Conjugacy Separability of Some One-Relator Groups

Abstract

Conjugacy separability of any group of the class of one-relator groups given by the presentation ⟨𝑎,𝑏;[𝑎^𝑚,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 finite (i.e., recall that for any nonidentity element 𝑔∈𝐺 there exists a homomorphism 𝜑 of group 𝐺 onto some finite group 𝑋 such that 𝑔𝜑≠1).

Conjugacy separability is related to the conjugacy problem, as the residual finiteness is related to the word problem in the study of groups. In fact, Mostowski [1] proved that finitely presented conjugacy separable groups have solvable conjugacy problem, just as finitely presented residually finite groups have solvable word problem.

Since 1962 when Baumslag and Solitar [2] discovered the first examples of nonresidually finite one-relator groups a lot of results establishing the residual finiteness of various one-relator groups have appeared. It was also shown that some of such groups are conjugacy separable. So, Dyer [2] has proved the conjugacy separability of any group that is free product of free groups amalgamating cycle. In the same paper she proved an unpublished result of M. Armstrong on conjugacy separability of any one-relator group with nontrivial center. The conjugacy separability of groups defined by relator of form (𝑎^𝑚𝑏^𝑛)^𝑡 and (𝑎⁻¹𝑏^𝑙𝑎𝑏^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 group 𝐺_𝑚𝑛=⟨𝑎,𝑏;[𝑎^𝑚,𝑏^𝑛]=1⟩ where integers 𝑚 and 𝑛 are 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 𝐺_𝑚𝑛 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 𝐺_𝑚𝑛 is well known; it follows, for example, from the result of paper [6]. Groups 𝐺_𝑚𝑛 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 𝐺_𝑚𝑛 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 𝐺_𝑚𝑛 are established and in Section 3 the proof of Theorem 1.1 will be completed.

We remind that conjugacy separability of groups 𝐺_𝑚𝑛 can also be obtained using [9]. However, the method in [9] presents any group 𝐺_𝑚𝑛 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 𝑔=𝑥₁𝑥₂⋯𝑥_𝑟,(𝑟⩾1), (2.1) where for any 𝑖=1,2,…,𝑟 element 𝑥_𝑖 belongs to one of the free factors 𝐴 or 𝐵 and if 𝑟>1 any successive 𝑥_𝑖 and 𝑥_𝑖+1 do not belong to the same factor 𝐴 or 𝐵 (and therefore for any 𝑖=1,2,…,𝑟 element 𝑥_𝑖 does not belong to amalgamated subgroup 𝐻). Such form of element 𝑔 is called reduced.

In general, element 𝑔∈𝐺 can have different reduced forms, but if 𝑔=𝑦₁𝑦₂⋯𝑦_𝑠 is one more reduced form of 𝑔, then 𝑟=𝑠 and, for any 𝑖=1,2,…,𝑟, 𝑥_𝑖 and 𝑦_𝑖 belong to the same factor 𝐴 or 𝐵. The length 𝑙(𝑔) of element 𝑔 is then defined as the number 𝑟 of the components in a reduced form of 𝑔. Element 𝑔 is called cyclically reduced if either 𝑙(𝑔)=1 or 𝑙(𝑔)>1 and the first and the last components of its reduced form do not belong to the same factor 𝐴 or 𝐵. If 𝑔=𝑥₁𝑥₂⋯𝑥_𝑟 is a reduced form of cyclically reduced element 𝑔 then cyclic permutations of element 𝑔 are elements of form 𝑥_𝑖𝑥_𝑖+1⋯𝑥_𝑟𝑥₁⋯𝑥_𝑖−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 𝑎=𝑥⁻¹𝑏𝑥, for some 𝑥∈𝑋.

The Solitar criterion of conjugacy of elements of amalgamated free product of two groups (Theorem 4.6 in [10]) can be formulated as follows.

Proposition 2.1. Let 𝐺=(𝐴∗𝐵;𝐻) be the free product of groups 𝐴 and 𝐵 with amalgamated subgroup 𝐻. Every element of 𝐺 is conjugate to a cyclically reduced element. If lengths of two cyclically reduced elements are nonequal then these elements are not conjugate in 𝐺. Let 𝑓 and 𝑔 be cyclically reduced elements of 𝐺 such that 𝑙(𝑓)=𝑙(𝑔). Then (1)if 𝑓∈𝐴 and 𝑓 is not conjugate in 𝐴 to any element of subgroup 𝐻, then 𝑓 and 𝑔 are conjugate in group 𝐺 if and only if 𝑔∈𝐴 and 𝑓 and 𝑔 are conjugate in 𝐴; similarly, if 𝑓∈𝐵 and 𝑓 is not conjugate in 𝐵 to any element of subgroup 𝐻 then 𝑓 and 𝑔 are conjugate in group 𝐺 if and only if 𝑔∈𝐵 and 𝑓 and 𝑔 are conjugate in 𝐵; (2)if 𝑓∈𝐻 then 𝑓 and 𝑔 are conjugate in group 𝐺 if and only if there exists a sequence of elements 𝑓=ℎ₀,ℎ₁,…,ℎ_𝑛,ℎ_𝑛+1=𝑔 (2.2) such that for any 𝑖=0,1,…,𝑛ℎ_𝑖∈𝐻 and elements ℎ_𝑖 and ℎ_𝑖+1 are 𝐴-conjugate or 𝐵-conjugate; (3)if 𝑙(𝑓)=𝑙(𝑔)>1, then 𝑓 and 𝑔 are conjugate in group 𝐺 if and only if element 𝑓 is 𝐻-conjugate to some cyclic permutation of 𝑔.

In the following special case assertion (2) of Proposition 2.1 became unnecessary.

Proposition 2.2. Let 𝐺=(𝐴∗𝐵;𝐻) be the free product of groups 𝐴 and 𝐵 with amalgamated subgroup 𝐻. Suppose that for any element 𝑓 of 𝐴 or 𝐵 and for any element ℎ of subgroup 𝐻, the inclusion 𝑓⁻¹ℎ𝑓∈𝐻 is valid if and only if elements 𝑓 and ℎ commute. If 𝑓 and 𝑔 are cyclically reduced elements of 𝐺 such that 𝑙(𝑓)=𝑙(𝑔)=1 then 𝑓 and 𝑔 are conjugate in group 𝐺 if and only if 𝑓 and 𝑔 belong to the same subgroup 𝐴 or 𝐵 and are conjugate in this subgroup.

Indeed, let 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 assertion (1) of Proposition 2.1. Otherwise, we can assume that 𝑓∈𝐻 and by assertion (2) of Proposition 2.1 there is a sequence of elements 𝑓=ℎ₀,ℎ₁,…,ℎ_𝑛,ℎ_𝑛+1=𝑔, (2.3) such that for any 𝑖=0,1,…,𝑛ℎ_𝑖∈𝐻, and for suitable element 𝑥_𝑖 belonging to one of the subgroups 𝐴 or 𝐵, the equality 𝑥_𝑖⁻¹ℎ_𝑖𝑥_𝑖=ℎ_𝑖+1 holds. Since for 𝑖=0,1,…,𝑛−1 the inclusion 𝑥⁻¹_𝑖+1ℎ_𝑖+1𝑥_𝑖+1∈𝐻 is valid, the hypothesis gives ℎ₀=ℎ_𝑛, that is, 𝑥_𝑛⁻¹𝑓𝑥_𝑛=𝑔. This means that elements 𝑓 and 𝑔 belong to that subgroup 𝐴 or 𝐵 which contains element 𝑥_𝑛 and also are conjugated in this subgroup.

Next, we describe more explicitly the situation arising in assertion (3) of Proposition 2.1.

Proposition 2.3. Let 𝑓=𝑥₁𝑥₂⋯𝑥_𝑟 and 𝑔=𝑦₁𝑦₂⋯𝑦_𝑟 be the reduced forms of elements 𝑓 and 𝑔 of group 𝐺=(𝐴∗𝐵;𝐻) where 𝑟>1. Then 𝑓 and 𝑔 are 𝐻-conjugate if and only if there exist elements ℎ₀,ℎ₁,ℎ₂,…,ℎ_𝑟=ℎ₀ in subgroup 𝐻 such that for any 𝑖=1,2,…,𝑟, one has 𝑥_𝑖=ℎ⁻¹_𝑖−1𝑦_𝑖ℎ_𝑖. (2.4)

Proof. If for some elements ℎ₀,ℎ₁,ℎ₂,…,ℎ_𝑟=ℎ₀ of subgroup 𝐻 equalities (2.4) hold, then 𝑓=𝑥₁𝑥₂⋯𝑥_𝑟=ℎ₀⁻¹𝑦₁ℎ₁ℎ₁⁻¹𝑦₂ℎ₂⋯ℎ⁻¹_𝑟−1𝑦₂ℎ_𝑟=ℎ₀⁻¹(𝑦₁𝑦₂⋯𝑦_𝑟)ℎ_𝑟=ℎ₀⁻¹𝑔ℎ₀. (2.5)
Conversely, by induction on 𝑟 we prove that if for some ℎ∈𝐻 the equality 𝑓=ℎ⁻¹𝑔ℎ holds then there exists a sequence of elements ℎ₀,ℎ₁,ℎ₂,…,ℎ_𝑟=ℎ₀ of subgroup 𝐻 satisfying the equalities (2.4) and such that ℎ₀=ℎ.
Rewriting the equality 𝑓=ℎ⁻¹𝑔ℎ in the form𝑥_𝑟⁻¹⋯𝑥₂⁻¹𝑥₁⁻¹ℎ⁻¹𝑦₁𝑦₂⋯𝑦_𝑟ℎ=1, (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 𝐴 or 𝐵 and element 𝑥₁⁻¹ℎ⁻¹𝑦₁ must belong to subgroup 𝐻. Denoting this element by ℎ₁⁻¹, we have 𝑥₁=ℎ⁻¹𝑦₁ℎ₁. If 𝑟=2 then the equality above takes the form 𝑥₂⁻¹ℎ₁⁻¹𝑦₂ℎ=1, whence 𝑥₂=ℎ₁⁻¹𝑦₂ℎ. Therefore, setting ℎ₀=ℎ₂=ℎ, we obtain in that case the desired sequence.
If 𝑟>2 let us rewrite the equality 𝑓=ℎ⁻¹𝑔ℎ in the formℎ⁻¹𝑦₁ℎ₁𝑥₂⋯𝑥_𝑟=ℎ⁻¹𝑦₁𝑦₂⋯𝑦_𝑟ℎ. (2.7) This implies that if we set 𝑓^′=ℎ⁻¹ℎ₁𝑥₂⋯𝑥_𝑟 and 𝑔^′=𝑦₂⋯𝑦_𝑟, then 𝑓^′=ℎ⁻¹𝑔^′ℎ. Since the length of elements 𝑓^′ and 𝑔^′ is equal to 𝑟−1, then by induction, there exists a sequence ℎ^′₁=ℎ, ℎ^′₂,…,ℎ^′_𝑟=ℎ^′₁ of elements of subgroup 𝐻, such that ℎ⁻¹ℎ₁𝑥₂=(ℎ^′₁)⁻¹𝑦₂ℎ^′₂ and for any 𝑖=2,3,…,𝑟𝑥_𝑖=(ℎ^′_𝑖−1)⁻¹𝑦_𝑖ℎ^′_𝑖. (2.8) Since 𝑥₂=ℎ₁⁻¹𝑦₂ℎ^′₂, then setting ℎ_𝑖=ℎ^′_𝑖 for 𝑖=2,3,…,𝑟, we obtain the desired sequence, and induction is completed. Proposition 2.3 is proven.

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

Proposition 2.4. Let 𝐺=(𝐴∗𝐵;𝐻) be the free product of groups 𝐴 and 𝐵 amalgamating subgroup 𝐻 where 𝐻 lies in the center of each of groups 𝐴 and 𝐵. Then for any element 𝑓∈𝐺 not belonging to subgroup 𝐴 one has 𝑓⁻¹𝐴𝑓∩𝐴=𝐻.

Proof. In fact, since subgroup 𝐻 coincides with the center of group 𝐺 (see [10, page 211]), the inclusion 𝐻⊆𝑓⁻¹𝐴𝑓∩𝐴 is obvious. Conversely, let element 𝑓⁻¹𝑎𝑓 belong to subgroup 𝐴 where 𝑎∈𝐴. Since 𝑓∉𝐴 then either 𝑓∈𝐵⧵𝐻 or 𝑙(𝑓)>1 and (without loss of generality) the first component of reduced form of 𝑓 belongs to 𝐵. In any case the assumption 𝑎∉𝐻 would imply that 𝑙(𝑓⁻¹𝑎𝑓)>1. Thus, 𝑎∈𝐻 and therefore 𝑓⁻¹𝑎𝑓=𝑎∈𝐻.

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

In what follows, our discussion will make use of the presentation of group 𝐺_𝑚𝑛 as an amalgamated free product of two groups. To describe such presentation, let 𝐻 be the subgroup of group 𝐺_𝑚𝑛, generated by elements 𝑐=𝑎^𝑚 and 𝑑=𝑏^𝑛. Also, let 𝐴 denote the subgroup of group 𝐺_𝑚𝑛 generated by element 𝑎 and subgroup 𝐻, and let 𝐵 denote the subgroup of group 𝐺_𝑚𝑛, 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;⟨𝑎^𝑚=𝑐⟩) of infinite cycle ⟨𝑎⟩ and group 𝐻 with amalgamation ⟨𝑎^𝑚⟩, group 𝐵 is the free product (⟨𝑏⟩∗𝐻;⟨𝑏^𝑛=𝑑⟩) of infinite cycle ⟨𝑏⟩ and group 𝐻 with amalgamation ⟨𝑏^𝑛⟩, 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 𝐺_𝑚𝑛. Namely, for any integer 𝑡>1 let 𝐺_𝑚𝑛(𝑡) be the group with presentation ⟨𝑎,𝑏;[𝑎^𝑚,𝑏^𝑛]=1,𝑎^𝑚𝑡=𝑏^𝑛𝑡=1⟩ (3.1) and 𝜌_𝑡 the natural homomorphism of groups 𝐺_𝑚𝑛 onto 𝐺_𝑚𝑛(𝑡). Then it is easy to verify that subgroup 𝐻(𝑡)=𝐻𝜌_𝑡 of group 𝐺_𝑚𝑛(𝑡) is isomorphic to the quotient 𝐻/𝐻^𝑡 (where, as usually, 𝐻^𝑡 consists of all elements ℎ^𝑡, ℎ∈𝐻), subgroup 𝐴(𝑡)=𝐴𝜌_𝑡 is the amalgamated free product (⟨𝑎;𝑎^𝑚𝑡=1⟩∗𝐻(𝑡);𝑎^𝑚=𝑐𝐻^𝑡) of cycle ⟨𝑎;𝑎^𝑚𝑡=1⟩ of order 𝑚𝑡 and group 𝐻(𝑡), subgroup 𝐵(𝑡)=𝐵𝜌_𝑡 is the amalgamated free product (⟨𝑏;𝑏^𝑛𝑡=1⟩∗𝐻(𝑡);𝑏^𝑛=𝑑𝐻^𝑡) of cycle ⟨𝑏;𝑏^𝑛𝑡=1⟩ of order 𝑛𝑡 and group 𝐻(𝑡), and group 𝐺_𝑚𝑛(𝑡) is the amalgamated free product (𝐴(𝑡)∗𝐵(𝑡);𝐻(𝑡)).

These decompositions of groups 𝐺_𝑚𝑛 and 𝐺_𝑚𝑛(𝑡) are assumed everywhere below, and such notions as free factor, reduced form, length of element and so on will refer to them.

Let us remark, at once, that since each of groups 𝐴(𝑡) and 𝐵(𝑡) is the amalgamated free product of two finite groups and group 𝐺_𝑚𝑛(𝑡) is the free product of groups 𝐴(𝑡) and 𝐵(𝑡) with finite amalgamation, it follows from results of [3] that for every 𝑡, group 𝐺_𝑚𝑛(𝑡) is conjugacy separable. So, to prove the conjugacy separability of group 𝐺_𝑚𝑛 it is enough, for any nonconjugate elements 𝑓 and 𝑔 of 𝐺_𝑚𝑛, to find an integer 𝑡 such that elements 𝑓𝜌_𝑡 and 𝑔𝜌_𝑡 are nonconjugate in group 𝐺_𝑚𝑛(𝑡).

Since in the decompositions of groups 𝐴 and 𝐵 as well, as of groups 𝐴(𝑡) and 𝐵(𝑡), into amalgamated free product stated above the amalgamated subgroups are contained in the centre of each free factor, by Proposition 2.4 we have the following proposition.

Proposition 3.1. For any element 𝑔 of group 𝐴 (or 𝐵) not belonging to subgroup 𝐻 the equality 𝑔⁻¹𝐻𝑔∩𝐻=⟨𝑐⟩ (resp., 𝑔⁻¹𝐻𝑔∩𝐻=⟨𝑑⟩) holds. In particular, for any element 𝑔 of group 𝐴 or 𝐵 and for any element ℎ of subgroup 𝐻 element 𝑔⁻¹ℎ𝑔 belongs to subgroup 𝐻 if and only if elements 𝑔 and ℎ commute.
Similarly, for any element 𝑔 of group 𝐴(𝑡) (or 𝐵(𝑡)) not belonging to subgroup 𝐻(𝑡) the equality 𝑔⁻¹𝐻(𝑡)𝑔∩𝐻(𝑡)=⟨𝑐𝐻^𝑡⟩ (resp., 𝑔⁻¹𝐻(𝑡)𝑔∩𝐻(𝑡)=⟨𝑑𝐻^𝑡⟩) holds. In particular, for any elements 𝑔 of group 𝐴(𝑡) or 𝐵(𝑡) and ℎ of subgroup 𝐻(𝑡), element 𝑔⁻¹ℎ𝑔 belongs to subgroup 𝐻(𝑡) if and only if elements 𝑔 and ℎ commute.

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

Proposition 3.2. Let 𝐺 be any of groups 𝐺_𝑚𝑛 and 𝐺_𝑚𝑛(𝑡). Every element of 𝐺 is conjugate to a cyclically reduced element. If lengths of two cyclically reduced elements are not equal then these elements are not conjugate in 𝐺. Let 𝑓 and 𝑔 be cyclically reduced elements of 𝐺 such that 𝑙(𝑓)=𝑙(𝑔). Then (1)If 𝑙(𝑓)=𝑙(𝑔)=1 then 𝑓 and 𝑔 are conjugate in group 𝐺 if and only if 𝑓 and 𝑔 belong to the same free factor and are conjugate in this factor. (2)If 𝑙(𝑓)=𝑙(𝑔)>1 then 𝑓 and 𝑔 are conjugate in group 𝐺 if and only if element 𝑓 is 𝐻-conjugate or 𝐻(𝑡)-conjugate to some cyclic permutation of 𝑔.

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

Proposition 3.3. For any homomorphism 𝜑 of group 𝐺_𝑚𝑛 onto a finite group 𝑋 there exist an integer 𝑡>1 and a homomorphism 𝜓 of group 𝐺_𝑚𝑛(𝑡) onto group 𝑋 such that 𝜑=𝜌_𝑡𝜓.
Also, for any integers 𝑡>1 and 𝑠>1 such that 𝑡 divides 𝑠 there exists homomorphism 𝜑∶𝐺_𝑚𝑛(𝑠)→𝐺_𝑚𝑛(𝑡) such that 𝜌_𝑡=𝜌_𝑠𝜑.
The same assertions are valid for groups 𝐴 and 𝐵.

Proposition 3.4. For any element 𝑔 of group 𝐴 or 𝐵, if 𝑔 does not belong to subgroup 𝐻 then there exists an integer 𝑡₀>1 such that for any positive integer 𝑡, divisible by 𝑡₀, element 𝑔𝜌_𝑡 does not belong to subgroup 𝐻𝜌_𝑡.

Proof. We will assume that 𝑔∈𝐴; the case when 𝑔∈𝐵 can be treated similarly.
So, let element 𝑔∈𝐴 do not belong to subgroup 𝐻 and let 𝑔=𝑥₁𝑥₂⋯𝑥_𝑟 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=𝑎^𝑘 for some integer 𝑘. Since 𝑔∉𝐻, integer 𝑘 is not divisible by 𝑚. Then for any integer 𝑡>0, in group 𝐴(𝑡), element 𝑔𝜌_𝑡 of subgroup ⟨𝑎;𝑎^𝑚𝑡=1⟩ cannot belong to the amalgamated subgroup (generated by element 𝑎^𝑚) of the decomposition of group 𝐴(𝑡) and consequently cannot belong to the free factor 𝐻(𝑡)=𝐻𝜌_𝑡.
Let 𝑟>1. Every component 𝑥_𝑖 of the reduced form of element 𝑔 is either of form 𝑎^𝑘, where integer 𝑘 is not divisible by 𝑚, or of form 𝑐^𝑘𝑑^𝑙 where 𝑙≠0. If integer 𝑡₀ is chosen such that the exponent 𝑙 of any component 𝑥_𝑖 of the second kind is not divisible by 𝑡₀, then for any 𝑡 divisible by 𝑡₀ the image 𝑥_𝑖𝜌_𝑡=𝑥_𝑖𝐻^𝑡 of such component will not belong to the amalgamated subgroup of group 𝐴(𝑡) (generated, let's remind, by element 𝑐𝐻^𝑡). Moreover, as above in case 𝑟=1, the images of components of the first kind will not belong to the amalgamated subgroup. Therefore, the form𝑔𝜌_𝑡=(𝑥₁𝜌_𝑡)(𝑥₂𝜌_𝑡)⋯(𝑥_𝑟𝜌_𝑡) (3.2) of element 𝑔𝜌_𝑡 is reduced in group 𝐴(𝑡) and since 𝑟>1, 𝑔𝜌_𝑡 does not belong to the free factor 𝐻(𝑡)=𝐻𝜌_𝑡. The proposition is proven.

Proposition 3.4 obviously implies the the following proposition.

Proposition 3.5. For any element 𝑔 of group 𝐺_𝑚𝑛 there exists an integer 𝑡₀>1 such that for all positive integer 𝑡, divisible by 𝑡₀, the length of element 𝑔𝜌_𝑡 in group 𝐺_𝑚𝑛(𝑡) coincides with the length of element 𝑔 in group 𝐺_𝑚𝑛.

Proposition 3.6. For any elements 𝑓 and 𝑔 of group 𝐴 (or 𝐵) such that element 𝑓 does not belong to the double coset 𝐻𝑔𝐻, there exists an integer 𝑡₀>1 such that for any positive integer 𝑡, divisible by 𝑡₀, element 𝑓𝜌_𝑡 does not belong to the double coset 𝐻(𝑡)(𝑔𝜌_𝑡)𝐻(𝑡).

Proof. We can again consider only the case when elements 𝑓 and 𝑔 belong to subgroup 𝐴. So, let us suppose that element 𝑓∈𝐴 does not belong to the double coset 𝐻𝑔𝐻. In view of Proposition 3.4, it is enough to prove that there exists a homomorphism 𝜑 of 𝐴 onto a finite group 𝑋 such that element 𝑓𝜑 of 𝑋 does not belong to the double coset (𝐻𝜑)(𝑔𝜑)(𝐻𝜑).
To this end let's consider the quotient group 𝐴=𝐴/𝐶 of group 𝐴 by its (central) subgroup 𝐶=⟨𝑐⟩. The image 𝑥𝐶 of an element 𝑥∈𝐴 in group 𝐴 will be denoted by 𝑥.
It is obvious that 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 𝑌𝑔𝑌. 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 𝑑^𝑘 for some integer 𝑘≠0. Since the set 𝑀 of all such exponents 𝑘 is finite, we can choose an integer 𝑡>0 such that for any 𝑘∈𝑀 the inequality 𝑡>2|𝑘| holds. Let's denote by 𝐴 the factor group of group 𝐴 by the normal closure of element 𝑑^𝑡. Group 𝐴 is the free product of groups 𝑋 and 𝑌/𝑌^𝑡 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/𝑌^𝑡𝑔𝑌/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, that𝑓∉(𝑌/𝑌^𝑡𝑔𝑌/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∉(𝐻𝑔𝐻)𝑁. Thus, the canonical homomorphism 𝜑 of group 𝐴 onto quotient group 𝐴/𝑁 has the required property and the proposition is proven.

4. Proof of Theorem 1.1

We prove first the following proposition.

Proposition 4.1. If elements 𝑓 and 𝑔 of group 𝐺_𝑚𝑛 such that 𝑙(𝑓)=𝑙(𝑔)>1 are not 𝐻-conjugate, then for some integer 𝑡>1, elements 𝑓𝜌_𝑡 and 𝑔𝜌_𝑡 of group 𝐺_𝑚𝑛(𝑡) are not 𝐻(𝑡)-conjugate.

Proof. Let 𝑓=𝑥₁𝑥₂⋯𝑥_𝑟 and 𝑔=𝑦₁𝑦₂⋯𝑦_𝑟 be the reduced forms in group 𝐺_𝑚𝑛=(𝐴∗𝐵;𝐻) of elements 𝑓 and 𝑔.
We remind (see Proposition 2.3) that elements 𝑓 and 𝑔 are 𝐻-conjugate if and only if there exist elements ℎ₀, ℎ₁, ℎ₂,…,ℎ_𝑟=ℎ₀ of 𝐻 such that for any 𝑖=1,2,…,𝑟, we have𝑥_𝑖=ℎ⁻¹_𝑖−1𝑦_𝑖ℎ_𝑖. (4.1) It then follows, in particular, that for each 𝑖=1,2,…,𝑟 elements 𝑥_𝑖 and 𝑦_𝑖 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 index 𝑖 elements 𝑥_𝑖 and 𝑦_𝑖 lie in different free factors 𝐴 and 𝐵 of group 𝐺_𝑚𝑛 (and, certainly, are not in subgroup 𝐻). It follows from Proposition 3.4 that there exists an integer 𝑡>1 such that elements 𝑥_𝑖𝜌_𝑡 and 𝑦_𝑖𝜌_𝑡 do not belong to the same free factor 𝐴(𝑡) or 𝐵(𝑡) of group 𝐺_𝑚𝑛(𝑡) (and, as above, lie in free factors of this group). Hence, by Proposition 2.3, in group 𝐺_𝑚𝑛(𝑡) elements 𝑓𝜌_𝑡 and 𝑔𝜌_𝑡 are not 𝐻(𝑡)-conjugate.Case 2. Let now for any 𝑖=1,2,…,𝑟 elements 𝑥_𝑖 and 𝑦_𝑖 belong to the same subgroup 𝐴 or 𝐵 and for some 𝑖 element 𝑥_𝑖 does not belong to the double coset 𝐻𝑦_𝑖𝐻. By Proposition 3.6, there exists an integer 𝑡₁>1 such that for any positive integer 𝑡, divisible by 𝑡₁, element 𝑥_𝑖𝜌_𝑡 is not in the double coset 𝐻(𝑡)(𝑦_𝑖𝜌_𝑡)𝐻(𝑡). From Proposition 3.5, there exist integers 𝑡₂>1 and 𝑡₃>1 such that for any positive integer 𝑡 divisible by 𝑡₂ the length of element 𝑓𝜌_𝑡 in group 𝐺_𝑚𝑛(𝑡) is equal to 𝑟 and for any positive integer 𝑡, divisible by 𝑡₃, the length of element 𝑔𝜏_𝑡 in group 𝐺_𝑚𝑛(𝑡) is equal to 𝑟. Thus, if 𝑡=𝑡₁𝑡₂𝑡₃ then in group 𝐺_𝑚𝑛(𝑡), elements 𝑓𝜌_𝑡 and 𝑔𝜌_𝑡 have the reduced forms (𝑥₁𝜌_𝑡)(𝑥₂𝜌_𝑡)⋯(𝑥_𝑟𝜌_𝑡),(𝑦₁𝜌_𝑡)(𝑦₂𝜌_𝑡)⋯(𝑦_𝑟𝜌_𝑡), (4.2) respectively, and element 𝑥_𝑖𝜌_𝑡 is not in the double coset 𝐻(𝑡)(𝑦_𝑖𝜌_𝑡)𝐻(𝑡). Again, by Proposition 2.3 these elements are not 𝐻(𝑡)-conjugate.Case 3. We now consider the case, when for any 𝑖=1,2,…,𝑟 elements 𝑥_𝑖 and 𝑦_𝑖 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 𝑥 and 𝑦 belong to one of the groups 𝐴 or 𝐵 and do not belong to the subgroup 𝐻 and 𝑥∈𝐻𝑦𝐻, that is, 𝑥=𝑐^𝛼𝑑^𝛽𝑦𝑐^𝛾𝑑^𝛿 (4.3) for some integers 𝛼, 𝛽, 𝛾 and 𝛿. If elements 𝑥 and 𝑦 belong to group 𝐴, then integers 𝛼+𝛾, 𝛽 and 𝛿 are uniquely determined by the equality (4.3). If elements 𝑥 and 𝑦 belong to group 𝐵, then integers 𝛽+𝛿, 𝛼 and 𝛾 are uniquely determined by equality (4.3).Proof. Let elements 𝑥 and 𝑦 belong to subgroup 𝐴 and let 𝑥=𝑐^𝛼1𝑑^𝛽1𝑦𝑐^𝛾1𝑑^𝛿1 and 𝑥=𝑐^𝛼2𝑑^𝛽2𝑦𝑐^𝛾2𝑑^𝛿2 for some integers 𝛼₁, 𝛽₁, 𝛾₁, 𝛿₁, 𝛼₂, 𝛽₂, 𝛾₂ and 𝛿₂. Rewriting the equality 𝑐^𝛼1𝑑^𝛽1𝑦𝑐^𝛾1𝑑^𝛿1=𝑐^𝛼2𝑑^𝛽2𝑦𝑐^𝛾2𝑑^𝛿2 as 𝑦⁻¹𝑐^{𝛼1−𝛼2}𝑑^{𝛽1−𝛽2}𝑦=𝑐^{𝛾2−𝛾1}𝑑^{𝛿2−𝛿1}, then, by Proposition 3.1, we conclude that 𝑐^{𝛼1−𝛼2}𝑑^{𝛽1−𝛽2}=𝑐^{𝛾2−𝛾1}𝑑^{𝛿2−𝛿1} and also that this element should belong to subgroup ⟨𝑐⟩, that is, 𝛽₁−𝛽₂=𝛿₂−𝛿₁=0. So, we have 𝑐^{𝛼1−𝛼2}=𝑐^{𝛾2−𝛾1} and since the order of element 𝑐 is infinite, then 𝛼₁−𝛼₂=𝛾₂−𝛾₁.
Thus, 𝛽₁=𝛽₂, 𝛿₁=𝛿₂ and 𝛼₁+𝛾₁=𝛼₂+𝛾₂ as it was required. The case when elements 𝑥 and 𝑦 belong to group 𝐵 is esteemed similarly.Lemma 4.3. Let elements 𝑥 and 𝑦 belong to one of groups 𝐴(𝑡) or 𝐵(𝑡) and do not belong to subgroup 𝐻(𝑡) and 𝑥∈𝐻(𝑡)𝑦𝐻(𝑡), that is, 𝑥=(𝑐𝐻^𝑡)^𝛼(𝑑𝐻^𝑡)^𝛽𝑦(𝑐𝐻^𝑡)^𝛾(𝑑𝐻^𝑡)^𝛿 (4.4) for some integers 𝛼, 𝛽, 𝛾 and 𝛿. If elements 𝑥 and 𝑦 belong to group 𝐴(𝑡), then integers 𝛼+𝛾, 𝛽 and 𝛿 are uniquely determined modulo 𝑡 by equality (4.4). If elements 𝑥 and 𝑦 belong to group 𝐵(𝑡), then integers 𝛽+𝛿, 𝛼 and 𝛾 are uniquely determined modulo 𝑡 by equality (4.4). The proof of Lemma 4.3 is completely similar to that of Lemma 4.2. Lemma 4.4. Let 𝑓=𝑥₁𝑥₂⋯𝑥_𝑟 and 𝑔=𝑦₁𝑦₂⋯𝑦_𝑟 be reduced elements of group 𝐺_𝑚𝑛, where 𝑟>1 and let for every 𝑖=1,2,…,𝑟 the equality 𝑥_𝑖=𝑢_𝑖𝑦_𝑖𝑣_𝑖 holds, for some elements 𝑢_𝑖 and 𝑣_𝑖 of subgroup 𝐻. Then there exists at most one sequence ℎ₀, ℎ₁, ℎ₂,…,ℎ_𝑟 of elements of subgroup 𝐻 such that for any 𝑖=1,2,…,𝑟𝑥_𝑖=ℎ⁻¹_𝑖−1𝑦_𝑖ℎ_𝑖. (4.5) Moreover, if 𝑢_𝑖=𝑐^𝛼𝑖𝑑^𝛽𝑖 and 𝑣_𝑖=𝑐^𝛾𝑖𝑑^𝛿𝑖 for some integers 𝛼_𝑖, 𝛽_𝑖, 𝛾_𝑖 and 𝛿_𝑖 (𝑖=1,2,…,𝑟) and 𝑥₁,𝑦₁∈𝐴, then such sequence exists if, and only if, for any 𝑖, 1<𝑖<𝑟, 𝛼_𝑖+𝛼_𝑖+1+𝛾_𝑖−1+𝛾_𝑖𝛽=0,if𝑖isodd,_𝑖+𝛽_𝑖+1+𝛿_𝑖−1+𝛿_𝑖=0,if𝑖iseven. (4.6)Proof. We suppose first that the sequence ℎ₀, ℎ₁, ℎ₂,…,ℎ_𝑟 of elements of subgroup 𝐻 satisfying equality (4.5) exists, and let's write ℎ_𝑖=𝑐^𝜇𝑖𝑑^𝜈𝑖, for some integers 𝜇_𝑖 and 𝜈_𝑖.
Then, since for any 𝑖=1,2,…,𝑟 the equality ℎ⁻¹_𝑖−1𝑦_𝑖ℎ_𝑖=𝑢_𝑖𝑦_𝑖𝑣_𝑖 holds, we have𝑦_𝑖⁻¹(ℎ_𝑖−1𝑢_𝑖)𝑦_𝑖=ℎ_𝑖𝑣_𝑖⁻¹. (4.7) Since 𝑟>1, then every element 𝑦_𝑖, belonging to one of the subgroups 𝐴 or 𝐵, does not lie in subgroup 𝐻, and consequently, from Proposition 3.1, for any 𝑖=1,2,…,𝑟, we have the equality ℎ_𝑖−1𝑢_𝑖=ℎ_𝑖𝑣_𝑖⁻¹. Substituting the expressions of elements ℎ_𝑖, 𝑢_𝑖 and 𝑣_𝑖, we have for every 𝑖=1,2,…,𝑟𝑐^{𝜇𝑖−1+𝛼𝑖}𝑑^{𝜈𝑖−1+𝛽𝑖}=𝑐^{𝜇𝑖−𝛾𝑖}𝑑^{𝜈𝑖−𝛿𝑖} and hence we obtain the system of numeric equations 𝜇_𝑖−1+𝛼_𝑖=𝜇_𝑖−𝛾_𝑖,𝜈_𝑖−1+𝛽_𝑖=𝜈_𝑖−𝛿_𝑖(𝑖=1,2,…,𝑟). (4.8) Since by (4.7) for every 𝑖=1,2,…,𝑟 element ℎ_𝑖𝑣_𝑖⁻¹ belongs to the intersection 𝑦_𝑖⁻¹𝐻𝑦_𝑖∩𝐻 and, by supposition, elements 𝑦₁, 𝑦₃,… belong to subgroup 𝐴 and elements 𝑦₂, 𝑦₄,… belong to subgroup 𝐵, then from Proposition 3.1, it follows that for odd 𝑖, we should have ℎ_𝑖−1𝑢_𝑖=ℎ_𝑖𝑣_𝑖⁻¹∈⟨𝑐⟩, and for even 𝑖, we should have ℎ_𝑖−1𝑢_𝑖=ℎ_𝑖𝑣_𝑖⁻¹∈⟨𝑑⟩. It means that for every odd 𝑖 (𝑖=1,2,…,𝑟) we have the equalities 𝜈_𝑖−1+𝛽_𝑖=0 and 𝜈_𝑖−𝛿_𝑖=0, and for every even 𝑖 (𝑖=1,2,…,𝑟) we have the equalities 𝜇_𝑖−1+𝛼_𝑖=0 and 𝜇_𝑖−𝛾_𝑖=0.
Hence, the values of the integers 𝜇_𝑖 and 𝜈_𝑖 are determined uniquely. Namely,𝜇_𝑖=𝛾_𝑖if𝑖isevenand2⩽𝑖⩽𝑟,−𝛼_𝑖+1if𝑖isoddand1⩽𝑖⩽𝑟−1, (4.9)𝜈_𝑖=−𝛽_𝑖+1𝛿if𝑖isevenand0⩽𝑖⩽𝑟−1,_𝑖if𝑖isoddand1⩽𝑖⩽𝑟. (4.10) Moreover, from equalities (4.8) it follows, that 𝜇₀=−(𝛼₁+𝛼₂+𝛾₁𝜈),_𝑟=𝛽_𝑟+𝛿_𝑟−1+𝛿_𝑟𝜇if𝑟iseven,_𝑟=𝛼_𝑟+𝛾_𝑟−1+𝛾_𝑟if𝑟isodd. (4.11)
Thus, the statement that there can exist at most one sequence of elements ℎ₀, ℎ₁, ℎ₂,…,ℎ_𝑟 of subgroup 𝐻 satisfying equalities (4.5) is demonstrated.
Substituting the value 𝜇_𝑖−1=𝛾_𝑖−1 and 𝜇_𝑖=−𝛼_𝑖+1 defined in (4.9) in the equalities 𝜇_𝑖−1+𝛼_𝑖=𝜇_𝑖−𝛾_𝑖 of system (4.8), where 1<𝑖<𝑟 and 𝑖 is odd, we obtain 𝛼_𝑖+𝛼_𝑖+1+𝛾_𝑖−1+𝛾_𝑖=0. Similarly, substituting the value 𝜈_𝑖−1=𝛿_𝑖−1 and 𝜈_𝑖=−𝛽_𝑖+1 defined in (4.10) in the equalities 𝜈_𝑖−1+𝛽_𝑖=𝜈_𝑖−𝛿_𝑖 of systems (4.8), where 1<𝑖<𝑟 and 𝑖 is even, we obtain 𝛽_𝑖+𝛽_𝑖+1+𝛿_𝑖−1+𝛿_𝑖=0.
Thus, under the existence in subgroup 𝐻 of sequence ℎ₀, ℎ₁, ℎ₂,…,ℎ_𝑟 of elements satisfying (4.5), conditions (4.6) are satisfied.
Conversely, suppose conditions (4.6) are satisfied. Letℎ₀=𝑐^{−(𝛼1+𝛼2+𝛾1)}𝑑^−𝛽1 (4.12) and for all indexes 𝑖 such that 1⩽𝑖<𝑟, we set ℎ_𝑖=𝑐^𝛾𝑖𝑑^{−𝛽𝑖+1}𝑐if𝑖iseven,_{−𝛼𝑖+1}𝑑^𝛿𝑖if𝑖isodd. (4.13) At last, for 𝑖=𝑟 we set ℎ_𝑟=𝑐^𝛾𝑟𝑑^{𝛽𝑟+𝛿𝑟−1+𝛿𝑟}𝑐if𝑟iseven,_{𝛼𝑟+𝛾𝑟−1+𝛾𝑟}𝑑^𝛿𝑟if𝑟isodd. (4.14)
Let us show that, the so-defined sequence of elements ℎ₀, ℎ₁, ℎ₂,…,ℎ_𝑟 really fits to equalities (4.5).
If 𝑖=1, using the permutability of elements 𝑐 and 𝑦₁ we haveℎ₀⁻¹𝑦₁ℎ₁=𝑐^{𝛼1+𝛼2+𝛾1}𝑑^𝛽1𝑦₁𝑐^−𝛼2𝑑^𝛿1=𝑐^𝛼1𝑑^𝛽1𝑦₁𝑐^𝛾1𝑑^𝛿1=𝑥₁. (4.15)
If 1<𝑖<𝑟 and integer 𝑖 is even, using the equality 𝛽_𝑖+𝛽_𝑖+1+𝛿_𝑖−1+𝛿_𝑖=0 and permutability of elements 𝑑 and 𝑦_𝑖, we haveℎ⁻¹_𝑖−1𝑦_𝑖ℎ_𝑖=𝑐^𝛼𝑖𝑑^{−𝛿𝑖−1}𝑦_𝑖𝑐^𝛾𝑖𝑑^{−𝛽𝑖+1}=𝑐^𝛼𝑖𝑦_𝑖𝑐^𝛾𝑖𝑑^{−(𝛿𝑖−1+𝛽𝑖+1)}=𝑐^𝛼𝑖𝑦_𝑖𝑐^𝛾𝑖𝑑^{𝛽𝑖+𝛿𝑖}=𝑐^𝛼𝑖𝑑^𝛽𝑖𝑦_𝑖𝑐^𝛾𝑖𝑑^𝛿𝑖=𝑥_𝑖. (4.16)
If 1<𝑖<𝑟 and integer 𝑖 is odd, using the equality 𝛼_𝑖+𝛼_𝑖+1+𝛾_𝑖−1+𝛾_𝑖=0 and permutability of elements 𝑐 and 𝑦_𝑖, we haveℎ⁻¹_𝑖−1𝑦_𝑖ℎ_𝑖=𝑐^{−𝛾𝑖−1}𝑑^𝛽𝑖𝑦_𝑖𝑐^{−𝛼𝑖+1}𝑑^𝛿𝑖=𝑐^{−(𝛾𝑖−1+𝛼𝑖+1)}𝑑^𝛽𝑖𝑦_𝑖𝑑^𝛿𝑖=𝑐^{𝛼𝑖+𝛾𝑖}𝑑^𝛽𝑖𝑦_𝑖𝑑^𝛿𝑖=𝑐^𝛼𝑖𝑑^𝛽𝑖𝑦_𝑖𝑐^𝛾𝑖𝑑^𝛿𝑖=𝑥_𝑖. (4.17)
If integer 𝑟 is even, thenℎ⁻¹_𝑟−1𝑦_𝑟ℎ_𝑟=𝑐^𝛼𝑟𝑑^{−𝛿𝑟−1}𝑦_𝑟𝑐^𝛾𝑟𝑑^{𝛽𝑟+𝛿𝑟−1+𝛿𝑟}=𝑐^𝛼𝑟𝑑^𝛽𝑟𝑦_𝑟𝑐^𝛾𝑟𝑑^𝛿𝑟=𝑥_𝑟, (4.18) and if 𝑟 is odd, then ℎ⁻¹_𝑟−1𝑦_𝑟ℎ_𝑟=𝑐^{−𝛾𝑟−1}𝑑^𝛽𝑟𝑦_𝑟𝑐^{𝛼𝑟+𝛾𝑟−1+𝛾𝑟}𝑑^𝛿𝑟=𝑐^𝛼𝑟𝑑^𝛽𝑟𝑦_𝑟𝑐^𝛾𝑟𝑑^𝛿𝑟=𝑥_𝑟. (4.19) So, Lemma 4.4 is completely demonstrated.
Similar argument gives the following lemma.
Lemma 4.5. Let 𝑓=𝑥₁𝑥₂⋯𝑥_𝑟 and 𝑔=𝑦₁𝑦₂⋯𝑦_𝑟 be reduced elements of group 𝐺_𝑚𝑛(𝑡), where 𝑟>1, and let for every 𝑖=1,2,…,𝑟 the equality 𝑥_𝑖=𝑢_𝑖𝑦_𝑖𝑣_𝑖 takes place, for some elements 𝑢_𝑖 and 𝑣_𝑖 of subgroup 𝐻(𝑡). Then there exists at most one sequence ℎ₀, ℎ₁, ℎ₂,…,ℎ_𝑟 of elements of subgroup 𝐻(𝑡) such that for any 𝑖=1,2,…,𝑟, one has 𝑥_𝑖=ℎ⁻¹_𝑖−1𝑦_𝑖ℎ_𝑖. (4.20) Moreover, if 𝑢_𝑖=(𝑐𝐻^𝑡)^𝛼𝑖(𝑑𝐻^𝑡)^𝛽𝑖 and 𝑣_𝑖=(𝑐𝐻^𝑡)^𝛾𝑖(𝑑𝐻^𝑡)^𝛿𝑖 for some integers 𝛼_𝑖, 𝛽_𝑖, 𝛾_𝑖 and 𝛿_𝑖 (𝑖=1,2,…,𝑟) and 𝑥₁,𝑦₁∈𝐴(𝑡), then such sequence exists if and only if for any integer 𝑖, 1<𝑖<𝑟, one has 𝛼_𝑖+𝛼_𝑖+1+𝛾_𝑖−1+𝛾_𝑖𝛽≡0(mod𝑡)if𝑖isodd,_𝑖+𝛽_𝑖+1+𝛿_𝑖−1+𝛿_𝑖≡0(mod𝑡)if𝑖iseven. (4.21)
We can now end the consideration of Case 3 and thus complete the proof of Proposition 4.1.
By Proposition 2.3, elements 𝑓 and 𝑔 are 𝐻-conjugate if and only if their cyclic permutation 𝑥₂⋯𝑥_𝑟𝑥₁ and 𝑦₂⋯𝑦_𝑟𝑦₁ are 𝐻-conjugate. Therefore, we can suppose, without loss of generality, that elements 𝑥₁ and 𝑦₁ belong to subgroup 𝐴.
By supposition, for every 𝑖=1,2,…,𝑟 there exist integers 𝛼_𝑖, 𝛽_𝑖, 𝛾_𝑖 and 𝛿_𝑖 such that𝑥_𝑖=𝑐^𝛼𝑖𝑑^𝛽𝑖𝑦_𝑖𝑐^𝛾𝑖𝑑^𝛿𝑖. (4.22)
If integer 𝑖, 1<𝑖<𝑟, is even, then 𝑦_𝑖∈𝐵, 𝑦_𝑖−1,𝑦_𝑖+1∈𝐴 and consequently, from Lemma 4.2, integers 𝛽_𝑖+𝛿_𝑖, 𝛿_𝑖−1 and 𝛽_𝑖+1 do not depend from the particular expressions of the elements 𝑥_𝑖 (in the form 𝑥_𝑖=𝑢𝑦_𝑖𝑣, where 𝑢,𝑣∈𝐻), and these integers are uniquely determined by the sequences 𝑥₁, 𝑥₂,…,𝑥_𝑟 and 𝑦₁, 𝑦₂,…,𝑦_𝑟. Similarly, for any odd 𝑖, 1<𝑖<𝑟, integers 𝛼_𝑖+𝛾_𝑖, 𝛾_𝑖−1 and 𝛼_𝑖+1 are uniquely determined by these sequences. It means, in turn, that the satisfiability of conditions (4.6) of Lemma 4.4 depends only on the sequences 𝑥₁, 𝑥₂,…,𝑥_𝑟 and 𝑦₁, 𝑦₂,…,𝑦_𝑟.
Let's now conditions (4.6) of Lemma 4.4 are not satisfied, that is, either for some even 𝑖, 1<𝑖<𝑟, the sum 𝛽_𝑖+𝛽_𝑖+1+𝛿_𝑖−1+𝛿_𝑖 is different from zero, or for some odd 𝑖, 1<𝑖<𝑟, the sum 𝛼_𝑖+𝛼_𝑖+1+𝛾_𝑖−1+𝛾_𝑖 is different from zero. Then it is possible to find an integer 𝑡₁>1, not dividing the respective sum. Let's also choose, according to Proposition 3.5, the integers 𝑡₂>1 and 𝑡₃>1 such that for all positive integer 𝑡, divisible by 𝑡₂, the length of element 𝑓𝜌_𝑡 in group 𝐺_𝑚𝑛(𝑡) is equal to 𝑟 and for all positive integer 𝑡, divisible by 𝑡₃, the length of element 𝑔𝜌_𝑡 in group 𝐺_𝑚𝑛(𝑡) is equal to 𝑟. Then if 𝑡=𝑡₁𝑡₂𝑡₃, in group 𝐺_𝑚𝑛(𝑡), elements 𝑓𝜌_𝑡 and 𝑔𝜌_𝑡 have reduced forms(𝑥₁𝜌_𝑡)(𝑥₂𝜌_𝑡)⋯(𝑥_𝑟𝜌_𝑡),(𝑦₁𝜌_𝑡)(𝑦₂𝜌_𝑡)⋯(𝑦_𝑟𝜌_𝑡), (4.23) respectively. Further, for every 𝑖=1,2,…,𝑟, 𝑥_𝑖𝜌_𝑡=(𝑐𝐻^𝑡)^𝛼𝑖(𝑑𝐻^𝑡)^𝛽𝑖(𝑦_𝑖𝜌_𝑡)(𝑐𝐻^𝑡)^𝛾𝑖(𝑑𝐻^𝑡)^𝛿𝑖. (4.24) From Lemma 4.3 and the selection of integer 𝑡₁ it follows that for elements 𝑓𝜌_𝑡 and 𝑔𝜌_𝑡, conditions (4.21) of Lemma 4.5 are not satisfied. Therefore from this lemma and Proposition 2.3, elements 𝑓𝜌_𝑡 and 𝑔𝜌_𝑡 are not 𝐻(𝑡)-conjugate in group 𝐺_𝑚𝑛(𝑡).
Let now the reduced forms of elements 𝑓 and 𝑔 satisfy conditions (4.6) of Lemma 4.4. Then according to this lemma, in subgroup 𝐻 there exists the only sequence of elements ℎ₀, ℎ₁, ℎ₂,…,ℎ_𝑟 such that for any 𝑖=1,2,…,𝑟, we have 𝑥_𝑖=ℎ⁻¹_𝑖−1𝑦_𝑖ℎ_𝑖. Since elements 𝑓 and 𝑔 are not 𝐻-conjugate, then elements ℎ₀ and ℎ_𝑟 should be different. Since group 𝐺_𝑚𝑛 is residually finite, by Proposition 3.3 there exists an integer 𝑡₁>1 such that ℎ₀𝜌_𝑡1≠ℎ_𝑟𝜌_𝑡1. Let's choose one more integer 𝑡₂ such that in group 𝐺_𝑚𝑛(𝑡₂) elements 𝑓𝜌_𝑡2 and 𝑔𝜌_𝑡2 have length 𝑟. Then if 𝑡=𝑡₁𝑡₂, in group 𝐺_𝑚𝑛(𝑡), ℎ₀𝜌_𝑡≠ℎ_𝑟𝜌_𝑡, elements 𝑓𝜌_𝑡 and 𝑔𝜌_𝑡 have reduced forms(𝑥₁𝜌_𝑡)(𝑥₂𝜌_𝑡)⋯(𝑥_𝑟𝜌_𝑡),(𝑦₁𝜌_𝑡)(𝑦₂𝜌_𝑡)⋯(𝑦_𝑟𝜌_𝑡), (4.25) respectively, and for every 𝑖=1,2,…,𝑟𝑥_𝑖𝜌_𝑡=(𝑐𝐻^𝑡)^𝛼𝑖(𝑑𝐻^𝑡)^𝛽𝑖(𝑦_𝑖𝜌_𝑡)(𝑐𝐻^𝑡)^𝛾𝑖(𝑑𝐻^𝑡)^𝛿𝑖. (4.26) Moreover, in group 𝐺_𝑚𝑛(𝑡), for any 𝑖=1,2,…,𝑟, the equality 𝑥_𝑖𝜌_𝑡=(ℎ_𝑖−1𝜌_𝑡)⁻¹(𝑦_𝑖𝜌_𝑡)(ℎ_𝑖𝜌_𝑡) (4.27) holds.
Since the sequence ℎ₀𝜌_𝑡, ℎ₁𝜌_𝑡, ℎ₂𝜌_𝑡,…,ℎ_𝑟𝜌_𝑡 is, by Lemma 4.5, the unique sequence of elements of subgroup 𝐻(𝑡) satisfying these equalities, then from Proposition 2.3, elements 𝑓𝜏_𝑡 and 𝑔𝜏_𝑡 are not 𝐻(𝑡)-conjugate in group 𝐺_𝑚𝑛(𝑡).
Hence Proposition 4.1 is completely demonstrated.

We now proceed directly to the proof of Theorem 1.1.

As remarked above, for any 𝑡>1, group 𝐺_𝑚𝑛(𝑡) is conjugacy separable. Therefore, for the proof of the Theorem it is enough to show that for any two nonconjugate in group 𝐺_𝑚𝑛 elements 𝑓 and 𝑔 of group 𝐺_𝑚𝑛, there exists an integer 𝑡>1 such that elements 𝑓𝜌_𝑡 and 𝑔𝜌_𝑡 are not conjugate in group 𝐺_𝑚𝑛(𝑡). To this end we make use of the conjugacy criterion of elements of groups 𝐺_𝑚𝑛 and 𝐺_𝑚𝑛(𝑡) given in Proposition 3.2.

So, let 𝑓=𝑥₁𝑥₂⋯𝑥_𝑟 and 𝑔=𝑦₁𝑦₂⋯𝑦_𝑟 be the reduced forms in group 𝐺_𝑚𝑛=(𝐴∗𝐵;𝐻) of two nonconjugate in group 𝐺_𝑚𝑛 elements 𝑓 and 𝑔. By Proposition 3.2 we can assume that elements 𝑓 and 𝑔 are cyclically reduced. In accordance with this proposition we must consider separately some cases.

Case 1 (𝑙(𝑓)≠𝑙(𝑔)). From Proposition 3.5, there exists an integer 𝑡₁>1 such that for any positive integer 𝑡, divisible by 𝑡₁, the length of element 𝑓𝜌_𝑡 in group 𝐺_𝑚𝑛(𝑡) coincides with the length of element 𝑓 in group 𝐺_𝑚𝑛. Similarly, there exists integer 𝑡₂>1 such that for any positive integer 𝑡, divisible by 𝑡₂, the length of element 𝑔𝜌_𝑡 in group 𝐺_𝑚𝑛(𝑡) coincides with the length of element 𝑔 in group 𝐺_𝑚𝑛. Thus, if 𝑡=𝑡₁𝑡₂ elements 𝑓𝜌_𝑡 and 𝑔𝜌_𝑡 of group 𝐺_𝑚𝑛(𝑡) are, as it is easy to see, cyclically reduced and have different length. Hence, by Proposition 3.2, these elements are not conjugate in this group. Thus integer 𝑡 is the required.

Case 2 (𝑙(𝑓)=𝑙(𝑔)=1). In this case each of these elements belongs to one of the 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 separable [3], then by Proposition 3.3 there exists an integer 𝑡>1 such that in group 𝐴(𝑡) elements 𝑓𝜌_𝑡 and 𝑔𝜌_𝑡 are not conjugate. Proposition 3.2 now implies that elements 𝑓𝜌_𝑡 and 𝑔𝜌_𝑡 are not conjugate in group 𝐺_𝑚𝑛(𝑡).

Let now element 𝑓 belongs to subgroup 𝐴, element 𝑔 belongs to subgroup 𝐵 and these elements do not belong to subgroup 𝐻. By Proposition 3.4, there exist integers 𝑡₁>1 and 𝑡₂>1 such that for any positive integer 𝑡, divisible by 𝑡₁, element 𝑓𝜌_𝑡 does not belong to subgroup 𝐻𝜌_𝑡, and for any positive integer 𝑡, divisible by 𝑡₂, element 𝑔𝜌_𝑡 does not belong to subgroup 𝐻𝜌_𝑡. Then if 𝑡=𝑡₁𝑡₂, Proposition 3.2 implies that elements 𝑓𝜌_𝑡 and 𝑔𝜌_𝑡 are not conjugate in group 𝐺_𝑚𝑛(𝑡).

Case 3 (𝑙(𝑓)=𝑙(𝑔)>1). Let 𝑔_𝑖=𝑦_𝑖𝑦_𝑖+1⋯𝑦_𝑟𝑦₁⋯𝑦_𝑖−1 (𝑖=1,2,…,𝑟) be all the cyclic permutations of element 𝑔. Since elements 𝑔 and 𝑔_𝑖 are conjugate and elements 𝑓 and 𝑔 are not conjugate, element 𝑓 is not 𝐻-conjugate to any of elements 𝑔₁, 𝑔₂,…,𝑔_𝑟. It follows from Proposition 4.1 that for every 𝑖=1,2,…,𝑟, there exists an integer 𝑡_𝑖>1 such that elements 𝑓𝜌_𝑡𝑖 and 𝑔_𝑖𝜌_𝑡𝑖 are not 𝐻(𝑡_𝑖)-conjugate in group 𝐺_𝑚𝑛(𝑡_𝑖). Let also integer 𝑡₀>1 be chosen such that for all positive integer 𝑡, divisible by 𝑡₀, elements 𝑓𝜌_𝑡 and 𝑔𝜌_𝑡 have length 𝑟 in group 𝐺_𝑚𝑛(𝑡). Then if 𝑡=𝑡₀𝑡₁⋯𝑡_𝑟 in group 𝐺_𝑚𝑛(𝑡), elements 𝑓𝜌_𝑡 and 𝑔_𝑖𝜌_𝑡 have reduced forms (𝑥₁𝜌_𝑡)(𝑥₂𝜌_𝑡)⋯(𝑥_𝑟𝜌_𝑡)(𝑦_𝑖𝜌_𝑡)(𝑦_𝑖+1𝜌_𝑡)⋯(𝑦_𝑟𝜌_𝑡)(𝑦₁𝜌_𝑡)⋯(𝑦_𝑖−1𝜌_𝑡), (4.28) respectively. Furthermore, elements 𝑓𝜌_𝑡 and 𝑔_𝑖𝜌_𝑡 are not 𝐻(𝑡)-conjugate in group 𝐺_𝑚𝑛(𝑡). Since an arbitrary cyclic permutation of element 𝑔𝜌_𝑡 coincides with some element 𝑔_𝑖𝜌_𝑡, then from Proposition 3.2 it follows that elements 𝑓𝜌_𝑡 and 𝑔𝜌_𝑡 are not conjugate in group 𝐺_𝑚𝑛(𝑡).

The proof of Theorem is now complete.