International Journal of Mathematics and Mathematical Sciences

Volume 2010 (2010), Article ID 803243, 14 pages

http://dx.doi.org/10.1155/2010/803243

## Conjugacy Separability of Some One-Relator Groups

^{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

Received 12 November 2009; Accepted 25 May 2010

Academic Editor: Howard Bell

Copyright © 2010 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.

#### Abstract

Conjugacy separability of any group of the class of one-relator groups given by the presentation 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 is conjugacy separable if for any two nonconjugate elements and of there exists a homomorphism of group onto some finite group such that the images and of elements and are not conjugate in group . It is clear that any conjugacy separable group is residually finite (i.e., recall that for any nonidentity element there exists a homomorphism of group onto some finite group such that ).

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 In the same paper she proved an unpublished result of . Armstrong on conjugacy separability of any one-relator group with nontrivial center. The conjugacy separability of groups defined by relator of form and , where , 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 group where integers and are greater than 1, is conjugacy separable.*

We note that the assertion of Theorem is also valid when or . Indeed, in this case group 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 groups is well known; it follows, for example, from the result of paper [6]. Groups 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 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 are established and in Section 3 the proof of Theorem 1.1 will be completed.

We remind that conjugacy separability of groups can also be obtained using [9]. However, the method in [9] presents any group 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 of groups and with amalgamated subgroup .

Every element can be written in a form
where for any element belongs to one of the free factors or and if any successive and do not belong to the same factor or (and therefore for any element does not belong to amalgamated subgroup ). Such form of element is called *reduced*.

In general, element can have different reduced forms, but if is one more reduced form of , then and, for any , and belong to the same factor or . The *length * of element is then defined as the number of the components in a reduced form of . Element is called *cyclically reduced* if either or and the first and the last components of its reduced form do not belong to the same factor or . If is a reduced form of cyclically reduced element then cyclic permutations of element are elements of form , where .

If is a subgroup of a group we will say that elements and of are *-conjugate* if , for some .

The Solitar criterion of conjugacy of elements of amalgamated free product of two groups (Theorem in [10]) can be formulated as follows.

Proposition 2.1. *Let be the free product of groups and with amalgamated subgroup . Every element of is conjugate to a cyclically reduced element. If lengths of two cyclically reduced elements are nonequal then these elements are not conjugate in . Let and be cyclically reduced elements of such that . Then **if and is not conjugate in to any element of subgroup , then and are conjugate in group if and only if and and are conjugate in ; similarly, if and is not conjugate in to any element of subgroup then and are conjugate in group if and only if and and are conjugate in ; **if then and are conjugate in group if and only if there exists a sequence of elements
such that for any and elements and are -conjugate or -conjugate; **if , then and are conjugate in group if and only if element is -conjugate to some cyclic permutation of .*

In the following special case assertion of Proposition 2.1 became unnecessary.

Proposition 2.2. *Let be the free product of groups and with amalgamated subgroup . Suppose that for any element of or and for any element of subgroup , the inclusion is valid if and only if elements and commute. If and are cyclically reduced elements of such that then and are conjugate in group if and only if and belong to the same subgroup or and are conjugate in this subgroup. *

Indeed, let elements and be conjugate in group and let belong to subgroup , say. If is not conjugate in to any element of subgroup , then the desired assertion is implied by assertion of Proposition 2.1. Otherwise, we can assume that and by assertion of Proposition 2.1 there is a sequence of elements such that for any , and for suitable element belonging to one of the subgroups or , the equality holds. Since for the inclusion is valid, the hypothesis gives , that is, . This means that elements and belong to that subgroup or which contains element and also are conjugated in this subgroup.

Next, we describe more explicitly the situation arising in assertion of Proposition 2.1.

Proposition 2.3. *Let and be the reduced forms of elements and of group where . Then and are -conjugate if and only if there exist elements in subgroup such that for any , one has
*

*Proof. *If for some elements of subgroup equalities (2.4) hold, then

Conversely, by induction on we prove that if for some the equality holds then there exists a sequence of elements of subgroup satisfying the equalities (2.4) and such that .

Rewriting the equality in the form
we see that since the expression in the left part of it cannot be reduced, elements and must be contained in the same subgroup or and element must belong to subgroup . Denoting this element by , we have . If then the equality above takes the form , whence . Therefore, setting , we obtain in that case the desired sequence.

If let us rewrite the equality in the form
This implies that if we set and , then . Since the length of elements and is equal to , then by induction, there exists a sequence , of elements of subgroup , such that and for any
Since , then setting for , 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. *Let be the free product of groups and amalgamating subgroup where lies in the center of each of groups and . Then for any element not belonging to subgroup one has . *

*Proof. *In fact, since subgroup coincides with the center of group (see [10, page 211]), the inclusion is obvious. Conversely, let element belong to subgroup where . Since then either or and (without loss of generality) the first component of reduced form of belongs to . In any case the assumption would imply that . Thus, and therefore .

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

In what follows, our discussion will make use of the presentation of group as an amalgamated free product of two groups. To describe such presentation, let be the subgroup of group , generated by elements and . Also, let denote the subgroup of group generated by element and subgroup , and let denote the subgroup of group , generated by element and subgroup . Then it can be immediately verified that is the free abelian group with base , , group is the free product of infinite cycle and group with amalgamation , group is the free product of infinite cycle and group with amalgamation , and group is the free product of groups and with amalgamation .

The same decomposition is satisfiable for certain quotients of groups . Namely, for any integer let be the group with presentation and the natural homomorphism of groups onto . Then it is easy to verify that subgroup of group is isomorphic to the quotient (where, as usually, consists of all elements , ), subgroup is the amalgamated free product of cycle of order and group , subgroup is the amalgamated free product of cycle of order and group , and group is the amalgamated free product .

These decompositions of groups and 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 and is the amalgamated free product of two finite groups and group is the free product of groups and with finite amalgamation, it follows from results of [3] that for every , group is conjugacy separable. So, to prove the conjugacy separability of group it is enough, for any nonconjugate elements and of , to find an integer such that elements and are nonconjugate in group .

Since in the decompositions of groups and as well, as of groups and , 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 element of group (or ) not belonging to subgroup the equality (resp., ) holds. In particular, for any element of group or and for any element of subgroup element belongs to subgroup if and only if elements and commute.**Similarly, for any element of group (or ) not belonging to subgroup the equality (resp., ) holds. In particular, for any elements of group or and of subgroup , element belongs to subgroup if and only if elements and commute.*

Propositions 2.1, 2.2 and 3.1 lead to the following criterion for conjugacy of elements of groups and .

Proposition 3.2. *Let be any of groups and . Every element of is conjugate to a cyclically reduced element. If lengths of two cyclically reduced elements are not equal then these elements are not conjugate in . Let and be cyclically reduced elements of such that . Then **If then and are conjugate in group if and only if and belong to the same free factor and are conjugate in this factor. **If then and are conjugate in group if and only if element is -conjugate or -conjugate to some cyclic permutation of . *

We now consider some further properties of groups and . The following assertion is easily checked.

Proposition 3.3. *For any homomorphism of group onto a finite group there exist an integer and a homomorphism of group onto group such that .**Also, for any integers and such that divides there exists homomorphism such that .**The same assertions are valid for groups and .*

Proposition 3.4. *For any element of group or , if does not belong to subgroup then there exists an integer such that for any positive integer , divisible by , element does not belong to subgroup .*

*Proof. *We will assume that ; the case when can be treated similarly.

So, let element do not belong to subgroup and let be a reduced form of (in the decomposition of group into amalgamated free product).

If then element belongs to subgroup that is, for some integer . Since , integer is not divisible by . Then for any integer , in group , element of subgroup cannot belong to the amalgamated subgroup (generated by element ) of the decomposition of group and consequently cannot belong to the free factor .

Let . Every component of the reduced form of element is either of form , where integer is not divisible by , or of form where . If integer is chosen such that the exponent of any component of the second kind is not divisible by , then for any divisible by the image of such component will not belong to the amalgamated subgroup of group (generated, let's remind, by element ). Moreover, as above in case , the images of components of the first kind will not belong to the amalgamated subgroup. Therefore, the form
of element is reduced in group and since , does not belong to the free factor . The proposition is proven.

Proposition 3.4 obviously implies the the following proposition.

Proposition 3.5. *For any element of group there exists an integer such that for all positive integer , divisible by , the length of element in group coincides with the length of element in group .*

Proposition 3.6. *For any elements and of group (or ) such that element does not belong to the double coset , there exists an integer such that for any positive integer , divisible by , element does not belong to the double coset . *

*Proof. *We can again consider only the case when elements and belong to subgroup . So, let us suppose that element does not belong to the double coset . In view of Proposition 3.4, it is enough to prove that there exists a homomorphism of onto a finite group such that element of does not belong to the double coset .

To this end let's consider the quotient group of group by its (central) subgroup . The image of an element in group will be denoted by .

It is obvious that group is the (ordinary) free product of the cyclic group of order , generated by element , and the infinite cycle , generated by . The canonical homomorphism of group onto group maps subgroup onto subgroup and consequently, the image of the double coset is the double coset . Since , the element does not belong to this coset.

We can assume, without loss of generality, that any element and , if it is different from identity, has reduced form the first and the last syllables of which do not belong to subgroup . Every -syllable of these reduced forms is of form for some integer . Since the set of all such exponents is finite, we can choose an integer such that for any the inequality holds. Let's denote by the factor group of group by the normal closure of element . Group is the free product of groups and and since different integers from are not relatively congruent and are not congruent to zero modulo , then the reduced forms of the images and of elements and in group are the same as in group . In particular, element does not belong to the double coset . Since this coset consists of a finite number of elements and group is residually finite, then there exists a normal subgroup of finite index of group such, that
If now is the product of the canonical homomorphisms of group onto group and of group onto group and is the full preimage by of subgroup , then is a normal subgroup of finite index of group and . Thus, the canonical homomorphism of group onto quotient group 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 and of group such that are not -conjugate, then for some integer , elements and of group are not -conjugate.*

*Proof. *Let and be the reduced forms in group of elements and .

We remind (see Proposition 2.3) that elements and are -conjugate if and only if there exist elements , , of such that for any , we have
It then follows, in particular, that for each elements and should lie in the same free factor or and define the same double coset modulo . So, we consider separately some cases.*Case 1. *Suppose that for some index elements and lie in different free factors and of group (and, certainly, are not in subgroup ). It follows from Proposition 3.4 that there exists an integer such that elements and do not belong to the same free factor or of group (and, as above, lie in free factors of this group). Hence, by Proposition 2.3, in group elements and are not -conjugate.*Case 2. *Let now for any elements and belong to the same subgroup or and for some element does not belong to the double coset . By Proposition 3.6, there exists an integer such that for any positive integer , divisible by , element is not in the double coset . From Proposition 3.5, there exist integers and such that for any positive integer divisible by the length of element in group is equal to and for any positive integer , divisible by , the length of element in group is equal to . Thus, if then in group , elements and have the reduced forms
respectively, and element is not in the double coset . Again, by Proposition 2.3 these elements are not -conjugate.*Case 3. *We now consider the case, when for any elements and lie in the same free factor or and also determine the same double coset modulo . We prove some lemmas.Lemma 4.2. *Let elements and belong to one of the groups or and do not belong to the subgroup and , that is,
**
for some integers , , and . If elements and belong to group , then integers , and are uniquely determined by the equality (4.3). If elements and belong to group , then integers , and are uniquely determined by equality (4.3).**Proof. *Let elements and belong to subgroup and let and for some integers , , , , , , and . Rewriting the equality as , then, by Proposition 3.1, we conclude that and also that this element should belong to subgroup , that is, . So, we have and since the order of element is infinite, then .

Thus, , and as it was required. The case when elements and belong to group is esteemed similarly.Lemma 4.3. *Let elements and belong to one of groups or and do not belong to subgroup and , that is,
**
for some integers , , and . If elements and belong to group , then integers , and are uniquely determined modulo by equality (4.4). If elements and belong to group , then integers , and are uniquely determined modulo by equality (4.4). *The proof of Lemma 4.3 is completely similar to that of Lemma 4.2. Lemma 4.4. *Let and be reduced elements of group , where and let for every the equality holds, for some elements and of subgroup . Then there exists at most one sequence , , of elements of subgroup such that for any **
Moreover, if and for some integers , , and () and , then such sequence exists if, and only if, for any , ,
**Proof. *We suppose first that the sequence , , of elements of subgroup satisfying equality (4.5) exists, and let's write , for some integers and .

Then, since for any the equality holds, we have
Since , then every element , belonging to one of the subgroups or , does not lie in subgroup , and consequently, from Proposition 3.1, for any , we have the equality . Substituting the expressions of elements , and , we have for every and hence we obtain the system of numeric equations
Since by (4.7) for every element belongs to the intersection and, by supposition, elements , belong to subgroup and elements , belong to subgroup , then from Proposition 3.1, it follows that for odd , we should have , and for even , we should have . It means that for every odd () we have the equalities and , and for every even () we have the equalities and .

Hence, the values of the integers and are determined uniquely. Namely,
Moreover, from equalities (4.8) it follows, that

Thus, the statement that there can exist at most one sequence of elements , , of subgroup satisfying equalities (4.5) is demonstrated.

Substituting the value and defined in (4.9) in the equalities of system (4.8), where and is odd, we obtain . Similarly, substituting the value and defined in (4.10) in the equalities of systems (4.8), where and is even, we obtain .

Thus, under the existence in subgroup of sequence , , of elements satisfying (4.5), conditions (4.6) are satisfied.

Conversely, suppose conditions (4.6) are satisfied. Let
and for all indexes such that , we set
At last, for we set

Let us show that, the so-defined sequence of elements , , really fits to equalities (4.5).

If , using the permutability of elements and we have

If and integer is even, using the equality and permutability of elements and , we have

If and integer is odd, using the equality and permutability of elements and , we have

If integer is even, then
and if is odd, then
So, Lemma 4.4 is completely demonstrated.

Similar argument gives the following lemma.

Lemma 4.5. *Let and be reduced elements of group , where , and let for every the equality takes place, for some elements and of subgroup . Then there exists at most one sequence , , of elements of subgroup such that for any , one has
**
Moreover, if and for some integers , , and () and , then such sequence exists if and only if for any integer , , one has
*

We can now end the consideration of Case 3 and thus complete the proof of Proposition 4.1.

By Proposition 2.3, elements and are -conjugate if and only if their cyclic permutation and are -conjugate. Therefore, we can suppose, without loss of generality, that elements and belong to subgroup .

By supposition, for every there exist integers , , and such that

If integer , , is even, then , and consequently, from Lemma 4.2, integers , and do not depend from the particular expressions of the elements (in the form , where ), and these integers are uniquely determined by the sequences , and , . Similarly, for any odd , , integers , and 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 , and , .

Let's now conditions (4.6) of Lemma 4.4 are not satisfied, that is, either for some even , , the sum is different from zero, or for some odd , , the sum is different from zero. Then it is possible to find an integer , not dividing the respective sum. Let's also choose, according to Proposition 3.5, the integers and such that for all positive integer , divisible by , the length of element in group is equal to and for all positive integer , divisible by , the length of element in group is equal to . Then if , in group , elements and have reduced forms
respectively. Further, for every ,
From Lemma 4.3 and the selection of integer it follows that for elements and , conditions (4.21) of Lemma 4.5 are not satisfied. Therefore from this lemma and Proposition 2.3, elements and are not -conjugate in group .

Let now the reduced forms of elements and satisfy conditions (4.6) of Lemma 4.4. Then according to this lemma, in subgroup there exists the only sequence of elements , , such that for any , we have . Since elements and are not -conjugate, then elements and should be different. Since group is residually finite, by Proposition 3.3 there exists an integer such that . Let's choose one more integer such that in group elements and have length . Then if , in group , , elements and have reduced forms
respectively, and for every
Moreover, in group , for any , the equality
holds.

Since the sequence , , is, by Lemma 4.5, the unique sequence of elements of subgroup satisfying these equalities, then from Proposition 2.3, elements and are not -conjugate in group .

Hence Proposition 4.1 is completely demonstrated.

We now proceed directly to the proof of Theorem 1.1.

As remarked above, for any , group is conjugacy separable. Therefore, for the proof of the Theorem it is enough to show that for any two nonconjugate in group elements and of group , there exists an integer such that elements and are not conjugate in group . To this end we make use of the conjugacy criterion of elements of groups and given in Proposition 3.2.

So, let and be the reduced forms in group of two nonconjugate in group elements and . By Proposition 3.2 we can assume that elements and are cyclically reduced. In accordance with this proposition we must consider separately some cases.

*Case 1 (). *From Proposition 3.5, there exists an integer such that for any positive integer , divisible by , the length of element in group coincides with the length of element in group . Similarly, there exists integer such that for any positive integer , divisible by , the length of element in group coincides with the length of element in group . Thus, if elements and of group 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 is the required.

*Case 2 (). *In this case each of these elements belongs to one of the subgroups or . Suppose first that both elements lie in the same of these subgroups; let it be, for instance, subgroup . Since elements and are not conjugate in group and group is conjugacy separable [3], then by Proposition 3.3 there exists an integer such that in group elements and are not conjugate. Proposition 3.2 now implies that elements and are not conjugate in group .

Let now element belongs to subgroup , element belongs to subgroup and these elements do not belong to subgroup . By Proposition 3.4, there exist integers and such that for any positive integer , divisible by , element does not belong to subgroup , and for any positive integer , divisible by , element does not belong to subgroup . Then if , Proposition 3.2 implies that elements and are not conjugate in group .

*Case 3 (). *Let () be all the cyclic permutations of element . Since elements and are conjugate and elements and are not conjugate, element is not -conjugate to any of elements , . It follows from Proposition 4.1 that for every , there exists an integer such that elements and are not -conjugate in group . Let also integer be chosen such that for all positive integer , divisible by , elements and have length in group . Then if in group , elements and have reduced forms
respectively. Furthermore, elements and are not -conjugate in group . Since an arbitrary cyclic permutation of element coincides with some element , then from Proposition 3.2 it follows that elements and are not conjugate in group .

The proof of Theorem is now complete.

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