Ascending HNN extensions of polycyclic groups have the same cohomology as their profinite completions
aa r X i v : . [ m a t h . G R ] N ov Ascending HNN extensions of polycyclic groups have thesame cohomology as their profinite completions
Karl Lorensen
Mathematics DepartmentPennsylvania State University, Altoona CollegeAltoona, PA 16601-3760USAe-mail: [email protected]
November 23, 2018
Abstract
Assume G is a polycyclic group and φ : G → G an endomorphism. Let G ∗ φ bethe ascending HNN extension of G with respect to φ ; that is, G ∗ φ is given by thepresentation G ∗ φ = h G, t | t − gt = φ ( g ) for all g ∈ G i . Furthermore, let d G ∗ φ be the profinite completion of G ∗ φ . We prove that, for any finite,discrete d G ∗ φ -module A , the map H ∗ ( d G ∗ φ , A ) → H ∗ ( G ∗ φ , A ) induced by the canonicalmap G ∗ φ → d G ∗ φ is an isomorphism. Mathematics Subject Classification (2010) : 20JO6, 20E18, 20E06 If φ : G → G is a group monomorphism, the ascending HNN extension of G with respect to φ , denoted G ∗ φ , is defined by G ∗ φ = h G, t | t − gt = φ ( g ) for all g ∈ G i . This paper is concerned with ascending HNN extensions of polycyclic groups. These typesof ascending HNN extensions merit study for an important reason: they comprise preciselythose finitely generated solvable groups whose finitely generated subgroups are all finitelypresented. This characterization is established in [ ], where the structure of these groups isexamined in detail. Another salient property manifested by every ascending HNN extensionof a polycyclic group is that of residual finiteness, proved in [ ].Our goal is to prove the following result about the profinite completion d G ∗ φ of G ∗ φ if G is polycyclic. 1 heorem. Let G be a polycyclic group and φ : G → G a monomorphism. Then, for anyfinite, discrete d G ∗ φ -module A , the map H n ( d G ∗ φ , A ) → H n ( G ∗ φ , A ) (1.1) induced by the canonical map G ∗ φ → d G ∗ φ is an isomorphism for all n ≥ . Such groups whose cohomology coincides with that of their profinite completions for all finitecoefficient modules were described as “good groups” by J-P. Serre in [ , Exercise 2, Chapter2], an appellation that has persisted to this day. These groups have sparked a great dealof interest recently, partly due to their applications to geometry; see, for instance, [ ], [ ],and [ ]. As reflected by the examples in these three references, the property of “goodness”often accompanies strong forms of residual finiteness, like subgroup separability or cyclicsubgroup separability.Among the most elementary examples of “good groups” are free groups and polycyclicgroups. On the surface, ascending HNN extensions of these types of groups would appear tobe unlikely candidates for Serre’s cohomological property, since, although residually finite,they are not, in general, cyclic subgroup separable. However, in [ ] it was established that,if G is a finitely generated free group and φ : G → G a monomorphism, then G ∗ φ is “good.”Moreover, strong evidence that this might also be true if G is polycyclic was adduced in [ ],where the author showed that, in this case, the map (1.1) is an isomorphism for n = 2.Our proof of the above theorem, presented in Section 3, begins with the observation that,for an arbitrary group G , G ∗ φ = G φ ⋊ Z , where G φ is the direct limit of the sequence G φ → G φ → G φ → · · · . Furthermore, if G is finitely generated, then d G ∗ φ = c G φ ⋊ ˆ Z , where c G φ is the profinitecompletion of G φ . From the Mayer-Vietoris sequences arising from these two semidirectproduct decompositions, we can see that, in order to prove that G ∗ φ is “good,” it sufficesto show that G φ is “good.” We prove the latter assertion for G polycyclic by induction onthe solvability length of G . Of pivotal importance for both the base case and the inductivestep is that G ’s being finitely generated ensures that G φ is finitely generated as a topologicalgroup with respect to its profinite topology. In fact, with the aid of this observation, theinductive step can be accomplished by a very routine argument. The base case, where G isabelian, on the other hand, presents considerably more difficulty, requiring a shift of focusfrom cohomology to homology via the universal coefficient theorem. The key ingredient inthe proof of this case turns out to be the fact that the profinite completion of H n ( G φ , Z ) isisomorphic to H n ( c G φ , ˆ Z ), provided G is free abelian of finite rank. In order to establish thisproperty, we employ the relation between homology and exterior powers, which forms thesubject of Section 2.The identification of this new class of residually finite, solvable groups that enjoy Serre’scohomological property invites the question whether there are perhaps other varieties of“good” solvable groups, still awaiting discovery. One natural class to consider next is that ofresidually finite, solvable, minimax groups, which includes all the ascending HNN extensionsof polycyclic groups. As our final result, we prove that every group in this larger class isindeed “good” as well. Theorem.
Let G be a residually finite, solvable, minimax group. Then, for any finite,discrete ˆ G -module A , the map H n ( ˆ G, A ) → H n ( G, A ) (1.2)2 nduced by the canonical map G → ˆ G is an isomorphism for all n ≥ . Semantic conventions.
When using the term “group,” we will mean an abstract group;profinite groups will always be identified with the adjective “profinite.”When employing the cohomology and homology of a profinite group, we will always meancontinuous cohomology and profinite homology, respectively. Moreover, the same conventionsapply to the Ext and Tor functors.The profinite topology on a group is the topology whose basis at the identity consists of allthe normal subgroups of finite index. A group G is finitely generated relative to its profinitetopology if it is finitely generated as a topological group, where the topology employed is theprofinite topology. This is equivalent to the assertion that there exist g , · · · , g n ∈ G suchthat, for every epimorphism ǫ from G onto a finite group F , ǫ ( g ) , · · · , ǫ ( g n ) generate F .When we refer to a “finitely generated profinite group,” we will always mean finitelygenerated in the topological sense.If G is a group, then ˆ G denotes its profinite completion and c G : G → ˆ G the completionmap.A graded ring R ∗ is a sequence ( R n ) ∞ n =0 of additive abelian groups such that R = Z ,together with bilinear product maps R i × R j → R i + j for all i, j ∈ N that obey the associativeproperty. If R ∗ is a graded ring such that, for each odd natural number n , x = 0 for all x ∈ R n , then R ∗ is strictly anticommutative .A profinite graded ring Ω ∗ is a sequence (Ω n ) ∞ n =0 of additive profinite abelian groupssuch that Ω = ˆ Z , together with continuous, bilinear product maps Ω i × Ω j → Ω i + j for all i, j ∈ N , obeying the associative property.For a group G we denote the homology group H n ( G, Z ), where the action of G on Z istrivial, by H n ( G ). Similarly, if Γ is a profinite group, then H n (Γ) represents H n (Γ , ˆ Z ), wherethe action of Γ on ˆ Z is trivial.Although in the introduction we only referred to ascending HNN extensions with respectto monomorphisms, in the body of the paper we will form these constructions with respectto arbitrary endomorphisms. Hence, if φ : G → G is a group endomorphism, G ∗ φ = h G, t | t − gt = φ ( g ) for all g ∈ G i . The notation G ∗ φ that we employ is borrowed from [ ]. This section is devoted to proving the formula \ H n ( G ) ∼ = H n ( ˆ G ) (2.1)if G is a torsion-free, abelian group that is finitely generated with respect to its profinitetopology. This formula will play an important role in the proof of the main theorem inSection 3. The proof of (2.1) is based on the connection between homology and exteriorpowers, for both abstract and profinite abelian groups.If G is an abelian group, then we denote the exterior power ring of G by V ∗ G . This is agraded ring that can be represented in each positive dimension n as the quotient of N ni =1 G by3he subgroup generated by all elements of the form g ⊗ · · · ⊗ g n such that g i = g i +1 for some i , with the multiplication defined by extending the tensoring operation linearly. The exteriorpower ring, then, is strictly anticommutative and enjoys the following universal property: forany strictly anticommutative graded ring R ∗ and group homomorphism θ : G → R , thereexists a unique graded ring homomorphism φ ∗ : V ∗ G → R ∗ such that φ = θ . H. Cartan [ ]established the following connection between the integral homology of a finitely generated,torsion-free abelian group and its exterior power; for a more contemporary proof, in English,see [ , p. 123]. Theorem 2.1. (Cartan) If G is a finitely generated, torsion-free abelian group and H ∗ ( G ) is regarded as a graded ring using the Pontryagin product, then H ∗ ( G ) ∼ = V ∗ G. For a profinite abelian group Γ we represent the profinite exterior power ring of Γ byˆ V ∗ Γ; it is a strictly anticommutative graded profinite ring with the following universalproperty: for any strictly anticommutative graded profinite ring Ω ∗ and any continuous grouphomomorphism θ : G → Ω , there exists a unique continuous homomorphism of graded rings φ ∗ : V ∗ G → Ω ∗ such that φ = θ . As described in [ , p. 131], ˆ V n Γ can be constructedas the completion of V n Γ with respect to the kernels of all the maps V n Γ → V n Γ /N for N (cid:2) o Γ. Alternatively, it may viewed as the quotient of the completed tensor product ˆ N n Γby the closed subgroup generated by all elements of the form g ˆ ⊗ · · · ˆ ⊗ g n such that g i = g i +1 for some i . As established below, the profinite exterior power coincides with the abstractexterior power for a finitely generated profinite abelian group. Proposition 2.2. If Γ is a finitely generated profinite abelian group, then the canonical map V ∗ Γ → ˆ V ∗ Γ is an isomorphism of graded rings.Proof. We will prove by induction that V n Γ ∼ = ˆ V n Γ for every n ∈ N . Consider the canonicalgroup epimomorphism φ : V n − Γ ⊗ Γ → V n Γ . By the inductive hypothesis, V n − Γ is a finitely generated profinite group. Hence, by [ ,Proposition 5.5.3(d)], the domain of φ is a profinite group. Thus V n Γ is compact in theprofinite topology, yielding the desired result.We wish to prove a profinite analogue of Cartan’s theorem. In order to do so, we requirethe following K¨unneth formula.
Theorem 2.3.
Let Γ and Γ be profinite groups. Then there is an exact sequence −−−−→ L i + j = n H i (Γ ) ˆ ⊗ H j (Γ ) −−−−→ H n (Γ × Γ ) −−−−→ L i + j = n − Tor( H i (Γ ) , H j (Γ )) −−−−→ . (2.2)Although well known, the above formula does not appear to be proven anywhere in theliterature. Nevertheless, the proof of the abstract version, as presented, for example, in [ ,Proposition 6.1.13], can be carried over with ease to the profinite realm.To prove our profinite version of Cartan’s theorem, we also need the following property ofprofinite exterior powers, which may be proved in the same manner as the analogous resultfor abstract groups; see [ , p. 122]. 4 emma 2.4. If Γ and Γ are profinite abelian groups, then ˆ V n (Γ ⊕ Γ ) ∼ = M i + j = n ˆ V i Γ ˆ ⊗ ˆ V j Γ for every n ≥ . Armed with the above two results, we can readily prove the desired formula for thehomology of a finitely generated profinite abelian group.
Theorem 2.5. If Γ is a torsion-free, finitely generated profinite abelian group, then H n (Γ) ∼ = V n Γ for all n ≥ .Proof. For n = 0 , n ≥
2. The profinite group Γcan be expressed as the direct sum of finitely many infinite procyclic groups. We prove theresult by induction on the number of procyclic groups in this decomposition. First assumeΓ is itself procyclic. Then Γ is a projective profinite group, making H n (Γ) = 0 for all n ≥ V n Γ is trivial if n ≥
2, thus confirming the result. Now assume Γ = Γ ⊕ Γ ,where Γ is infinite procyclic. Applying Theorem 2.3 to Γ, we have that the fourth term insequence (2.2) is trivial. Therefore, by the inductive hypothesis together with Lemma 2.4,we have that H n (Γ) ∼ = ˆ V n Γ = V n Γ.In order to use the above theorem to prove formula (2.1), we need to establish that [ V n G ∼ = V n ˆ G. (2.3)if G is an abelian group that is finitely generated relative to its profinite topology. The proofof (2.3) is based on the following lemma about graded rings. Lemma 2.6.
Let R ∗ be a strictly anticommutative graded ring such that R n is finitelygenerated in its profinite topology for each n ∈ N . Then the following two statements hold.(i) The family { c R n | n ≥ } of profinite abelian groups can be made into a strictly anticom-mutative graded profinite ring so that the completion maps c n : R n → c R n constitute a gradedring homomorphism.(ii) For any strictly anticommutative graded profinite ring Ω ∗ and graded ring homomorphism φ ∗ : R ∗ → Ω ∗ , there exists a unique continuous graded ring homomorphism ψ ∗ : c R ∗ → Ω ∗ such that ψ ∗ c ∗ = φ ∗ .Proof. (i). Let f ij : R i × R j → R i + j be the function arising from the multiplication in R ∗ .We may obtain a function g ij : c R i × R j → [ R i + j extending f ij such that, for each y ∈ R j , g ij ( , y ) is a continuous homomorphism c R i → [ R i + j . We claim that g ij is continuous, where R j is given the profinite topology. This will follow if we can show that g − ij ( ¯ N + a ) is open in c R i × R j for any ¯ N (cid:2) o [ R i + j and a ∈ [ R i + j . For each y ∈ R j , set ¯ M y = { x ∈ c R i | g ij ( x, y ) ∈ ¯ N } .5ince the map x ¯ N + g ij ( x, y ) from c R i to [ R i + j / ¯ N is a continuous homomorphism withkernel ¯ M y , we have that [ c R i : ¯ M y ] divides \ [ R i + j : ¯ N ] for every y ∈ R j . However, c R i ,being a finitely generated profinite group, possesses only finitely many open subgroups ofany given index, which means that the set { ¯ M y : y ∈ R j } is finite. Thus ¯ M = T y ∈ R j ¯ M y isan open subgroup of b R i . Next take N to be the preimage of ¯ N under the completion map R i + j → [ R i + j . Proceeding in a fashion similar to above, we let P x = { y ∈ R j | f ij ( x, y ) ∈ N } for each x ∈ R i , obtaining that P = T x ∈ R i P x is an open subgroup of R j .Now assume ( b, c ) ∈ g − ij ( ¯ N + a ). Let b ′ be an element of the image of R i in c R i such that¯ M + b = ¯ M + b ′ . For any m ∈ ¯ M and p ∈ P , g ij ( m + b ′ , p + c ) = g ij ( m, p + c ) + g ij ( b ′ , p + c ) = g ij ( m, p + c ) + g ij ( b ′ , p ) + g ij ( b ′ , c ) . Since g ij ( m, p + c ) ∈ ¯ N , g ij ( b ′ , p ) ∈ ¯ N , and g ij ( b ′ , c ) ∈ ¯ N + a , we have that g ij ( m + b ′ , p + c ) ∈ ¯ N + a . Thus ( ¯ M + b ) × ( P + c ) ≤ g − ij ( ¯ N + a ). Therefore, g − ij ( ¯ N + a ) is open in c R i × R j .It follows, then, that g ij is continuous. As a consequence, we can deduce that g ij is linearin the second component by virtue of its being linear there on a dense subset. This allowsus to extend g ij to a function h ij : c R i × c R j → [ R i + j such that, for each x ∈ c R i , h ij ( x, )is a continuous homomorphism c R j → [ R i + j . By reasoning like we did above for g ij , wecan conclude that h ij is continuous and, therefore, bilinear. The maps h ij , then, furnishthe desired product on the family of groups { R n | n ∈ N } , the associativity and strictanticommutativity following from the fact that these properties hold on dense subsets.(ii). The universal property of the profinite completion yields a family { ψ n : c R n → Ω n | n ∈ N } of continuous group homomorphisms such that ψ n c n = φ n . Moreover, for all( x, y ) ∈ c R i × c R j , ψ i + j ( xy ) = ψ i ( x ) ψ j ( y ), since this identity holds on a dense subset.Now we are prepared to prove formula (2.3). Proposition 2.7.
Assume G is an abelian group that is finitely generated with respect to itsprofinite topology. Then the following two statements hold.(i) The family of profinite abelian groups { [ V n G | n ≥ } can be made into a strictly anti-commutative graded profinite ring so that the completion maps c n : V n G → [ V n G constitutea graded ring homomorphism.(ii) The graded ring homomorphism V ∗ G → V ∗ ˆ G arising from c G : G → ˆ G induces acontinuous graded profinite ring isomorphism [ V ∗ G → V ∗ ˆ G. Proof.
Throughout the proof, we will make repeated use of the fact that V ∗ ˆ G = ˆ V ∗ ˆ G ,which follows from Proposition 2.2. Since G is finitely generated relative to its profinitetopology, the same is true for V n G for n >
1. Hence statement (i) follows by the precedingproposition. Moreover, by the universal property of ˆ V ∗ ˆ G , there is a continuous graded ringhomomorphism φ ∗ : V ∗ ˆ G → [ V ∗ G such that φ is just the identity map ˆ G → ˆ G . In addition,we have a graded ring homomorphism V ∗ G → V ∗ ˆ G , which, according to Lemma 2.6(ii),induces a continuous graded ring homomorphism ψ ∗ : [ V ∗ G → V ∗ ˆ G such that ψ is the6dentity map ˆ G → ˆ G . We claim that ψ ∗ φ ∗ is the identity map V ∗ ˆ G → V ∗ ˆ G , and that φ ∗ ψ ∗ is the identity map [ V ∗ G → [ V ∗ G . The first assertion follows immediately from the universalproperty of ˆ V ∗ ˆ G since ψ φ is the identity map ˆ G → ˆ G . To verify the second, consider thecomposition V ∗ G c ∗ −−−−→ [ V ∗ G φ ∗ ψ ∗ −−−−→ [ V ∗ G. The universal property of the exterior power ensures that this composition is the completionmap since that is its form in dimension one. Consequently, by the universal property of theprofinite completion, φ ∗ ψ ∗ can only be the identity map.In order to deduce formula (2.1) from the above proposition, we still require the fact thatthe profinite completion of a torsion-free abelian group is itself torsion-free. This propertymay be established by the same argument used to prove [ , Proposition 2.1]. Notice thatfor our result, as opposed to the formulation in [ ], it is not necessary to assume that G isresidually finite. Lemma 2.8. (P. Kropholler and J. Wilson) If G is a torsion-free abelian group, then ˆ G isalso torsion-free.Proof. Assume ¯ x ∈ ˆ G such that n ¯ x = 0 for some integer n = 0. Let N be a subgroup offinite index in G , and let x ∈ G such that N + x is the image of ¯ x in G/N . Then nx ∈ N .We claim that, in fact, nx ∈ nN . To show this, suppose otherwise. As an abelian groupwith finite exponent, N/nN is a direct sum of finite cyclic groups and thus residually finite.Hence N contains a subgroup M of finite index such that nN ≤ M and nx / ∈ M . Now let y ∈ G such that M + y is the image of ¯ x in G/M . Hence ny ∈ M and x − y ∈ N . It followsthat n ( x − y ) ∈ nN ≤ M , so that nx ∈ M , a contradiction. Therefore, nx ∈ nN , whichmeans, since G is torsion-free, that x ∈ N . Since N was an arbitrary subgroup of finiteindex in G , we can conclude that ¯ x = 0. Consequently, ˆ G is torsion-free.Now we are prepared to prove formula (2.1). Theorem 2.9.
Let G be a torsion-free abelian group that is finitely generated with respectto its profinite topology. Then, for each n ≥ , the map \ H n ( G ) → H n ( ˆ G ) induced by c G : G → ˆ G is an isomorphism.Proof. By Lemma 2.8, ˆ G is torsion-free. In view of this, the result follows immediately fromTheorem 2.1, Theorem 2.5 and Proposition 2.7. We begin by defining the class of groups that is the focus of this section.
Definition.
Define G to be the class of groups G such that, for each n ≥ G -module A , the following two properties hold:(i) the group H n ( G, A ) is finite;(ii) the map c G : G → ˆ G induces an isomorphism H n ( ˆ G, A ) → H n ( G, A ).7ur objective is to prove that every ascending HNN extension of a polycyclic group isin G . In our proof we will make use of the fact that G is closed under the formation of thefollowing type of group extension. Proposition 3.1.
Let −→ N −→ G −→ Q −→ be a group extension such that N isfinitely generated in its profinite topology. If N and Q are both in G , then G belongs to G . The essential details of the proof of the above proposition are provided by Serre [ ,Exercise 2, Chapter 2], though with one important difference between the hypotheses: inplace of our condition on N , Serre assumes that N is finitely generated as an abstract group.An examination of Serre’s argument, however, reveals that all that is really required is that N has only finitely many subgroups of any given finite index, a property that also holds inthe presence of our weaker condition on N .In analysing ascending HNN extensions, the following species of direct limit shall playan important role. Definition. If G is a group and φ : G → G an endomorphism, then G φ is the direct limitof the sequence G φ → G φ → G φ → · · · . The above variety of direct limit enjoys the following property, which will be highlysignificant for our proof that every ascending HNN extension of a polycyclic group is in G . Lemma 3.2. If G is a finitely generated group and φ : G → G an endomorphism, then G φ is finitely generated relative to its profinite topology.Proof. We have that G φ is the direct limit of the sequence G φ → G φ → G φ → · · · . (3.1)Let ǫ : G φ → F be an epimorphism, where F is a finite group. For each i ∈ N , let N i be thesubgroup of G formed by intersecting Ker ǫ with the copy of G occupying the i -th spot inthe sequence (3.1). We have, then, that φ ( N i ) ≤ N i +1 for all i ∈ N , and that φ induces amonomorphism G/N i → G/N i +1 for all i ∈ N . It follows from the finiteness of F that thereexists k ∈ N such that the map φ : G → G induces an isomorphism G/N i → G/N i +1 forall i ≥ k . Moreover, invoking the fact that G has only finitely many subgroups of any givenfinite index, we can conclude that there is an l ≥ k such that N i = N l for infinitely many i ≥ k . To simplify the notation, we let N = N l .For each nonnegative integer j , let φ − j ( N ) = { x ∈ G : φ j ( x ) ∈ N } , where φ is understood to be the identity map from G to G . Now set M = T ∞ j =0 φ − j ( N ).It is easy to see that M ≤ N i for all i ∈ N , M (cid:2) G , and φ ( M ) ≤ M . We claim that, inaddition, [ G : M ] < ∞ and the map G/M → G/M induced by φ is an isomorphism. Toestablish the former assertion, we first observe that, for each j ≥
0, [ G : φ − j ( N )] ≤ [ G : N ] . Since G has only finitely many subgroups with index ≤ [ G : N ], it follows that there are onlyfinitely many subgroups of the form φ − j ( N ) for j ≥
0. Therefore, M , as the intersection offinitely many subgroups with finite index, has finite index. Turning now to prove our second8ssertion about M , we let x ∈ G such that φ ( x ) ∈ M . From the definition of N , we havethat, for some n > φ n ( N ) ≤ N and φ n induces an isomorphism G/N → G/N . Also, forany j ≥ φ n + j ( x ) ∈ N , implying that φ j ( x ) ∈ N . Hence x ∈ M . Therefore, the map G/M → G/M induced by φ is an isomorphism.Now let φ ′ be the map M → M induced by φ . Treating M φ ′ as a subgroup of G φ , wehave M φ ′ ≤ Ker ǫ . Moreover, any element of G φ is congruent modulo M φ ′ to an elementof the first G in the sequence (3.1). It follows, then, that the image of the first copy of G under ǫ is the entire group F . Consequently, we can conclude that G φ is finitely generatedrelative to its profinite topology.Below we establish the connection between the groups G ∗ φ and G φ . Lemma 3.3. If G is a group and φ : G → G an endomorphism, then G ∗ φ ∼ = G φ ⋊ Z . Proof.
Each element of G ∗ φ can be written in the form t i gt − j , where i and j are nonnegativeintegers and g ∈ G . Thus G ∗ φ is the product of the normal subgroup S ∞ i =0 t i Gt − i with thesubgroup h t i . Moreover, the commutative diagram G φ −−−−→ G φ −−−−→ G φ −−−−→ · · · y θ y θ θ y G ⊂ −−−−→ tGt − ⊂ −−−−→ t Gt − ⊂ −−−−→ · · · , where θ i ( g ) = t i gt − i , reveals that G φ ∼ = S ∞ i =0 t i Gt − i . Hence the result follows.The above decomposition, combined with Lemma 3.2 and Proposition 3.1, yields thefollowing corollary. Corollary 3.4.
Assume G is a finitely generated group and φ : G → G is an endomorphism.If G φ is in G , then G ∗ φ is also in G . Before proving our main theorem, we state a universal coefficient theorem for profinitegroups. Rather than provide a proof, we refer the reader to the proof of the abstract versionin [ , Theorem 3.6.5, Exercise 6.1.5], as it can easily be translated to the profinite context. Theorem 3.5.
Let Γ be a profinite group and A a trivial discrete Γ -module. Then, for any n ≥ , there is an exact sequence −−−−→ Ext( H n − (Γ) , A ) −−−−→ H n (Γ , A ) −−−−→ Hom( H n (Γ) , A ) −−−−→ Theorem 3.6. If G is a polycyclic group and φ : G → G an endomorphism, then G ∗ φ is inthe class G .Proof. By Corollary 3.4, it suffices to show that G φ belongs to G . We prove this assertionby induction on the length of the derived series of G . First assume G is abelian. Taking A to be a finite, discrete c G φ -module, we wish to establish the following two properties:9i) H n ( G φ , A ) is finite;(ii) H n ( c G φ , A ) ∼ = H n ( G φ , A ) for all n ≥ G is torsion-free andthe action of G φ on A is trivial. The two properties are clearly true for n = 0, so we willassume that n ≥
1. In this case, the universal coefficient formulas yield the commutativediagram0 −−−−→
Ext( H n − ( c G φ ) , A ) −−−−→ H n ( c G φ , A ) −−−−→ Hom( H n ( c G φ ) , A ) −−−−→ y y y −−−−→ Ext( H n − ( G φ ) , A ) −−−−→ H n ( G φ , A ) −−−−→ Hom( H n ( G φ ) , A ) −−−−→ c G φ must be torsion-free and topologicallyfinitely generated. Hence c G φ is a direct sum of finitely many infinite procyclic groups, whichmeans, by the K¨unneth formula, that the same property holds for H n − ( c G φ ). As a result,Ext( H n − ( c G φ ) , A ) = 0. Furthermore, H n − ( G ) is torsion-free, which implies, since homologycommutes with direct limits, that H n − ( G φ ) is also torsion-free. Because A is finite, thisyields that Ext( H n − ( G φ ) , A ) = 0 (see Lemma 3.8 below). In addition, it follows fromTheorem 2.9 that the third vertical map in (3.2) is an isomorphism. Therefore, property (ii)holds. Also, since H n ( c G φ ) is a finitely generated profinite group, property (i) is true.Next we establish properties (i) and (ii) without the restriction that A is a trivial G φ -module, still assuming, however, that G is free abelian of finite rank. Let ω : G φ → Aut( A )be the homomorphism arising from the action of G φ on A . Arguing just as we did for themap ǫ in the proof of Lemma 3.2, we can find a subgroup M in G of finite index such that φ ( M ) ≤ M and M φ ′ ≤ Ker ω , where φ ′ : M → M is the map induced by φ . Now set Q = G/M , and let φ ′′ : Q → Q be the map induced by φ . Then there is an exact sequence1 → M φ ′ → G φ → Q φ ′′ → . (3.3)Noticing that Q φ ′′ is finite, we obtain from (3.3) an exact sequence1 → d M φ ′ → c G φ → Q φ ′′ → H n ( M φ ′ , A ) is finite and H n [ ( M φ ′ , A ) ∼ = H n ( M φ ′ , A ) for all n ≥
0. Thus, invoking theLyndon-Hochschild-Serre spectral sequences for (3.3) and (3.4), we can conclude that bothproperties (i) and (ii) hold. Therefore, G φ lies in G whenever G is free abelian of finite rank.We now treat the case where G is a finitely generated, abelian group that may containtorsion. In this case, G contains a torsion-free subgroup N such that G/N is finite. Taking M = T ∞ j =0 φ − j ( N ), we have M ≤ N , [ G : M ] < ∞ , and φ ( M ) ≤ M . Hence, letting Q , φ ′ and φ ′′ be exactly as in the previous paragraph, we have the exact sequences (3.3) and(3.4) in this case, too. Also, by the torsion-free case proved above, H n ( M φ ′ , A ) is finiteand H n [ ( M φ ′ , A ) ∼ = H n ( M φ ′ , A ) for all n ≥
0. Properties (i) and (ii), then, follow as above.Therefore, G φ belongs to G .Finally, we assume that the solvability length of G exceeds 1. Let N be the commutatorsubgroup of G and Q = G/N . Then φ ( N ) ≤ N . Let φ ′ : N → N and φ ′′ : Q → Q be the10aps induced by φ . Then we have an exact sequence1 → N φ ′ → G φ → Q φ ′′ → . By the base case, we have that Q φ ′′ is in G , and, by the inductive hypothesis, N φ ′ belongs to G . Moreover, by Lemma 3.2, N φ ′ is finitely generated with respect to its profinite topology.Therefore, by Proposition 3.1, G φ belongs to G .A scrutiny of the above proof, particularly its third paragraph, reveals that the argumentcan be easily extended to prove that every ascending HNN extension of a virtually polycyclicgroup is in G . Theorem 3.7. If G is a virtually polycyclic group and φ : G → G an endomorphism, then G ∗ φ is in the class G . It still remains to prove the following elementary result about abelian groups, which weinvoked in the proof of Theorem 3.6.
Lemma 3.8. If A is a torsion-free abelian group, then Ext(
A, B ) = 0 for any finite abeliangroup B .Proof. We will prove the conclusion by showing that Ext( A, Z /p ) = 0 for every prime p .Let A = A ⊗ Q , and consider the monomorphism A → A . This map induces an epimor-phism Ext( A , Z /p ) → Ext( A, Z /p ). Hence the result will follow if we can establish thatExt( A , Z /p ) = 0. To accomplish this, we employ the exact sequence0 → Z × p → Z → Z /p → , which gives rise to an exact sequenceExt( A , Z ) × p → Ext( A , Z ) → Ext( A , Z /p ) → . Moreover, since multiplication by p induces an isomorphism A → A , the first map in theabove sequence is an isomorphism, forcing the third group to be trivial.In [ , Theorem 1.4] it is shown that every ascending HNN extension of a polycyclicgroup is a minimax group. In light of this, the question poses itself whether the result inTheorem 3.6 extends to all residually finite, solvable, minimax groups. Applying the sameapproach employed in the proof of Theorem 3.6, we provide a positive answer to this questionin the following theorem. Theorem 3.9. If G is a residually finite, solvable, minimax group, then G is in G .Proof. We begin by treating the case where G is a residually finite, abelian, minimax group.As a minimax group, G decomposes as G = H ⊕ T , where T is the torsion subgroup of G and H is torsion-free– this is established in [ , Lemma 10.31(i)]. Moreover, since G isresidually finite, T must be finite. According to [ , Lemma 10.31(ii)], H is an extension ofa free abelian group of finite rank by a direct sum of finitely many quasicyclic groups. Sinceboth kernel and quotient in this extension are finitely generated relative to their profinitetopologies, it follows that H , too, enjoys this property. Thus, invoking the same argument11hat was employed in the first two paragraphs of the proof of Theorem 3.6 in order toestablish that G φ is in G , we can deduce that H must belong to the class G . Hence G residesin this class as well.Now we consider the general case. Since G is residually finite, it possesses an abeliannormal series consisting solely of subgroups that are closed in the profinite topology of G .We will prove that G is in G by inducting on the length of this series, the case of a seriesof length one having been disposed of above. Let A be the first subgroup in the series.Then A is in G by our base case. Since A is closed, G/A is residually finite, which means,in view of the inductive hypothesis, that it belongs to G . In addition, by our reasoning inthe first paragraph, A is finitely generated with regard to its profinite topology. Therefore,Proposition 3.1 yields that G is in G . Acknowledgements.
The author benefited from discussions with Pavel Zalesskii aboutprofinite completions of ascending HNN extensions. In addition, he is indebted to PeterSymonds for enlightening him regarding profinite homology.
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