Volume 2010, Article ID 803243,14pages doi:10.1155/2010/803243

*Research Article*

**Conjugacy Separability of Some**

**One-Relator Groups**

**D. Tieudjo**

**1**

_{and D. I. Moldavanskii}

_{and D. I. Moldavanskii}

**2**

*1 _{Department of Mathematics and Computer Science, University of Ngaoundere, P.O. Box 454,}*

*Ngaoundere, Cameroon*

*2 _{Department of Algebra and Mathematical Logic, Ivanovo State University, 39,}*

*Ermak St, Ivanovo 153025, Russia*

Correspondence should be addressed to D. Tieudjo,tieudjo@yahoo.com

Received 12 November 2009; Accepted 25 May 2010

Academic Editor: Howard Bell

Copyrightq2010 D. Tieudjo and D. I. Moldavanskii. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Conjugacy separability of any group of the class of one-relator groups given by the presentation

*a, b*;*am _{, b}n*

_{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.

**1. Introduction**

A group*G*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 finitei.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, Mostowski1proved that finitely presented conjugacy separable groups have solvable conjugacy problem, just as finitely presented residually finite groups have solvable word problem.

unpublished result of M. Armstrong on conjugacy separability of any one-relator group with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 *am _{b}n*

_{}

*t*

and*a*−1_{b}l_{ab}m_{}*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 groupGmna, b*;

*am, bn*1

*where integersmandnare greater than 1, is*

*conjugacy separable.*

We note that the assertion of Theorem is also valid when*m* 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 of3.
The residual finiteness of groups*Gmn*is well known; it follows, for example, from the

result of paper6. Groups*Gmn* are of some particular interest: they are centerless,

torsion-free,*. . .*Some properties of these groups were considered in7,8where, 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, inSection 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 2some properties of separability of groups *Gmn* are established and in

Section 3the proof ofTheorem 1.1will be completed.

We remind that conjugacy separability of groups *Gmn* can also be obtained using

9. However, the method in 9presents 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 product

*G* *A*∗*B*; *H*of groups*A*and*B*with amalgamated subgroup*H*.
Every element*g*∈*G*can be written in a form

*gx*1*x*2· · ·*xr,* *r*1*,* 2.1

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

*gis called reduced.*

In general, element*g*∈*G*can have diﬀerent reduced forms, but if*gy*1*y*2· · ·*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

If*g* *x*1*x*2· · ·*xr* is a reduced form of cyclically reduced element*g* then cyclic permutations

of element*g*are elements of form*xixi*1· · ·*xrx*1· · ·*xi*−1, where*i*1*,*2*, . . . , r*.

If*X*is a subgroup of a group*Y* we will say that elements*a*and*b*of*Y*are*X-conjugate*

if*ax*−1_{bx}_{, for some}_{x}_{∈}_{X}_{.}

The Solitar criterion of conjugacy of elements of amalgamated free product of two
groupsTheorem 4*.*6 in10can be formulated as follows.

**Proposition 2.1. Let**G*A*∗*B*;*H* *be the free product of groupsAand* *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 ofGsuch thatlf* *lg. Then*

1*if* *f* ∈ *Aand* *f* *is not conjugate in* *Ato any element of subgroupH, then* *f* *andg* *are*
*conjugate in groupGif and only ifg* ∈ *Aandf* *and* *g* *are conjugate inA; similarly, if*

*f* ∈*Bandfis not conjugate inBto any element of subgroupHthenfandgare conjugate*
*in groupGif and only ifg*∈*Bandfandgare conjugate inB;*

2*iff* ∈*H* *thenf* *andg* *are conjugate in groupGif and only if there exists a sequence of*
*elements*

*f* *h*0*, h*1*, . . . , hn, hn*1*g* 2.2

*such that for any* *i* 0*,*1*, . . . , n hi* ∈ *H* *and elements* *hi* *and* *hi*1 *areA-conjugate or*

*B-conjugate;*

3*if* *lf* *lg* *>* *1, thenf* *and* *g* *are conjugate in groupG* *if and only if elementf* *is*

*H-conjugate to some cyclic permutation ofg.*

In the following special case assertion2ofProposition 2.1became unnecessary.

**Proposition 2.2. Let**G*A*∗*B*;*H* *be the free product of groupsAand* *B* *with amalgamated*
*subgroupH. Suppose that for any elementf* *ofAor* *Band for any elementhof subgroupH, the*
*inclusionf*−1_{hf}_{∈} _{H}_{is valid if and only if elements}_{f}_{and}_{h}_{commute. If}_{f}_{and}_{g}_{are cyclically}

*reduced elements ofGsuch thatlf* *lg* *1 thenfandgare conjugate in groupGif and only if*

*fandgbelong to the same subgroupAorBand are conjugate in this subgroup.*

Indeed, let elements*f* 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 assertion1ofProposition 2.1. Otherwise, we can assume that*f* ∈ *H*and by
assertion2ofProposition 2.1there is a sequence of elements

*fh*0*, h*1*, . . . , hn, hn*1*g,* 2.3

such that for any*i* 0*,*1*, . . . , n hi* ∈ *H*, and for suitable element*xi*belonging to one of the

subgroups*A*or*B*, the equality*x*−1

*i* *hixi* *hi*1holds. Since for*i*0*,*1*, . . . , n*−1 the inclusion

*x*−_{i}_{}1_{1}*hi*1*xi*1 ∈ *H*is valid, the hypothesis gives*h*0 *hn*, that is,*x*−*n*1*fxn* *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.

* Proposition 2.3. Letfx*1

*x*2· · ·

*xrandgy*1

*y*2· · ·

*yr*

*be the reduced forms of elementsfandgof*

*groupG* *A*∗*B*;*Hwherer >1. ThenfandgareH-conjugate if and only if there exist elements*

*h*0*, h*1*, h*2*, . . . , hr* *h*0*in subgroupHsuch that for anyi*1*,*2*, . . . , r, one has*

*xih*−*i*−11*yihi.* 2.4

*Proof. If for some elementsh*0*, h*1*, h*2*, . . . , hr* *h*0of subgroup*H*equalities2.4hold, then

*fx*1*x*2· · ·*xr* *h*−01*y*1*h*1 *h*1−1*y*2*h*2· · · *hr*−−11*y*2*hr* *h*−01

*y*1*y*2· · ·*yr*

*hr* *h*−01*gh*0*.* 2.5

Conversely, by induction on *r* we prove that if for some *h* ∈ *H* the equality *f*
*h*−1*gh*holds then there exists a sequence of elements*h*0*, h*1*, h*2*, . . . , hr* *h*0 of subgroup*H*

satisfying the equalities2.4and such that*h*0*h*.

Rewriting the equality*fh*−1_{gh}_{in the form}

*x*−* _{r}*1· · ·

*x*−

_{2}1

*x*−

_{1}1

*h*−1

*y*1

*y*2· · ·

*yrh*1

*,*2.6

we see that since the expression in the left part of it cannot be reduced, elements *x*1 and

*y*1 must be contained in the same subgroup *A*or *B*and element *x*_{1}−1*h*−1*y*1 must belong to

subgroup*H*. Denoting this element by*h*−_{1}1, we have*x*1 *h*−1*y*1*h*1. If*r* 2 then the equality

above takes the form*x*−1

2 *h*−11*y*2*h*1, whence*x*2 *h*1−1*y*2*h*. Therefore, setting*h*0 *h*2 *h*, we

obtain in that case the desired sequence.

If*r >*2 let us rewrite the equality*fh*−1*gh*in the form

*h*−1*y*1*h*1*x*2· · ·*xr* *h*−1*y*1*y*2· · ·*yrh.* 2.7

This implies that if we set*f* *h*−1_{h}

1*x*2· · ·*xr* and*g* *y*2· · ·*yr*, then*f* *h*−1*gh*. Since the

length of elements*f*and*g*is equal to*r*−1, then by induction, there exists a sequence*h*_{1}*h*,

*h*_{2}*, . . . , h _{r}*

*h*

_{1}of elements of subgroup

*H*, such that

*h*−1

_{h}1*x*2 *h*_{1}−1*y*2*h*_{2} and for any

*i*2*,*3*, . . . , r*

*xi*

*h _{i}*

_{−}

_{1}−1

*yihi.*2.8

Since*x*2 *h*−11*y*2*h*2, then setting*hi* *hi*for*i*2*,*3*, . . . , r*, we obtain the desired sequence, and

induction is completed.Proposition 2.3is proven.

We conclude this section with one more property of amalgamated free product.

**Proposition 2.4. Let**G*A*∗*B*;*Hbe the free product of groupsAandBamalgamating subgroup*

*HwhereHlies in the center of each of groupsAandB. Then for any elementf* ∈*Gnot belonging to*
*subgroupAone hasf*−1*Af*∩*AH.*

*Proof. In fact, since subgroupH*coincides with the center of group*G*see10, page 211, the
inclusion*H* ⊆ *f*−1_{Af}_{∩}_{A}_{is obvious. Conversely, let element}* _{f}*−1

_{af}_{belong to subgroup}

_{A}first component of reduced form of*f* belongs to*B*. In any case the assumption*a /*∈*H*would
imply that*lf*−1_{af}_{}_{>}_{1. Thus,}_{a}_{∈}_{H}_{and therefore}* _{f}*−1

_{af}_{}

_{a}_{∈}

_{H}_{.}

**3. Some Properties of Groups**

*G*

*mn*

**and of Their Certain Quotients**

In what follows, our discussion will make use of the presentation of group *Gmn* 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*;*amc*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}_{G}

*mn*is the free product*A*∗*B*;*H*of groups*A*and*B*with amalgamation*H*.

The same decomposition is satisfiable for certain quotients of groups*Gmn*. Namely, for

any integer*t >*1 let*Gmnt*be the group with presentation

*a, b*;*am, bn* 1*, amtbnt*1 3.1

and*ρt*the natural homomorphism of groups*Gmn*onto*Gmnt*. Then it is easy to verify that

subgroup*Ht* *Hρt*of group*Gmnt*is isomorphic to the quotient*H/Ht*where, as usually,

*Ht*_{consists of all elements}_{h}t_{,}_{h}_{∈}_{H}_{}_{, subgroup}_{A}_{}_{t}_{Aρ}

*t*is the amalgamated free product

*a*;*amt* 1 ∗*Ht*;*am* *cHt*of cycle*a*;*amt* 1of order*mt*and group*Ht*, subgroup

*Bt* *Bρt*is the amalgamated free product*b*;*bnt* 1 ∗*Ht*;*bn* *dHt*of cycle*b*;*bnt*

1 of order*nt*and group *Ht*, and group *Gmnt*is the amalgamated free product*At*∗

*Bt*;*Ht*.

These decompositions of groups*Gmn*and*Gmnt*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 groups*At*and*Bt* is the amalgamated
free product of two finite groups and group*Gmnt*is the free product of groups*At*and

*Bt*with finite amalgamation, it follows from results of3that for every*t*, group*Gmnt*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*Gmnt*.

Since in the decompositions of groups*A*and*B*as well, as of groups *At*and*Bt*,
into amalgamated free product stated above the amalgamated subgroups are contained in
the centre of each free factor, byProposition 2.4we have the following proposition.

**Proposition 3.1. For any element**g*of groupA(or* *B) not belonging to subgroupH* *the equality*

*g*−1_{Hg}_{∩}_{H}_{}_{}_{c}_{}* _{(resp.,}_{g}*−1

_{Hg}_{∩}

_{H}_{}

_{}

_{d}_{}

_{) holds. In particular, for any element}_{g}_{of group}_{A}_{or}_{B}*and for any elementhof subgroupHelementg*−1_{hg}_{belongs to subgroup}_{H}_{if and only if elements}_{g}

*andhcommute.*

*Similarly, for any element* *g* *of group* *At* *(or* *Bt) not belonging to subgroup* *Htthe*
*equalityg*−1_{H}_{}_{t}_{}_{g}_{∩}_{H}_{}_{t}_{}_{cH}t_{}* _{(resp.,}_{g}*−1

_{H}_{}

_{t}_{}

_{g}_{∩}

_{H}_{}

_{t}_{}

_{dH}t_{}

_{) holds. In particular, for any}*elementsg* *of groupAtorBtandhof subgroupHt, elementg*−1_{hg}_{belongs to subgroup}_{H}_{}_{t}_{}

Propositions2.1,2.2and3.1lead to the following criterion for conjugacy of elements
of groups*Gmn*and*Gmnt*.

**Proposition 3.2. Let***G* *be any of groups* *Gmn* *and* *Gmnt. 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 inG. Letfandgbe cyclically reduced elements ofGsuch thatlf* *lg.*
*Then*

1*Iflf* *lg* *1 thenf* *andgare conjugate in groupGif and only iff* *andg* *belong to*
*the same free factor and are conjugate in this factor.*

2*If* *lf* *lg* *>* *1 thenf* *and* *g* *are conjugate in group* *G* *if and only if element* *f* *is*

*H-conjugate orHt-conjugate to some cyclic permutation ofg.*

We now consider some further properties of groups*Gmn* and *Gmnt*. The following

assertion is easily checked.

**Proposition 3.3. For any homomorphism**ϕof groupGmnonto a finite groupXthere exist an integer

*t >1 and a homomorphismψof groupGmntonto groupXsuch thatϕρtψ.*

*Also, for any integers* *t >* *1 and* *s >* *1 such that* *t* *divides* *s* *there exists homomorphism*

*ϕ*:*Gmns* → *Gmntsuch thatρtρsϕ.*

*The same assertions are valid for groupsAandB.*

**Proposition 3.4. For any element**gof groupAorB, ifgdoes not belong to subgroupHthen there*exists an integert*0*>1 such that for any positive integert, divisible byt*0*, elementgρtdoes not belong*

*to subgroupHρt.*

*Proof. We will assume thatg* ∈*A*; the case when*g*∈*B*can be treated similarly.

So, let element*g*∈*A*do not belong to subgroup*H*and let*gx*1*x*2· · ·*xr*be a reduced

form of*g*in the decomposition of group*A*into amalgamated free product.

If*r*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*At*, 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}

*Ht* *Hρt*.

Let*r >* 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*ck _{d}l*

_{where}

_{l /}_{}

_{0. If integer}

_{t}0is chosen such

that the exponent*l*of any component*xi*of the second kind is not divisible by*t*0, then for any

*t*divisible by*t*0the image*xiρtxiHt*of such component will not belong to the amalgamated

subgroup of group*At* 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

*gρt*

*x*1*ρt*

*x*2*ρt*

· · ·*xrρt*

3.2

of element*gρt*is reduced in group*At*and since*r >*1,*gρt*does not belong to the free factor

*Ht* *Hρt*. The proposition is proven.

**Proposition 3.5. For any element**g*of groupGmn* *there exists an integert*0 *>* *1 such that for all*

*positive integert, divisible byt*0*, the length of elementgρtin groupGmntcoincides with the length*

*of elementgin groupGmn.*

**Proposition 3.6. For any elements**fandgof groupA(orB) such that elementfdoes not belong to*the double cosetHgH, there exists an integert*0 *>1 such that for any positive integert, divisible by*

*t*0*, elementfρtdoes not belong to the double cosetHtgρtHt.*

*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*HϕgϕHϕ*.

To this end let’s consider the quotient group*A* *A/C* of group*A*by its central
subgroup*Cc*. The image*xC*of an element*x*∈*A*in group*A*will be denoted by*x*.

It is obvious that group *A* 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*Y gY*. Since*C* *H*,
the element*f*does not belong to this coset.

We can assume, without loss of generality, that any element*f* 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*dk* 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*A*the factor group of group*A*by the
normal closure of element*dt*. Group*A*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*f*and*g*of elements*f*and*g*in group*A*are the same
as in group*A*. In particular, element*f*does not belong to the double coset*Y/Yt _{g Y/Y}*

_{}

*t*

_{. Since}

this coset consists of a finite number of elements and group*A*is residually finite, then there
exists a normal subgroup*N*of finite index of group*A*such, that

*f /*∈*Y/YtgY/Y* *t*·*N.* 3.3

If now*θ*is the product of the canonical homomorphisms of group*A*onto group*A*and of
group*A*onto group*A*and*N*is the full preimage by*θ*of subgroup*N*, then*N*is a normal
subgroup of finite index of group*A*and*f /*∈*HgHN*. 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**

**Theorem 1.1**

We prove first the following proposition.

**Proposition 4.1. If elements**fandgof groupGmnsuch thatlf*lg>1 are notH-conjugate,*

*Proof. Letf* *x*1*x*2· · ·*xr*and*gy*1*y*2· · ·*yr* be the reduced forms in group*Gmn* *A*∗*B*;*H*

of elements*f*and*g*.

We remindseeProposition 2.3that elements*f* and*g* are*H*-conjugate if and only if
there exist elements*h*0,*h*1,*h*2*, . . . , hr* *h*0of*H*such that for any*i*1*,*2*, . . . , r*, we have

*xih*−_{i}_{−}1_{1}*yihi.* 4.1

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 fromProposition 3.4that

there exists an integer*t >*1 such that elements*xiρt*and*yiρt*do not belong to the same free

factor*At*or*Bt*of group*Gmnt* and, as above, lie in free factors of this group. Hence, by

Proposition 2.3, in group*Gmnt*elements*fρt*and*gρt*are not*Ht*-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*. ByProposition 3.6,

there exists an integer*t*1 *>*1 such that for any positive integer*t*, divisible by*t*1, element*xiρt*

is not in the double coset*HtyiρtHt*. FromProposition 3.5, there exist integers*t*2*>*1 and

*t*3 *>* 1 such that for any positive integer*t*divisible by*t*2 the length of element*fρt*in group

*Gmnt*is equal to*r*and for any positive integer*t*, divisible by*t*3, the length of element*gτt*in

group*Gmnt*is equal to*r*. Thus, if*tt*1*t*2*t*3then in group*Gmnt*, elements*fρt*and*gρt*have

the reduced forms

*x*1*ρt*

*x*2*ρt*

· · ·*xrρt*

*,* *y*1*ρt*

*y*2*ρt*

· · ·*yrρt*

*,* 4.2

respectively, and element *xiρt* is not in the double coset *HtyiρtHt*. Again, by

Proposition 2.3these elements are not*Ht*-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 elements***xand* *y* *belong to one of the groupsA* *or* *B* *and do not belong to the*
*subgroupHandx*∈*HyH, that is,*

*xcαdβycγdδ* 4.3

*for some integersα,β,γandδ. If elementsxandybelong to groupA, then integersαγ,βandδare*
*uniquely determined by the equality*4.3*. If elementsxandybelong to groupB, then integersβδ,*

*αandγare uniquely determined by equality*4.3*.*

*Proof. Let elements* *x* and *y* belong to subgroup *A* and let *x* *cα*1* _{d}β*1

*1*

_{yc}γ*1*

_{d}δ_{and}

_{x}*cα*2* _{d}β*2

*2*

_{yc}γ*2*

_{d}δ_{for some integers}

_{α}1, *β*1, *γ*1, *δ*1, *α*2, *β*2, *γ*2 and *δ*2. Rewriting the equality

conclude that*cα*1−*α*2* _{d}β*1−

*β*2

*2−*

_{c}γ*γ*1

*2−*

_{d}δ*δ*1

_{and also that this element should belong to subgroup}

*c*, that is,*β*1−*β*2*δ*2−*δ*10. So, we have*cα*1−*α*2 *cγ*2−*γ*1and since the order of element*c*is

infinite, then*α*1−*α*2*γ*2−*γ*1.

Thus,*β*1*β*2,*δ*1*δ*2and*α*1*γ*1*α*2*γ*2as it was required. The case when elements

*x*and*y*belong to group*B*is esteemed similarly.

**Lemma 4.3. Let elements**xandybelong to one of groupsAtorBtand do not belong to subgroup

*Htandx*∈*HtyHt, that is,*

*xcHtαdHtβycHtγdHtδ* 4.4

*for some integersα,β,γandδ. If elementsxandybelong to groupAt, then integersαγ,βand*

*δare uniquely determined modulotby equality*4.4*. If elementsxandybelong to groupBt, then*
*integersβδ,αandγare uniquely determined modulotby equality*4.4*.*

The proof ofLemma 4.3is completely similar to that ofLemma 4.2.

* Lemma 4.4. Letfx*1

*x*2· · ·

*xr*

*andg*

*y*1

*y*2· · ·

*yr*

*be reduced elements of groupGmn, wherer >*1

*and let for everyi*1*,*2*, . . . , rthe equalityxiuiyiviholds, for some elementsuiandviof subgroup*

*H. Then there exists at most one sequenceh*0*,h*1*,h*2*, . . . , hrof elements of subgroupHsuch that for*

*anyi*1*,*2*, . . . , r*

*xih*−*i*−11*yihi.* 4.5

*Moreover, ifui* *cαidβi* *and* *vi* *cγidδi* *for some integers* *αi,* *βi,γi* *and* *δi* *(i* 1*,*2*, . . . , r) and*

*x*1*, y*1∈*A, then such sequence exists if, and only if, for anyi, 1< i < r,*

*αiαi*1*γi*−1*γi*0*,* *ifiis odd,*

*βiβi*1*δi*−1*δi*0*,* *ifiis even.*

4.6

*Proof. We suppose first that the sequence* *h*0, *h*1, *h*2*, . . . , hr* of elements of subgroup *H*

satisfying equality4.5exists, and let’s write*hicμidνi*, for some integers*μi*and*νi*.

Then, since for any*i*1*,*2*, . . . , r*the equality*h*−_{i}_{−}1_{1}*yihiuiyivi*holds, we have

*y*−* _{i}*1

*hi*−1

*uiyihiv*−

*i*1

*.*4.7

Since*r >* 1, then every element*yi*, belonging to one of the subgroups*A*or*B*, does not lie

in subgroup*H*, and consequently, fromProposition 3.1, for any*i* 1*,*2*, . . . , r*, we have the
equality*hi*−1*ui* *hiv _{i}*−1. Substituting the expressions of elements

*hi*,

*ui*and

*vi*, we have for

every *i* 1*,*2*, . . . , rcμi*−1*αi _{d}νi*−1

*βi*

_{c}μi−γi_{d}νi−δi_{and hence we obtain the system of numeric}

equations

*μi*−1*αi* *μi*−*γi,* *νi*−1*βiνi*−*δi* *i*1*,*2*, . . . , r.* 4.8

Since by4.7for every*i* 1*,*2*, . . . , r* element*hiv _{i}*−1 belongs to the intersection

*y*−

*1*

_{i}*Hyi*∩

*H*

to subgroup*B*, then fromProposition 3.1, it follows that for odd*i*, we should have*hi*−1*ui*

*hiv _{i}*−1∈

*c*, and for even

*i*, we should have

*hi*−1

*ui*

*hiv*−

*1 ∈*

_{i}*d*. It means that for every odd

*ii* 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 integers*μi*and*νi*are determined uniquely. Namely,

*μi*
⎧
⎨
⎩

*γi* if*i*is even and 2*ir,*

−*αi*1 if*i*is odd and 1*ir*−1*,*

4.9

*νi*
⎧
⎨
⎩

−*βi*1 if*i*is even and 0*ir*−1*,*

*δi* if*i*is odd and 1*ir.*

4.10

Moreover, from equalities4.8it follows, that

*μ*0−

*α*1*α*2*γ*1

*,*

*νr* *βrδr*−1*δr* if*r* is even*,*

*μr* *αrγr*−1*γr* if *r* is odd*.*

4.11

Thus, the statement that there can exist at most one sequence of elements *h*0, *h*1,

*h*2*, . . . , hr* of subgroup*H*satisfying equalities4.5is demonstrated.

Substituting the value*μi*−1*γi*−1and*μi*−*αi*1defined in4.9in the equalities*μi*−1

*αi* *μi*−*γi*of system4.8, where 1*< i < r* and*i*is odd, we obtain*αiαi*1*γi*−1*γi* 0.

Similarly, substituting the value*νi*−1 *δi*−1 and*νi* −*βi*1defined in4.10in the equalities

*νi*−1*βiνi*−*δi*of systems4.8, where 1*< i < r*and*i*is even, we obtain*βiβi*1*δi*−1*δi*0.

Thus, under the existence in subgroup*H* of sequence*h*0,*h*1,*h*2*, . . . , hr* of elements

satisfying4.5, conditions4.6are satisfied.

Conversely, suppose conditions4.6are satisfied. Let

*h*0*c*−*α*1*α*2*γ*1*d*−*β*1 4.12

and for all indexes*i*such that 1*i < r*, we set

*hi*
⎧
⎨
⎩

*cγi _{d}*−

*βi*1

_{if}

_{i}_{is even}

_{,}*c*−*αi*1_{d}δi_{if} _{i}_{is odd}_{.}

4.13

At last, for*ir*we set

*hr*
⎧
⎨
⎩

*cγr _{d}βrδr*−1

*δr*

_{if}

_{r}_{is even}

_{,}*cαrγr*−1*γr _{d}δr*

_{if}

_{r}_{is odd}

_{.}Let us show that, the so-defined sequence of elements*h*0,*h*1,*h*2*, . . . , hr* really fits to

equalities4.5.

If*i*1, using the permutability of elements*c*and*y*1we have

*h*−_{0}1*y*1*h*1*cα*1*α*2*γ*1*dβ*1*y*1*c*−*α*2*dδ*1*cα*1*dβ*1*y*1*cγ*1*dδ*1*x*1*.* 4.15

If 1 *< 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

*h*−_{i}_{−}1_{1}*yihicαid*−*δi*−1*yicγid*−*βi*1*cαiyicγid*−*δi*−1*βi*1*cαiyicγidβiδi* *cαidβiyicγidδi* *xi.* 4.16

If 1*< i < r*and integer*i*is odd, using the equality*αiαi*1*γi*−1*γi*0 and permutability

of elements*c*and*yi*, we have

*h*−1

*i*−1*yihi* *c*−*γi*−1*dβiyic*−*αi*1*dδi* *c*−*γi*−1*αi*1*dβiyidδi* *cαiγidβiyidδi* *cαidβiyicγidδi* *xi.* 4.17

If integer*r*is even, then

*h*−_{r}_{−}1_{1}*yrhr* *cαrd*−*δr*−1*yrcγrdβrδr*−1*δr* *cαrdβryrcγrdδr* *xr,* 4.18

and if*r*is odd, then

*h*−_{r}_{−}1_{1}*yrhr* *c*−*γr*−1*dβryrcαrγr*−1*γrdδr* *cαrdβryrcγrdδr* *xr.* 4.19

So,Lemma 4.4is completely demonstrated.

Similar argument gives the following lemma.

**Lemma 4.5. Let**f*x*1*x*2· · ·*xr* *andg* *y*1*y*2· · ·*yr* *be reduced elements of groupGmnt, where*

*r >1, and let for everyi*1*,*2*, . . . , rthe equalityxiuiyivitakes place, for some elementsuiandvi*

*of subgroupHt. Then there exists at most one sequenceh*0*,h*1*,h*2*, . . . , hr* *of elements of subgroup*

*Htsuch that for anyi*1*,*2*, . . . , r, one has*

*xih*−_{i}_{−}1_{1}*yihi.* 4.20

*Moreover, ifui* *cHtαidHtβi* *andvi* *cHtγidHtδi* *for some integersαi,βi,γi* *andδi* *(i*

1*,*2*, . . . , r) andx*1*, y*1 ∈*At, then such sequence exists if and only if for any integeri, 1< i < r, one*

*has*

*αiαi*1*γi*−1*γi*≡0mod*t* *ifiis odd,*

*βiβi*1*δi*−1*δi*≡0mod*t* *ifiis even.*

4.21

By Proposition 2.3, elements *f* and *g* are *H*-conjugate if and only if their cyclic
permutation*x*2· · ·*xrx*1and*y*2· · ·*yry*1 are*H*-conjugate. Therefore, we can suppose, without

loss of generality, that elements*x*1and*y*1belong to subgroup*A*.

By supposition, for every*i*1*,*2*, . . . , r*there exist integers*αi*,*βi*,*γi*and*δi*such that

*xicαidβiyicγidδi.* 4.22

If integer *i*, 1 *< i < r*, is even, then *yi* ∈ *B*,*yi*−1*, yi*1 ∈ *A* and consequently, from

Lemma 4.2, integers*βiδi*,*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*x*1,*x*2*, . . . , xr*and*y*1,*y*2*, . . . , yr*. Similarly, for any odd*i*, 1*< i < r*,

integers*αiγi*,*γi*−1 and*αi*1are uniquely determined by these sequences. It means, in turn,

that the satisfiability of conditions4.6 of Lemma 4.4depends only on the sequences*x*1,

*x*2*, . . . , xr* and*y*1,*y*2*, . . . , yr*.

Let’s now conditions4.6ofLemma 4.4are not satisfied, that is, either for some even

*i*, 1*< i < r*, the sum*βiβi*1*δi*−1*δi*is diﬀerent from zero, or for some odd*i*, 1*< i < r*,

the sum*αiαi*1*γi*−1*γi*is diﬀerent from zero. Then it is possible to find an integer*t*1 *>*1,

not dividing the respective sum. Let’s also choose, according toProposition 3.5, the integers

*t*2*>*1 and*t*3*>*1 such that for all positive integer*t*, divisible by*t*2, the length of element*fρt*in

group*Gmnt*is equal to*r*and for all positive integer*t*, divisible by*t*3, the length of element

*gρt*in group *Gmnt*is equal to*r*. Then if*t* *t*1*t*2*t*3, in group*Gmnt*, elements*fρt*and*gρt*

have reduced forms

*x*1*ρt*

*x*2*ρt*

· · ·*xrρt*

*,* *y*1*ρt*

*y*2*ρt*

· · ·*yrρt*

*,* 4.23

respectively. Further, for every*i*1*,*2*, . . . , r*,

*xiρt*

*cHtαi*

*dHtβi*
*yiρt*

*cHtγi*
*dHtδi*

*.* 4.24

From Lemma 4.3 and the selection of integer *t*1 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*Ht*-conjugate in group*Gmnt*.

Let now the reduced forms of elements*f*and*g*satisfy conditions4.6ofLemma 4.4.
Then according to this lemma, in subgroup*H*there exists the only sequence of elements*h*0,

*h*1,*h*2*, . . . , hr* such that for any*i*1*,*2*, . . . , r*, we have*xih*−_{i}_{−}1_{1}*yihi*. Since elements*f*and*g*are

not*H*-conjugate, then elements*h*0and*hr*should be diﬀerent. Since group*Gmn*is residually

finite, by Proposition 3.3there exists an integer*t*1 *>* 1 such that *h*0*ρt*1*/hrρt*1. Let’s choose

one more integer*t*2such that in group*Gmnt*2elements*fρt*2and*gρt*2have length*r*. Then if

*tt*1*t*2, in group*Gmnt*,*h*0*ρt/hrρt*, elements*fρt*and*gρt*have reduced forms

*x*1*ρt*

*x*2*ρt*

· · ·*xrρt*

*,* *y*1*ρt*

*y*2*ρt*

· · ·*yrρt*

*,* 4.25

respectively, and for every*i*1*,*2*, . . . , r*

*xiρt*

*cHtαi*

*dHtβi*
*yiρt*

*cHtγi*
*dHtδi*

Moreover, in group*Gmnt*, for any*i*1*,*2*, . . . , r*, the equality

*xiρt*

*hi*−1*ρt*
_{−}1

*yiρt*

*hiρt*

4.27

holds.

Since the sequence*h*0*ρt*,*h*1*ρt*,*h*2*ρt, . . . , hrρt*is, byLemma 4.5, the unique sequence of

elements of subgroup *Ht* satisfying these equalities, then from Proposition 2.3, elements

*fτt*and*gτt*are not*Ht*-conjugate in group*Gmnt*.

HenceProposition 4.1is completely demonstrated.

We now proceed directly to the proof ofTheorem 1.1.

As remarked above, for any*t >*1, group*Gmnt*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*Gmnt*. To this end we make use of the conjugacy criterion of elements

of groups*Gmn*and*Gmnt*given inProposition 3.2.

So, let *f* *x*1*x*2· · ·*xr* and *g* *y*1*y*2· · ·*yr* be the reduced forms in group *Gmn*

*A*∗*B*; *H*of two nonconjugate in group*Gmn*elements*f*and*g*. ByProposition 3.2we can

assume that elements*f*and*g*are cyclically reduced. In accordance with this proposition we
must consider separately some cases.

*Case 1* *lf/lg*. FromProposition 3.5, there exists an integer *t*1 *>* 1 such that for any

positive integer*t*, divisible by*t*1, the length of element*fρt* in group*Gmnt*coincides with

the length of element*f* in group*Gmn*. Similarly, there exists integer*t*2 *>*1 such that for any

positive integer*t*, divisible by*t*2, the length of element*gρt*in group*Gmnt*coincides with the

length of element*g*in group*Gmn*. Thus, if*tt*1*t*2elements*fρt*and*gρt*of group*Gmnt*are,

as it is easy to see, cyclically reduced and have diﬀerent length. Hence, byProposition 3.2,
these elements are not conjugate in this group. Thus integer*t*is the required.

*Case 2lf* *lg* 1. In this case each of these elements belongs to one of the subgroups*A*

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
separable3, then byProposition 3.3there exists an integer*t >* 1 such that in group*At*

elements*fρt*and *gρt*are not conjugate.Proposition 3.2now implies that elements*fρt*and

*gρt*are not conjugate in group*Gmnt*.

Let now element*f*belongs to subgroup*A*, element*g*belongs to subgroup*B*and these
elements do not belong to subgroup*H*. ByProposition 3.4, there exist integers *t*1 *>* 1 and

*t*2 *>* 1 such that for any positive integer*t*, divisible by *t*1, element*fρt* does not belong to

subgroup*Hρt*, and for any positive integer*t*, divisible by*t*2, element*gρt* does not belong

to subgroup*Hρt*. Then if*t* *t*1*t*2,Proposition 3.2implies that elements*fρt*and*gρt*are not

conjugate in group*Gmnt*.

*Case 3* *lf* *lg* *>* 1. Let *gi* *yiyi*1· · ·*yry*1· · ·*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*g*1,*g*2*, . . . , gr*. It follows from

Proposition 4.1that for every*i* 1*,*2*, . . . , r*, there exists an integer*ti* *>* 1 such that elements

that for all positive integer*t*, divisible by*t*0, elements *fρt* and*gρt* have length *r* in group

*Gmnt*. Then if*tt*0*t*1· · ·*tr*in group*Gmnt*, elements*fρt*and*giρt*have reduced forms

*x*1*ρt*

*x*2*ρt*

· · ·*xrρt*

*yiρt*

*yi*1*ρt*

· · ·*yrρt*

*y*1*ρt*

· · ·*yi*−1*ρt*

*,* 4.28

respectively. Furthermore, elements *fρt* and *giρt* are not *Ht*-conjugate in group *Gmnt*.

Since an arbitrary cyclic permutation of element*gρt*coincides with some element*giρt*, then

fromProposition 3.2it follows that elements*fρt*and*gρt*are not conjugate in group*Gmnt*.

The proof of Theorem is now complete.

**References**

1 *A. W. Mostowski, “On the decidability of some problems in special classes of groups,” Fundamenta*

*Mathematicae, vol. 59, pp. 123–135, 1966.*

2 *G. Baumslag and D. Solitar, “Some two-generator one-relator non-Hopfian groups,” Bulletin of the*

*American Mathematical Society, vol. 68, pp. 199–201, 1962.*

3 *J. L. Dyer, “Separating conjugates in amalgamated free products and HNN extensions,” Australian*

*Mathematical Society A, vol. 29, no. 1, pp. 35–51, 1980.*

4 *C. Y. Tang, “Conjugacy separability of certain 1-relator groups,” Proceedings of the American*

*Mathematical Society, vol. 86, no. 3, pp. 379–384, 1982.*

5 R. B. J. T. Allenby and C. Y. Tang, “Conjugacy separability of certain 1-relator groups with torsion,”

*Journal of Algebra, vol. 103, no. 2, pp. 619–637, 1986.*

6 *G. Baumslag, “Free subgroups of certain one-relator groups defined by positive words,” Mathematical*

*Proceedings of the Cambridge Philosophical Society, vol. 93, no. 2, pp. 247–251, 1983.*

7 D. Tieudjo and D. I. Moldavanski, “Endomorphisms of the group*Gmn* *a, b*;*am, bn* 1*m, n >*

1*,” Journal of the African Mathematical Union, vol. 9, pp. 11–18, 1998.*

8 D. Tieudjo and D. I. Moldavanskii, “On the automorphisms of some one-relator groups,”

*Communications in Algebra, vol. 34, no. 11, pp. 3975–3983, 2006.*

9 *G. Kim, J. McCarron, and C. Y. Tang, “Adjoining roots to conjugacy separable groups,” Journal of*

*Algebra, vol. 176, no. 2, pp. 327–345, 1995.*