Conjugacy separability of some one-relator groups

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.


Introduction
A group G is conjugacy separable if for any two non-conjugate elements f and g of G there exists a homomorphism ϕ of group G onto some finite group X such that the images f ϕ 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, for any non-identity element g ∈ G there exists a homomorphism ϕ of group G onto some finite group X such that gϕ = 1).
Since 1962 when G. Baumslag and D. Solitar [3] 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, J. Dyer [4] have proved the conjugacy separability of any group being free product of free groups amalgamating cycle. In the same paper she proved an unpublished result of M. Amstrong on conjugacy separability of any one-relator group with nontrivial center. The conjugacy separability of groups defined by relator of form (a m b n ) t and (a −1 b l ab m ) s , where s > 1, was proved in [6] and [1] 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 Theorem. 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 [4].
The residual finiteness of groups G mn is well known; it follows, for example, from the result of paper [2]. Some properties of these groups were considered in [7] where, in particular, the description of their endomorphisms was given. The proofs in this paper made use of presentation of group G mn as amalgamated free product. The same presentation will be the crucial tool in the proof of Theorem above 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 will be completed.

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 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 = y 1 y 2 · · · 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 = x 1 x 2 · · · x r is a reduced form of cyclically reduced element g then cyclic permutations of element g are elements of form If X is subgroup of a group Y we shall say that elements a and b of Y are X-conjugate if a = x −1 bx, for some x ∈ X.
The Solitar criterion of conjugacy of elements of amalgamated free product of two groups (Theorem 4.6 in [5]) can be formulated as follows. Proposition 1.1. 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 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 such that for any i = 0, 1, . . . , n h i ∈ H and elements h i and h i+1 are Aconjugate 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 1.1 became unnecessary. In fact, 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 desired assertion is implied by assertion (1) of Proposition 1.1. Otherwise, we can assume that f ∈ H and by assertion (2) of Proposition 1.1 there is a sequence of elements f = h 0 , h 1 , . . . , h n , h n+1 = g, such that for any i = 0, 1, . . . , n h i ∈ H and for suitable element x i belonging to one of the subgroups A or B, the equality x −1 i h i x i = h i+1 holds. Since for i = 0, 1, . . . , n − 1 the inclusion f −1 h i+1 f ∈ H is valid, the hypothesis gives h 0 = h n , i. e. x −1 n f x 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 explicit the situation arising in assertion (3) of Proposition 1.1. Proposition 1.3. Let f = x 1 x 2 · · · x r and g = y 1 y 2 · · · 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 h 0 , h 1 , h 2 , . . . , h r = h 0 in subgroup H such that for any i = 1, 2, . . . , r, we have Proof. If for some elements h 0 , h 1 , h 2 , . . . , h r = h 0 of subgroup H the equalities (1) hold then Conversely, by induction on r we prove that if for some h ∈ H the equality f = h −1 gh holds then there exists a sequence of elements h 0 , h 1 , h 2 , . . . , h r = h 0 of subgroup H satisfying the equalities (1) and such that h 0 = h.
Rewriting the equality f = h −1 gh in the form we see that since the expression in the left part of it cannot be reduced, elements x 1 and y 1 must be contained in the same subgroup A or B and element x −1 1 h −1 y 1 must belong to subgroup H. Denoting this element by h −1 1 , we have x 1 = h −1 y 1 h 1 . If r = 2 then the equality above takes the form Since the length of elements f ′ and g ′ is equal to r − 1, then by induction, there exists a sequence h ′ . . , r, we obtain the desired sequence, and induction is complete. Proposition 1.3 is proven.
We conclude this section with one more property of amalgamated free product. In fact, since subgroup H coincides with the center of group G (see [5], page 211), 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 −1 af ) > 1. Thus, a ∈ H and therefore f −1 af = a ∈ H.

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 4 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 and ρ t be the natural homomorphism of group 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 paper [4] 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 non-conjugate elements f and g of G mn , to find an integer t such that elements f ρ t and gρ t are non-conjugate 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 1.4 we have Proposition 2.1. For any element g of group A (or B) not belonging to subgroup H the equality g −1 Hg ∩ H = c (respectively, g −1 Hg ∩ H = d ) holds. In particular, for any element g of group A or B and for any element h of subgroup H element g −1 hg 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 −1 H(t)g ∩ H(t) = cH t (respectively, g −1 H(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 −1 hg belongs to subgroup H(t) if and only if elements g and h commute. Proposition 2.2. 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 5 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 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 2.3. 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 t ψ.
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 s ϕ.
The same assertions are valid for groups A and B.
Proposition 2.4. For any element g of group A or B, if g does not belong to subgroup H then there exists an integer t 0 > 1 such that for any positive integer t, divisible by t 0 , element gρ t does not belong to subgroup Hρ t .
Proof. We shall 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 = x 1 x 2 · · · 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 i. e. 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 k d l where l = 0. If integer t 0 is chosen such that the exponent l of any component x i of the second kind is not divisible by t 0 , then for any t divisible by t 0 the image x i ρ t = x i H 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 form 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 2.4 obviously implies the
Proposition 2.5. For any element g of group G mn there exists an integer t 0 > 1 such that for all positive integer t, divisible by t 0 , the length of element gρ t in group G mn (t) coincides with the length of element g in group G mn . Proposition 2.6. For any elements f and g of group A (or B) such that element f does not belong to the double coset HgH, then there exists an integer t 0 > 1 such that for any positive integer t, divisible by t 0 , element f ρ 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 2.4, it is enough to prove that there exists a homomorphism ϕ of A onto a finite group X such that element f ϕ 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 cyclic group X of order m, generated by element a, and the infinite cycle Y , generated by d. The canonical homomorphism of group A onto group A maps subgroup H onto subgroup Y and consequently, the image of the double coset HgH is the double coset Y gY . Since C H, the element f does not belong to this coset.
We can assume, without loss of generality, that any element f and g, if it is different from identity, has reduced form the first and the last syllables of which do not belong to subgroup Y . Every Y -syllable of these reduced forms is of form 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 t g Y /Y t . Since this coset consists of a finite number of elements and group A is residually finite, then there exists a normal subgroup N of finite index of group A such, that 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 pre-image 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.

Proof of Theorem
We prove first the following proposition.
Proposition 3.1. 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 f ρ t and gρ t of group G mn (t) are not H(t)-conjugate. 7 Proof. Let f = x 1 x 2 · · · x r and g = y 1 y 2 · · · y r be the reduced forms in group G mn = (A * B; H) of elements f and g.
We remind (see Proposition 1.3) that elements f and g are H-conjugate if and only if there exist elements h 0 , h 1 , h 2 , . . . , h r = h 0 of H such that for any i = 1, 2, . . . , r, we have 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 2.4 that there exists an integer t > 1 such that elements x i ρ t and y i ρ t do not belong to the same free factor A(t) or B(t) of group G mn (t) (and, as before, lie in free factors of this group). Hence, by Proposition 1.3, in group G mn (t) elements f ρ 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 i H. By Proposition 2.6, there exists an integer t 1 > 1 such that for any positive integer t, divisible by t 1 , element x i ρ t is not in the double coset H(t)(y i ρ t )H(t). From Proposition 2.5, there exist integers t 2 > 1 and t 3 > 1 such that for any positive integer t divisible by t 2 the length of element f ρ t in group G mn (t) is equal to r and for any positive integer t, divisible by t 3 , the length of element gτ t in group G mn (t) is equal to r. Thus, if t = t 1 t 2 t 3 then in group G mn (t), elements f ρ t and gρ t have the reduced forms (x 1 ρ t )(x 2 ρ t ) · · · (x r ρ t ) and (y 1 ρ t )(y 2 ρ t ) · · · (y r ρ t ) respectively and element x i ρ t is not in the double coset H(t)(y i ρ t )H(t). Again, by Proposition 1.3 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 1.
Let elements x and y of one of groups A or B do not belong to subgroup H and x ∈ HyH, i. e.
x = c α d β yc γ d δ 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 (2). If elements x and y belong to group B, then integers β + δ, α and γ are uniquely determined by equality (2).

Lemma 2.
Let elements x and y of one of groups A(t) or B(t) do not belong to subgroup H(t) and x ∈ H(t)yH(t), i. e.
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 (3). If elements x and y belong to group B(t), then integers β + δ, α and γ are uniquely determined modulo t by equality (3).
The proof of lemma 2 is completely similar to that of lemma 1.
Lemma 3. Let f = x 1 x 2 · · · x r and g = y 1 y 2 · · · y r be reduced forms of elements f and g of group G mn , where r > 1 and let for every i = 1, 2, . . . , r the equality x i = u i y i v i holds, for some elements u i and v i of subgroup H. Then there exists at most one sequence h 0 , h 1 , h 2 , . . . , h r of elements of subgroup H such that for any i = 1, 2, . . . , r Moreover, if u i = c α i d β i and v i = c γ i d δ i for some integers α i , β i , γ i and δ i (i = 1, 2, . . . , r) and x 1 , y 1 ∈ A, then such sequence exists if, and only if, for any i, 1 < i < r, Proof. We suppose first that the sequence h 0 , h 1 , h 2 , . . . , h r of elements of subgroup H, satisfying equality (4), exists and let's write its elements as h i = c µ i d ν i , for some integers µ i and ν i .
Then, since for any i = 1, 2, . . . , r the equality As 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 2.1, for any i = 1, 2, . . . , r, we have the equality h i−1 u i = h i v −1 i . Substituting the expressions of elements h i , u i and v i , we have for every i = 1, 2, . . . , r c µ i−1 +α i d ν i−1 +β i = c µ i −γ i d ν i −δ i and hence we obtain the system of numeric equations Since by (6) 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, and Moreover, from the equalities (7) it follows, that µ 0 = −(α 1 + α 2 + γ 1 ), and ν r = β r + δ r−1 + δ r , if r is even, Thus, the statement that there can exist at most one sequence of elements h 0 , h 1 , h 2 , . . . , h r of subgroup H satisfying the equalities (4) is demonstrated.
Conversely, suppose conditions (5) are satisfied. Let and for all indexes i such that 1 i < r, we set At last, for i = r we set h r = c γ r d β r +δ r−1 +δ r , if r is even, c α r +γ r−1 +γ r d δ r , if r is odd.
Let's show that, the so defined sequence of elements h 0 , h 1 , h 2 , . . . , h r really fits to the equalities (4).
If i = 1, using the permutability of elements c and y 1 we have 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 have 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 have If integer r is even, then h −1 r−1 y r h r = c α r d −δ r−1 y r c γ r d β r +δ r−1 +δ r = c α r d β r y r c γ r d δ r = x r , and if r is odd, then h −1 r−1 y r h r = c −γ r−1 d β r y r c α r +γ r−1 +γ r d δ r = c α r d β r y r c γ r d δ r = x r .

Lemma 3 is completely demonstrated.
Similar argument gives the Lemma 4. Let f = x 1 x 2 · · · x r and g = y 1 y 2 · · · y r be reduced forms of elements f and g of group G mn (t), where r > 1, and let for every i = 1, 2, . . . , r the equality x i = u i y i v i takes place, for some elements u i and v i of subgroup H(t). Then there exists at most one sequence h 0 , h 1 , h 2 , . . . , h r of elements of subgroup H(t) such that for any i = 1, 2, . . . , r, we have Moreover, if u i = (cH t ) α i (dH t ) β i and v i = (cH t ) γ i (dH t ) δ i for some integers α i , β i , γ i and δ i (i = 1, 2, . . . , r) and x 1 , y 1 ∈ A(t), then such sequence exist if and only if for any integer i, 1 < i < r, we have α i + α i+1 + γ i−1 + γ i ≡ 0 (mod t), if i is odd, and β i + β i+1 + δ i−1 + δ i ≡ 0 (mod t), if i is even.
gρ t does not belong to subgroup Hρ t . Then if t = t 1 t 2 , Proposition 2.2 implies that elements f ρ t and gρ t are not conjugate in group G mn (t).
3. l(f ) = l(g) > 1. Let g i = y i y i+1 · · · y r y 1 · · · 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 g 1 , g 2 , . . . , g r . It follows from Proposition 3.1 that for every i = 1, 2, . . . , r, there exists an integer t i > 1 such that elements f ρ t i and g i ρ t i are not H(t i )-conjugate in group G mn (t i ). Let also integer t 0 > 1 be chosen such that for all positive integer t, divisible by t 0 , elements f ρ t and gρ t have length r in group G mn (t). Then if t = t 0 t 1 · · · t r in group G mn (t), elements f ρ t and g i ρ t have reduced forms (x 1 ρ t )(x 2 ρ t ) · · · (x r ρ t ) and (y i ρ t )(y i+1 ρ t ) · · · (y r ρ t )(y 1 ρ t ) · · · (y i−1 ρ t ) respectively. Furthermore, elements f ρ t and g i ρ 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 i ρ t , then from Proposition 2.2 it follows that elements f ρ t and gρ t are not conjugate in group G mn (t).
The proof of Theorem is now complete.