχ(Γ) = iχ(Γ), by (3.25).

Take Γ in (3.29) with π1(Γ) = Fn and χ(Γ) = 1 − n. For H of index i in Fn with

π1(Γ) = H , we have χ(Γ) = i(1− n).

Now in (3.28), note χ(∆) = 1, so the set of edges {eα} for Γ \∆ numbers i(n− 1) + 1,so this is the rank of H . 2

Notes.

1. F2 6 F3 and F3 6 F2.

2. If N � F (X), it can be shown that if N has infinite index (and N 6= I) then Nhas infinite rank.

4. Presentations of groupsLecture 6

By (3.7), we can write any group G as F (X)/N , where (the image of X) is a generating setfor G.

Definition 4.1. A presentation 〈X |R〉 for a group G is a set X and a subset R of F (X)such that G ∼= G(X)/〈〈R〉〉.

Elements of X are generators, and elements of R are relators.

Theorem 4.2 (van Dyck, ‘quotient = more relators’). If G = 〈X |R〉, then Q is a quo-tient of G ⇐⇒ Q ∼= 〈X |R ∪ S〉.

Proof.

(⇐) Have that F (X)π−→ 〈X |R ∪ S〉 factors through G as 〈〈R〉〉 6 〈〈R ∪ S〉〉.

16

(⇒) Now say Q = G/L, for G = F (X)/M , where M = 〈〈R〉〉. Then L = NM/M forN � F (X), so Q ∼= F (X)/NM .

Take any S with 〈〈S〉〉 = N , then 〈〈R ∪ S〉〉 = 〈〈N ∪M〉〉 = NM , so Q ∼= 〈X |R ∪ S〉. 2

Definition 4.3. G is finitely presented (f.p.) if G = 〈x1, . . ., xn|r1, . . ., rm〉. I.e., finitelymany generators and finitely many relators.

Note. If N 6= I is of infinite index in F (X), then N is not finitely generated.

Proposition 4.4 (B. Neumann, 1937). If G = 〈x1, . . ., xn | r1, . . ., rm〉 and also G =〈y1, . . ., yk | s1, s2, . . .〉, then G = 〈y1, . . ., yk | s1, . . ., sℓ〉 for some ℓ.

Proof. We set yi = vi(x1, . . ., xn) and xi = wi(y1, . . ., yn), as we think of the xi and yi aselements of G.

Thus yi = vi(w1(yj), . . ., wn(yj)) and ri(w1(yj), . . ., wn(yj)) = e in G.

Let N = 〈〈s1, s2, . . .〉〉� F ({yi}) and form

G = 〈y1, . . ., yk | yi = vi(w1(yj), . . ., wn(yj)), ri( ¯. . .)〉

So there exists θ : G։ F ({yi})/N , as each relator holds in G, as F ({yi}) ։ F ({yi})/Nfactors through G.

Now take ϕ : 〈x1, . . ., xn|r1, . . ., rm〉։ G by ϕ(xi) = xi ≡ wi(y1, . . ., yk).

Then θϕ(xi) = xi, with ϕ surjective, so θ is an isomorphism. So N = 〈〈finite set〉〉.

But N =∞⋃i=1

〈〈s1, . . ., sℓ〉〉. 2

Proposition 4.5. G = 〈a, b∣∣[a2n+1ba−(2n+1), b], n ∈ N〉 is not finitely presented.

Proof. Notation: let cn = [a2n+1ba−(2n+1), b].

In Aj , j odd, let α = (12. . .j) and β = (123). Then αkβα−k commutes with β if3 6 k 6 j − 3, but not if k = j − 2.

Thus in A2ℓ+3, we have that relations c1, . . ., cℓ−1 = e hold, but cℓ 6= e. So cℓ /∈〈〈c1, . . ., cℓ−1〉〉 6 F2 (via ‘a 7→ α, b 7→ β’).

Now use (4.4). 2

Aside. The same Neumann paper shows that there are uncountably many finitely generatedgroups up to isomorphism, but only countably many finitely presented groups.

Examples 4.6.

0. Finite groups are f.p.

1. Fn = 〈x1, . . ., xn|〉.

2. Cn = 〈x|xn〉, but 〈x, y|x14, y21, x3 = y4〉 = C7.

17

Proposition 4.7. If N = 〈ni|rj〉 and H = 〈hk|sℓ〉, then G = N ⋊ϕ H has presentation

P = 〈ni, hk|rj , sℓ, hknin−1k ϕ(hk)(ni)〉

Proof. As hkni = (element of N)hk, any p ∈ P can be written as v(some ni)w(some hk).

Define θ : P ։ G via extending F ({ni} ∪ {hk}) ։ G.

If θ(p) = e, then θ(v(ni))︸ ︷︷ ︸∈N

θ(w(hk))︸ ︷︷ ︸∈H

= e. But N ∩H = I, so v(ni), w(hk) = e in P .

So θ is an isomorphism. 2

If G = 〈x1, . . ., xn|r1, r2, . . .〉 and p is prime, let rj ∈ Fnp be the exponential sum vector modulo

p of rj . That is, (#x1’s−#x−11 ’s) + . . . mod p.

Then Fn ։ Cnp , r 7→ r, is a homomorphism.

Proposition 4.8. If S = span{rj} 6= Fnp then G 6= I.

Proof. There exists a linear functional f : Fnp ։ Fp with S 6 ker f .

So Fn ։ Fnp

f։ Fp factors through G. So G։ Fp. 2

Example 4.9. If #relators < #generators in a finite presentation, then G ։ Fp for all p,so it’s infinite.

Proposition 4.10. If G = 〈xi|rk〉 and 〈yj |sℓ〉, then G ∗H = 〈xi, yj |rk, sℓ〉.

Proof. Call the RHS G|H , and let G|Hθ։ G ∗H (where θ fixes xi, yj).

Also, there is a homomorphism from G into G|H fixing each xi and from H similarly.Extend to ϕ : G ∗H → G|H by (3.20). Now compose with θ, and extend. Uniquenesssays θϕ = id, but ϕ is surjective, so θ is an isomorphism 2

We can now extend the table on page 5.

Free FPSubgroups y n (e.g., F2)Quotients n n (e.g., (4.5))Extensions n y

Theorem 4.11 (P. Hall). If H �G and N,G/N are f.p., then G is f.p.

Proof. Let N = 〈x1, . . ., xn|r1, . . ., rm〉 and G/N = 〈y1, . . ., yℓ|s1, . . ., sℓ〉.

Take g1, . . ., gℓ with giN = yi, so have a generating set. Now take relations1 ri = e,si(g1, . . ., gℓ) = ti(x1, . . ., xn) ∈ N , and gjxig

−1j = uij(x1, . . ., xn) ∈ N , and g−1

j xigj =vij(x1, . . ., xm).

Take G = 〈x1, . . ., xm, g1, . . ., gℓ〉, with above relations.Lecture 7

1A relation is an expression of the form w = u, compared with a relator which is of the form wu−1.

18

Then Gθ։ G. Let K = ker θ, and let N = 〈x1, . . ., xn〉. Then N �G by uij , vij . Now

restrict θ to N where it is an isomorphism. So K ∩N = I.

Now as θ(N) = N , we obtain θ0 : G/N → G/N which is also injective as above. Butker θ0 = KN/N ∼= K/K ∩N , so K = I. 2

Free products with amalgamation; HNN extensions

Definition 4.12. Let G, H be groups with A 6 G, B 6 H such that ϕ : A → B is anisomorphism.

The free product with amalgamation is the group G∗H/〈〈a = ϕ(a) for all a ∈ A〉〉.(Note it’s enough to use a generating set for A in the normal closure.)

Definition 4.13. If G is any group with A,B 6 G such that ϕ : A→ B is an isomorphism,then the HNN extension G∗ϕ is the group G ∗ 〈t〉/〈〈tat−1 = ϕ(a)〉〉.

(So 〈t〉 ∼= Z, and t is called a stable letter.)

Note that ta = ϕ(a)t and t−1b = ϕ−1(b)t−1. So we can move a’s (resp. b’s) to the left of t(resp. t−1) in G∗ϕ.

Choose right transversals TA and TB for A and B in G such that both include e. A normalform is a sequence g0t

ε1g1tε2 . . .tεngn (for n > 0 and εi = ±1) such that each gi ∈ G, with

εi = +1⇒ gi ∈ TA and a εi = −1⇒ gi ∈ TB, and there is no subsequence tεet−ε.

Every element of G∗ϕ can be put into normal form (work from right to left).

Theorem 4.14 (Normal form for HNN extensions). Every element inG∗ϕ has a uniquenormal form.

Proof. Define ρ : G∗ϕ → S(N ), where N is the set of normal forms as follows.

1. ρ(g)(g0tε1 . . .gn) = gg0t

ε1 . . .gn for g ∈ G.

2. If ε = −1 and g0 ∈ A then ρ(t)(g0t−1. . .gn) = (ϕ(g0)g1)t

ε2 . . .gn.

Otherwise, ρ(t)(g0tε1 . . .gn) = ϕ(a)tg0t

ε1 . . .gn, where g0 = ag0 for g0 ∈ TA.

This has inverse ρ(t−1)(g0tg1. . .gn) = (ϕ−1(g0)g1)tε2 . . .gn (if g0 ∈ B), and otherwise

g0tε1 . . .gn 7→ ϕ−1(b)t−1g0t

ε1 . . .gn (where g0 = bg0 for g0 ∈ TB).

Also, ρ(a) = ρ(t−1)ρ(ϕ(a))ρ(t), so ρ is a well-defined homomorphism on G∗ϕ, andρ(g0t

ε1 . . .gn)(e) = g0tε1 . . .gn for normal forms. 2

Say that a sequence g0tε1 . . .gn is reduced is there is no subsequence tgit

−1 for gi ∈ A, ort−1git for gi ∈ B. (Such a sequence is sometimes called a pinch.)

Corollary 4.15 (Britton’s Lemma). G embeds in G∗ϕ by g 7→ g, and if for n > 1 theform g0t

ε1 . . .gn is reduced then it 6= e in G∗ϕ.

Proof. g (6= e) and e are both in normal form, so g 6= e in G∗ϕ by (4.14). On changing areduced sequence into a normal form, no t±ε cancel, so it becomes g′0t

ε1 . . .g′n. 2

19

Corollary 4.16 (Torsion in HNN extensions). If γ ∈ G∗ϕ has finite order then γ isconjugate to some g ∈ G.

Proof. If γ = g0tε1 . . .tεngn (with n > 1) is a reduced sequence and tεngng0t

ε1 is not a pinch,then γk is reduced, so 6= e by (4.15).

Otherwise, replace γ by tεngnγ(tεngn)

−1 = gtε2 . . .tεn−1gn−1, which is either reduced orin G. Now repeat. 2

Imre Leader moment 2. There is an infinite group where every non-identity element isconjugate.

Proof. Take G to be countably infinite, torsion-free – e.g., G = Z.

Let {g0, g1, . . ., } be an enumeration of the non-identity elements.

Form HNN extensions. First, G1 = 〈G, t1 : t1g0t−11 = g1〉. (Taking A = 〈g0〉 and

B = 〈g1〉.)

Note that G embeds in G1 by (4.15), so g0 and g2 still have infinite order in G1.

So we can form G2 = 〈G1, t2 : t2g0t−12 = g2〉, . . . , Gn = 〈Gn−1, tn : tng0t

−1n = gn〉.

We have G1 6 G2 6 . . ., so we can let Γ(G) =⋃Gn.

Then Γ is countably infinite and torsion-free by (4.16).

Note any two non-identity elements of G are conjugate in Γ(G) = Γ1.

New set Γ2 = Γ(Γ1), . . . , Γm = Γ(Γm−1), . . . . And finally form ∆ =⋃Γm.

Now, if non-identity γ, δ ∈ ∆ then γ, δ ∈ Γm for some m, so are conjugate in Γm+1 andhence are ∆. 2

Note. D. Osin (2010) showed that there are finitely-generated examples. It is unknownwhether there exist finitely-presented examples.

For free products with amalgamation, G ∗ϕ H , say element c1. . .cn ∈ G ∗ H is a-reduced(for ‘amalgamated-reduced’) if for n > 1 no ci is in A or B. We can turn any c ∈ G ∗H intoan a-reduced element by ‘absorbing’ elements of A or B.

Corollary 4.17. If c1. . .cn is a-reduced then it equals e in G∗ϕH . (Using G,H → G∗ϕH .)

Proof. Let F = (G ∗H) ∗ 〈t〉/〈〈tat−1 = ϕ(a)〉〉, an HNN extension of G ∗H .

Let ψ : G ∗ϕ H → F be defined by ψ(g) = tgt−1 and ψ(h) = h. This is well-defined.

For an a-reduced element, if c1 ∈ A then c1 7→ ϕ(a) 6= e (normal form). Otherwise, itmaps to a reduced sequence in F (HNN), so done by (4.15). 2

20

5. Soluble, polycyclic, nilpotent groupsLecture 8

See D.J.S Robinson, ‘A course in the theory of groups’, Springer (GTM 80)

Definition 5.1. The group G is soluble/solvable if there exists a sequence of subgroups

I = Gn �Gn−1 � . . .�G1 �G0 = G

such that Gi/Gi+1 is abelian.

Theorem 5.2. The soluble property is preserved by subgroups, quotients and extensions.

Proof. Let G be soluble.

(i) If H 6 G, then as Gi+1 �Gi, we have Gi+1 ∩ (H ∩Gi) 6 H ∩ Gi by (1.15), and(H ∩Gi)/(H ∩Gi+1) ∼= (H ∩Gi)Gi+1/Gi+1 6 Gi/Gi+1, so it’s abelian.

(ii) If H�G, then Gi+1N/N �GiN/N with quotient Gi(Gi+1N)/Gi+1N ∼= Gi/(Gi∩Gi+1N), which is a quotient of Gi/Gi+1, so it’s abelian.

(iii) G/N soluble implies we have N 6 Gi+1 6 Gi 6 G with Gi+1/N � Gi/N , soGi+1�Gi andGi/Gi+1 abelian. Now put together with I = Nn�Nn−1�. . .�N0 =N , with Ni/Ni+1 abelian. 2

Definition 5.3. For any group G, the derived series is the sequence of subgroups

G = G(0)> G(1)

> G(2)> . . ., where Gi+1 = (G(i))′.

Each G(i) is characteristic in G.

Proposition 5.4. G is soluble ⇐⇒ its derived series terminates at I. If so, then it haslength that of a smallest series with abelian quotients.

Proof. G(i)/G(i+1) = G(i)/(G(i))′, so abelian. Now say Gi is as in (5.1) and assume thatG(i) 6 Gi. As H 6 G implies H ′ 6 G′, we have G(i+1) 6 (Gi)

′. But Gi/Gi+1 isabelian, so (Gi)

′ 6 Gi+1. 2

So in (5.1), there does exist a series where each Gi �G.

We say a group G is perfect if G = G′. A perfect group G(6= I) is not soluble, and if aquotient has G։ Q։ A with A abelian then A = I, so Q is also perfect.

E.g., if G is simple, non-abelian, then G is perfect. E.g., A5.

Corollary 5.5. If G contains a non-abelian free subgroup then G is not soluble.

Proof. If G is soluble and F2 6 G then there is N such that F2/N ∼= A5, which would besoluble. //\\ 2

Polycyclic groups

Definition 5.6. The group G is polycyclic if if there exists a sequence of subgroups

I = Gn �Gn−1 � . . .�G1 �G0 = G

such that Gi/Gi−1 is cyclic.

21

So polycyclic =⇒ soluble.

Theorem 5.7. The polycyclic property is preserved by subgroups, quotients and extensions.

Proof. Exactly as (5.2), with ‘abelian’ replaced by ‘cyclic’ throughout. 2

In fact, if property P is ‘soluble: yes; quotients: yes; extensions: no’, then (5.2) shows that‘poly-P ’ has all three.

Corollary 5.8. Finitely-generated abelian groups A are polycyclic.

Proof. (2.1) expresses A as a direct product of cyclic groups, so build out of extensions. 2

Theorem 5.9. G polycyclic ⇐⇒ G soluble and has max.

Proof. (⇒) A cyclic group is soluble and has max, and both of these are preserved byextensions.

(⇐) Each Gi in (5.1) is finitely generated, so each Gi/Gi+1 is polycyclic by (5.8), andthis is preserved by extensions. 2

Corollary 5.10. If G is polycyclic and H 6 G, then H is finitely presented.

Proof. A cyclic group is finitely presented and this is preserved by extensions by (4.11), soG is finitely presented. And H 6 G implies H polycyclic. 2

Example 5.11. The subgroup D 6 Q of dyadic rationals {n/2i : n ∈ Z, i > 0} is not cyclic,hence infinitely generated as Q is locally cyclic.

Let B = D ⋊ϕ Z, where for t ∈ Z we define ϕ(t) to be the automorphism d 7→ 2d ford ∈ D. Then B is metabelian (‘abelian-by-abelian’) and generated by 1 ∈ D and t ∈ Z.

Write the Z part as 〈t〉. Then all elements of B have the form (n/2i, tj), with

(0, tk)(n/2i, 1)(0, t−k) = (2kn/2i, 1),

so D 6 〈1, t〉 by taking k > i. (Where ‘1’= (1, 1) and ‘t’ = (0, t).)

So this is a group which is finitely generated but doesn’t have max, so isn’t polycyclic.

What about a finitely presented example?

Let G = 〈a, b : bab−1 = a2〉. Then send a 7→ (1, 1) and b 7→ (0, t), and extend to asurjective homomorphism θ : G→ B.

Now, G/G′ ∼= Z, generated by bG′.

As ba = a2b and ba−1 = a−2b, and ab−1 = b−1a2 and a−1b−1 = b−1a−2, given anyword in a, b, we can move the positive powers of b past a±1 to the right, and of b−1 tothe left.

Thus any g ∈ G will have the form b−maℓbn for ℓ ∈ Z and m,n ∈ N.

Now θ(g) = (ℓ/2m, tn−m), and if this equals eB = (0, 1), then ℓ = 0, m = n. Hence θis also injective, and so G ∼= B.

22

We can also do this with matrices. Let a =

(1 10 1

)and b =

(2 00 1

)∈ GL(2,Q).

Then b−maℓbn =

(2n−m ℓ2−m

0 1

)and bab−1 = a2.

Proposition 5.12. We have infinite finitely-presented groups G with finitely-presented H 6

G and g ∈ G \H such that gHg−1 � H .

Proof. Take G as in (5.11), with H = 〈a〉. Then bHb−1 = 〈a2〉 � H . 2

Note. In G we have . . . � g2Hg−1 � gHg−1 � H � g−1Hg � g−2Hg2 � . . ..

So any example in (5.12) cannot have max (or min).

Nilpotent groupsLecture 9

Definition 5.13. A group G is nilpotent if there exists a sequence of subgroups

I = Gn 6 Gn−1 6 . . . 6 G0 = G

such that Gi �G and Gi/Gi+1 is in the centre of G/Gi+1.

Such a series (if it exists) is called a central series.

Note. Abelian =⇒ nilpotent =⇒ soluble, but G(6= I) nilpotent =⇒ Z(G) 6= I.

So S3 is polycyclic, not nilpotent, so ‘nilpotent’ is not preserved by extensions.

Theorem 5.14. The nilpotent property is preserved by subgroups, quotients and directproducts.

Proof.

(i) As in (5.2), we want (H ∩Gi)/(H ∩Gi+1) to be in the centre of H/(H ∩Gi+1) ∼=HGi+1/Gi+1. The former becomes (H ∩ Gi)Gi+1/Gi+1, which is in Gi/Gi+1, socommutes with G/Gi+1.

(ii) We need GiN/Gi+1H to be in the centre of G/Gi+1N , but xgx−1g−1 ∈ Gi+1 forx ∈ Gi, g ∈ G, so xng(xn)−1g−1 ∈ Gi+1N .

(iii) Given Gi 6 G and Hj 6 H as in (5.13), we have Gi × G 6 G × H with (Gi ×H)/(Gi+1 ×H) in the centre of (G×H)/(Gi+1 ×H), so we use

I × I 6 I ×Hn 6 I ×Hn−1 6 . . . 6 I ×H 6 Gn ×H 6 Gn−1 ×H 6 . . . 6 G×H

For H1, . . ., Hn 6 G, define [H1, H2] = 〈[h1, h2]〉 and [H1, . . ., Hn] =[[H1, . . ., Hn−1], Hn

].

For any G, the lower central series of G is

γ1(G) = G > γ2(G) = [G,G] > γ3(G) = [[G,G], G] > . . . > γi+1(G) = [γi(G), G], . . .

Lemma 5.15. G is nilpotent ⇐⇒ the lower central series terminates at I.

Proof. (⇐) Each γi(G) is characteristic in G, and γi(G)/γi+1(G) is in Z(G/γi+1G), so weget a central series.

23

(⇒) Assume γi(G) 6 Gi−1 (some central series) for i > 1, then γi+1(G) = [γi(G), G] 6[Gi−1, G] which (as a central series) is in Gi. 2

Theorem 5.16 (Baer). If G is nilpotent and finitely generated then γi(G) is finitely gen-erated.

Proof. Suppose first that γ3(G) = I and G = 〈x1, . . ., xn〉 is a symmetric generatingset (i.e., closed under inverses). In any group we have [y, x]x[y, z]x−1 = [y, xz] andz[x, y]z−1[z, y] = [zx, y]. So, in G we get [g, xh] = [g, x][g, h] and [gx, y] = [x, y][g, y],so any element of γ2(G) is a finite product

∏[xj , xk].

This also shows for any G that γ2(G)/γ3(G) is generated by {[xj , xk]} modulo γ3(G).

Now assume that anything in γi(G)/γi+1(G) is a product of {[xj1 , . . ., xji ] ≡ i[x]}.

In γi+1(G)/γi+2(G), suppose g = xih, then[i[g], g

]=

[i[g], xℓg

]=

[i[g], xℓ

][i[g], h

], as

[· · · ]︸︷︷︸i+1

is in the centre.

So, for [g1, . . ., gi+1] ∈ γi+1(G), we can write it as(∏

r

[[g1, . . ., gi], xr

])γ for γ ∈

γi+2(G).

This equals(∏

r

(∏s

[i[xs]β, xr

]))γ′, for β ∈ γi+1(G), γ

′ ∈ γi+2(G).

E.g.. [i[xs]β, xr ] = [β, xr ][i[xs], xr]δ, some δ ∈ γi+2(G).

And this equals∏

r

∏s

[i[xs], xr

]modulo γi+2(G).

So if γn+1(G) = I then γn(G) is finitely generated, and γn−1(G)/γn(G) finitely gener-ated implies γn−1(G) finitely generated, etc. 2

Corollary 5.13. Finitely-generated nilpotent groups G are polycyclic and have max.

Proof. γi(G)/γi+1(G) is abelian and finitely generated, so have max and polycyclic, exten-sions. 2

6. Finite index subgroups and virtual properties

If a subgroup H has finite index in G (i.e., finite number of left cosets), we write H 6f G,with index [G : H ].

Lemma 6.1.

(i) If H 6f G and H 6 J 6 G then H 6f J 6 fG.

(ii) If J 6f H 6f G then J 6f G with [G : J ] = [G : H ][H : J ].

(iii) If H 6f G and S 6 G then H ∩ S 6f S with index 6 [G : H ], with equality iffSH = G, and dividing [G : H ] if SH 6 G.

(iv) If H 6f G and J 6f G then H ∩ J 6f G with [G : H ∩ J ] 6 [G : H ][G : J ].

Proof.

24

(i) Check

(ii) Suppose g1, . . ., gk is a (left) transversal for H in G, and h1, . . ., hℓ is for J in H .Then G =

⋃i,j gihjJ . Check that if gihjJ = gi′hj′J then same coset.

(iii) Take G = g1H ∪ . . . ∪ gkH and throw away any giH with giH ∩ S = ∅. (Thishappens ⇐⇒ SH 6= G.) Then g1(H∩S), . . ., gk(H∩S) are disjoint. Check unionU .

Replace gi with si ∈ S as above to form si(H∩S), etc. Then U ⊂ S and if s ∈ giHthen s = sih ∈ si(H ∩ S), so if SH is a subgroup then H 6 SH 6 G, so use (i),(ii) and we’ve shown that [SH : H ] = [S : H ∩ S].

(iv) [G : H ∩ J ](ii)= [G : J ][J : H ∩ J ]

(iii)

6 [G : J ][G : H ]. 2

A corollary of (iv) above is the following.Lecture 10

Theorem 6.2 (Poincare). A finite intersection of finite index subgroups has finite index.

Lemma 6.3. If [G : H ] = k then for any g ∈ G, there exists i with 1 6 i 6 k such thatgi ∈ H .

If H �G then we can take i | k (or even i = k).

Proof. The left cosets H, gH, . . ., gkH cannot all be distinct, so giH = gjH for some 1 6

i < j 6 k, and then gj−i ∈ H with 1 6 j − i 6 k.

If H �G then gH has order dividing k = |G/H |. 2

Proposition 6.4. Let θ : G։ H be a surjective homomorphism.

(i) If B 6f H then θ−1(B) 6f G. In fact, [G : θ−1(B)] = [H : B].

(ii) If A 6f G then θ(A) 6f H with [H : θ(A)] | [G : A].

Proof.

(i) Have H = h1B ∪ . . . ∪ hkB, and can take gi with θ(gi) = hi to get a transversalfor C = θ−1(B) in G. For g ∈ G, say θ(g) = hib, then g

−1g ∈ C, and pulling backsends disjoint sets to disjoint sets with θ−1(hiB) = giC.

(ii) θ−1(θ(A)) = KA 6f G for K = ker θ with [G : KA] = [H : θ(A)] by (i). But[G : KA] divides [G : A] by (6.1)(ii). 2

The Regular Representation

Any group G acts on itself by (left) multiplication. Now let H be any subgroup and L be theset of left cosets of H in G. The (left) regular representation ρ of G on L is the actionof G given by ρ(g)(xH) = gxH .

Note orb(H) = L and the stabiliser of H ∈ L is H 6 G.

Lemma 6.5. ker ρ =⋂

x∈G

xHx−1

25

Proof. xH = gxH for all x ∈ G ⇐⇒ x−1gx ∈ H for all x ∈ G. 2

Definition 6.6. For H 6 G, the core of H in G is ker ρ.

Proposition 6.7. Core(H) � G and is the largest normal subgroup of G that is containedin H .

Proof. The core is normal as it is a kernel.

If N �G and N 6 H then x−1Nx 6 H for all x ∈ G. 2

Theorem 6.8 (‘The useful theorem’). If H 6f G with [G : H ] = n then there existsN �f G with N 6 H and [G : N ] | n!.

Proof. L has n elements, so ρ : G→ Sn. Then |G/ kerρ| = |im ρ| which divides |Sn|. 2

Theorem 6.9. If G is finitely generated then for any n ∈ N, there are only finitely manysubgroups of index n in G.

Proof. If G = 〈g1, . . ., gk〉 then there are only finitely many homomorphisms θ : G→ Sn asθ would be determined by θ(gi), so only finitely many homomorphisms are transitiveon {1, . . ., n}, where stab(1) has index n (by Orbit-Stabiliser).

Now say H 6f G with [G : H ] = n, then on ordering L as {H, . . ., } we have that theregular representation of G is a transitive homomorphism from G to Sn with stab(1) =H . 2

Corollary 6.10. If G is finitely generated and H 6f G then there exists C, characteristicin G, with C 6f H 6f G.

Proof. Let C =⋂

α∈Aut(G)

α(H).

Then H and α(H) have the same index in G by (6.4), with C 6f G by (6.9) and (6.2).

If β ∈ Aut(G) then β(C) =⋂

α∈Aut(G)

βα(H) =⋂

γ∈Aut(G)

γ(H). 2

Theorem 6.11. If H 6f G then G finitely generated (resp. presented) ⇐⇒ H finitelygenerated (resp. presented).

Proof. G finitely generated means that there exists θ : Fk ։ G. Now H 6f G impliesθ−1(H) 6f Fk by (6.4), so it is some Fi by the Nielsen-Schreier index formula (3.31).Now restrict θ to Fi ։ H .

Now say G = Fk/N is finitely presented, so N = 〈〈r1, . . ., rm〉〉 (in Fk). Then H = Fℓ/Nfor N 6 Fℓ 6f Fk by above.

Take a right transversal t1, . . ., tn for Fℓ in Fk where [G : H ] = n.

By 1.18, N consists of all elements of the form

g1ra1

i1g−11 . . . gjr

aj

ijg−1j

for g1, . . ., gj ∈ Fk, which need not be in Fℓ.

26

But as any g ∈ Fk is htj for h ∈ Fℓ, we do have that the normal closure of {tjrit−1j :

1 6 i 6 m} in Fℓ in N .

Finally, finitely generated/presented groups are preserved by extensions by (1.29),(4.11), so ifH is finitely generated/presented then takeN�G with N 6f H by (6.8). Nis finitely generated/presented and G/N is finite, so G is finitely generated/presented.2

Note. If G = 〈k gens |m rels〉 then the deficiency of the presentation is k−m. If k−m > 0then G is infinite.

(6.11) shows that if existsH with [G : H ] = n then H has a presentation with deficiencyn(k − 1−m) + 1. So def(presn for H)− 1 = [G : H ] (def(presn for G)− 1).

How do we know a group has proper finite index subgroups?

E.g., if H 6f Q with index n then for each q ∈ Q, we have q = n(qn

)∈ H by (6.3).

Theorem 6.12 (Higman, 1951). The group

G = 〈a1, a2, a3, a4 | a1a2a−11 = a22, a2a3a

−12 = a23, a3a4a

−13 = a24, a4a1a

−14 = a21〉

has no proper finite index subgroups.

Proof. If H <f G then (6.8) gives a non-trivial finite quotient G/N . Note that for n > 1and prime p dividing 2n − 1, the least prime factor of n is less than p.

Take r the order of 2 mod p, then r | n and r | p − 1 (by Fermat’s Little Theorem).Now say ni is the order of ai in G/N . Then an1a2a

−n1 = a2

n

2 , so n2 | 2n1 − 1, and so on.

Let p be the smallest prime dividing n1n2n3n4, with (wlog) p | n2. Then n1 has asmaller prime factor. This is a contradiction unless n1n2n3n4 = 1. 2

Proposition 6.13. G in (6.12) is infinite.Lecture 11

Proof. If H = 〈x, y | yxy−1 = x2〉 then x, y have infinite order, by (5.11). Let H ′ be anisomorphic copy of H . Then H ∗ϕ H ′, where ϕ(x) = y′ is infinite and is (via x = y′,z = x′)

〈x, y, z | yxy−1 = x2, xzx−1 = z2〉.

Note that y, z freely generate F2 by (4.17) as a word w(y, x′) with powers gathered isa-reduced.

Now take four copies of H , say Hi = 〈ai, bi | biaib−1i = ai〉. Form

K = H1 ∗ϕ H2 = 〈a1(= b2), b1, a2 | b1a1b−11 = a21, a1a2a

−11 = a22〉,

infinite with 〈b1, a2〉 free, and L = H3 ∗ϕ H4 (send 1 7→ 3, 2 7→ 4).

Finally, make K ∗θ L for θ(b1) = a4, θ(a2) = b3. This is G. 2

Virtual Properties

Here we regard groups H and G as ‘basically the same’ if H 6f G. Say a group property Pis ‘okay’ if when G has P and H 6f G then H has P .

27

But what about finite index ‘supergroups’? If H has P and H 6f G then G might not haveP .

Definition 6.14. If property P is ‘okay’, we say that a group G is virtually P if thereexists H 6f G where H was P .

By (6.8), this is the same as G being P-by-finite. For finitely-generated groups, cyclic =⇒abelian =⇒ nilpotent =⇒ polycyclic =⇒ soluble.

Example 6.15.

(i) Z× Z is abelian but not cyclic.

(ii) A nilpotent group which is not virtually abelian.

Let G = 〈a, b, t | ab = ba, tat−1 = ab, tbt−1 = b〉 be the semidirect product Z2 ⋊ Z

(〈a, b〉⋊ 〈t〉). Then b ∈ Z(G) and G/〈b〉 ∼= Z2.

Now, if there is abelian A 6f G then ti, aj ∈ A for i, j > 0 by (6.3). Buttiajt−i = (abi)j = ajaij 6= aj in G.

(iii) A polycyclic group which is not virtually nilpotent.

Let G = 〈a, b, t | ab = ba, tat−1 = a2b, tbt−1 = ab〉.

If H 6f G with H nilpotent then take h 6= e in Z(H) and set h = akbℓtm. Now,there exists i > 0 with ti ∈ H , and thus tiht−i = h. But this changes ℓ if k 6= 0,and adds a’s if ℓ 6= 0.

** Non-examinable aside: Word Growth **

If G = 〈X〉 for X finite, then the growth function γX : N→ N is defined by γX(n) = #{g ∈G : g = w(X±1) for word length 6 n}.

If S = X ∪X−1 ∪ {e}, then γX(n) = |Sn|.

Say that finitely-generated G has polynomial word growth if ∃/∀ finite generating set X ,there exists c, d > 0 such that for all n, γX(n) 6 cnd.

Theorem (Gromov, 1981). G has polynomial word growth ⇐⇒ G is not nilpotent.

** End of aside **

Virtually Polycyclic Groups

Theorem 6.16. The following are equivalent:

(i) G is virtually polycyclic.

(ii) G is polycyclic by finite.

(iii) G is poly-Z by finite.

(iv) G is poly-(Z or finite).

Proof. We have (i) = (ii) and (iii) ⇒ (ii) ⇒ (iv).

28

So, (iv) ⇒ (iii). Say N is (. . .(Z by Z). . . by Z) by F (finite), and G/N = Z.

Now P 6f N is finitely generated, so by (6.10) we have P0 6f N and P0 characteristicin N with P0 also poly-Z (preserved by subgroups).

Now P0 � G with G/N ∼= (G/P0)/(N/P0) so G is (poly-Z) by (finite by Z), but thelatter is Z by finite, using (3.8). So by (1.28), G is (poly-Z by Z) by finite.

So in general we can push all finite factors ‘to the right’ and can gather as (P by finite)by finite, which is P by finite. 2

Corollary 6.17. Virtually polycyclic groups are preserved by subgroups, quotients, exten-sions; it is the smallest such class containing all finite groups and Z; and all have maxand are finitely presented.

Proof. Property P = (Z or finite) is preserved by subgroups and quotients, so poly-P ispreserved by all three, by (5.7)

Any poly-P group is contained in any class with the above properties, and max andfinitely-presented are preserved by extensions. 2

Virtually Soluble Groups

Corollary 6.18. The virtually soluble groups with max are precisely the virtually polycyclicgroups.

Proof. (5.9) and (6.17). 2

Question. Are these all the groups with max? (See the final lecture.)

Theorem 6.19. Virtually soluble groups are preserved by subgroups, quotients and exten-sions.

Proof.

(i) If H 6f G with H soluble and S 6 G, then H ∩ S 6f S by (6.1)(iii), butH ∩H 6 H so it’s soluble by (5.2)(i).

(ii) Use (6.4)(ii) and (5.2)(ii).

(iii) Let G/N , N be virtually soluble, and takeM�fN soluble and normal, of minimalindex. Then if soluble S �f N , we have S 6 M as SM soluble, normal, finiteindex.

Thus M is the unique normal soluble subgroup of that index, so it’s characteristicin N , and normal in G.

But G/N ∼= (G/M)/(N/M), which we’ll call Q/R. This R is finite with no non-Lecture 12trivial soluble normal subgroups.

Now, the centraliser C = CQ(R) =⋂r∈R

CQ(r) �f Q by (6.2) and Orbit-Stabiliser,

so abelian C∩R�R, so equals I. Thus in Q, C ∼= CR/R 6 Q/R wirtually soluble,so G/M = Q is too.

Now take H 6f G with H/M soluble, so H is by (5.2)(iii). 2

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Corollary 6.20. Virtually soluble groups are the smallest class preserved by subgroups,quotients, extensions, that contains all abelian and finite groups.

Proof. Like (6.17), using (6.19). 2

As in (5.5), if F2 6 G then G is not virtually soluble. For if soluble S 6f G and F2 6 Gthen F2 ∩ S 6f F2 with F2 ∩ S 6 S non-abelian, free. So a contradiction by (5.5).

On Sheet 2, we see that not containing F2 is preserved by subgroups, quotients, exten-sions. Finitely-generated example (1.41) is not virtually soluble, no F2 subgroup. Finitelypresented? There are examples, but only 5 known constructions.

(Tits Alternative (1972). A finitely-generated linear group (i.e., a group of n × n matricesover a field F), or even any linear group in characteristic 0, is either virtually soluble orcontains a non-abelian free group.)

7. Maximal (normal) subgroups

Definition 7.1. A proper subgroup H < G is maximal if H 6 J 6 G implies H = J orH = G.

Example 7.2. Q has no maximal subgroups. For if M < Q is maximal then M � Q andQ/M has no proper normal subgroups (correspondence theorem), so it’s Cp. But Q

has no proper finite index subgroups.

Zorn’s Lemma. A poset X is a set with a relation 6 which is reflexive, antisymmetric,transitive. In a total order we always have either x 6 y or y 6 x. A subset S of X isa chain if S is totally ordered.

Then, if every chain S of X has an upper bound in X (i.e., some b ∈ X with s 6 b forall s ∈ S) then X has a maximal element m (i.e., if m 6 x then m = x).

Proposition 7.3 (Neumann, 1937). IfH < G and g ∈ G\H , then there exists a maximalsubgroup M containing H relative to g: i.e., H 6 M < G and g /∈ M , such that ifM 6 L and g /∈ L then L =M .

Proof. Poset X = {J < G : H 6 J and g /∈ J}, ordered by 6. Then for chain S = {Ji} weget

⋃Ji a subgroup which contains H but not g. So there is a maximal element M .2

Corollary 7.4. If G is finitely generated and H < G then H 6M for M maximal.

Proof. G = 〈g1, . . ., gk, h1, . . ., hℓ〉 for gi /∈ H , hj ∈ H . Take M1 maximal with H 6 M1

relative to g1.

If M1 < L < G then g1 ∈ L but not all gi are, so wlog g2 /∈ L, and if there is no L thendone. Take M2 maximal with L 6M2 relative to g2. This must stop at or before k: ifg1, . . ., gk−1 ∈Mk maximal relative to gk and Mk < L then gk ∈ L and so L = G. 2

Note. This all works exactly the same if we replace ‘subgroup’ with ‘normal subgroup’.

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Infinite Simple Groups

G(6= I) is simple if N �G implies N = I or N = G. Note that G simple and soluble impliesG = Cp, some prime p.

Other simple groups include An (n > 5) and PSL(n,F) (for n > 2 or |F| > 3 – and the proofdoes work for infinite fields).

Imre moment 3. H =⋃An 6 S(N) (as in (1.5)(ii)) is simple.

Suppose N�G. Then N ∩An �An. If N 6= I then take k with N ∩Ak 6= I, so An 6 Nfor all n > k.

This H is not finitely generated. If G is an infinite simple group then G is not virtuallysoluble, as S 6f G =⇒ N 6 S by (6.8), with N �f G. So N = S = G and so G has nofinite index subgroups.

Theorem 7.6 (Higman, 1951). Infinite finitely-generated simple groups exist.

Proof. Take Higman’s G and N � G a maximal normal subgroup (containing I), so G/Nsimple (correspondence theorem). But N 6= G, N 66f G, so G/N infinite and finitelygenerated. 2

What about infinite finitely-presented simple groups? R. Thompson (60s/70s/80s?): Thomp-son groups F, T, V , finitely presented, infinite. F not simple, not virtually soluble, no F2

subgroup. T, V simple. (Cannon, Floyd, Parry – something Math (42) 1996 p215.)

Open: does there exist an infinite finitely-presented simple group with no F2 subgroup?

Burger/Mozes (1998): there exist k, ℓ > 2 such that Γ = Fk ∗FℓFk for Fℓ 6f Fk is finitely

presented, torsion-free, and simple.

8. Residual Finiteness

Definition 8.1. A group G is residually finite if⋂

H6fG

H = I.

Proposition 8.2. The following are equivalent:Lecture 13

(i) G is residually finite

(ii)⋂

N�fG

= I

(iii) For all g ∈ G \ {e}, there exists a homomorphism θ onto a finite group F suchthat θ(g) 6= e

(iv) For all g1, . . ., gn ∈ G \ {e}, there exists a homomorphism θ onto a finite group Fsuch that θ(gi) 6= e for all i.

Proof.

(ii)⇒(i) is clear, and (i)⇒(ii) is by (6.8).

(ii)⇔(iii), as for g ∈ G \ {e}, take N = ker θ, or θ : G։ G/N where g /∈ N .

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(ii)⇒(iv), via θ : G։ G/⋂n

i=1Ni and Poincare, whereas (iv)⇒(iii). 2

So finite groups and Z are residually finite, but Q and the Higman group are not.

An infinite residually finite group G has infinitely many finite index subgroups (Poincare),of arbitrarily high index by (6.1): if both H =

⋂ni=1Hi and J 6f G, then [G : H ] = [G :

H ∩ J ] ⇐⇒ H = J , so take Hn+1 = J with h ∈ H \ J , continue, and index →∞.

Aside. Given finitely-generated group G, we might expect the discrete topology on G. Butdefine basic open sets as the cosets of N �f G, giving the profinite topology. It is:Hausdorff ⇐⇒ G is residually finite; indiscrete ⇐⇒ no proper finite index subgroup;discrete ⇐⇒ G finite.

Lemma 8.3. If RG =⋂

N�fG

N then G/RG is residually finite.

Proof. Normal finite index subgroups of G/RG are N/RG for RG 6 N�f G, but for g /∈ RG

we have RG 6 N �f G with g /∈ N . 2

Note that if Gθ։ Q, residually finite, then θ factors through G/RG as θ(RG) ⊂ RQ.

Proposition 8.4.

(i) If G is residually finite and H 6 G, then H is residually finite

(ii) If H is residually finite and H 6f G, then G is residually finite

(iii) If G, H are residually finite, then G×H is residually finite, and if G is also finitelygenerated then G⋊H is residually finite.

Proof.

(i) RG =⋂

L6fG

L from (8.2), and RG ∩H > RH by (6.1).

(ii) For H 6f G, we have L 6f H =⇒ L 6f G, so RG 6 RH but RG > RH by (i).

(iii) For (g, h) 6= id, take θ1 : G → F1 and θ2 : H → F2 with θ1(g), θ2(g) not both e.Then θ1 × θ2 : G×H → F1 × F2 finite and (g, h) 67→ (e, e).

For G ⋊ H , we have θ : G ⋊ H ։ H with θ(gh) = h, so there exists ϕ withϕθ(gh) = ϕ(h) 6= e in a finite group, unless h = e. Take g ∈ G \ {e}. Now takeL 6f G with g /∈ L and C 6f L characteristic in G by (6.10). Then C � G⋊Hso CH 6 G⋊H . As G ∩H = I, we have g /∈ CH 6f G⋊H . 2

Corollary 8.5. G virtually polycyclic =⇒ G residually finite.

Proof. We have H 6f G with H poly-Z by (6.16). Now if M/N = Z with N finitelygenerated and residually finite, then M ∼= N × Z by (3.8), so M is residually finite by(8.4)(iii). Thus H is residually finite, and so G is too by (8.4)(ii). 2

Theorem 8.6. Free groups are residually finite.

Proof 1. Recall (3.23), that F =

(1 20 1

), G =

(1 02 1

)∈ SL(2,Z) freely generate F2.

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Given a reduced word w (6= ∅) in F2, we have w(F,G) =

(a bc d

)6=

(1 00 1

).

Take prime p > max{|a − 1|, |d − 1|, |b|, |c|}, then θ : SL(2,Z) → SL(2,Fp), a finitegroup with θ(w(F,G)) 6= e. So F2 is residually finite.

Now, Fn 6 F2, so there are residually finite, and for w(6= ∅) ∈ F (X), where {xi : i ∈ I},only xi1 , . . ., xik appear in w. So we have θ : F (X) ։ Fk = F (xi1 , . . ., xik) given by“rest of X goes to e and extend”. Now θ(w) 6= e, so now have ϕ : Fk → finite group,with ϕθ(w) 6= e. 2

Proof 2. F2 free on a, b create a reduced word w using a, b, A,B (formal inverses).

Let w = AbABBBAAbA, labelled: 11A10b9A8B7B6B5A4A4b2A1.

Let f : F2 → S11 (or n+ 1 if length n) be given by

1 2 3 4 5 6 7 8 9 10 11a 1 3 4 8 10b 3 5 6 7 10

Partial functions, each mapping i 7→ i + 1. Injective? E.g., maybe f(b)(9) = 10 =f(b)(11)? But then 11B10b9 reduced. Fill in:

1 2 3 4 5 6 7 8 9 10 11a 11 1 2 3 4 5 6 7 8 9 10b 11 3 1 2 4 5 6 7 10 8 9

f(a), f(b) ∈ S11. Now universal property extends to a homomorphism f : F2 → S11.And f(w)(1) = 11, so f(w) 6= id. 2

Theorem 8.6a. If G1, G2 are residually finite, then G1 ∗G2 is residually finite.Lecture 14

Proof. First say G1, G2 are finite. Given a reduced sequence g1g2. . .gn in G1 ∗G2 of lengthn > 1, let Xn = {g ∈ G1 ∗G2 : 0 6 length(g) 6 n}, a finite set.

Define an action on Xn: if γ ∈ G2 then γ(g1. . .gk) =

{γg1. . .gk if length(RHS) 6 ng1. . .gk otherwise

This gives a homomorphism G2 → S(Xn), and similarly G1 → S(Xn), which we extendto a homomorphism G1 ∗G2 → S(Xn), which is a finite group.

Now g1. . .gn(∅) = g1. . .gn 6= ∅, so g1. . .gn is non-trivial in S(Xn).

For g1. . .gn in general G1 ∗G2, choose N1 �f G1 and N2 �f G2 such that g1, . . ., gn /∈N1 ∪N2 by (8.2)(iv).

Then Gi ։ Gi/Ni → (G1/N1) ∗ (G2/N2) extends to G1 ∗G2to(G1/N1) ∗ (G2/N2) andthe image of g1. . .gn is reduced and so has length n. 2

Hopfian Groups

33

Definition 8.7. A group G is Hopfian if every surjective endomorphism θ : G ։ G isinjective.

If not, then G/ ker θ ∼= G, so G is isomorphic to a proper quotient of itself. Finite groupsare Hopfian, as are Z. The free group F (N) is not Hopfian, as we can send x1 7→ x1 andxi+1 7→ xi for all i > 2 – but it is residually finite.

Theorem 8.8 (Malcev, 1940). A finitely-generated, residually-finite group G is Hopfian.

Proof. For θ : G→ G and H 6f G with index n, θ−1(H) has index n too, and if θ−1(H1) =θ−1(H2) then θθ−1(H1) = H1 = H2. So the pullback map is injective on {index nsubgroups of G}. But this is a finite set by (6.9), so is a permutation.

However, ker θ 6 θ−1(H) for all H 6f G, so ker θ is in⋂

H6fG

θ−1(H) = RG = I. 2

Corollary 8.9. If g1, . . ., gn generate the free group Fn then they freely generate Fn.

Proof. Fn is Hopfian, so if w(g1, . . ., gn) = e in F (x1, . . ., xn) then θ : F (g1, . . ., gn) →F (x1, . . ., xn) by: given symbol gi, send it to its image in F (x1, . . ., xn) and extend.This is surjective (it hits a generating set) and hence injective by (8.6), (8.8). Sow = ∅. 2

An infinite simple group is Hopfian but not residually finite.

Baumslag-Solitar Groups

These are very useful for counterexamples.

Definition 8.10. The Baumslag-Solitar group Bm,n = 〈a, t | tamt−1 = an〉 for m,n 6= 0.

These include B1,1 = Z× Z and B1,−1 = the Klein bottle group. Can also change m,n butkeep the group the same: Bm,n

∼= B−m,−n∼= Bn,m. They are HNN extensions: Z = 〈a〉∗ϕ

with ϕ : 〈am〉 → 〈an〉.

Proposition 8.11. Bm,n is soluble if |m| or |n| equals 1, and it contains F2 otherwise.

Proof. If |m| or |n| is 1, then wlog we have B1,n and this is soluble, just as in (5.11) (forB1,2). Otherwise, a is not in the domain or image of ϕ. So for any reduced wordw(x, y) in F2 we have w(t, ata−1) a reduced sequence in an HNN extension, so is not eby Britton’s Lemma.

Theorem 8.12. B2,3 is not Hopfian.

Proof. Let θ(t) = t, θ(a) = a2. This is a homomorphism as it preserves the relation:θ(ta2t−1) = (ta2t−1)2, θ(a3) = a6. It is surjective: tat−1a−1 7→ ta2t−1a−2 = a.

What is the kernel? θ([tat−1, a]) = [a3, a2] = e, but tat−1ata−1t−1a−1 is a reducedsequence, so 6= e by Britton’s Lemma. 2

Theorem 8.13. There exists a finite-generated soluble group which is not finitely presented.

Proof. Consider G = B2,3 and θ as above. Set Ki = ker θi �G. We have Ki < Ki+1, sincefor y 6= e with θ(y) = e we have y = θi(x) as θi is surjective, so x ∈ Ki+1 \Ki.

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So G/⋃Ki = Q is not finitely presented by (4.4): if it is, then Q = 〈a, s | S〉 with

S = 〈〈s1, . . ., sk〉〉 in G, so all in KN but x ∈ KN+1 \KN is e in Q but not in S 6 KN .

Now, G′ is generated by tiat−1 for i ∈ Z. But θi(tiat−i) = tia2i

t−i = a3i

(for i > 0),which commutes which θi(a). This gives [tiat−i, a] 7→ e in Q as it’s in ker θi, as well as[tjat−j , tkat−k] by conjugacy (set j = i + k).

Thus θ(G′) = Q′ is abelian, so Q′′ = I. 2

9. The Generalised Burnside ProblemLecture 15

Examples of torsion groups: finite groups F , infinite F × F × F × . . ., example (1.5), etc.But finitely generated?

In 1902, W. Burnside asked:

(1). The Generalised Burnside problem. Do there exist finitely-generated infinite tor-sion groups?

Lemma 9.1. If G is finitely-generated infinite torsion, then

(i) For G։ Q, we have Q finite or finitely-generated infinite torsion.

(ii) For H 6f G, we have H finitely-generated infinite torsion by (6.11).

(iii) G is not virtually soluble.

Proof.

(iii) If H is soluble then H/H ′ is finitely-generated torsion and abelian, so finite, thusH ′ is finitely-generated infinite torsion by (ii) and soluble. Continue until H(n) =I.//\\ 2

Burnside also asked:

(2). The Burnside problem. If G is finitely generated and there is k such that for allg ∈ G we have gk = e (i.e., bounded torsion), then can G be infinite?

Let FB(n, k) = 〈x1, . . ., xn | wk = e for all w ∈ Fn〉. Then a group G is n-generated andgk = e for all g ∈ G ⇐⇒ FB(n, k) ։ G, so (2) says: do there exist n, k for which FB(n, k)is infinite?

And also:

(3). The Restricted Burnside Problem. Can G in (2) be infinite and residually finite?Yes ⇐⇒ FB(n, k)/R infinite by (8.3).

In 1964, Golod answered (1) with ‘yes’: there are finitely-generated infinite p-groups (i.e.,every element has order pk for some k).

Let p be any prime.

Definition 9.2. In Fn, the p-value vp(w) of non-identity w ∈ Fn is max{k : w = upk

foru ∈ Fn}.

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Definition 9.3. The p-deficiency, written p-def, of a presentation 〈x1, . . ., xn | r1, r2, . . .〉

is n−∞∑i=1

p−vp(ri), it it converges.

Lemma 9.4. Say F acts on X and S � F with [F : S] = p. For x ∈ X , if there isg ∈ stabF (x) \ S, then orbF (x) = orbS(x).

Proof. Have SstabF (x) = F , so for f(x) ∈ orbF (x) set f = st with s ∈ S and t ∈ stabF (x).Then f(x) = s(x) ∈ orbS(x). 2

Theorem 9.5. For any prime p and n > 2, there exists an infinite n-generated p-groupwhich is residually finite.

Proof. Suppose 〈x1, . . ., xn | r1, r2, . . .〉 is a presentation P with p-def(P ) >. Define G =Fn/R.

(i) G ։ Cp. We must have vp(ri) = 0 for at most n − 1 relators, or else p-def(P ) 6n− n.

By (4.8), if span{ri} 6= Fnp , then G ։ Cp, but if vp(ri) > 1 then ri = 0 ∈ Fn

p . Sodim span {ri} < n.

So let N = ker and set N = S/R for R 6 S � Fn with [Fn : S] = p.

By the proof of (6.11), N is generated by p(n−1)+1 elements and R = 〈〈tjrit−j :i ∈ N, 0 6 j 6 p − 1〉〉S gives relators, where {e, t, . . ., tp−1} is a transversal for Sin Fn if t /∈ S.

What is the p-deficiency? Take one relator r = ri in P and set k = vp(r) so

r = wpk

for w ∈ Fn.

Case (a). w /∈ S.

By (9.4) with conjugacy and x = r, f = w (these commute), so get cclF (r) =cclS(r) so 〈〈r, . . ., tp−1rt1−p〉〉S = 〈〈r〉〉S , since the tirt−i are conjugate in S (al-though not by t!).

Now, r = (wp)pk−1

for wp ∈ S.

Case (b). w ∈ S.

So tjrt−j = (tjwt−j)pk

.

Thus R = 〈〈

{ri if vp(ri) in S equals vp(ri)− 1 in Fn

tjrt−j , 0 6 j 6 p− 1 if the vp are equal

}〉〉.

So this presentation Q for N as p-deficiency

p(n− 1) + 1−∞∑

i=1

p

pvp(ri)= p(p-def(P )− 1) + 1

(ii) If p-def(P ) > 1 then G is infinite, as we get p-def(Q) > 1, so N ։ Cp. So repeatto get G > N1 > N2 > . . .

Now list the non-identity elements of Fn as {w1, w2, . . .} and set P = 〈x1, . . ., xn |

wp1 , w

p2

2 , wp3

3 , . . .〉. Then p-def(P ) > 1. So G is an infinite p-group, finitely generated.

Residually finite? G/RG is a residually-finite p-group, and N1, N2, . . . > RG, so G/RG

infinite. 2

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What about (2), the Burnside Problem? FB(n, 2) is abelian. FB(n, 3), FB(n, 4), FB(n, 6)are finite (1940s-1950s). FB(n, 5) is an open problem.

Novikov, Adyan (1970s) showed that FB(n, k) are infinite for all odd k > 665.

Ol’shansky (1982): for all large primes p, there exists a finitely-generated infinite group Gsuch that if I < H < G then then H ∼= Cp. So G has max!

What about (3), the Restricted Burnside Problem? The answer is no, by Zelmanov (1994 –Fields medal).

Question. Do there exist finitely-presented infinite torsion groups?

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