On the profinite topology on solvable groups

Question 1.1.

If G is a f.g. (or finitely presented) solvable group which is RF but not LERF, then must it contain a strictly ascending HNN-extension subgroup H=〈t,B〉 where B is a f.g. subgroup of G?

Lemma 2.1 ([8, Proposition 2.1]).

The wreath product W=A≀B of the groups A and B is f.g. if and only if both A and B are f.g.

Theorem 2.2 ([9, Theorem 3.2]).

The wreath product A≀B is RF if and only if A and B are RF and either A is abelian or B is finite.

Lemma 2.3 ([15, Statement 22.14]).

Let A and Q be two groups and N≤Q. Let T be a right transversal to N in Q. Then there is an isomorphism Θ between A(Q)⋉N and A(T)≀N which is defined as follows: For every element (f,y) in A(Q)⋉N, Θ⁢(f,y)=(F,y) where, for every x∈N and t∈T, F⁢(x)⁢(t)=f⁢(x⁢t).

Lemma 2.4 ([9, Lemma 3.2]).

Let Q be a group and A be an abelian group. If N is a normal subgroup of Q, then there is a natural homomorphism φ from A≀Q onto A≀(Q/N) which is defined by

for every (∏i=1nδxi,yi,y)∈A≀Q. In particular, if A1≤A and R is a subgroup of Q, then we have

Remark 2.5.

Let F⁢(n) be the free group of rank n≥2. Denote by Fd⁢(n) the dth derived subgroup, that is Fd⁢(n)=[Fd-1⁢(n),Fd-1⁢(n)] where F0⁢(n)=F⁢(n). The free solvable group of rank n and solvable class d is Sn,d=F⁢(n)/Fd⁢(n). Let T be the free left module of rank n over the ring ℤ⁢[Sn,d]. In [12], Magnus showed that the free solvable group Sn,d+1 of solvable class d+1 can be embedded into the group of matrices M of the form (ft01), where f∈Sn,d and t∈T. The group M is isomorphic to the wreath product of the free abelian group of rank n with Sn,d.

Theorem 2.6 ([9, Theorem 6.3]).

Every f.g. free solvable group of class d≥1 is an RF group.

Definition 2.7.

Let B be a group and let A1 and A2 be subgroups of B with φ:A1→A2 an isomorphism. Let 〈t〉 be the infinite cyclic group generated by a new element t. The HNN-extension H is formed by adjoining to the free product B*〈t〉 the relations t⁢a⁢t-1=φ⁢(a) for all a∈A1. Hence, H admits the presentation

The group B is called the base ofH, t is called the stable letter, and A1 and A2 are called the associated groups.

An HNN-extension H is called an ascending HNN-extension if at least one of the subgroups A1 and A2 is equal to the base B and it is called a strictly ascending HNN-extension, if at least one of the subgroups A1 and A2 is equal to the base B and the other one is strictly contained in B, so that conjugation by t induces an injective but not surjective endomorphism of B.

Remark 2.8.

If H is a subgroup of the group G, we can consider two topologies on H: its own profinite topology as a group and the restriction to H of the profinite topology of G. It is clear that the restriction to H of the profinite topology of G is contained in the profinite topology of H. In particular, if H1 is a subgroup of H which is closed in the profinite topology of G, then H1 is closed in the profinite topology of H. If H is a finite index subgroup of G, then the induced topology on H from G coincides with its own profinite topology of H.

Lemma 2.9 ([16, Lemma 3.1.5]).

Let G=N⋉H be an RF group which is a semidirect product of its normal subgroup N with a subgroup H. Then the induced topology on H coincides with its own profinite topology and H is closed in the profinite topology of G.

Remark 2.10.

Let N be a normal subgroup of a group G. Then N is closed in the profinite topology of G if and only if G/N is RF.

Theorem 2.11 ([5]).

Let G be a group and H a subgroup of G that is closed in the profinite topology of G. Then H is not conjugate in G to a proper subgroup of itself. In other words, if G contains a (respectively f.g.) strictly ascending HNN-extension H=〈B,t∣t⁢B⁢t-1⊂B〉, then B is not closed in the profinite topology of G.

Theorem 3.1 ([11, Theorem 3]).

Every f.g. abelian-by-polycyclic group is RF.

Lemma 3.2.

Let G be a group with a normal subgroup of finite index N. If N is ERF (respectively LERF), then so is G.

Proof.

In the case ERF, the result is [18, Lemma 4.2]. In the LERF case, the proof is essentially the same; one needs only to observe that if H is a f.g. subgroup of G, then H∩N is a f.g. subgroup of N: since N has finite index in G, the group

is finite and so, H∩N is a finite index subgroup of a f.g. group H. Hence, by Schreier’s theorem [17, statement 1.6.11], H∩N is f.g. ∎

Corollary 3.3.

Let G be a group with a finite normal subgroup N. If G/N is ERF (respectively LERF) and G is an RF group, then G is ERF (respectively LERF).

Lemma 3.4.

Let G be a group and N be a normal subgroup of G. Let H be a subgroup of G such that H⁢N/N is closed in the profinite topology of G/N. Then ClG⁢(H)=H⁢(ClG⁢(H)∩N). In particular, if for every n∈N∖H, n is not in the closure of H in G, then H is closed in the profinite topology of G.

Proof.

Let Φ:G→G/N be natural projection. Since H⁢N/N is closed in the profinite topology of G/N, the subgroup HN is closed in the profinite topology of G. Hence, we have

So, by Dedekind’s Modular Law [17, Statement 1.3.14], we have

If for every n∈N∖H, n is not in the closure of H in G, then we have

Lemma 3.5.

Let G be a group with normal subgroup N such that G/N is an ERF (respectively a LERF) group. Let H be a (respectively f.g.) subgroup of G. Then H is closed in the profinite topology of G if and only if H∩N is closed in the profinite topology of G.

Proof.

Since G/N is an ERF (respectively LERF) group, this quotient group is RF, so, by Remark 2.10, N is closed in the profinite topology of G.

Suppose that H is a closed (respectively f.g.) subgroup of G. Hence, H∩N is closed in the profinite topology of G.

Conversely, suppose that H∩N is a closed subgroup in G. By the preceding lemma, it is enough to show that for every n∈N∖H, n does not belong to ClG⁢(H). Since n∉H∩N and this subgroup is closed in the profinite topology of G, there is a finite index normal subgroup M of G such that n∉(H∩N)⁢M. The subgroup L=M∩N is a closed normal subgroup of G such that L is of finite index in N and n∉L.

Now we claim that the subgroup LH is a closed subgroup of G containing H but not n, which entails that n is not in the closure of H in G. First if n is in LH, then n=l⁢h with l∈L and h∈H. Then l-1⁢n=h∈H∩N≤(H∩N)⁢M, whence n is in (H∩N)⁢M, which is a contradiction with the choice of M. So, n is not in LH. Finally, we show that LH is a closed subgroup of G. As L is closed in the profinite topology of G, by Remark 2.10, G/L is RF. The RF group G/L contains the finite normal subgroup N/L such that (G/L)/(N/L)≃G/N is ERF (respectively LERF). By Corollary 3.3, G/L is ERF (respectively LERF). Hence (H⁢L)/L is closed in the profinite topology of G/L; so HL is closed in the profinite topology of G. ∎

Corollary 3.6.

Let G be a f.g. group which has a surjective morphism φ onto a polycyclic group Q, with abelian kernel K. Let H be a subgroup of G such that φ⁢(H) has finite index in Q. Then the induced topology on H coincides with its own profinite topology and H is closed in the profinite topology of G.

Proof.

Let H1 be a clopen subgroup of H. By the definition of the profinite topology, H1 is a finite index subgroup of H. Hence, the assumptions of the corollary imply that H1⁢K is a finite index subgroup of G containing H1. So, by Remark 2.8, it is enough to show that H1 is closed in the profinite topology of G1=H1⁢K. By Schreier’s theorem [17, statement 1.6.11], G1 is a f.g. group and as the class of abelian-by-polycyclic groups is closed under taking subgroups, G1 is a f.g. abelian-by-polycyclic group. Since K is an abelian group, H1∩K is a normal subgroup of G1. By Theorem 3.1, the group G1/(H1∩K) is RF. Hence, by Remark 2.10, H1∩K is closed in the profinite topology of G1. By the preceding lemma, H1 is closed in G1. ∎

Lemma 3.7.

Let A and Q be arbitrary groups, W=A≀Q and let Φ:W→Q be the natural projection. Let H be a f.g. subgroup of W such that R=Φ⁢(H) is closed in the profinite topology of Q. Then there are a finite index subgroup N and a finite subset T of Q such that H≤A(T)≀R≤A(T)≀N and A(T)≀N is a finite index subgroup of W.

Proof.

Let

For every s∈S, let ts∈Q∖R and rs∈R such that s=rs⁢ts. Let T be a complete set of representatives of R-right cosets of the set {ts∣s∈S}. Since R is closed in the profinite topology of Q, for all distinct ts,ts′∈T, there is a subgroup Ns,s′ of finite index of Q containing R such that ts⁢Ns,s′≠ts′⁢Ns,s′. Since the set T is a finite subset of Q, the subgroup

is a subgroup of finite index of Q containing R such that for all distinct ts,ts′∈T, ts⁢N≠ts′⁢N. The subgroup H is contained in B⋉N. We can choose a right transversal T′ to N in Q such that T⊆T′. By Lemma 2.3, the subgroup B⋉N is isomorphic to A(T′)≀N and it is clear that this subgroup has finite index in W. The group A(T′)≀R is a subgroup of A(T′)≀N. We claim that every generator of H is in A(T′)≀R. First, note that for every generator (f,r) of H and s∈Supp⁢(f)⊆S, we have s=rs⁢ts with rs∈R and ts∈Q∖R. By our definition of T, there is t∈T such that ts⁢t-1∈R, and so ts=r′⁢t for some r′∈R. Therefore, by Lemma 2.3, we have

Hence, F belongs to A(T′)⁢(R). So H is contained in A(T′)≀R. ∎

Theorem 3.8.

Let W=A≀Q where A is a f.g. abelian group and Q is a polycyclic group. Then W is a LERF group.

Proof.

Let B=A(Q), H be a f.g. subgroup of W and let Φ:W→Q be the natural projection, and R=Φ⁢(H)(≃H⁢B/B). By the Mal’cev theorem [13], Q is an ERF group, whence R is closed in the profinite topology of Q. Hence, by Lemma 3.7, there is a finite index subgroup A(T)≀N of W such that

By Remark 2.8, it is enough to show that H is closed in the subgroup of finite index A(T)≀N. Hence, by replacing W with its finite index subgroup A(T)≀N, we can assume that

So, we have H∩A(Q)=H∩A(R), where we are identifying A(Q) with A(Q)≀{1} and A(R) with A(R)≀{1}.

To show that H is closed in the profinite topology of W, by Lemma 3.5, it is enough to show that A(Q)∩H=A(R)∩H is closed in the profinite topology of W. First we show that A(R) is closed in the profinite topology of W. As W/B≃Q is an ERF group and therefore W/B is RF, by Remark 2.10, B is a closed subgroup of W. Hence, it is enough to show that for every g∈B∖A(R), g is not in the closure of A(R). Since g∉A(R), there is y∈Supp⁢(g) such that y is not in R. As Q is an ERF group, there exist a subgroup N of finite index of Q containing R such that y∉N. The subgroup A≀N contains A(R) but not g. Since N is of finite index in Q, by Corollary 3.6, the subgroup A≀N is closed in the profinite topology of W. So, g is not in the closure of A(R). This establishes that A(R) is a closed subgroup of W, which contains H∩A(R). Hence, we have

Take g∈A(R)∖(H∩A(R)). We have the natural projection π:A≀R→R. By the second isomorphism theorem and the fact that R=π⁢(H) we have

Hence, H⁢A(R)=A≀R. Since A(R) is abelian, the subgroup H∩A(R) is normal in H⁢A(R)=A≀R. Since the class of abelian-by-polycyclic groups is closed under taking quotients, (A≀R)/(H∩A(R)) is a f.g. abelian-by-polycyclic group and so RF, by Theorem 3.1. Since g∉H∩A(R), there is a normal subgroup M of finite index of A≀R containing H∩A(R) but not g. The subgroup L=M∩A(R) is a normal subgroup of A≀R containing H∩A(R) but not g.

Let LW be the normal closure of L in W. By Theorem 3.1, the f.g. abelian-by-polycyclic group W/LW is an RF group, and so, by Remark 2.10, LW is closed in the profinite topology of W. We claim that L=LW∩A(R). It follows from this claim that L is a closed subgroup of W containing H∩A(R) but not g, whence g is not in ClW⁢(H∩A(R)). Hence, H∩A(R) is closed in the profinite topology of W, and so by Lemma 3.5, H is closed in the profinite topology of W.

It remains to establish the preceding claim. As B is an abelian normal subgroup of W and L≤A(R)≤B, B normalizes L and LW=LB⋉Q=LQ⊆B where

Let (f,1)∈LQ∩A(R) and

where (fi,1)∈L and qi∈Q for every i with 1≤i≤m. For every element x∈Supp⁢(f)⊆R, there exists an i∈{1,…,m} such that fiqi⁢(x)=fi⁢(qi-1⁢x) is nontrivial. As (fi,1) is an element of L⊆A(R), we must have qi∈R. Hence, (fiqi,1) is in L.

Since B is an abelian group, we can write (f,1)=(g1,1)⁢(g2,1) such that (g1,1)∈L and (g2,1)=(fi1qi1,1)⁢…⁢(fikqik,1) where qij∉R. For each element x∈Supp⁢(f) if g2⁢(x) is nontrivial, then by similar arguments as above there is j∈{1,…,k} such that qij∈R, which is in contradiction with the choice of g2; whence, we have g2⁢(x)=1. For each x∉Supp⁢(f), we have two cases:

1. g1⁢(x)=g2⁢(x)=1,

2. g1⁢(x)=g2-1⁢(x)≠1.

In case (ii), since Supp⁢(g1)⊆R, by a similar argument as above there is an index j∈{1,…,k} such that qij∈R, which again contradicts the choice of g2. Therefore, g2 is trivial and f∈L. So, LQ∩A(R) is contained in L. The reverse inclusion is trivial. ∎

Corollary 3.9.

A f.g. free metabelian group is LERF.

Proof.

Via the Magnus embedding, the free metabelian group can be embedded in a group A≀Q where A and Q are f.g. abelian groups. The corollary follows now since the class of LERF groups is closed under taking subgroups. ∎

Theorem 3.10 ([17, Statement 6.1.11]).

Every f.g. residually finite group is Hopfian.

Lemma 3.11.

Let A be a f.g. abelian group and let Q be an RF group. Then every f.g. subgroup H of A(Q) is closed in the profinite topology of W=A≀Q.

Proof.

By Lemma 3.7, there is a finite subset T of Q and a finite index subgroup N of Q such that H≤A(T)≀{1}≤A(T)≀N and A(T)≀N is a finite index subgroup of W. By Remark 2.8, it is enough to show that H is closed in the profinite topology of AT≀N.

By Lemma 2.4, there is a projection φ:A(T)≀N→A(T)≀{1} such that

By Theorem 3.10, the f.g. and RF group A(T)≀{1} is Hopfian. Hence,

which entails that AT≀N=ker⁡φ⋉(A(T)≀{1}). By Theorem 2.2, W is RF and so, the subgroup AT≀N is RF. Hence, by Lemma 2.9, the induced topology on A(T)≀{1} from AT≀N coincides with its own profinite topology and A(T)≀{1} is closed in the profinite topology of AT≀N. Since A(T)≀{1} is a f.g. abelian group and so, a LERF group, it follows that H is closed in A(T)≀N. ∎

Lemma 4.1.

Let G be a group with a normal subgroup N such that G/N is an ERF (respectively a LERF) group. Let H=〈B,t〉 be a subgroup of G which is a strictly ascending HNN-extension with (respectively f.g.) base B and stable letter t. Then the subgroup 〈B∩N,t〉 is also a strictly ascending HNN-extension.

Proof.

Let C=B∩N. Since N is a normal subgroup of G,

Let us suppose that t⁢C⁢t-1=C. Since G/N is an ERF (respectively a LERF) group, the subgroup B⁢N/N is closed in the profinite topology of G/N and so by Theorem 2.11, we have

Hence, for every b∈B, there is an element b0∈B such that t⁢b0⁢t-1⁢N=b⁢N and so t⁢b0-1⁢t-1⁢b∈N∩B=C as t⁢B⁢t-1<B. Since we assumed that t⁢C⁢t-1=C, there is c0∈C such that t⁢b0-1⁢t-1⁢b=t⁢c0⁢t-1, whence b=t⁢b0⁢c0⁢t-1∈t⁢B⁢t-1, which entails that B is equal to t⁢B⁢t-1 in contradiction with the assumption that t⁢B⁢t-1<B. Thus, we have t⁢C⁢t-1<C. ∎

Theorem 4.2.

The free solvable group Sn,3(n≥2) is not LERF but it does not contain a strictly ascending HNN-extension H=〈t,B〉 where B is a f.g. subgroup.

Proof.

Let A be the f.g. free abelian group of rank n, Sn,2 be the free metabelian group of rank n, and let W=A≀Sn,2. By the Magnus embedding, Sn,3 can be embedded in W. We will show that W does not contain a strictly ascending HNN-extension with a f.g. base. We assume by contradiction that the group H=〈B,t∣t⁢B⁢t-1⊂B〉 is a strictly ascending HNN-extension with f.g. base B. Let π:W→Sn,2 be the natural projection, N=ker⁡π, R=π⁢(B), t1=π⁢(t). Since, by Corollary 3.9, Sn,2 is a LERF group, by Lemma 4.1, 〈B∩N,t〉 is a strictly ascending HNN-extension. If t belongs to N, then as N is an abelian group, B∩N=(B∩N)t which contradicts the fact that 〈B∩N,t〉 is a strictly ascending HNN-extension. Hence, we have that π⁢(t) is nontrivial. Since Sn,2 is a LERF group, R is a closed subgroup of Sn,2. So, Theorem 2.11 implies that the equalities R=π⁢(B)=π⁢(Bt)=Rt1 hold. Therefore, R is a normal subgroup of π⁢(H)=〈R,t1〉. The f.g. subgroup 〈R,t1〉 is closed in the profinite topology of the LERF group Sn,2. Hence, by Lemma 3.7 there is a finite set T of Sn,2 such that H≤A(T)≀〈R,t1〉. Since π⁢(B)=R, it follows that B is contained in A(T)⁢(〈R,t1〉)⋉R. Let A1=A(T) and W1=A1≀〈R,t1〉.

Suppose that there is a positive integer j such that t1j∈R. Then A1〈R,t1〉⋉R is a finite index subgroup of the group W1 containing B. Let

be the natural projection. We observe that π1 is a restriction of π to the subgroup A1〈R,t1〉⋉R. Since π1⁢(B)=R, we have

Since A1〈R,t1〉 is an abelian group, B∩A1〈R,t1〉 is a normal subgroup of A1〈R,t1〉⋉R. We have

So, we obtain the following equalities:

where the last equality holds because t=(f,t1) for some f∈N and N is an abelian group. Since (B∩N)t is contained in (B∩N)tj, we have

which contradicts the assumption that 〈B∩N,t〉 is a strictly ascending HNN-extension. We thus conclude that R∩〈t1〉={1} and so 〈R,t1〉=R⋉〈t1〉.

Let B=〈(f1,(r1,1)),…,(fm,(rm,1))〉 and

for some elements f1,…,fm∈A1(R⋉〈t1〉), (r1′,t1α1),…,(rk′,t1αk)∈R⋉〈t1〉, and r1,…,rm∈R. Let

The subset A1(T′) is a subgroup of A1(R⋉〈t1〉) because for every f and g in A1(R⋉〈t1〉) we always have

1. Supp⁢(f)=Supp⁢(f-1),

2. Supp⁢(f⁢g)⊆Supp⁢(f)∪Supp⁢(g).

The group A1(T′)≀R is a subgroup of W1 and every generator of B is in this subgroup, whence the group B is contained in A1(T′)≀R. Let α be an element of the set {α1,…,αk} such that for every 1≤i≤k, |α|≥|αi|. We take (g,(1,1)) in B∩N=B∩A1(T′)⁢(R). Since N is an abelian group and t=(f,(1,t1)) for some f∈N, we have

Hence, for every positive integer i, the element

is in B∩N. Let (r,t1αj)∈Supp⁢(g) with r∈R and t1αj∈T′. For every positive integer i, we have that the element (rt1i,t1αj+i) is in the support of g(1,t1i) because

We can take the positive integer i large enough so that |αj+i|>|α|. Hence, the element (g(1,t1i),1) is not in B∩N=B∩A1(T′)⁢(R), whence ti⁢(B∩N)⁢t-i is not contained in B∩N. This contradicts the fact that 〈B∩N,t〉 is a strictly ascending HNN-extension. ∎