A Note on the continuous extensions of injective morphisms between free groups to relatively free profinite groups
Abstract
Let V be a pseudovariety of finite groups such that free groups are residually V, and let ϕ: F(A) → F(B) be an injective morphism between finitely generated free groups. We characterize the situations where the continuous extension ˆϕ of ϕ between the pro-V completions of F(A) and F(B) is also injective. In particular, if V is extension-closed, this is the case if and only if ϕ(F(A)) and its pro-V closure in F(B) have the same rank. We examine a number of situations where the injectivity of ˆϕ can be asserted, or at least decided, and we draw a few corollaries.
Full text
Publ. Mat. 47 (2003), 477–487 A NOTE ON THE CONTINUOUS EXTENSIONS OF INJECTIVE MORPHISMS BETWEEN FREE GROUPS TO RELATIVELY FREE PROFINITE GROUPS Thierry Coulbois, Mark Sapir and Pascal Weil Abstract Let Vbeapseudovariety of finite groups such that free groups are residually V, and let ϕ:F(A)→F(B)beaninjective morphism between finitely generated free groups. We characterize the situations where the continuous extension ˆϕof ϕbetween the pro-V completions of F(A) and F(B)isalso injective. In particular, if V is extension-closed, this is the case if and only if ϕ(F(A)) and its pro-Vclosure in F(B)have the same rank. We examine a number of situations where the injectivity of ˆϕcan be asserted, or at least decided, and we draw a few corollaries. In this paper, we are interested in the pro-Vtopologies on finitely generated free groups, where Vis a pseudovariety of groups (a class of finite groups closed under taking subgroups, quotients and finite direct products). These topologies were introduced in the 1950s by Hall. When Vis the class of all finite groups, the finite index subgroups are exactly the open subgroups, and Hall proved [6] that every finitely generated subgroup is closed. More recent papers (Ribes and Zalesski˘ı[9], Margolis, Sapir and Weil [7], Weil [12]) focused on the problem of effectively computing the pro-Vclosure ClV(H)ofagiven finitely generated subgroup Hof a free group. It is known for instance that if V is extension-closed, then ClV(H) has finite rank, at most equal to the rank of H[9]. In general, a finite rank subgroup may have an infinite rank closure (e.g. if Vis the pseudovariety of finite abelian groups), or it may be the case that if His a finite rank subgroup, then its closure 2000 Mathematics Subject Classification. 20E18, 20E05. Key words. Profinite topology, rank of subgroups of free groups, monomorphisms between free groups. This paper was prepared while the second author was an invited Professor at Universit´e Bordeaux-1. The third author also acknowledges partial support from 1999 INTAS grant number 1224 Combinatorial and geometric theory of groups and semigroups and its applications to computer science.
478 T. Coulbois, M. Sapir, P. Weil always has finite rank, possibly greater than the rank of H(e.g. if Vis the pseudovariety of finite nilpotent groups [7]). It is interesting to note that deciding whether a given subgroup is pro-V-closed is equivalent to deciding an extension property for a certain set of partial isomorphisms of a finite set [7]. If Vconsists of all finite p-groups, for some fixed prime p, the closure of a given finite rank subgroup can be effectively computed [9], in polynomial time [7]. On the other hand, it is not known whether the pro-solvable closure of a finite rank subgroup is effectively computable; a positive solution for this difficult open question would have interesting consequences in finite monoid theory [7] and in computational complexity (Straubing and Th´erien [10]). It is also known that finite rank closed subgroups are free factors of clopen subgroups, and that the converse holds if Vis extensionclosed [9]. Moreover, again in the extension-closed case, if His a finite rank pro-V-closed subgroup of a free group F, then the pro-Vtopology of Hcoincides with the topology it inherits from F. The central result in this paper (Theorem 1.1) characterizes the situations where this property of coincidence of topologies holds: it is equivalent to another extension property, namely to the fact that a certain injective morphism between two free groups Fand Fadmits an injective continuous extension between the pro-Vcompletions of Fand F.Itturns out that in the extension-closed case, this is equivalent to the fact that Hand its pro-Vclosure have equal rank. After the proof of the main result, we list a number of immediate consequences: for instance, it follows from our result that if Vis extensionclosed, the continuous extension of an injective endomorphism of the free group of rank 2 is always injective. In the last section, we illustrate our result by considering a simple example of an injective morphism ϕ:F→Fbetween finitely generated free groups whose continuous extension ˆϕto the pro-pcompletions is not injective, and we exhibit a sequence (tn)nof elements of Fwhose limit points are non-trivial elements of ker ˆϕ. 1. Injective extendability If Ais an alphabet (that is, a finite non-empty set), then F(A) denotes the free group on A. Let Vbeapseudovariety of finite groups: the pro-V topology on a group Gis the least topology which makes every morphism from Ginto an element of Vcontinuous. A basis of neighborhoods of 1 in this topology is given by the finite-index normal subgroups Kof G such that G/K ∈V.
Continuous Extensions of Injective Morphisms 479 The pro-Vtopology on Gis Hausdorff if and only if Gis residually V. In that case, the pro-Vtopology on Gcan be defined by an ultrametric distance function. This situation arises in particular if Gis a free group and Vis a non-trivial extension-closed pseudovariety. In the sequel, we consider only pseudovarieties Vsuch that free groups are residually V. If Gis a group, we denote by ˆ Gthe pro-Vcompletion of G:itis compact and totally disconnected. If G=F(A), we write ˆ FV(A) for ˆ G; this is also the free pro-Vgroup on A.IfH⊆G,wewrite ClV(H) (or simply Cl(H)) for the closure of Hin G.IfH⊆ˆ G,wewrite Hfor the closure of Hin ˆ G.Inparticular, if H⊆G,H= Cl(H) and, if Gis residually V, then Cl(H)=H∩G. We note that every morphism ϕ:F→Fbetween free groups is uniformly continuous when both groups are equipped with their respective pro-Vtopologies. In particular, ϕadmits a (uniquely defined) continuous extension between the pro-Vcompletions, written ˆϕ:ˆ F→ˆ F. Forajustification of these assertions, we refer the readers, for instance, to [7]. We now consider injective morphisms, and we state our main result. Theorem 1.1. Let ϕ:F(A)→F(B)be an injective morphism and let H=ϕ(F(A)).LetVbeapseudovariety of groups such that free groups are residually V. The following conditions are equivalent: •The continuous extension of ϕ,ˆϕ:ˆ FV(A)→ˆ FV(B)is one-to-one. •The pro-Vtopology on Hcoincides with the topology on Hinduced by the pro-Vtopology on F(B). If, in addition, Vis extension-closed, these properties are equivalent to: •Hand Cl(H)have the same rank. The proof of Theorem 1.1 follows directly from Propositions 1.4, 1.6 and 1.8 below. 1.1. Comparing the pro-V topologies on a subgroup. We will use the following elementary remark. Lemma 1.2. Let ϕ:F(A)→F(B)beamorphism between finitely generated free groups, let H=ϕ(F(A)) be the range of ϕand let ˆϕ:ˆ FV(A)→ ˆ FV(B)be the continuous extension of ϕbetween the pro-Vcompletions of F(A)and F(B). Then the range of ˆϕis H. Proof: By continuity, we have ˆϕ(ˆ FV(A)) = ˆϕ(F(A)) ⊆ϕ(F(A)) = H. Of course, we also have H⊆ˆϕ(ˆ FV(A)). Finally, as ˆ FV(A)iscompact and ˆϕis continuous, the group ˆϕ(ˆ FV(A)) is closed, so ˆϕ(ˆ FV(A))=H.
480 T. Coulbois, M. Sapir, P. Weil Next we consider the following property of a finitely generated subgroup Hof a free group F(B): Property Coinc(V). The pro-Vtopology on Hcoincides with the topology on Hinduced by the pro-Vtopology on F(B). This property translates as follows. Lemma 1.3. Let Hbe afinitely generated subgroup of a free group F(B), let ı:H→F(B)be the natural injection of Hinto F(B), and let ˆı:ˆ H→ˆ FV(B)be the continuous extension of ıbetween the pro-Vcompletions of Hand F(B). Then Hhas Property Coinc(V)if and only if ˆı is injective. In particular, His homeomorphic to ˆ H. Proof: This is immediate once we observe that ˆıhas range H(by Lemma 1.2). We are now ready to prove the first equivalence in Theorem 1.1. Proposition 1.4. Let Vbeapseudovariety of groups such that free groups are residually V.Letϕ,ˆϕand Hbe as in the statement of Theorem 1.1. Then ˆϕis injective if and only if Hhas Property Coinc(V). Proof: Let ψ:F(A)→Hbe the restriction of ϕto an isomorphism between F(A) and H, and let ˆ ψ:ˆ FV(A)→ˆ Hbe the continuous extension of ψ.Asψis an isomorphism, ˆ ψis a homeomorphism. Let ı:H→F(B) and ˆı:ˆ H→ˆ FV(B)beasinLemma 1.3. We observe that ϕ=ı◦ψ,sothat ˆϕ=ˆı◦ˆ ψ.Asˆ ψis a homeomorphism, ˆϕ is one-to-one if and only if ˆıis. By Lemma 1.3, this is equivalent to H having Property Coinc(V), as we wanted to prove. 1.2. The extension-closed case. The following sufficient condition for a finitely generated subgroup to have Property Coinc(V)wasproved in[7, Proposition 2.17]: Proposition 1.5. If Vis extension-closed, then every finitely generated, closed subgroup of F(A)has Property Coinc(V). We will see that this sufficient condition is not necessary (Proposition 1.7 below). However, we can immediately use this property to prove one half of the remaining equivalence. Proposition 1.6. Let Vbe a non-trivial extension-closed pseudovariety of groups. Let ϕ:F(A)→F(B)be a morphism between free groups, let H=ϕ(F(A)), and let ˆϕbe the continuous extension of ϕbetween the pro-Vcompletions of F(A)and F(B).Ifϕand ˆϕare injective, then H and Cl(H)have equal ranks.
Continuous Extensions of Injective Morphisms 481 Proof: By Lemma 1.2, the range of ˆϕis H. Since ˆϕis injective between two compact spaces, it is a homeomorphism onto its image, so His homeomorphic to the free pro-Vgroup of rank |A|= rank(H). On the other hand, we know from Proposition 1.5 that Cl(H) has Property Coinc(V). Applying Lemma 1.3 to Cl(H), we find that Cl(H) is homeomorphic to the free pro-Vgroup of rank rank(Cl(H)). But H= Cl(H), so we have proved that rank(H)=rank(Cl(H)). Before we prove the reverse implication, we show the following result, which does not require the hypothesis that Vis extension-closed. Proposition 1.7. Let Hbeafinitely generated subgroup of the free group F(A).IfHis dense in the pro-Vtopology of F(A)and if rank(H)=rank(F(A)), then Hhas Property Coinc(V). Proof: As Hand F(A)have the same rank, we may consider an injective endomorphism ψof F(A) with range H. Let ˆ ψ:ˆ FV(A)→ˆ FV(A)bethe continuous extension of ψ.ByLemma 1.2, ˆ ψis an onto endomorphism of ˆ FV(A). But every onto continuous endomorphism of a finitely generated profinite group is injective [5, Proposition 15.3]. So ˆϕis injective: by Proposition 1.4, this implies that Hhas Property Coinc(V). We can now give the last element in the proof of Theorem 1.1. Proposition 1.8. Let Hbeafinitely generated subgroup of the free group F(A).IfVis extension-closed and rank(H)=rank(Cl(H)), then Hhas Property Coinc(V). Proof: By Proposition 1.5, the pro-Vtopology on Cl(H) coincides with the topology on Cl(H) induced by the pro-Vtopology on F(A). Therefore His dense in the pro-Vtopology on Cl(H). Now Proposition 1.7 implies that the pro-Vtopology on Hcoincides with the topology on Hinduced by the pro-Vtopology on Cl(H), and this concludes the proof. 2. Corollaries The following collection of remarks is immediately deduced from Theorem 1.1. Throughout this section, Vdenotes a pseudovariety of groups such that free groups are residually V,ϕ:F(A)→F(B)isaninjective morphism between free groups, ˆϕ:ˆ FV(A)→ˆ FV(B)isthe continuous extension of ϕbetween the pro-Vcompletions of F(A) and F(B), and H=ϕ(F(A)). Corollary 2.1. Whether ˆϕis injective depends only on H, not on ϕ.
482 T. Coulbois, M. Sapir, P. Weil If V=G, the pseudovariety of all finite groups, the pro-Vcompletion of a group is called its profinite completion. It is well-known that for the pro-Gtopology, every finitely generated subgroup of the free group is closed [6]. As a result, we have: Corollary 2.2. Every injective morphism between free groups of finite rank admits an injective continuous extension to the profinite completions of these groups. Let pbeaprime number and let Gpbe the pseudovariety of finite p-groups. The pro-Gpcompletion of a group is called its pro-pcompletion. It is shown in [9] that if pis a prime number, one can effectively compute the pro-pclosure of a finitely generated subgroup of the free group (see [7] for a polynomial time algorithm). It follows that: Corollary 2.3. Given a prime number p, one can decide whether the continuous extension of ϕto the pro-pcompletions is injective. Let Gsol be the pseudovariety of finite solvable groups; the pro-Gsol completion of a group is called its pro-solvable completion. It is also shown in [12] that one can compute the rank of the pro-solvable closure of a finite index subgroup: Corollary 2.4. If Hhas finite index, one can decide whether the continuous extension of ϕto the pro-solvable completions is injective. For the general case however, we do not know whether one can effectively compute the rank of the pro-solvable closure of a given finitely generated subgroup (see the conclusion of [7]or[12] for a discussion). In particular, we do not know whether the injectivity of the continuous extension of ϕto the pro-solvable completions is decidable. In [7], the pro-Vtopology is considered also when Vis the pseudovariety Gnil of finite nilpotent groups, a pseudovariety which is not extensionclosed. An example is given of a finitely generated subgroup which is closed in that topology yet does not have Property Coinc(V)[7, Example 1.10]. This shows that the extension-closed assumption in Proposition 1.8 cannot be dispensed with. However, we also know [7] that the pro-nilpotent completion of the free group (its pro-Gnil completion) is a subdirect product of its pro-pcompletions (pprime). Therefore we have: Corollary 2.5. The continuous extension of ϕto the pro-nilpotent completions is injective if and only if, for each prime p, the continuous extension of ϕto the pro-pcompletions is injective.
Continuous Extensions of Injective Morphisms 483 In view of Theorem 1.1, deciding the injectivity of the pro-nilpotent extension of ϕis equivalent to deciding whether every p-closure of H has the same rank as H. But there are only finitely many subgroups of F(A)ofthe form Clp(H), and they are effectively computable [7], [12]. It follows that: Corollary 2.6. It is decidable whether the extension of ϕto the pronilpotent completions is injective. Returning to extension-closed pseudovarieties, it is shown in [9] that rank(Cl(H)) ≤rank(H). As free groups and free pro-Vgroups of rank 1 are commutative, it follows that if Hhas rank 1 or 2, then rank(H)= rank(Cl(H)). This translates into the following result. Corollary 2.7. If Vis extension-closed and ϕis defined on the free group of rank 1or 2, then ˆϕis injective. This last result leads to the following consequences. Let Gbe the pseudovariety of all finite groups, Bbeafinite alphabet, and u∈ˆ FG(B). We say that a finite group Gsatisfies the pseudo-identity u=1if, for every continuous morphism ψ:ˆ FG(B)→G,wehave ψ(u)=1. The class of finite groups which satisfy a given set of pseudo-identities is a pseudovariety, and every pseudovariety can be defined in this fashion (Reiterman’s theorem, see [1]). Let A={a, b},Bbe a finite alphabet, ϕ:F(A)→F(B)bean injective morphism (that is, ϕ(a)ϕ(b)=ϕ(b)ϕ(a)) and ˆϕ:ˆ FG(A)→ ˆ FG(B)bethe continuous extension of ϕto the free profinite groups over Aand B. Corollary 2.8. Let Vbeanon-trivial extension-closed pseudovariety and let (ui)i∈Ibe acollection of elements of ˆ FG(A).IfVsatisfies the pseudo-identities ˆϕ(ui)=1, then Vsatisfies the pseudo-identities ui=1. To build from this result, let us observe that, if pis a prime number, it is immediate that a finite group is a p-group if and only if every one of its cyclic subgroup is a p-group. That is equivalent to saying that Gp is defined by a set of one-variable pseudo-identities. In fact, it is even the case that there exists a single element up∈ˆ FG(a) such that a finite group is a p-group if and only if it satisfies the pseudo-identity up=1 (upis the limit in the profinite topology of the sequence apn!, denoted up=apω[2, Example 2.6(1)]). It is also known that a finite group is nilpotent (resp. solvable) if and only if each of its 2-generated subgroups is nilpotent (resp. solvable), so that the pseudovarieties Gnil and Gsol are both defined by a set of
484 T. Coulbois, M. Sapir, P. Weil 2-variable pseudo-identities. In the nilpotent case, this is a result of Neumann and Taylor [8] and in the solvable case, it was proved by Thompson [11], see also Flavell [4]. In fact, it is known that there exists an element unil(a, b) (resp. usol(a, b)) of ˆ FG(A) such that the single 2-variable pseudo-identity unil(a, b)=1defines exactly Gnil, Almeida [2, Example 2.7(1)] (resp. usol(a, b)=1defines exactly Gsol, Bandman et al. [3]). Thus, Corollary 2.8 implies the following. Corollary 2.9. Let Bbe a finite alphabet, and let x, y ∈F(B)such that xy =yx.LetVbe an extension-closed pseudovariety of groups. •If Vsatisfies the pseudo-identity xpω=1, then V=Gp. •If Vsatisfies the pseudo-identity unil(x, y)=1, then V=Gpfor some prime p. •If Vsatisfies the pseudo-identity usol(x, y)=1, then V=Gsol. This can be rewritten in the, perhaps more readable, following form (without actually using the subtle results of the existence of a single pseudo-identity defining Gp,Gnil or Gsol). Corollary 2.10. Let Bbeafinite alphabet, and let x, y ∈F(B)such that xy =yx.LetVbe an extension-closed pseudovariety of groups. •If for every morphism ψ:F(B)→Ginto an element G∈V,ψ(x) has exponent a power of p(for some fixed prime p), then V=Gp. •If for every morphism ψ:F(B)→Ginto an element G∈V,ψ(x) and ψ(y)generate a nilpotent subgroup of G, then V=Gpfor some prime p. •If for every morphism ψ:F(B)→Ginto an element G∈V,ψ(x) and ψ(y)generate a solvable subgroup of G, then V=Gsol. 3. An example We now give an explicit example of an injective morphism ϕbetween finitely generated free groups whose continuous extension ˆϕbetween the corresponding free pro-pgroups is not injective, and we exhibit a sequence of words (tn)nsuch that lim ϕ(tn)=1,yet 1 is not a limit point of(tn)n:thus the limit points of (tn)nare non-trivial elements of ker ˆϕ. Put differently, this means that p-groups ultimately satisfy ϕ(tn)=1,yet there exists a p-group that does not satisfy any of the identities tn=1.
Continuous Extensions of Injective Morphisms 485 Let qbeafixed odd prime, let B={x, y}, and let Hbe the kernel of the morphism from F(B)into the additive group Z/qZwhich maps letters xand yto 1. Then Hhas rank q+1and if A={a0,a 1,...,a q}, then His the range of the injective morphism ϕ:F(A)→F(B) given by ϕ(ai)=xiyx−(i+1),i=0,...,q−2 ϕ(aq−1)=xq−1y ϕ(aq)=xq. As F(B)/H is a q-group, His closed in the pro-qtopology: by Theorem 1.1, the continuous extension of ϕto the pro-qcompletions of F(A) and F(B)isinjective. On the other hand, one can show that for every other prime number p,His dense in the pro-ptopology (see [7, Section 3.1]), and by Theorem 1.1 again, the continuous extension of ϕto the pro-pcompletions is not injective. Let u0=x,v0=y, and for each n≥0, un+1 =unvnand vn+1 = vnun.Wenow fix a prime number p=q.Itiswell-known that the identities un=vnare ultimately verified by every finite p-group (Engel identities, [8]). This means that the sequence (unv−1 n)nconverges to 1 in the pro-ptopology. It is also easily verified that, for each n≥0, the word unv−1 nis reduced and lies in H.Thus there exists a (unique) word tn∈F(A) such that ϕ(tn)=unv−1 nfor each n≥0. Let Sbe the q-dimensional vector space over the p-element field Fp with basis e0,...,e q−1. Let πbe the projection of F(A)ontoSdefined by π(ai)=eifor i=0,...,q−1 and π(aq)=0.Weprove that π(tn) is never 0 in S,sothat the additive group S(an abelian p-group) does not satisfy tn=1. We consider the morphism π◦ϕ−1from Hto S.Foreach n≥0, let sn=π(tn)=π◦ϕ−1(unv−1 n). Then sn+1 =π◦ϕ−1(un+1v−1 n+1)=π◦ϕ−1(unvnu−1 nv−1 n). Since His normal, unvnu−1 nu−1 n∈Hand we have sn+1 =π◦ϕ−1(unvnu−1 nu−1 n)+π◦ϕ−1(unv−1 n)=sn−π◦ϕ−1(ununv−1 nu−1 n). Let σbe the linear isomorphism of Sgiven by σ(ei)=ei+1 where the indices iand i+1 are taken modulo q.Weleave it to the reader to verify that, for each i=0,...,q−1, we have π◦ϕ−1(xϕ(ai)x−1)=π◦ϕ−1(yϕ(ai)y−1)=σ◦π(ai).