# Conjugacy Separability of Some One-Relator Groups

Authors: D. Tieudjo; D. I. Moldavanskii
Journal: International Journal of Mathematics and Mathematical Sciences
 (2010)

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## Abstract

Conjugacy separability of any group of the class of one-relator groups given by the presentation〈a,b;[am,bn]=1〉(m,n>1) is proven. The proof made used of theoretical combinatorial group methods, namely the structure of amalgamated free
products and some properties of the subgroups and quotients of any group of the class of
one-relator groups given above.


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## Body

## 1. Introduction

A groupG is conjugacy separable if for any two nonconjugate elements f and g of G there exists a homomorphism φ of group G onto some finite group X such that the images fφ and gφ of elements f and g are not conjugate in group X. It is clear that any conjugacy separable group G is residually finite (i.e., recall that for any nonidentity element g∈G there exists a homomorphism φ of group G onto some finite group X such that gφ≠1).Conjugacy separability is related to the conjugacy problem, as the residual finiteness is related to the word problem in the study of groups. In fact, Mostowski [1] proved that finitely presented conjugacy separable groups have solvable conjugacy problem, just as finitely presented residually finite groups have solvable word problem.Since 1962 when Baumslag and Solitar [2] discovered the first examples of nonresidually finite one-relator groups a lot of results establishing the residual finiteness of various one-relator groups have appeared. It was also shown that some of such groups are conjugacy separable.  So, Dyer [2] has proved the conjugacy separability of any group that is free product of free groups amalgamating cycle. In the same paper she proved an unpublished result of M. Armstrong on conjugacy separability of any one-relator group with nontrivial center. The conjugacy separability of groups defined by relator of form (ambn)t and (a-1blabm)s, where s>1, was proved in [4] and [5], respectively. It should mention that up to now it is not known whether there exists one-relator group that is residually finite but not conjugacy separable.In this paper we enlarge the class of conjugacy separable one-relator groups. Namely, we prove here the following theorem.Theorem 1.1.
Any groupGmn=〈a,b;[am,bn]=1〉 where integers m and n are greater than 1, is conjugacy separable.We note that the assertion of Theorem is also valid whenm=1 or n=1. Indeed, in this case group Gmn is the generalized free product of two finitely generated abelian groups with cyclic amalgamation and its conjugacy separability follows from the result of [3].The residual finiteness of groupsGmn is well known; it follows, for example, from the result of paper [6]. Groups Gmn are of some particular interest: they are centerless, torsion-free,… Some properties of these groups were considered in [7, 8] where, in particular, the description of their endomorphisms was given and their automorphisms group were studied, respectively. In this paper, the proofs made use of presentation of group Gmn as amalgamated free product. This presentation will be the crucial tool in the proof of Theorem 1.1 and because of this, in Section 1, we recall the Solitar theorem on the conjugacy of elements of amalgamated free products and derive, for our special case, the criterion which is somewhat simpler. In Section 2 some properties of separability of groups Gmn are established and in Section 3 the proof of Theorem 1.1 will be completed.We remind that conjugacy separability of groupsGmn can also be obtained using [9]. However, the method in [9] presents any group Gmn as the result of adjoining roots to conjugacy separable groups.


## 2. Preliminary Remarks on the Conjugacy in  Amalgamated Free Products

Let us recall some notions and properties concerned with the construction of free productG=(A*B;H) of groups A and B with amalgamated subgroup H.Every elementg∈G can be written in a form (2.1)g=x1x2⋯xr,(r⩾1),
where for any i=1,2,…,r element xi belongs to one of the free factors A or B and if r>1 any successive xi and xi+1 do not belong to the same factor A or B (and therefore for any i=1,2,…,r element xi does not belong to amalgamated subgroup H). Such form of element g is called reduced.In general, elementg∈G can have different reduced forms, but if g=y1y2⋯ys is one more reduced form of g, then r=s and, for any i=1,2,…,r, xi and yi belong to the same factor A or B. The lengthl(g) of element g is then defined as the number r of the components in a reduced form of g. Element g is called cyclically reduced if either l(g)=1 or l(g)>1 and the first and the last components of its reduced form do not belong to the same factor A or B. If g=x1x2⋯xr is a reduced form of cyclically reduced element g then cyclic permutations of element g are elements of form xixi+1⋯xrx1⋯xi-1, where i=1,2,…,r.IfX is a subgroup of a group Y we will say that elements a and b of Y are  X-conjugate if a=x-1bx, for some x∈X.The Solitar criterion of conjugacy of elements of amalgamated free product of two groups (Theorem4.6 in [10]) can be formulated as follows.Proposition 2.1.
LetG=(A*B;H) be the free product of groups A and B with amalgamated subgroup H. Every element of G is conjugate to a cyclically reduced element. If lengths of two cyclically reduced elements are nonequal then these elements are not conjugate in G. Let f and g be cyclically reduced elements of G such that l(f)=l(g). Then  (1)
iff∈A and f is not conjugate in A to any element of subgroup H, then f and g are conjugate in group G if and only if g∈A and f and g are conjugate in A; similarly, if f∈B and f is not conjugate in B to any element of subgroup H then f and g are conjugate in group G if and only if g∈B and f and g are conjugate in B;(2)
iff∈H then f and g are conjugate in group G if and only if there exists a sequence of elements
(2.2)f=h0,h1,…,hn,hn+1=g
such that for any i=0,1,…,nhi∈H and elements hi and hi+1 are A-conjugate or B-conjugate;(3)
ifl(f)=l(g)>1, then f and g are conjugate in group G if and only if element f is H-conjugate to some cyclic permutation of g.In the following special case assertion(2)  of Proposition 2.1 became unnecessary.Proposition 2.2.
LetG=(A*B;H) be the free product of groups A and B with amalgamated subgroup H. Suppose that for any element f of A or B and for any element h of subgroup H, the inclusion f-1hf∈H is valid if and only if elements f and h commute. If f and g are cyclically reduced elements of G such that l(f)=l(g)=1 then f and g are conjugate in group G if and only if f and g belong to the same subgroup A or B and are conjugate in this subgroup.Indeed, let elementsf and g be conjugate in group G and let f belong to subgroup A, say. If f is not conjugate in A to any element of subgroup H, then the desired assertion is implied by assertion  (1)  of Proposition 2.1. Otherwise, we can assume that f∈H and by assertion  (2)   of Proposition 2.1 there is a sequence of elements (2.3)f=h0,h1,…,hn,hn+1=g,
such that for any i=0,1,…,nhi∈H, and for suitable element xi belonging to one of the subgroups A or B, the equality xi-1hixi=hi+1 holds. Since for i=0,1,…,n-1 the inclusion xi+1-1hi+1xi+1∈H is valid, the hypothesis gives h0=hn, that is, xn-1fxn=g. This means that elements f and g belong to that subgroup A or B which contains element xn and also are conjugated in this subgroup.Next, we describe more explicitly the situation arising in assertion(3)  of Proposition 2.1.Proposition 2.3.
Letf=x1x2⋯xr and g=y1y2⋯yr be the reduced forms of elements f and g of group G=(A*B;H) where r>1. Then f and g are H-conjugate if and only if there exist elements h0,h1,h2,…,hr=h0 in subgroup H such that for any i=1,2,…,r, one has
(2.4)xi=hi-1-1yihi.Proof.
If for some elementsh0,h1,h2,…,hr=h0 of subgroup H equalities (2.4) hold, then
(2.5)f=x1x2⋯xr=h0-1y1h1h1-1y2h2⋯hr-1-1y2hr=h0-1(y1y2⋯yr)hr=h0-1gh0.
Conversely, by induction onr we prove that if for some h∈H the equality f=h-1gh holds then there exists a sequence of elements h0,h1,h2,…,hr=h0 of subgroup H satisfying the equalities (2.4) and such that h0=h.
Rewriting the equalityf=h-1gh in the form(2.6)xr-1⋯x2-1x1-1h-1y1y2⋯yrh=1,
we see that since the expression in the left part of it cannot be reduced, elements x1 and y1 must be contained in the same subgroup A or B and element x1-1h-1y1 must belong to subgroup H. Denoting this element by h1-1, we have x1=h-1y1h1. If r=2 then the equality above takes the form x2-1h1-1y2h=1, whence x2=h1-1y2h. Therefore, setting h0=h2=h, we obtain in that case the desired sequence.
Ifr>2 let us rewrite the equality f=h-1gh in the form(2.7)h-1y1h1x2⋯xr=h-1y1y2⋯yrh.
This implies that if we set f'=h-1h1x2⋯xr and g'=y2⋯yr, then f'=h-1g'h. Since the length of elements f' and g' is equal to r-1, then by induction, there exists a sequence h1′=h, h2′,…,hr′=h1′ of elements of subgroup H, such that h-1h1x2=(h1′)-1y2h2′ and for any i=2,3,…,r(2.8)xi=(hi-1′)-1yihi′.
Since x2=h1-1y2h2′, then setting hi=hi′ for i=2,3,…,r, we obtain the desired sequence, and induction is completed. Proposition 2.3 is proven.We conclude this section with one more property of amalgamated free product.Proposition 2.4.
LetG=(A*B;H) be the free product of groups A and B amalgamating subgroup H where H lies in the center of each of groups A and B. Then for any element f∈G not belonging to subgroup A one has f-1Af∩A=H.Proof.
In fact, since subgroupH coincides with the center of group G (see [10, page 211]), the inclusion H⊆f-1Af∩A is obvious. Conversely, let element f-1af belong to subgroup A where a∈A. Since f∉A then either f∈B∖H or l(f)>1 and (without loss of generality) the first component of reduced form of f belongs to B. In any case the assumption a∉H would imply that l(f-1af)>1. Thus, a∈H and therefore f-1af=a∈H.


## 3. Some Properties of GroupsGmn   and of Their Certain Quotients

In what follows, our discussion will make use of the presentation of groupGmn as an amalgamated free product of two groups. To describe such presentation, let H be the subgroup of group Gmn, generated by elements c=am and d=bn. Also, let A denote the subgroup of group Gmn generated by element a and subgroup H, and let B denote the subgroup of group Gmn, generated by element b and subgroup H. Then it can be immediately verified that H is the free abelian group with base c, d, group A is the free product (〈a〉*H;〈am=c〉) of infinite cycle 〈a〉 and group H with amalgamation 〈am〉, group B is the free product (〈b〉*H;〈bn=d〉) of infinite cycle 〈b〉 and group H with amalgamation 〈bn〉, and group Gmn is the free product (A*B;H) of groups A and B with amalgamation H.The same decomposition is satisfiable for certain quotients of groupsGmn. Namely, for any integer t>1 let Gmn(t) be the group with presentation (3.1)〈a,b;[am,bn]=1,amt=bnt=1〉
and ρt the natural homomorphism of groups Gmn onto Gmn(t). Then it is easy to verify that subgroup H(t)=Hρt of group Gmn(t) is isomorphic to the quotient H/Ht (where, as usually, Ht consists of all elements ht, h∈H), subgroup A(t)=Aρt is the amalgamated free product (〈a;amt=1〉*H(t);am=cHt) of cycle 〈a;amt=1〉 of order mt and group H(t), subgroup B(t)=Bρt is the amalgamated free product (〈b;bnt=1〉*H(t);bn=dHt) of cycle 〈b;bnt=1〉 of order nt and group H(t), and group Gmn(t) is the amalgamated free product (A(t)*B(t);H(t)).These decompositions of groupsGmn and Gmn(t) are assumed everywhere below, and such notions as free factor, reduced form, length of element and so on will refer to them.Let us remark, at once, that since each of groupsA(t) and B(t) is the amalgamated free product of two finite groups and group Gmn(t) is the free product of groups A(t) and B(t) with finite amalgamation, it follows from results of [3] that for every t, group Gmn(t) is conjugacy separable. So, to prove the conjugacy separability of group Gmn it is enough, for any nonconjugate elements f and g of Gmn, to find an integer t such that elements fρt and gρt are nonconjugate in group Gmn(t).Since in the decompositions of groupsA and B as well, as of groups A(t) and B(t), into amalgamated free product stated above the amalgamated subgroups are contained in the centre of each free factor, by Proposition 2.4 we have the following proposition.Proposition 3.1.
For any elementg of group A (or B) not belonging to subgroup H the equality g-1Hg∩H=〈c〉 (resp., g-1Hg∩H=〈d〉) holds. In particular, for any element g of group A or B and for any element h of subgroup H element g-1hg belongs to subgroup H if and only if elements g and h commute.
Similarly, for any elementg of group A(t) (or B(t)) not belonging to subgroup H(t) the equality g-1H(t)g∩H(t)=〈cHt〉 (resp., g-1H(t)g∩H(t)=〈dHt〉) holds. In particular, for any elements g of group A(t) or B(t) and h of subgroup H(t), element g-1hg belongs to subgroup H(t) if and only if elements g and h commute.Propositions2.1, 2.2 and 3.1 lead to the following criterion for conjugacy of elements of groups Gmn and Gmn(t).Proposition 3.2.
LetG be any of groups Gmn and Gmn(t). Every element of G is conjugate to a cyclically reduced element. If lengths of two cyclically reduced elements are not equal then these elements are not conjugate in G. Let f and g be cyclically reduced elements of G such that l(f)=l(g). Then   (1)
Ifl(f)=l(g)=1 then f and g are conjugate in group G if and only if f and g belong to the same free factor and are conjugate in this factor.(2)
Ifl(f)=l(g)>1 then f and g are conjugate in group G if and only if element f is H-conjugate or H(t)-conjugate to some cyclic permutation of g.We now consider some further properties of groupsGmn and Gmn(t). The following assertion is easily checked.Proposition 3.3.
For any homomorphismφ of group Gmn onto a finite group X there exist an integer t>1 and a homomorphism ψ of group Gmn(t) onto group X such that φ=ρtψ.
Also, for any integerst>1 and s>1 such that t divides s there exists homomorphism φ:Gmn(s)→Gmn(t) such that ρt=ρsφ.
The same assertions are valid for groupsA and B.Proposition 3.4.
For any elementg of group A or B, if g does not belong to subgroup H then there exists an integer t0>1 such that for any positive integer t, divisible by t0, element gρt does not belong to subgroup Hρt.Proof.
We will assume thatg∈A; the case when g∈B can be treated similarly.
So, let elementg∈A do not belong to subgroup H and let g=x1x2⋯xr be a reduced form of g (in the decomposition of group A into amalgamated free product).
Ifr=1 then element g belongs to subgroup 〈a〉 that is, g=ak for some integer k. Since g∉H, integer k is not divisible by m. Then for any integer t>0, in group A(t), element gρt of subgroup 〈a;amt=1〉 cannot belong to the amalgamated subgroup (generated by element am) of the decomposition of group A(t) and consequently cannot belong to the free factor H(t)=Hρt.
Letr>1. Every component xi of the reduced form of element g is either of form ak, where integer k is not divisible by m, or of form ckdl where l≠0. If integer t0 is chosen such that the exponent l of any component xi of the second kind is not divisible by t0, then for any t divisible by t0 the image xiρt=xiHt of such component will not belong to the amalgamated subgroup of group A(t) (generated, let's remind, by element cHt). Moreover, as above in case r=1, the images of components of the first kind will not belong to the amalgamated subgroup. Therefore, the form(3.2)gρt=(x1ρt)(x2ρt)⋯(xrρt)
of element gρt is reduced in group A(t) and since r>1, gρt does not belong to the free factor H(t)=Hρt. The proposition is proven.Proposition3.4 obviously implies the the following proposition.Proposition 3.5.
For any elementg of group Gmn there exists an integer t0>1 such that for all positive integer t, divisible by t0, the length of element gρt in group Gmn(t) coincides with the length of element g in group Gmn.Proposition 3.6.
For any elementsf and g of group A (or B) such that element f does not belong to the double coset HgH, there exists an integer t0>1 such that for any positive integer t, divisible by t0, element fρt does not belong to the double coset H(t)(gρt)H(t).Proof.
We can again consider only the case when elementsf and g belong to subgroup A. So, let us suppose that element f∈A does not belong to the double coset HgH. In view of Proposition 3.4, it is enough to prove that there exists a homomorphism φ of A onto a finite group X such that element fφ of X does not belong to the double coset (Hφ)(gφ)(Hφ).
To this end let's consider the quotient groupA¯=A/C of group A by its (central) subgroup C=〈c〉. The image xC of an element x∈A in group A¯ will be denoted by x¯.
It is obvious that groupA¯ is the (ordinary) free product of the cyclic group X of order m, generated by element a¯, and the infinite cycle Y, generated by d¯. The canonical homomorphism of group A onto group A¯ maps subgroup H onto subgroup Y and consequently, the image of the double coset HgH is the double coset Yg¯Y. Since C⩽H, the element f¯ does not belong to this coset.
We can assume, without loss of generality, that any elementf¯ and g¯, if it is different from identity, has reduced form the first and the last syllables of which do not belong to subgroup Y. Every Y-syllable of these reduced forms is of form d¯k for some integer k≠0. Since the set M of all such exponents k is finite, we can choose an integer t>0 such that for any k∈M the inequality t>2|k| holds. Let's denote by Ã the factor group of group A¯ by the normal closure of element d¯t. Group Ã is the free product of groups X and Y/Yt and since different integers from M are not relatively congruent and are not congruent to zero modulo t, then the reduced forms of the images f̃ and g̃ of elements f¯ and g¯ in group Ã are the same as in group A¯. In particular, element f̃ does not belong to the double coset Y/Ytg̃Y/Yt. Since this coset consists of a finite number of elements and group Ã is residually finite, then there exists a normal subgroup Ñ of finite index of group Ã such, that(3.3)f̃∉(Y/Ytg̃Y/Yt)·Ñ.
If now θ is the product of the canonical homomorphisms of group A onto group A¯ and of group A¯ onto group Ã and N is the full preimage by θ of subgroup Ñ, then N is a normal subgroup of finite index of group A and f∉(HgH)N. Thus, the canonical homomorphism φ of group A onto quotient group A/N has the required property and the proposition is proven.


## 4. Proof of Theorem1.1

We prove first the following proposition.Proposition 4.1.
If elementsf and g of group Gmn such that l(f)=l(g)>1 are not H-conjugate, then for some integer t>1, elements fρt and gρt of group Gmn(t) are not H(t)-conjugate.Proof.
Letf=x1x2⋯xr and g=y1y2⋯yr be the reduced forms in group Gmn=(A*B;H) of elements f and g.
We remind (see Proposition2.3) that elements f and g are H-conjugate if and only if there exist elements h0, h1, h2,…,hr=h0 of H such that for any i=1,2,…,r, we have(4.1)xi=hi-1-1yihi.
It then follows, in particular, that for each i=1,2,…,r elements xi and yi should lie in the same free factor A or B and define the same double coset modulo (H,H). So, we consider separately some cases.Case 1.
Suppose that for some indexi elements xi and yi lie in different free factors A and B of group Gmn (and, certainly, are not in subgroup H). It follows from Proposition 3.4 that there exists an integer t>1 such that elements xiρt and yiρt do not belong to the same free factor A(t) or B(t) of group Gmn(t) (and, as above, lie in free factors of this group). Hence, by Proposition 2.3, in group Gmn(t) elements fρt and gρt are not H(t)-conjugate.Case 2.
Let now for anyi=1,2,…,r elements xi and yi belong to the same subgroup A or B and for some i element xi does not belong to the double coset HyiH. By Proposition 3.6, there exists an integer t1>1 such that for any positive integer t, divisible by t1, element xiρt is not in the double coset H(t)(yiρt)H(t). From Proposition 3.5, there exist integers t2>1 and t3>1 such that for any positive integer t divisible by t2 the length of element fρt in group Gmn(t) is equal to r and for any positive integer t, divisible by t3, the length of element gτt in group Gmn(t) is equal to r. Thus, if t=t1t2t3 then in group Gmn(t), elements fρt and gρt have the reduced forms
(4.2)(x1ρt)(x2ρt)⋯(xrρt),(y1ρt)(y2ρt)⋯(yrρt),
respectively, and element xiρt is not in the double coset H(t)(yiρt)H(t). Again, by Proposition 2.3 these elements are not H(t)-conjugate.Case 3.
We now consider the case, when for anyi=1,2,…,r elements xi and yi lie in the same free factor A or B and also determine the same double coset modulo (H,H). We prove some lemmas.Lemma 4.2.
Let elementsx and y belong to one of the groups A or B and do not belong to the subgroup H and x∈HyH, that is,
(4.3)x=cαdβycγdδ
for some integers α, β, γ and δ. If elements x and y belong to group A, then integers α+γ, β and δ are uniquely determined by the equality (4.3). If elements x and y belong to group B, then integers β+δ, α and γ are uniquely determined by equality (4.3).Proof.
Let elementsx and y belong to subgroup A and let x=cα1dβ1ycγ1dδ1 and x=cα2dβ2ycγ2dδ2 for some integers α1, β1, γ1, δ1, α2, β2, γ2 and δ2. Rewriting the equality cα1dβ1ycγ1dδ1=cα2dβ2ycγ2dδ2 as y-1cα1-α2dβ1-β2y=cγ2-γ1dδ2-δ1, then, by Proposition 3.1, we conclude that cα1-α2dβ1-β2=cγ2-γ1dδ2-δ1 and also that this element should belong to subgroup 〈c〉, that is, β1-β2=δ2-δ1=0. So, we have cα1-α2=cγ2-γ1 and since the order of element c is infinite, then α1-α2=γ2-γ1.
Thus,β1=β2, δ1=δ2 and α1+γ1=α2+γ2 as it was required. The case when elements x and y belong to group B is esteemed  similarly.Lemma 4.3.
Let elementsx and y belong to one of groups A(t) or B(t) and do not belong to subgroup H(t) and x∈H(t)yH(t), that is,
(4.4)x=(cHt)α(dHt)βy(cHt)γ(dHt)δ
for some integers α, β, γ and δ. If elements x and y belong to group A(t), then integers α+γ, β and δ are uniquely determined modulo t by equality (4.4). If elements x and y belong to group B(t), then integers β+δ, α and γ are uniquely determined modulo t by equality (4.4).The proof of Lemma 4.3 is completely similar to that of Lemma 4.2. Lemma 4.4.
Letf=x1x2⋯xr and g=y1y2⋯yr be reduced elements of group Gmn, where r>1 and let for every i=1,2,…,r the equality xi=uiyivi holds, for some elements ui and vi of subgroup H. Then there exists at most one sequence h0, h1, h2,…,hr of elements of subgroup H such that for any i=1,2,…,r(4.5)xi=hi-1-1yihi.
Moreover, if ui=cαidβi and vi=cγidδi for some integers αi, βi, γi and δi (i=1,2,…,r) and x1,y1∈A, then such sequence exists if, and only if, for any i, 1<i<r,
(4.6)αi+αi+1+γi-1+γi=0,ifiisodd,βi+βi+1+δi-1+δi=0,ifiiseven.Proof.
We suppose first that the sequenceh0, h1, h2,…,hr of elements of subgroup H satisfying equality (4.5) exists, and let's write hi=cμidνi, for some integers μi and νi.
Then, since for anyi=1,2,…,r the equality hi-1-1yihi=uiyivi holds, we have(4.7)yi-1(hi-1ui)yi=hivi-1.
Since r>1, then every element yi, belonging to one of the subgroups A or B, does not lie in subgroup H, and consequently, from Proposition 3.1, for any i=1,2,…,r, we have the equality hi-1ui=hivi-1. Substituting the expressions of elements hi, ui and vi, we have for every i=1,2,…,rcμi-1+αidνi-1+βi=cμi-γidνi-δi and hence we obtain the system of numeric equations
(4.8)μi-1+αi=μi-γi,νi-1+βi=νi-δi(i=1,2,…,r).
Since by (4.7) for every i=1,2,…,r element hivi-1 belongs to the intersection yi-1Hyi∩H and, by supposition, elements y1, y3,…  belong to subgroup A and elements y2, y4,…  belong to subgroup B, then from Proposition 3.1, it follows that for odd i, we should have hi-1ui=hivi-1∈〈c〉, and for even i, we should have hi-1ui=hivi-1∈〈d〉. It means that for every odd i (i=1,2,…,r) we have the equalities νi-1+βi=0 and νi-δi=0, and for every even i (i=1,2,…,r) we have the equalities μi-1+αi=0 and μi-γi=0.
Hence, the values of the integersμi and νi are determined uniquely. Namely,(4.9)μi={γiifiisevenand2⩽i⩽r,-αi+1ifiisoddand1⩽i⩽r-1,(4.10)νi={-βi+1ifiisevenand0⩽i⩽r-1,δiifiisoddand1⩽i⩽r.
Moreover, from equalities (4.8) it follows, that
(4.11)μ0=-(α1+α2+γ1),νr=βr+δr-1+δrifriseven,μr=αr+γr-1+γrifrisodd.
Thus, the statement that there can exist at most one sequence of elementsh0, h1, h2,…,hr of subgroup H satisfying equalities (4.5) is demonstrated.
Substituting the valueμi-1=γi-1 and μi=-αi+1 defined in (4.9) in the equalities μi-1+αi=μi-γi of system (4.8), where 1<i<r and i is odd, we obtain αi+αi+1+γi-1+γi=0. Similarly, substituting the value νi-1=δi-1 and νi=-βi+1 defined in (4.10) in the equalities νi-1+βi=νi-δi of systems (4.8), where 1<i<r and i is even, we obtain βi+βi+1+δi-1+δi=0.
Thus, under the existence in subgroupH of sequence h0, h1, h2,…,hr of elements satisfying (4.5), conditions (4.6) are satisfied.
Conversely, suppose conditions (4.6) are satisfied. Let(4.12)h0=c-(α1+α2+γ1)d-β1
and for all indexes i such that 1⩽i<r, we set
(4.13)hi={cγid-βi+1ifiiseven,c-αi+1dδiifiisodd.
At last, for i=r we set
(4.14)hr={cγrdβr+δr-1+δrifriseven,cαr+γr-1+γrdδrifrisodd.
Let us show that, the so-defined sequence of elementsh0, h1, h2,…,hr really fits to equalities (4.5).
Ifi=1, using the permutability of elements c and y1 we have(4.15)h0-1y1h1=cα1+α2+γ1dβ1y1c-α2dδ1=cα1dβ1y1cγ1dδ1=x1.
If1<i<r and integer i is even, using the equality βi+βi+1+δi-1+δi=0 and permutability of elements d and yi, we have(4.16)hi-1-1yihi=cαid-δi-1yicγid-βi+1=cαiyicγid-(δi-1+βi+1)=cαiyicγidβi+δi=cαidβiyicγidδi=xi.
If1<i<r and integer i is odd, using the equality αi+αi+1+γi-1+γi=0 and permutability of elements c and yi, we have(4.17)hi-1-1yihi=c-γi-1dβiyic-αi+1dδi=c-(γi-1+αi+1)dβiyidδi=cαi+γidβiyidδi=cαidβiyicγidδi=xi.
If integerr is even, then(4.18)hr-1-1yrhr=cαrd-δr-1yrcγrdβr+δr-1+δr=cαrdβryrcγrdδr=xr,
and if r is odd, then
(4.19)hr-1-1yrhr=c-γr-1dβryrcαr+γr-1+γrdδr=cαrdβryrcγrdδr=xr.
So, Lemma 4.4 is completely demonstrated.
Similar argument gives the following lemma.
Lemma 4.5.
Letf=x1x2⋯xr and g=y1y2⋯yr be reduced elements of group Gmn(t), where r>1, and let for every i=1,2,…,r the equality xi=uiyivi takes place, for some elements ui and vi of subgroup H(t). Then there exists at most one sequence h0, h1, h2,…,hr of elements of subgroup H(t) such that for any i=1,2,…,r, one has
(4.20)xi=hi-1-1yihi.
Moreover, if ui=(cHt)αi(dHt)βi and vi=(cHt)γi(dHt)δi for some integers αi, βi, γi and δi (i=1,2,…,r) and x1,y1∈A(t), then such sequence exists if and only if for any integer i, 1<i<r, one has
(4.21)αi+αi+1+γi-1+γi≡0(modt)ifiisodd,βi+βi+1+δi-1+δi≡0(modt)ifiiseven.
We can now end the consideration of Case3 and thus complete the proof of Proposition 4.1.
By Proposition2.3, elements f and g are H-conjugate if and only if their cyclic permutation x2⋯xrx1 and y2⋯yry1 are H-conjugate. Therefore, we can suppose, without loss of generality, that elements x1 and y1 belong to subgroup A.
By supposition, for everyi=1,2,…,r there exist integers αi, βi, γi and δi such that(4.22)xi=cαidβiyicγidδi.
If integeri, 1<i<r, is even, then yi∈B, yi-1,yi+1∈A and consequently, from Lemma 4.2, integers βi+δi, δi-1 and βi+1 do not depend from the particular expressions of the elements xi (in the form xi=uyiv, where u,v∈H), and these integers are uniquely determined by the sequences x1, x2,…,xr and y1, y2,…,yr. Similarly, for any odd i, 1<i<r, integers αi+γi, γi-1 and αi+1 are uniquely determined by these sequences. It means, in turn, that the satisfiability of conditions (4.6) of Lemma 4.4 depends only on the sequences x1, x2,…,xr and y1, y2,…,yr.
Let's now conditions (4.6) of Lemma 4.4 are not satisfied, that is, either for some even i, 1<i<r, the sum βi+βi+1+δi-1+δi is different from zero, or for some odd i, 1<i<r, the sum αi+αi+1+γi-1+γi is different from zero. Then it is possible to find an integer t1>1, not dividing the respective sum. Let's also choose, according to Proposition 3.5, the integers t2>1 and t3>1 such that for all positive integer t, divisible by t2, the length of element fρt in group Gmn(t) is equal to r and for all positive integer t, divisible by t3, the length of element gρt in group Gmn(t) is equal to r. Then if t=t1t2t3, in group Gmn(t), elements fρt and gρt have reduced forms(4.23)(x1ρt)(x2ρt)⋯(xrρt),(y1ρt)(y2ρt)⋯(yrρt),
respectively. Further, for every i=1,2,…,r,
(4.24)xiρt=(cHt)αi(dHt)βi(yiρt)(cHt)γi(dHt)δi.
From Lemma 4.3 and the selection of integer t1 it follows that for elements fρt and gρt, conditions (4.21) of Lemma 4.5 are not satisfied. Therefore from this lemma and Proposition 2.3, elements fρt and gρt are not H(t)-conjugate in group Gmn(t).
Let now the reduced forms of elementsf and g satisfy conditions (4.6) of Lemma 4.4. Then according to this lemma, in subgroup H there exists the only sequence of elements h0, h1, h2,…,hr such that for any i=1,2,…,r, we have xi=hi-1-1yihi. Since elements f and g are not H-conjugate, then elements h0 and hr should be different. Since group Gmn is residually finite, by Proposition 3.3 there exists an integer t1>1 such that h0ρt1≠hrρt1. Let's choose one more integer t2 such that in group Gmn(t2) elements fρt2 and gρt2 have length r. Then if t=t1t2, in group Gmn(t), h0ρt≠hrρt, elements fρt and gρt have reduced forms(4.25)(x1ρt)(x2ρt)⋯(xrρt),(y1ρt)(y2ρt)⋯(yrρt),
respectively, and for every i=1,2,…,r(4.26)xiρt=(cHt)αi(dHt)βi(yiρt)(cHt)γi(dHt)δi.
Moreover, in group Gmn(t), for any i=1,2,…,r, the equality
(4.27)xiρt=(hi-1ρt)-1(yiρt)(hiρt)
holds.
Since the sequenceh0ρt, h1ρt, h2ρt,…,hrρt is, by Lemma 4.5, the unique sequence of elements of subgroup H(t) satisfying these equalities, then from Proposition 2.3, elements fτt and gτt are not H(t)-conjugate in group Gmn(t).
Hence Proposition4.1 is completely demonstrated.We now proceed directly to the proof of Theorem1.1.As remarked above, for anyt>1, group Gmn(t) is conjugacy separable. Therefore, for the proof of the Theorem it is enough to show that for any two nonconjugate in group Gmn elements f and g of group Gmn, there exists an integer t>1 such that elements fρt and gρt are not conjugate in group Gmn(t). To this end we make use of the conjugacy criterion of elements of groups Gmn and Gmn(t) given in Proposition 3.2.So, letf=x1x2⋯xr and g=y1y2⋯yr be the reduced forms in group Gmn=(A*B;H) of two nonconjugate in group Gmn elements f and g. By Proposition 3.2 we can assume that elements f and g are cyclically reduced. In accordance with this proposition we must  consider separately some cases.Case 1 (l(f)≠l(g)).
From Proposition3.5, there exists an integer t1>1 such that for any positive integer t, divisible by t1, the length of element fρt in group Gmn(t) coincides with the length of element f in group Gmn. Similarly, there exists integer t2>1 such that for any positive integer t, divisible by t2, the length of element gρt in group Gmn(t) coincides with the length of element g in group Gmn. Thus, if t=t1t2 elements fρt and gρt of group Gmn(t) are, as it is easy to see, cyclically reduced and have different length. Hence, by Proposition 3.2, these elements are not conjugate in this group. Thus integer t is the required.Case 2 (l(f)=l(g)=1).
In this case each of these elements belongs to one of the subgroupsA or B. Suppose first that both elements lie in the same of these subgroups; let it be, for instance, subgroup A. Since elements f and g are not conjugate in group A and group A is conjugacy separable [3], then by Proposition 3.3 there exists an integer t>1 such that in group A(t) elements fρt and gρt are not conjugate. Proposition 3.2 now implies that elements fρt and gρt are not conjugate in group Gmn(t).Let now elementf belongs to subgroup A, element g belongs to subgroup B and these elements do not belong to subgroup H. By Proposition 3.4, there exist integers t1>1 and t2>1 such that for any positive integer t, divisible by t1, element fρt does not belong to subgroup Hρt, and for any positive integer t, divisible by t2, element gρt does not belong to subgroup Hρt. Then if t=t1t2, Proposition 3.2 implies that elements fρt and gρt are not conjugate in group Gmn(t).Case 3 (l(f)=l(g)>1).
Letgi=yiyi+1⋯yry1⋯yi-1 (i=1,2,…,r) be all the cyclic permutations of element g. Since elements g and gi are conjugate and elements f and g are not conjugate, element f is not H-conjugate to any of elements g1, g2,…,gr. It follows from Proposition 4.1 that for every i=1,2,…,r, there exists an integer ti>1 such that elements fρti and giρti are not H(ti)-conjugate in group Gmn(ti). Let also integer t0>1 be chosen such that for all positive integer t, divisible by t0, elements fρt and gρt have length r in group Gmn(t). Then if t=t0t1⋯tr in group Gmn(t), elements fρt and giρt have reduced forms
(4.28)(x1ρt)(x2ρt)⋯(xrρt)(yiρt)(yi+1ρt)⋯(yrρt)(y1ρt)⋯(yi-1ρt),
respectively. Furthermore, elements fρt and giρt are not H(t)-conjugate in group Gmn(t). Since an arbitrary cyclic permutation of element gρt coincides with some element giρt, then from Proposition 3.2 it follows that elements fρt and gρt are not conjugate in group Gmn(t).The proof of Theorem is now complete.



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*Source: 803243-2010-10-03.xml*