The twisted conjugacy problem for endomorphisms of

J. Group Theory 13 (2010), 355–364 DOI 10.1515/JGT.2009.054

Journal of Group Theory ( de Gruyter 2010

The twisted conjugacy problem for endomorphisms of polycyclic groups V. Roman’kov (Communicated by E. I. Khukhro)

Abstract. An algorithm is constructed that, when given an explicit presentation of a polycyclic group G, decides for any endomorphism c A EndðGÞ and any pair of elements u; v A G, whether or not the equation ðxcÞu ¼ vx has a solution x A G. Thus it is shown that the problem of the title is decidable. Also we present an algorithm that produces a finite set of generators of the subgroup Fixc ðGÞ c G of all c-invariant elements of G.

1 Introduction Let G be a group, and u; v A G. Given an endomorphism c A EndðGÞ, one says that u and v are c-twisted conjugate, and one writes u @c v, if and only if there exists x A G such that u ¼ ðxcÞ1 vx, or equivalently ðxcÞu ¼ vx. This notion has important applications in modern Nielsen fixed-point theory. The recognition of twisted conjugacy classes with respect to a given endomorphism c A EndðGÞ in the case of any polycyclic group G is the main concern of this paper. The class P of all polycyclic groups has a satisfactory algorithmic theory. A basic fact about polycyclic groups due to Hall is that every such group is finitely presented (see [18] or [23]). Polycyclic groups are residually finite by a result of Hirsch [10]. Hence the word problem is decidable in any polycyclic group G. By a known theorem of Mal’cev [14] (for an elementary proof due to J. S. Wilson see [18]) every subgroup of a polycyclic group is closed in the profinite topology. Hence the generalized word problem (also known as the membership problem) is also decidable in polycyclic groups. Every polycyclic group is conjugacy separable, i.e. if two elements of a polycyclic group G are not conjugate in G, they fail to be conjugate in some finite quotient. This deep result was proved by Formanek [4] and Remeslennikov [16]. So the conjugacy problem is decidable in any polycyclic group.

The author was partially supported by RFBR, Grant 07-01-00392.

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V. Roman’kov

It follows (see [3]) that the twisted conjugacy problem is decidable for any automorphism c A AutðGÞ of any polycyclic group G. Indeed, the split extension Gc ¼ G z hci is polycyclic, and the twisted conjugacy problem for c reduces to the ordinary conjugacy problem for Gc , which is decidable from above. However we cannot apply this method when c A EndðGÞ is not automorphism. Here we develop a new approach based on the free Fox calculus. We want to determine whether or not the equation ðxcÞu ¼ vx;

ð1Þ

where c A EndðGÞ; u; v A G, has a solution x in an arbitrary polycyclic group G. This question can be reduced (in any group G, not only a polycyclic group) to the case where one of the elements is trivial. To do this we change c to j ¼ c su , where su A AutðGÞ is the inner automorphism g 7! u1 gu. Hence ðxcÞu ¼ vx if and only if xj ¼ wx;

ð2Þ

1

where w ¼ u v. Note that every polycyclic group is Noetherian, i.e. satisfies to the maximal condition for subgroups (equivalently, every subgroup is finitely generated). The class P can be characterized as the class of all solvable matrix groups over the integers Z (see [11]). Thus, by the famous Mal’cev–Kolchin theorem, any polycyclic group is virtually nilpotent-by-abelian. This last description will be used in the paper. The main result of this paper is the following Theorem. Let G be any e‰ciently presented polycyclic group. Then there is an algorithm that decides the twisted conjugacy problem in G for any endomorphism j A EndðGÞ. Section 2 is devoted to preliminary results and Section 3 contains a proof of the main theorem.

2 Preliminary results Let G be any group and u; v A G. We consider the question of the decidability of the equation (1), equivalently (2). The following lemma shows that the question can under some assumptions be transformed to a similar question about a subgroup H of finite index in G. Lemma 2.1. Let G be any group and H c G be any EndðGÞ-invariant subgroup of finite index. Suppose that the membership problem in H is decidable in G. Then, if the twisted conjugacy problem is decidable in H, it is also decidable in G. Proof. Let G ¼ t1 H U U tk H


Twisted conjugacy in polycyclic groups

357

be a representation of G as a union of distinct cosets. Let (2) have a solution x ¼ ti h for some i A f1; . . . ; kg and h A H. Then the equation yi j ¼ ðti jÞ1 wti yi ¼ wi yi

ð3Þ

has the solution yi ¼ h A H. The necessary condition that wi ¼ ðti jÞ1 wti A H can be verified e¤ectively by our assumption. Obviously, any solution yi A H of (3) gives a solution x ¼ ti yi A G of (2). Hence (2), and so also (1), is solvable in G if and only if (3) has a solution in H for some i A f1; . . . ; kg, and this can be decided e¤ectively. r Let Fn be the free group of rank n with basis f f1 ; . . . ; fn g. In [5], Fox developed the di¤erential calculus in the free group ring ZFn . The partial Fox derivatives may be defined as the mappings q=qfj : ZFn ! ZFn ;

for 1 c j c n;

satisfying the following conditions for all a; b A Z and all u; v A Fn : qfi =qfj ¼ dij

ðwhere dij is the Kronecker deltaÞ;

qðau þ bvÞ=qfj ¼ aqu=qfj þ bqv=qfj ; qðuvÞ=qfj ¼ qu=qfj þ uqv=qfj : It is easy to verify that these derivatives have another property. Every element b A ZFn can be uniquely written in the form b ¼ b e þ qb=qf1 ð f1 1Þ þ þ qb=qfn ð fn 1Þ; where e : ZFn ! Z is the augmentation homomorphism, defined by fj e ¼ 1 for j ¼ 1; . . . ; n. It follows from the definitions that qa=qfj ¼ 0 for all a A Z and for each j. The following formula which holds for every element u A Fn is called the main identity for the Fox derivatives: n X

qu=qfi ð fi 1Þ ¼ u 1:

ð4Þ

i¼1

Let G ¼ Fn =R be a group presented as a quotient of Fn . For brevity we keep the notation f1 ; . . . ; fn for the images in G of the generators of Fn . So we have G ¼ h f1 ; . . . ; fn i. For any word w in these generators we define formal derivatives qw=qfi for i ¼ 1; . . . ; n as elements of the group ring ZG. We simply compute their values as sums of words (i.e. elements of ZFn ) and apply the epimorphism ZFn ! ZG. These values depend on the choice of the word w representing an element of G. But in any case property (4) is valid in ZG.


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V. Roman’kov

Let C be an abelian normal subgroup of G. We regard C as a module over ZG via conjugation in G extended by linearity to ZG, and also as a module over ZðG=CÞ; this is possible because C acts trivially on itself by conjuation. Then for any word c representing an element c A C and linear combinations of words a representing an element a A ZG one has qc a =qfi ¼ aqc=qfi ;

for i ¼ 1; . . . ; n:

Our notation for conjugates and commutators is as follows: g f ¼ fgf 1 and ½g; f ¼ gfg1 f 1 . Let j A EndðGÞ be any endomorphism that acts trivially on G=C, where C is as before an abelian normal subgroup of G. Then j is uniquely determined by the images f i j ¼ ci f i

ði ¼ 1; . . . ; nÞ;

ð5Þ

where ci are some words representing elements ci A C. Then for any element g A G the image under j can be written as qg=qf1

gj ¼ c1

. . . cnqg=qfn g:

ð6Þ

Lemma 2.2. Let G ¼ Fn =R be presented as a quotient of Fn , and let j A EndðGÞ be an endomorphism that acts trivially on G=C, where C is an abelian normal subgroup of G. Assume that j is determined by (5). Also assume that the derived subgroup G 0 acts trivially on C, and G 0 c Fixj ðGÞ, where Fixj ðGÞ ¼ fg A G j gj ¼ gg is the subgroup of j-invariant elements of G. Then Fixj ðGÞ lies in CG ðc1 ; . . . ; cn Þ, the centralizer of the elements c1 ; . . . ; cn in G. If the center z1 ðGÞ is trivial then Fixj ðGÞ ¼ CG ðc1 ; . . . ; cn Þ. Proof. Recall that G ¼ h f1 ; . . . ; fn i. Since j acts trivially on G 0 , it maps every commutator ½ fi ; fj to itself. Since j is determined by (5) with c1 ; . . . ; cn A C we can use (6). We obtain 1fj

½ fi ; fj ¼ ½ fi ; fj j ¼ ½ci fi ; cj fj ¼ ci

cjfi 1 ½ fi ; fj ;

or equivalently f 1

ci j

¼ cjfi 1 ;

for i; j ¼ 1; . . . ; n:

ð7Þ

So we can use the formula (6) again to conclude that gj ¼ g if and only if qg=qf1

c1

. . . cnqg=qfn ¼ 1:

ð8Þ

Here both sides of (8) are regarded as elements of the module C. Since G 0 acts trivially on C by hypothesis, G acts as an abelian group on G, so we can permute elements



359

acting on elements of C. Multiplying both sides of (8) by fi 1 for i ¼ 1; . . . ; n, we deduce from (7) and (4) that gj ¼ g is equivalent to n

Tj¼1 qg=qfj ð fj 1Þ

ci

¼ cig1 ¼ 1;

for i ¼ 1; . . . ; n

and so equivalent to cig ¼ ci for i ¼ 1; . . . ; n. Hence g A CG ðc1 ; . . . ; cn Þ. Conversely, suppose that g A CG ðc1 ; . . . ; cn Þ. Suppose that qg=qf1

c1

. . . cnqg=qfn ¼ v:

Then using the same argument as before we obtain that v fi 1 ¼ cig1 ¼ 1;

for i ¼ 1; . . . ; n:

Thus v A z1 ðGÞ, so if the additional assumption that z1 ðGÞ ¼ 1 holds then v ¼ 1, hence g A Fixj ðGÞ. We conclude that if z1 ðGÞ ¼ 1 then Fixj ðGÞ ¼ CG ðc1 ; . . . ; cn Þ. r Corollary 2.3. Let G be a centerless polycyclic group satisfying the assumptions of Lemma 2.2. Then there is an algorithm which finds a finite set fg1 ; . . . ; gk g of generators of Fixj ðGÞ. Proof. By Lemma 2.2, Fixj ðGÞ is the centralizer of a finite set of elements of G. It is known (see [19] or [23]) that a finite generating set of such a centralizer can be constructed e¤ectively. r The statement of Corollary 2.3 can be improved by elimination of the condition that z1 ðGÞ ¼ 1. Lemma 2.4. Let G be a polycyclic group satisfying the assumptions of Lemma 2.2. Then there is an algorithm which finds a finite set fg1 ; . . . ; gk g of generators of Fixj ðGÞ. Proof. It was proved in Lemma 2.2 that G1 ¼ CG ðc1 ; . . . ; cn Þ d Fixj ðGÞ. This group G1 can be constructed e¤ectively. Moreover, the proof of Lemma 2.2 shows that for any g A G1 one has gj ¼ cðgÞg, with cðgÞ A z1 ðGÞ. In particular G1 is invariant under j, so we may consider j as an endomorphism of G1 . Define a homomorphism m : G1 ! z1 ðGÞ by g 7! cðgÞ. Obviously we have Fixj ðGÞ ¼ kerðmÞ. Hence a set of generators of Fixj ðGÞ can be e¤ectively constructed by the standard procedure. r Lemma 2.5. Let G be any group, and let j A EndðGÞ be an endomorphism that acts trivially on G=U, where U is an abelian normal subgroup of G. Then every term Ci ¼ zi ðGÞ of the upper central series of G is j-invariant, so j induces an automorphism of every quotient G=Ci .


360

V. Roman’kov

Proof. It is enough to prove the statement for the center C ¼ C1 . Let c A C. Since j acts trivially modulo U and U is abelian, cj acts trivially on U. For each g A G one has gj ¼ uðgÞg, with uðgÞ A U. It follows that if ½g; c ¼ 1 then ½gj; cj ¼ ½uðgÞg; cj ¼ ½g; cj ¼ 1: Hence cj A C.

r

Lemma 2.6. Let G be a polycyclic group with nilpotent derived subgroup G 0 . Let j A EndðGÞ be any endomorphism. Then there is an algorithm which finds a finite set fg1 ; . . . ; gk g of generators of the subgroup Fixj ðGÞ. Proof. In the case when G 0 is abelian the statement was proved in [22]. So we can assume that j acts trivially on G modulo the second derived subgroup G 00 . Clearly Fixj ðGÞ c Fixj; G 00 ðGÞ ¼ fg A G : gj 1 g mod G 00 g: Obviously Fixj; G 00 ðGÞ is the full preimage of Fixj ðG=G 00 Þ and so can be constructed e¤ectively. Suppose by induction that we have found the subgroup GðiÞ ¼ Fixj; gi ðG 0 Þ ðGÞ ¼ fg A G : gj 1 g mod gi ðG 0 Þg; for some i d 2. Consider the quotient Giþ1 ¼ GðiÞ=giþ1 ðG 0 Þ and let Ciþ1 ¼ gi ðGðiÞ 0 Þ=giþ1 ðG 0 Þ 0 the be an abelian EndðGiþ1 Þ-invariant subgroup of Giþ1 . Since Ciþ1 is central in Giþ1 0 derived subgroup Giþ1 acts trivially on Ciþ1 . Moreover j A EndðGiþ1 Þ acts trivially 0 on Giþ1 =Ciþ1 and fixes every element g A Giþ1 . Then by Lemma 2.4 we can find e¤ectively a finite set of generators of Gði þ 1Þ ¼ Fixj; giþ1 ðG 0 Þ ðGÞ. Continuing this process we will eventually obtain a finite set of generators of Fixj ðGÞ. r

Lemma 2.7. Suppose that the twisted conjugacy problem is decidable for any endomorphism of any polycyclic group H with nilpotent derived subgroup H 0 . Then there is an algorithm which finds the subgroup Fixj ðGÞ in any polycyclic group G for any endomorphism j A EndðGÞ. Proof. It is known (see for example [11]) that each polycyclic group G contains an EndðGÞ-invariant subgroup H such that H 0 is nilpotent. Such a subgroup H can be constructed e¤ectively. Let G ¼ t1 H U U tk H



361

be a representation of G as a union of cosets with t1 ¼ 1. If a coset ti H intersects Fixj ðGÞ one has yj ¼ ðti jÞ1 ti y;

ð9Þ

for some y ¼ h A H. Since j induces an endomorphism of H and we can e¤ectively verify the inclusion ðti jÞ1 ti A H which is necessary for the solvability of (9), we can check this by our assumption. If (9) has a solution y ¼ h we change ti to ti h. If a coset ti H does not intersect Fixj ðGÞ, we do not need to consider it further. In this way we obtain representatives ti1 ; . . . ; tiq A Fixj ðGÞ of all cosets with non-empty intersection with Fixj ðGÞ. From this we obtain by Lemma 2.7 the subgroup Fixj ðHÞ, and we conclude that Fixj ðGÞ ¼ ti1 Fixj ðHÞ U U tiq Fixj ðHÞ:

r

3 Proof of the theorem Now G is a polycyclic group, and j A EndðGÞ is an arbitrary endomorphism. We must decide whether or not the equation (2) has a solution x A G. As a first step we reduce the problem to the case when G is nilpotent-by-abelian. It is well known (see [11] or [23]) that every polycyclic group admits a characteristic series 1 c N c A c G; where N is nilpotent, A=N is abelian, and G=A is finite. By Lemma 2.1 it is enough to find an algorithm deciding the twisted conjugacy problem in this subgroup A. To simplify notation we may therefore replace G by A and assume that G itself is polycyclic group with nilpotent derived group N ¼ G 0 . It was proved in [22] that the twisted conjugacy problem is decidable for any endomorphism in any metabelian polycyclic group. So we shall apply induction on the nilpotency class c of N, and we may assume that c d 2. Let C be the last non-trivial member of the lower central series of N. We consider the equation induced by (2) in the quotient G1 ¼ G=C. By the inductive assumption we have an algorithm that determines whether or not the equation (2) has a solution in G1 . If this equation is not solvable in G1 it is not solvable in G. So suppose that there is a solution of (2) in G1 . Then there is an element x1 A G for which x1 j ¼ wx1 c1 ; where c A C. Then (2) is solvable in G if and only if the equation zj ¼ cz


ð10Þ

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V. Roman’kov

is solvable in G. Indeed, if x A G is a solution of (2) then z ¼ x1 1 x is a solution of (10), and vice versa. Any possible solution of (10) belongs to the subgroup Fixj; C ðGÞ which is the full preimage of Fixj ðG1 Þ. Thus it can be found e¤ectively by Lemma 2.6. For simplicity we assume that G ¼ Fixj; C ðGÞ. Then j acts trivially on G 0 and on G=C. Moreover, we can assume that the center z1 ðGÞ is trivial. By Lemma 2.5 every member of the upper central series of G is j-admissible. Since G is polycyclic and so Noetherian this series is finite. Suppose that (10) is solvable in the quotient G=z1 ðGÞ. Then there is an element z1 A G for which z1 j ¼ cz1 d 1 ; where d A z1 ðGÞ. Using an argument similar to the one above we conclude that (10) is solvable if and only if the equation yj ¼ dy

ð11Þ

is solvable. We replace G by Fixj; z1 ðGÞ ðGÞ and look for a solution of (11) in this new group. Now for every element g A Fixj; z1 ðGÞ ðGÞ one has gj ¼ dðgÞg; where dðgÞ A z1 ðGÞ. The equation (11) has a solution if and only if d belongs to the image of Fixj; z1 ðGÞ ðGÞ under the homomorphism g 7! dðgÞ. This is a homomorphism of abelian groups and hence we can answer this question by the standard procedure. Continuing this process we reduce the question to the case of a centerless group. Let j be defined by (5). Then for any z A G we have qz=qf1

zj ¼ c1

. . . cnqz=qfn z:

If z is a solution of (10) then qz=qf1

c1

. . . cnqz=qfn ¼ c:

Regarding C as a module and multiplying both sides of this equation by fi 1 for i ¼ 1; . . . ; n, and using (7) and (4) we obtain the set of equations ciz ¼ c fi 1 ci ;

for i ¼ 1; . . . ; n:

ð12Þ

Since the conjugacy problem is decidable in any polycyclic group we first find a solution of (12) for c1 . We can assume that such solution z1 exists, or (2) has no solutions in G. Conjugating by z1 we reduce the problem to a similar question for CG ðc1 Þ. This subgroup can be constructed e¤ectively. We continue this process with the new group CG ðc1 Þ replacing G and n 1 equations arising from (12). Suppose



363

that (12) is solvable for some z. Let qz=qf1

zjz1 ¼ c1

. . . cnqz=qfn ¼ v:

Multiplying both sides by fi 1 for i ¼ 1; . . . ; n, and applying the same argument as above we obtain that ðcv1 Þ fi 1 ¼ 1;

for i ¼ 1; . . . ; n:

Since z1 ðGÞ ¼ 1, we conclude that c ¼ v, and the theorem is proved.

r

Corollary 3.1. Let G be a polycyclic group and j any endomorphism of G. Then there is an algorithm that produces generators of the fixed-point group Fixj ðGÞ of all jinvariant elements of G. Proof. The statement follows from the Theorem and Lemma 2.7.

r

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[16] V. N. Remeslennikov. Conjugacy in polycyclic groups. Algebra and Logic 8 (1969), 404–411. [17] V. N. Remeslennikov and V. A. Roman’kov. Algorithmic and model theoretic problems in groups. J. Soviet Math. 21 (1983), 2887–2939. [18] D. J. S. Robinson. A course in the theory of groups (Springer-Verlag, 1982). [19] D. J. S. Robinson. Reflections on the constructive theory of polycyclic groups. In Groups– Korea 1988 (Pusan, 1988), Lecture Notes in Math. 1398 (Springer-Verlag, 1989), pp. 163–169. [20] V. A. Roman’kov. About unsolvability of problem of endomorphic reducibility in free nilpotent groups and free rings. Algebra and Logic 16 (1977), 310–320. [21] V. Roman’kov and E. Ventura. On the twisted conjugacy problem for endomorphisms of nilpotent groups. (In preparation.) [22] V. Roman’kov and E. Ventura. The twisted conjugacy problem for endomorphisms of metabelian groups. Algebra and Logic 48 (2009), 89–98. [23] D. Segal. Polycyclic groups (Cambridge University Press, 1983). [24] A. Seidenberg. Constructions in a polynomial ring over the ring of integers. Amer. J. Math. 100 (1978), 685–703. [25] C. C. Sims. Computations with finitely presented groups. Encyclopedia of Math. and its Appl. (Cambridge University Press, 1994). Received 21 January, 2009 V. Roman’kov, Omsk State Dostoevskii University Omsk, Russia E-mail: [email protected]
