Conjugacy Separability of Some One Relator Groups

14  Download (0)

Full text

(1)

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

2

1Department of Mathematics and Computer Science, University of Ngaoundere, P.O. Box 454,

Ngaoundere, Cameroon

2Department 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, bn 1m, 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 groupGis conjugacy separable if for any two nonconjugate elementsf andg ofGthere exists a homomorphismϕof groupGonto some finite groupXsuch that the imagesand

of elementsfandgare not conjugate in groupX. It is clear that any conjugacy separable groupGis residually finitei.e., recall that for any nonidentity elementgGthere exists a homomorphismϕof groupGonto some finite groupXsuch thatgϕ /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.

(2)

unpublished result of M. Armstrong on conjugacy separability of any one-relator group with nontrivial center.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 ambnt

anda−1blabms, wheres >1, was proved in4and5, 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 1where integersmandnare greater than 1, is

conjugacy separable.

We note that the assertion of Theorem is also valid whenm 1 orn 1. Indeed, in this case groupGmnis 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 groupsGmnis well known; it follows, for example, from the

result of paper6. GroupsGmn 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 groupGmnas 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 groupGmn 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 AB; Hof groupsAandBwith amalgamated subgroupH. Every elementgGcan be written in a form

gx1x2· · ·xr, r1, 2.1

where for anyi1,2, . . . , r elementxibelongs to one of the free factorsAorBand ifr >1

any successivexi and xi1 do not belong to the same factor AorB and therefore for any

i1,2, . . . , relementxidoes not belong to amalgamated subgroupH. Such form of element

gis called reduced.

In general, elementgGcan have different reduced forms, but ifgy1y2· · ·ysis one

more reduced form ofg, thenr sand, for anyi 1,2, . . . , r,xiandyibelong to the same

(3)

Ifg x1x2· · ·xr is a reduced form of cyclically reduced elementg then cyclic permutations

of elementgare elements of formxixi1· · ·xrx1· · ·xi−1, wherei1,2, . . . , r.

IfXis a subgroup of a groupY we will say that elementsaandbofYareX-conjugate

ifax−1bx, for somexX.

The Solitar criterion of conjugacy of elements of amalgamated free product of two groupsTheorem 4.6 in10can be formulated as follows.

Proposition 2.1. LetG AB;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

1if fAand f is not conjugate in Ato any element of subgroupH, then f andg are conjugate in groupGif and only ifgAandf and g are conjugate inA; similarly, if

fBandfis not conjugate inBto any element of subgroupHthenfandgare conjugate in groupGif and only ifgBandfandgare conjugate inB;

2iffH thenf andg are conjugate in groupGif and only if there exists a sequence of elements

f h0, h1, . . . , hn, hn1g 2.2

such that for any i 0,1, . . . , n hiH and elements hi and hi1 areA-conjugate or

B-conjugate;

3if 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. LetG AB;H be the free product of groupsAand B with amalgamated subgroupH. Suppose that for any elementf ofAor Band for any elementhof subgroupH, the inclusionf−1hf His valid if and only if elementsf and hcommute. Iff 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 elementsf and g be conjugate in groupGand letf belong to subgroup

A, say. Iff is not conjugate inAto any element of subgroupH, then the desired assertion is implied by assertion1ofProposition 2.1. Otherwise, we can assume thatfHand by assertion2ofProposition 2.1there is a sequence of elements

fh0, h1, . . . , hn, hn1g, 2.3

such that for anyi 0,1, . . . , n hiH, and for suitable elementxibelonging to one of the

subgroupsAorB, the equalityx−1

i hixi hi1holds. Since fori0,1, . . . , n−1 the inclusion

xi11hi1xi1 ∈ His valid, the hypothesis givesh0 hn, that is,xn1fxn g. This means that

elementsf and g belong to that subgroupAor Bwhich contains elementxn and also are

conjugated in this subgroup.

(4)

Proposition 2.3. Letfx1x2· · ·xrandgy1y2· · ·yr be the reduced forms of elementsfandgof

groupG AB;Hwherer >1. ThenfandgareH-conjugate if and only if there exist elements

h0, h1, h2, . . . , hr h0in subgroupHsuch that for anyi1,2, . . . , r, one has

xihi−11yihi. 2.4

Proof. If for some elementsh0, h1, h2, . . . , hr h0of subgroupHequalities2.4hold, then

fx1x2· · ·xr h−01y1h1 h1−1y2h2· · · hr−−11y2hr h−01

y1y2· · ·yr

hr h−01gh0. 2.5

Conversely, by induction on r we prove that if for some hH the equality f h−1ghholds then there exists a sequence of elementsh0, h1, h2, . . . , hr h0 of subgroupH

satisfying the equalities2.4and such thath0h.

Rewriting the equalityfh−1ghin the form

xr1· · ·x21x11h−1y1y2· · ·yrh1, 2.6

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 Aor Band element x1−1h−1y1 must belong to

subgroupH. Denoting this element byh11, we havex1 h−1y1h1. Ifr 2 then the equality

above takes the formx−1

2 h−11y2h1, whencex2 h1−1y2h. Therefore, settingh0 h2 h, we

obtain in that case the desired sequence.

Ifr >2 let us rewrite the equalityfh−1ghin the form

h−1y1h1x2· · ·xr h−1y1y2· · ·yrh. 2.7

This implies that if we setf h−1h

1x2· · ·xr andg y2· · ·yr, thenf h−1gh. Since the

length of elementsfandgis equal tor−1, then by induction, there exists a sequenceh1h,

h2, . . . , hr h1 of elements of subgroup H, such that h−1h

1x2 h1−1y2h2 and for any

i2,3, . . . , r

xi

hi1−1yihi. 2.8

Sincex2 h−11y2h2, then settinghi hifori2,3, . . . , 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. LetG AB;Hbe the free product of groupsAandBamalgamating subgroup

HwhereHlies in the center of each of groupsAandB. Then for any elementfGnot belonging to subgroupAone hasf−1AfAH.

Proof. In fact, since subgroupHcoincides with the center of groupGsee10, page 211, the inclusionHf−1Af Ais obvious. Conversely, let elementf−1af belong to subgroupA

(5)

first component of reduced form off belongs toB. In any case the assumptiona /Hwould imply thatlf−1af>1. Thus,aHand thereforef−1af aH.

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, groupA is the free productaH;amcof infinite cycleaand groupHwith amalgamationam, group

Bis the free productbH;bn dof infinite cycleband groupHwith amalgamation

bn, and groupG

mnis the free productAB;Hof groupsAandBwith amalgamationH.

The same decomposition is satisfiable for certain quotients of groupsGmn. Namely, for

any integert >1 letGmntbe the group with presentation

a, b;am, bn 1, amtbnt1 3.1

andρtthe natural homomorphism of groupsGmnontoGmnt. Then it is easy to verify that

subgroupHt Hρtof groupGmntis isomorphic to the quotientH/Htwhere, as usually,

Htconsists of all elementsht,hH, subgroupAt

tis the amalgamated free product

a;amt 1 ∗Ht;am cHtof cyclea;amt 1of ordermtand groupHt, subgroup

Bt Bρtis the amalgamated free productb;bnt 1 ∗Ht;bn dHtof cycleb;bnt

1 of orderntand group Ht, and group Gmntis the amalgamated free productAt

Bt;Ht.

These decompositions of groupsGmnandGmntare 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 groupsAtandBt is the amalgamated free product of two finite groups and groupGmntis the free product of groupsAtand

Btwith finite amalgamation, it follows from results of3that for everyt, groupGmntis

conjugacy separable. So, to prove the conjugacy separability of groupGmn it is enough, for

any nonconjugate elementsf andgofGmn, to find an integertsuch that elementsfρtand

gρtare nonconjugate in groupGmnt.

Since in the decompositions of groupsAandBas well, as of groups AtandBt, 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 elementg of groupA(or B) not belonging to subgroupH the equality

g−1HgHc(resp.,g−1HgHd) holds. In particular, for any elementgof groupAorB

and for any elementhof subgroupHelementg−1hgbelongs to subgroupHif and only if elementsg

andhcommute.

Similarly, for any element g of group At (or Bt) not belonging to subgroup Htthe equalityg−1HtgHt cHt(resp.,g−1HtgHt dHt) holds. In particular, for any

elementsg of groupAtorBtandhof subgroupHt, elementg−1hgbelongs to subgroupHt

(6)

Propositions2.1,2.2and3.1lead to the following criterion for conjugacy of elements of groupsGmnandGmnt.

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

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

2If 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 groupsGmn and Gmnt. The following

assertion is easily checked.

Proposition 3.3. For any homomorphismϕof 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:GmnsGmntsuch thatρtρsϕ.

The same assertions are valid for groupsAandB.

Proposition 3.4. For any elementgof groupAorB, ifgdoes not belong to subgroupHthen there exists an integert0>1 such that for any positive integert, divisible byt0, elementgρtdoes not belong

to subgroupHρt.

Proof. We will assume thatgA; the case whengBcan be treated similarly.

So, let elementgAdo not belong to subgroupHand letgx1x2· · ·xrbe a reduced

form ofgin the decomposition of groupAinto amalgamated free product.

Ifr1 then elementgbelongs to subgroupathat is,g akfor some integerk. Since

g /H, integerkis not divisible bym. Then for any integert >0, in groupAt, elementgρt

of subgroupa;amt1cannot belong to the amalgamated subgroupgenerated by element

amof the decomposition of groupAtand consequently cannot belong to the free factor

Ht Hρt.

Letr > 1. Every componentxi of the reduced form of elementg is either of formak,

where integerkis not divisible bym, or of formckdlwherel /0. If integert

0is chosen such

that the exponentlof any componentxiof the second kind is not divisible byt0, then for any

tdivisible byt0the imagexiρtxiHtof such component will not belong to the amalgamated

subgroup of groupAt generated, let’s remind, by elementcHt. Moreover, as above in

caser 1, the images of components of the first kind will not belong to the amalgamated subgroup. Therefore, the form

gρt

x1ρt

x2ρt

· · ·xrρt

3.2

of elementgρtis reduced in groupAtand sincer >1,gρtdoes not belong to the free factor

Ht Hρt. The proposition is proven.

(7)

Proposition 3.5. For any elementg of groupGmn there exists an integert0 > 1 such that for all

positive integert, divisible byt0, the length of elementgρtin groupGmntcoincides with the length

of elementgin groupGmn.

Proposition 3.6. For any elementsfandgof groupA(orB) such that elementfdoes not belong to the double cosetHgH, there exists an integert0 >1 such that for any positive integert, divisible by

t0, elementfρtdoes not belong to the double cosetHtgρtHt.

Proof. We can again consider only the case when elementsf and g belong to subgroupA. So, let us suppose that elementfAdoes not belong to the double cosetHgH. In view of Proposition 3.4, it is enough to prove that there exists a homomorphismϕofAonto a finite groupXsuch that elementofXdoes not belong to the double cosetHϕgϕHϕ.

To this end let’s consider the quotient groupA A/C of groupAby its central subgroupCc. The imagexCof an elementxAin groupAwill be denoted byx.

It is obvious that group A is the ordinary free product of the cyclic group X of order m, generated by elementa, and the infinite cycle Y, generated byd. The canonical homomorphism of group A onto group A maps subgroup H onto subgroup Y and consequently, the image of the double cosetHgH is the double cosetY gY. SinceC H, the elementfdoes not belong to this coset.

We can assume, without loss of generality, that any elementf andg, if it is different from identity, has reduced form the first and the last syllables of which do not belong to subgroupY. EveryY-syllable of these reduced forms is of formdk for some integer k /0. Since the setMof all such exponentskis finite, we can choose an integert >0 such that for anykMthe inequalityt >2|k|holds. Let’s denote byAthe factor group of groupAby the normal closure of elementdt. GroupAis the free product of groups XandY/Ytand since

different integers fromMare not relatively congruent and are not congruent to zero modulo

t, then the reduced forms of the imagesfandgof elementsfandgin groupAare the same as in groupA. In particular, elementfdoes not belong to the double cosetY/Ytg Y/Y t. Since

this coset consists of a finite number of elements and groupAis residually finite, then there exists a normal subgroupNof finite index of groupAsuch, that

f /Y/YtgY/Y t·N. 3.3

If nowθis the product of the canonical homomorphisms of groupAonto groupAand of groupAonto groupAandNis the full preimage byθof subgroupN, thenNis a normal subgroup of finite index of groupAandf /HgHN. Thus, the canonical homomorphism

ϕ of groupA onto quotient groupA/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 elementsfandgof groupGmnsuch thatlf lg>1 are notH-conjugate,

(8)

Proof. Letf x1x2· · ·xrandgy1y2· · ·yr be the reduced forms in groupGmn AB;H

of elementsfandg.

We remindseeProposition 2.3that elementsf andg areH-conjugate if and only if there exist elementsh0,h1,h2, . . . , hr h0ofHsuch that for anyi1,2, . . . , r, we have

xihi11yihi. 4.1

It then follows, in particular, that for eachi 1,2, . . . , r elementsxi andyi should lie in the

same free factorAorBand define the same double coset moduloH, H. So, we consider separately some cases.

Case 1. Suppose that for some indexielementsxi andyi lie in different free factorsAand

Bof groupGmnand, certainly, are not in subgroupH. It follows fromProposition 3.4that

there exists an integert >1 such that elementsxiρtandyiρtdo not belong to the same free

factorAtorBtof groupGmnt and, as above, lie in free factors of this group. Hence, by

Proposition 2.3, in groupGmntelementsfρtandgρtare notHt-conjugate.

Case 2. Let now for anyi 1,2, . . . , r elementsxi andyi belong to the same subgroupAor

Band for someielementxi does not belong to the double cosetHyiH. ByProposition 3.6,

there exists an integert1 >1 such that for any positive integert, divisible byt1, elementxiρt

is not in the double cosetHtyiρtHt. FromProposition 3.5, there exist integerst2>1 and

t3 > 1 such that for any positive integertdivisible byt2 the length of elementfρtin group

Gmntis equal torand for any positive integert, divisible byt3, the length of elementgτtin

groupGmntis equal tor. Thus, iftt1t2t3then in groupGmnt, elementsfρtandgρthave

the reduced forms

x1ρt

x2ρt

· · ·xrρt

, y1ρt

y2ρt

· · ·yrρt

, 4.2

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

Proposition 2.3these elements are notHt-conjugate.

Case 3. We now consider the case, when for anyi 1,2, . . . , r elementsxi andyi lie in the

same free factorAorBand also determine the same double coset moduloH, 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 subgroupHandxHyH, that is,

xcαdβycγdδ 4.3

for some integersα,β,γandδ. If elementsxandybelong to groupA, then integersαγ,βandδare uniquely determined by the equality4.3. If elementsxandybelong to groupB, then integersβδ,

αandγare uniquely determined by equality4.3.

Proof. Let elements x and y belong to subgroup A and let x 1dβ1ycγ1dδ1 and x

2dβ2ycγ2dδ2 for some integers α

1, β1, γ1, δ1, α2, β2, γ2 and δ2. Rewriting the equality

(9)

conclude that1−α2dβ1−β2 cγ2−γ1dδ2−δ1and also that this element should belong to subgroup

c, that is,β1−β2δ2−δ10. So, we have1−α2 2−γ1and since the order of elementcis

infinite, thenα1−α2γ2−γ1.

Thus,β1β2,δ1δ2andα1γ1α2γ2as it was required. The case when elements

xandybelong to groupBis esteemed similarly.

Lemma 4.3. Let elementsxandybelong to one of groupsAtorBtand do not belong to subgroup

HtandxHtyHt, that is,

xcHtαdHtβycHtγdHtδ 4.4

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

δare uniquely determined modulotby equality4.4. If elementsxandybelong to groupBt, then integersβδ,αandγare uniquely determined modulotby equality4.4.

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

Lemma 4.4. Letfx1x2· · ·xr andg y1y2· · ·yr be reduced elements of groupGmn, wherer >1

and let for everyi1,2, . . . , rthe equalityxiuiyiviholds, for some elementsuiandviof subgroup

H. Then there exists at most one sequenceh0,h1,h2, . . . , hrof elements of subgroupHsuch that for

anyi1,2, . . . , r

xihi−11yihi. 4.5

Moreover, ifui 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 anyi, 1< i < r,

αiαi1γi−1γi0, ifiis odd,

βiβi1δi−1δi0, ifiis even.

4.6

Proof. We suppose first that the sequence h0, h1, h2, . . . , hr of elements of subgroup H

satisfying equality4.5exists, and let’s writehicμidνi, for some integersμiandνi.

Then, since for anyi1,2, . . . , rthe equalityhi11yihiuiyiviholds, we have

yi1hi−1uiyihivi1. 4.7

Sincer > 1, then every elementyi, belonging to one of the subgroupsAorB, does not lie

in subgroupH, and consequently, fromProposition 3.1, for anyi 1,2, . . . , r, we have the equalityhi−1ui hivi−1. Substituting the expressions of elements hi,ui and vi, we have for

every i 1,2, . . . , rcμi−1αidνi−1βi cμi−γidνi−δi and hence we obtain the system of numeric

equations

μi−1αi μiγi, νi−1βiνiδi i1,2, . . . , r. 4.8

Since by4.7for everyi 1,2, . . . , r elementhivi−1 belongs to the intersectionyi1HyiH

(10)

to subgroupB, then fromProposition 3.1, it follows that for oddi, we should havehi−1ui

hivi−1∈ c, and for eveni, we should havehi−1ui hivi1 ∈ d. It means that for every odd

ii 1,2, . . . , rwe have the equalities νi−1βi 0 andνiδi 0, and for every eveni

i1,2, . . . , rwe have the equalitiesμi−1αi0 andμiγi0.

Hence, the values of the integersμiandνiare determined uniquely. Namely,

μi ⎧ ⎨ ⎩

γi ifiis even and 2ir,

αi1 ifiis odd and 1ir−1,

4.9

νi ⎧ ⎨ ⎩

βi1 ifiis even and 0ir−1,

δi ifiis odd and 1ir.

4.10

Moreover, from equalities4.8it follows, that

μ0−

α1α2γ1

,

νr βrδr−1δr ifr is even,

μr αrγr−1γr if r is odd.

4.11

Thus, the statement that there can exist at most one sequence of elements h0, h1,

h2, . . . , hr of subgroupHsatisfying equalities4.5is demonstrated.

Substituting the valueμi−1γi−1andμiαi1defined in4.9in the equalitiesμi−1

αi μiγiof system4.8, where 1< i < r andiis odd, we obtainαiαi1γi−1γi 0.

Similarly, substituting the valueνi−1 δi−1 andνiβi1defined in4.10in the equalities

νi−1βiνiδiof systems4.8, where 1< i < randiis even, we obtainβiβi1δi−1δi0.

Thus, under the existence in subgroupH of sequenceh0,h1,h2, . . . , hr of elements

satisfying4.5, conditions4.6are satisfied.

Conversely, suppose conditions4.6are satisfied. Let

h0cα1α2γ1dβ1 4.12

and for all indexesisuch that 1i < r, we set

hi ⎧ ⎨ ⎩

cγidβi1 if iis even,

cαi1dδi if iis odd.

4.13

At last, forirwe set

hr ⎧ ⎨ ⎩

cγrdβrδr−1δr ifr is even,

cαrγr−1γrdδr ifr is odd.

(11)

Let us show that, the so-defined sequence of elementsh0,h1,h2, . . . , hr really fits to

equalities4.5.

Ifi1, using the permutability of elementscandy1we have

h01y1h11α2γ11y1cα2111y111x1. 4.15

If 1 < i < r and integeri is even, using the equality βi βi1 δi−1δi 0 and

permutability of elementsdandyi, we have

hi11yihicαidδi−1yicγidβi1cαiyicγidδi−1βi1cαiyicγidβiδi cαidβiyicγidδi xi. 4.16

If 1< i < rand integeriis odd, using the equalityαiαi1γi−1γi0 and permutability

of elementscandyi, we have

h−1

i−1yihi cγi−1dβiyicαi1dδi cγi−1αi1dβiyidδi cαiγidβiyidδi cαidβiyicγidδi xi. 4.17

If integerris even, then

hr11yrhr cαrdδr−1yrcγrdβrδr−1δr cαrdβryrcγrdδr xr, 4.18

and ifris odd, then

hr11yrhr cγr−1dβryrcαrγr−1γrdδr cαrdβryrcγrdδr xr. 4.19

So,Lemma 4.4is completely demonstrated.

Similar argument gives the following lemma.

Lemma 4.5. Letf x1x2· · ·xr andg y1y2· · ·yr be reduced elements of groupGmnt, where

r >1, and let for everyi1,2, . . . , rthe equalityxiuiyivitakes place, for some elementsuiandvi

of subgroupHt. Then there exists at most one sequenceh0,h1,h2, . . . , hr of elements of subgroup

Htsuch that for anyi1,2, . . . , r, one has

xihi11yihi. 4.20

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

1,2, . . . , r) andx1, y1 ∈At, then such sequence exists if and only if for any integeri, 1< i < r, one

has

αiαi1γi−1γi≡0modt ifiis odd,

βiβi1δi−1δi≡0modt ifiis even.

4.21

(12)

By Proposition 2.3, elements f and g are H-conjugate if and only if their cyclic permutationx2· · ·xrx1andy2· · ·yry1 areH-conjugate. Therefore, we can suppose, without

loss of generality, that elementsx1andy1belong to subgroupA.

By supposition, for everyi1,2, . . . , rthere exist integersαi,βi,γiandδisuch that

xicαidβiyicγidδi. 4.22

If integer i, 1 < i < r, is even, then yiB,yi−1, yi1 ∈ A and consequently, from

Lemma 4.2, integersβiδi,δi−1 andβi1 do not depend from the particular expressions of

the elementsxi in the formxi uyiv, whereu, vH, and these integers are uniquely

determined by the sequencesx1,x2, . . . , xrandy1,y2, . . . , yr. Similarly, for any oddi, 1< i < r,

integersαiγi,γi−1 andαi1are uniquely determined by these sequences. It means, in turn,

that the satisfiability of conditions4.6 of Lemma 4.4depends only on the sequencesx1,

x2, . . . , xr andy1,y2, . . . , yr.

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

i, 1< i < r, the sumβiβi1δi−1δiis different from zero, or for some oddi, 1< i < r,

the sumαiαi1γi−1γiis different from zero. Then it is possible to find an integert1 >1,

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

t2>1 andt3>1 such that for all positive integert, divisible byt2, the length of elementfρtin

groupGmntis equal torand for all positive integert, divisible byt3, the length of element

gρtin group Gmntis equal tor. Then ift t1t2t3, in groupGmnt, elementsfρtandgρt

have reduced forms

x1ρt

x2ρt

· · ·xrρt

, y1ρt

y2ρt

· · ·yrρt

, 4.23

respectively. Further, for everyi1,2, . . . , r,

xiρt

cHtαi

dHtβi yiρt

cHtγi dHtδi

. 4.24

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, elementsfρtandgρtare notHt-conjugate in groupGmnt.

Let now the reduced forms of elementsfandgsatisfy conditions4.6ofLemma 4.4. Then according to this lemma, in subgroupHthere exists the only sequence of elementsh0,

h1,h2, . . . , hr such that for anyi1,2, . . . , r, we havexihi11yihi. Since elementsfandgare

notH-conjugate, then elementsh0andhrshould be different. Since groupGmnis residually

finite, by Proposition 3.3there exists an integert1 > 1 such that h0ρt1/hrρt1. Let’s choose

one more integert2such that in groupGmnt2elementsfρt2andgρt2have lengthr. Then if

tt1t2, in groupGmnt,h0ρt/hrρt, elementsfρtandgρthave reduced forms

x1ρt

x2ρt

· · ·xrρt

, y1ρt

y2ρt

· · ·yrρt

, 4.25

respectively, and for everyi1,2, . . . , r

xiρt

cHtαi

dHtβi yiρt

cHtγi dHtδi

(13)

Moreover, in groupGmnt, for anyi1,2, . . . , r, the equality

xiρt

hi−1ρt 1

yiρt

hiρt

4.27

holds.

Since the sequenceh0ρt,h1ρt,h2ρt, . . . , hrρtis, byLemma 4.5, the unique sequence of

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

fτtandgτtare notHt-conjugate in groupGmnt.

HenceProposition 4.1is completely demonstrated.

We now proceed directly to the proof ofTheorem 1.1.

As remarked above, for anyt >1, groupGmntis conjugacy separable. Therefore, for

the proof of the Theorem it is enough to show that for any two nonconjugate in groupGmn

elementsfandgof groupGmn, there exists an integert >1 such that elementsfρtandgρtare

not conjugate in groupGmnt. To this end we make use of the conjugacy criterion of elements

of groupsGmnandGmntgiven inProposition 3.2.

So, let f x1x2· · ·xr and g y1y2· · ·yr be the reduced forms in group Gmn

AB; Hof two nonconjugate in groupGmnelementsfandg. ByProposition 3.2we can

assume that elementsfandgare cyclically reduced. In accordance with this proposition we must consider separately some cases.

Case 1 lf/lg. FromProposition 3.5, there exists an integer t1 > 1 such that for any

positive integert, divisible byt1, the length of elementfρt in groupGmntcoincides with

the length of elementf in groupGmn. Similarly, there exists integert2 >1 such that for any

positive integert, divisible byt2, the length of elementgρtin groupGmntcoincides with the

length of elementgin groupGmn. Thus, iftt1t2elementsfρtandgρtof groupGmntare,

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 integertis the required.

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

orB. Suppose first that both elements lie in the same of these subgroups; let it be, for instance, subgroupA. Since elementsfandgare not conjugate in groupAand groupAis conjugacy separable3, then byProposition 3.3there exists an integert > 1 such that in groupAt

elementsfρtand gρtare not conjugate.Proposition 3.2now implies that elementsfρtand

gρtare not conjugate in groupGmnt.

Let now elementfbelongs to subgroupA, elementgbelongs to subgroupBand these elements do not belong to subgroupH. ByProposition 3.4, there exist integers t1 > 1 and

t2 > 1 such that for any positive integert, divisible by t1, elementfρt does not belong to

subgroupHρt, and for any positive integert, divisible byt2, elementgρt does not belong

to subgroupHρt. Then ift t1t2,Proposition 3.2implies that elementsfρtandgρtare not

conjugate in groupGmnt.

Case 3 lf lg > 1. Let gi yiyi1· · ·yry1· · ·yi−1 i 1,2, . . . , r be all the cyclic

permutations of elementg. Since elementsg andgiare conjugate and elementsfandgare

not conjugate, elementfis notH-conjugate to any of elementsg1,g2, . . . , gr. It follows from

Proposition 4.1that for everyi 1,2, . . . , r, there exists an integerti > 1 such that elements

(14)

that for all positive integert, divisible byt0, elements fρt andgρt have length r in group

Gmnt. Then iftt0t1· · ·trin groupGmnt, elementsfρtandgiρthave reduced forms

x1ρt

x2ρt

· · ·xrρt

yiρt

yi1ρt

· · ·yrρt

y1ρt

· · ·yi−1ρt

, 4.28

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

Since an arbitrary cyclic permutation of elementgρtcoincides with some elementgiρt, then

fromProposition 3.2it follows that elementsfρtandgρtare not conjugate in groupGmnt.

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 groupGmn a, b;am, bn 1m, 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.

Figure

Updating...

References