Countable locally virtually abelian groups

Proposition 5.1. LetGbe a torsion-free group andA an abelian subgroup of finite index inG. Then CG(A) is abelian.

Proof. Since A is abelian,A ≤Z(CG(A)), and hence Z(CG(A)) has finite index in

CG(A). It follows from Lemma 1.10 that CG(A)0 is finite and thus in fact trivial, sinceGis torsion-free.

A subgroupH of a groupGischaracteristic inGifH is preserved by all automor- phisms ofG.

Proposition 5.2. Let G be a countable torsion-free locally virtually abelian group and suppose thatG does not contain free abelian subgroups of arbitrarily high finite rank. ThenG has a finite index normal abelian subgroup that is characteristic.

Proof. Let k ∈ N be maximal such that G has a free abelian subgroup of rank k. We can choose elements gi ∈ G for i ∈ N such that G = hgi|i∈Ni and

hg1, . . . , gki ∼=Zk.

For i ≥ k, let Hi = hg1, . . . , gii and let Ai be a subgroup of Hi that is maximal subject to having finite index inHiand being isomorphic toZk. By Proposition 5.1, CHi(Ai) is abelian, and so the maximality ofAi implies CHi(Ai) =Ai.

SupposeBi is another rankkabelian subgroup ofHi. ThenAi∩Bi has finite index in Bi, so has rank k, so has finite index in Ai and hence in Hi. Proposition 5.1 implies that CHi(Ai ∩Bi) is abelian. But Ai, Bi ≤ CHi(Ai ∩Bi), and so, by the

maximality of Ai, we have CHi(Ai∩Bi) = Ai, so Bi ≤ Ai. This proves that Ai

is the unique maximal abelian subgroup of rank k in Hi. It follows that Ai EHi and Ai ≤Ai+1 for all i≥k. Since Ai is self-centralising, Hi/Ai is isomorphic to a subgroup of Aut(Ai)∼= GL(k,Z).

We will now show that for each i ∈ _N, Hi/Ai is isomorphic to a subgroup of

Hi+1/Ai+1. For i≤kthis is obvious, since Ai =Hi ≤Hi+1 =Ai+1.

Fori≥k, let ιi be the natural embedding of Hi in Hi+1,φi the natural homomor- phism fromHi toHi/Ai and letψi :Hi/Ai →Hi+1/Ai+1be given byxAi7→xAi+1.

Thenψi◦φi =φi+1◦ιi and so kerψi◦φi = kerφi+1◦ιi. Now kerφi+1◦ιi =Hi∩kerφi+1=Hi∩Ai+1 =Ai.

The last equality is due to the maximality of Ai and the fact that Hi ∩Ai+1 is

abelian and contains Ai. Thus kerψi ◦φi = Ai. Since already kerφi = Ai, this implies thatψi is injective and soHi/Ai is isomorphic to imψi≤Hi+1/Ai+1.

Hence, since each Hi/Ai is isomorphic to a finite subgroup of GL(m,Z), Proposi-

tion 1.11 implies that there existsm∈Nsuch thatHi/Ai is isomorphic toHm/Am for alli≥m.

We have a commutative diagram:

Hm ιm −−−−→ Hm+1 ιm+1 −−−−→ . . . −−−−→ιi−1 Hi ιi −−−−→ Hi+1   yφm   yφm+1   y   yφi   yφi+1 Hm/Am ψm −−−−→ Hm+1/Am+1 ψm+1 −−−−→ . . . −−−−→ψi−1 Hi/Ai ψi −−−−→ Hi+1/Ai+1

Fori≥m, defineθi:Hi →Hm/Am by θm=φm and fori > m

θi=ψm−1◦ψm−1+1. . .◦ψ

−1

i−1◦φi.

Since eachφi is an epimorphism and all the ψ−_j1 (j≥m) are isomorphisms, each θi is an epimorphism. Moreover, sinceφi=ψi−1◦φi+1◦ιi, we have

θi+1◦ιi = ψ−m1◦ψ −1 m+1. . .◦ψ −1 i ◦φi+1◦ιi = ψ−_m1◦ψ_m−1₊₁. . .◦ψ_i−₋1₁◦φi =θi, which means thatθi+1(x) =θi(x) for x∈Hi.

Defineθ:G→Hm/Am by θ(x) =θi(x) wherei= min{j≥m|x∈Hj}. Then θ is a well-defined epimorphism with kernelA:=∪i∈NAi. SoAis abelian of finite index

inG.

Finally, we show that Ais characteristic in G. Let B be an abelian subgroup of G

containing a subgroup isomorphic to Zk, with generating set {hi | i ∈ N}, where

hh1, . . . , hki ∼=Zk. LetBi =hh1, . . . , hiifori∈N. EachBi, being finitely generated, is contained in someHji, and hence in Aji. Now sinceBi ≤Bi+1 for all i∈N,

B =∪i∈NBi ≤ ∪i∈NAji ≤ ∪i∈NAi =A.

ThusAis the unique maximal abelian subgroup ofGcontaining aZk subgroup and

so is characteristic inG.

Next we deal with groups which do have subgroups isomorphic toZk for all k∈N.

Observation 5.3. An automorphism of a free abelian group of finite rank that centralises a subgroup of finite index is trivial.

Proof. LetGbe a free abelian group of rankkand let Hbe a finite index subgroup of G. Letα be an automorphism of Gwhich centralises H. We can regardG and

H as subgroups of Qk, so α can thus be considered as an element of GL(k,Q). A

finite index subgroup of a free abelian group of rank k is also free abelian of rank

k, so any generating set for H is a basis for Qk over Q. Since α centralises H, it

fixes a basis ofQk and thus also fixes the whole ofQk and in particular is trivial on G.

Lemma 5.4. Let G be a countable torsion-free locally virtually abelian group and let N be an abelian normal subgroup of G such that G/N ∼= Z∞. Then G has a

subgroup isomorphic toZ∞.

Proof. IfN containsZkfor allk∈Nthen we are done, sinceN is abelian; so assume tis maximal withZt≤N. WriteG=Si∈NGi, with eachGi finitely generated (and

hence virtually abelian),Gi< Gi+1 for alli∈N, and eachNi :=Gi∩N isomorphic toZt.

LetAi =CGi(Ni) and letBi be an abelian subgroup of finite index inGi. ThenBi

centralises the finite index subgroup Bi∩Ni of Ni. Hence, by Observation 5.3, Bi must centraliseNi. ThusBi ≤Ai, so|Gi :Ai|is finite.

Each Ni has finite index in Ni+1; so, by Observation 5.3, for all i ∈ N we have Ai≤CGi+1(Ni+1) =Ai+1.

IfAi is not abelian, then there existx, y∈Ai with [x, y] =z∈Ni\ {1}. But then, sincez∈Z(Ai), the subgroup ofAi generated byx, y and z is

hx, y, z |[x, y] =z,[x, z] = [y, z] = 1i,

which is the Heisenberg group, which is not virtually abelian. But this contra- dicts the hypothesis that all finitely generated subgroups ofGare virtually abelian. ThusAi must be abelian. Hence S_i∈_NAi is abelian and contains finitely generated subgroups of arbitrarily high rank; so it must containZ∞.

Lemma 5.5. Let G be a locally virtually abelian group with a normal torsion sub- group T such that G/T ∼=Z∞. Then G has a subgroup isomorphic to Z∞.

Proof. Let g1, g2, . . . be an irredundant generating set for G modulo T. Then, for

all i ∈ N, Gi := hg1, . . . , gii is virtually abelian; so Ti := Gi ∩T is finite. Also

Gi/Ti ∼=Zi.

We shall construct a chain of subgroups H1 < H2 < · · · < Hi < · · · such that

Hi ≤Gi with|Gi:Hi|finite and Hi ∼=Zi. Suppose that we have definedH1, . . . , Hi for some i≥0. Then gi+1 ∈Gi+1 centralises HiTi+1/Ti+1, while some powergik+1

ofgi+1 centralises Ti+1. If l=k|Ti+1|, then, for anyh ∈Hi, there is some t∈Ti+1

such thathgi+1 _{=ht, and so}

hgli+1 _{= htt}gi+1_tgi2+1_{. . . t}gil+1 = h ttgi+1_{. . . t}gik−+11 |Ti+1| =h,

hence g_il₊₁ centralises Hi. Thus Hi+1 :=

Hi, gli+1

has finite index in Gi+1 and is

isomorphic toZi+1.

So we can construct the chain as claimed, andS

i∈NHi

∼

=Z∞.

Proposition 5.6. Let G be a soluble group with all finitely generated subgroups virtually abelian, and suppose thatG has subgroups isomorphic to Zk for all k∈N.

ThenG has a subgroup isomorphic to Z∞.

Proof. The proof is by induction on the derived length ofG. The statement is true for abelian groups. Now suppose it is true for groups of derived length at mostn, and letG be locally virtually abelian of derived length n+ 1, with Zk ≤G for all k∈_N.

Let N = G(n). If Zk ≤ N for all k ∈ N then we are done, since N is abelian.

So suppose there exists t ∈ _N maximal such that Zt ∈ N. This means that G/N

must contain Zk for all k∈N. By the induction hypothesis, this means that G/N

has a subgroup isomorphic to Z∞. We can assume without loss of generality that G/N ∼=Z∞.

Let T be the torsion subgroup of N. Then N/T is torsion-free, and so G/T and

N/T satisfy the hypothesis of Lemma 5.4. Thus we can replace G by a subgroup and assume G/T ∼= Z∞. Then, by Lemma 5.5, G has a subgroup isomorphic to