Volume 2010, Article ID 803243,14pages doi:10.1155/2010/803243
Conjugacy Separability of Some
and D. I. Moldavanskii2
1Department of Mathematics and Computer Science, University of Ngaoundere, P.O. Box 454,
2Department of Algebra and Mathematical Logic, Ivanovo State University, 39,
Ermak St, Ivanovo 153025, Russia
Correspondence should be addressed to D. Tieudjo,email@example.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.
A groupGis conjugacy separable if for any two nonconjugate elementsf andg ofGthere exists a homomorphismϕof groupGonto some finite groupXsuch that the imagesfϕand
gϕof elementsfandgare not conjugate in groupX. It is clear that any conjugacy separable groupGis residually finitei.e., recall that for any nonidentity elementg ∈Gthere 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.
unpublished result of M. Armstrong on conjugacy separability of any one-relator group with nontrivial center.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
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 A∗B; Hof groupsAandBwith amalgamated subgroupH. Every elementg∈Gcan 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, elementg∈Gcan have diﬀerent reduced forms, but ifgy1y2· · ·ysis one
more reduced form ofg, thenr sand, for anyi 1,2, . . . , r,xiandyibelong to the same
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 somex∈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. LetG 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
1if 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;
2iff ∈H 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 hi ∈ H and elements hi and hi1 areA-conjugate or
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 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−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 thatf ∈ Hand by assertion2ofProposition 2.1there is a sequence of elements
fh0, h1, . . . , hn, hn1g, 2.3
such that for anyi 0,1, . . . , n hi ∈ H, and for suitable elementxibelonging to one of the
subgroupsAorB, the equalityx−1
i hixi hi1holds. Since fori0,1, . . . , n−1 the inclusion
x−i11hi1xi1 ∈ His valid, the hypothesis givesh0 hn, that is,x−n1fxn g. This means that
elementsf and g belong to that subgroupAor Bwhich contains elementxn and also are
conjugated in this subgroup.
Proposition 2.3. Letfx1x2· · ·xrandgy1y2· · ·yr be the reduced forms of elementsfandgof
groupG A∗B;Hwherer >1. ThenfandgareH-conjugate if and only if there exist elements
h0, h1, h2, . . . , hr h0in subgroupHsuch that for anyi1,2, . . . , r, one has
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 h ∈ H 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
x−r1· · ·x−21x−11h−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 byh−11, 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
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 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−1Af∩AH.
Proof. In fact, since subgroupHcoincides with the center of groupGsee10, page 211, the inclusionH ⊆ f−1Af ∩Ais obvious. Conversely, let elementf−1af belong to subgroupA
first component of reduced form off belongs toB. In any case the assumptiona /∈Hwould imply thatlf−1af>1. Thus,a∈Hand thereforef−1af a∈H.
3. Some Properties of Groups
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 producta∗H;amcof infinite cycleaand groupHwith amalgamationam, group
Bis the free productb ∗H;bn dof infinite cycleband groupHwith amalgamation
bn, and groupG
mnis the free productA∗B;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,h∈H, subgroupAt Aρ
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∗
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−1Hg∩Hc(resp.,g−1Hg∩Hd) holds. In particular, for any elementgof groupAorB
and for any elementhof subgroupHelementg−1hgbelongs to subgroupHif and only if elementsg
Similarly, for any element g of group At (or Bt) not belonging to subgroup Htthe equalityg−1Htg∩Ht cHt(resp.,g−1Htg∩Ht dHt) holds. In particular, for any
elementsg of groupAtorBtandhof subgroupHt, elementg−1hgbelongs to subgroupHt
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:Gmns → Gmntsuch 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
Proof. We will assume thatg ∈A; the case wheng∈Bcan be treated similarly.
So, let elementg∈Ado 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
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
· · ·xrρt
of elementgρtis reduced in groupAtand sincer >1,gρtdoes not belong to the free factor
Ht Hρt. The proposition is proven.
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 elementf ∈Adoes 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 elementfϕofXdoes 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 elementx∈Ain 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 anyk∈Mthe 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
We prove first the following proposition.
Proposition 4.1. If elementsfandgof groupGmnsuch thatlf lg>1 are notH-conjugate,
Proof. Letf x1x2· · ·xrandgy1y2· · ·yr be the reduced forms in groupGmn A∗B;H
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
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
· · ·xrρt
· · ·yrρt
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 subgroupHandx∈HyH, that is,
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 cα1dβ1ycγ1dδ1 and x
cα2dβ2ycγ2dδ2 for some integers α
1, β1, γ1, δ1, α2, β2, γ2 and δ2. Rewriting the equality
conclude thatcα1−α2dβ1−β2 cγ2−γ1dδ2−δ1and also that this element should belong to subgroup
c, that is,β1−β2δ2−δ10. So, we havecα1−α2 cγ2−γ1and since the order of elementcis
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
Htandx∈HtyHt, that is,
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
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.
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 equalityh−i−11yihiuiyiviholds, we have
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
μi−1αi μi−γi, νi−1βiνi−δi i1,2, . . . , r. 4.8
Since by4.7for everyi 1,2, . . . , r elementhivi−1 belongs to the intersectiony−i1Hyi∩H
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 hiv−i1 ∈ d. It means that for every odd
ii 1,2, . . . , rwe have the equalities νi−1βi 0 andνi−δi 0, and for every eveni
i1,2, . . . , rwe have the equalitiesμi−1αi0 andμi−γi0.
Hence, the values of the integersμiandνiare determined uniquely. Namely,
μi ⎧ ⎨ ⎩
γi ifiis even and 2ir,
−αi1 ifiis odd and 1ir−1,
νi ⎧ ⎨ ⎩
−βi1 ifiis even and 0ir−1,
δi ifiis odd and 1ir.
Moreover, from equalities4.8it follows, that
νr βrδr−1δr ifr is even,
μr αrγr−1γr if r is odd.
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−1andμi−αi1defined in4.9in the equalitiesμi−1
αi μi−γiof system4.8, where 1< i < r andiis odd, we obtainαiαi1γi−1γi 0.
Similarly, substituting the valueνi−1 δi−1 andνi −βi1defined in4.10in the equalities
νi−1βiνi−δiof 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
and for all indexesisuch that 1i < r, we set
hi ⎧ ⎨ ⎩
cγid−βi1 if iis even,
c−αi1dδi if iis odd.
At last, forirwe set
hr ⎧ ⎨ ⎩
cγrdβrδr−1δr ifr is even,
cαrγr−1γrdδr ifr is odd.
Let us show that, the so-defined sequence of elementsh0,h1,h2, . . . , hr really fits to
Ifi1, using the permutability of elementscandy1we have
If 1 < i < r and integeri is even, using the equality βi βi1 δi−1δi 0 and
permutability of elementsdandyi, we have
h−i−11yihicαid−δi−1yicγid−βi1cαiyicγid−δi−1βi1cαiyicγidβiδi cαidβiyicγidδi xi. 4.16
If 1< i < rand integeriis odd, using the equalityαiαi1γi−1γi0 and permutability
of elementscandyi, we have
i−1yihi c−γi−1dβiyic−αi1dδi c−γi−1αi1dβiyidδi cαiγidβiyidδi cαidβiyicγidδi xi. 4.17
If integerris even, then
h−r−11yrhr cαrd−δr−1yrcγrdβrδr−1δr cαrdβryrcγrdδr xr, 4.18
and ifris odd, then
h−r−11yrhr c−γr−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
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
αiαi1γi−1γi≡0modt ifiis odd,
βiβi1δi−1δi≡0modt ifiis even.
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
If integer i, 1 < i < r, is even, then yi ∈ B,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, v ∈ H, 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
· · ·xrρt
· · ·yrρt
respectively. Further, for everyi1,2, . . . , r,
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 havexih−i−11yihi. 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
· · ·xrρt
· · ·yrρt
respectively, and for everyi1,2, . . . , r
Moreover, in groupGmnt, for anyi1,2, . . . , r, the equality
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
A∗B; 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
that for all positive integert, divisible byt0, elements fρt andgρt have length r in group
Gmnt. Then iftt0t1· · ·trin groupGmnt, elementsfρtandgiρthave reduced forms
· · ·xrρt
· · ·yrρt
· · ·yi−1ρt
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.
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