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A Criterion for the Conjugacy Separability of Amalgamated Free Products of Conjugacy Separable Groups
JOURNAL OF ALGEBRA ARTICLE NO.
184, 1052]1072 Ž1996.
0298
A Criterion for the Conjugacy Separability of Amalgamated Free Products of Conjugacy Separable Groups Goansu KimU Department of Mathematics, Kangnung National Uni¨ ersity, Kangnung, 210-702 Korea
and C. Y. Tang† Uni¨ ersity of Waterloo, Waterloo, Ontario, N2L 3G1, Canada Communicated by Walter Feit Received August 4, 1995
We first prove a criterion for the conjugacy separability of generalized free products of two conjugacy separable groups amalgamating a cyclic subgroup. Applying this result, we show that tree products of a finite number of conjugacy separable, residually finitely generated torsion-free nilpotent groups amalgamating cyclic subgroups are conjugacy separable. From this we derive that tree products of finitely generated torsion-free nilpotent groups, free groups, or surface groups Q 1996 Academic Press, Inc. amalgamating cyclic subgroups are conjugacy separable.
1. INTRODUCTION Let S be a subset of a group G. Then G is said to be S-separable if for each g f S there exists a finite index normal subgroup N of G with g f NS. This is equivalent to S being a closed subset in the profinite topology on G. If G is S-separable for every finitely generated subgroup S of G, then G is said to be subgroup separable Žor LERF.. In the special case of S s 14 , G is S-separable means G is residually finite. If G is U
Supported by GARC-KOSEF and NDRF, Korea Research Foundation, 1995. E-mail: [email protected]. † Partially supported by the Natural Science and Engineering Research Council of Canada, Grant No. A-6064. E-mail: [email protected]. 1052 0021-8693r96 $18.00 Copyright Q 1996 by Academic Press, Inc. All rights of reproduction in any form reserved.
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S-separable for every conjugacy class S of G, then G is called conjugacy separable. Separability properties of groups are of great interest to both group theorists and three-manifold topologists. In particular, Mostowski w16x showed that: Ž1. finitely presented residually finite groups have solvable word problem, Ž2. finitely presented conjugacy separable groups have solvable conjugacy problem, and Ž3. finitely presented subgroup separable groups have solvable generalized word problem. It is well known that free groups and surface groups are both subgroup separable and conjugacy separable ŽHall w10x, Scott w19x, Stebe w21x.. Thurston w24x showed that the fundamental groups of Haken manifolds are residually finite. Scott asked whether these groups are subgroup separable. Burns, Karrass and Solitar w5x showed that these groups needs not be subgroup separable. The question of whether these groups are always conjugacy separable is still unknown. In particular, whether knot groups are subgroup separable or conjugacy separable is still unknown. Blackburn w4x showed that finitely generated nilpotent groups are conjugacy separable. Subsequently, Formanek w8x Žalso Remeslennikov w18x. showed that polycyclic-by-finite groups are conjugacy separable. Stebe w21, 22x first considered the conjugacy separability of generalized free products of free groups and that of Fuchsian groups. The former was completed by Dyer w6x, who showed that generalized free products of two free groups or finitely generated nilpotent groups amalgamating a cyclic subgroup are conjugacy separable. The latter was completed by Fine and Rosenberger w7x using some results of Stebe w22x and Allenby and Tang w3x. This naturally leads us to ask whether generalized free products of finitely generated Fuchsian groups amalgamating a cyclic subgroup are conjugacy separable. These groups are proved to be conjugacy separable by Kim and Tang w13x. In the same paper we asked whether generalized free products of polycyclic groups amalgamating a cyclic subgroup are conjugacy separable. L. Ribes recently informed us that together with D. Segal and P. A. Zalesskii, they have proved the conjugacy separability of these groups. Thus it is natural to ask whether tree products of conjugacy separable groups amalgamating cyclic subgroups are conjugacy separable. In this paper we show that tree products of a finite number of conjugacy separable residually nilpotent groups are conjugacy separable. The main difficulty in the proof lies in proving the double coset separability of these groups. Other results on conjugacy separability and double coset separability can be found in Allenby w1x, Shirvani w20x, Niblo w17x, and Gitik and Rips w9x. In Section 2 we prove one of our main results, a criterion for the generalized free products of two conjugacy separable groups amalgamating a cyclic subgroup to be conjugacy separable; this is Theorem 2.3. The main body of our proof is in Section 3, where we prove that tree products of
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residually nilpotent groups amalgamating cyclic subgroups are double coset separable with respect to the subgroups ² h: and ² k :, where h and k are elements of some vertex groups. This involves a detailed analysis of the various cases. Finally, in Section 4 we prove our main result, Theorem 4.3, that tree products of a finite number of conjugacy separable residually nilpotent groups, amalgamating cyclic subgroups, are conjugacy separable. From this it follows that tree products of a finite number of free groups, finitely generated nilpotent groups, and surface groups, amalgamating cyclic subgroups, are conjugacy separable. At this point it is natural to ask whether tree products of Fuchsian groups amalgamating cyclic subgroups are conjugacy separable. In view of Ribes, Segal, and Zalesskii’s result we also ask whether tree products of polycyclic groups amalgamating cyclic subgroups are conjugacy separable. Throughout this paper we use standard notations and terminology. For convenience we list the following: The letter G always denotes a group. x 4G denotes the set of all conjugates of x in G. x ;G y means x, y are conjugate in G. If x g G s A)H B, then 5 x 5 denotes the free product length of x in G. N 1f G means N is a normal subgroup of finite index in G. A group G is pc if, for every cyclic subgroup H of G and every x g G _ H, there exists N 1f G such that in G s GrN, x f H. A group G is conjugacy separable if, for each pair of nonconjugate elements x, y g G, there exists N 1f G such that in G s GrN, x ¤G y. CH Ž x . is the centralizer of x g G in the subgroup H of G. Let H s ² h: be a infinite cyclic subgroup of G. If, for every positive integer n, there exists Nn 1f G such that Nn l H s ² h n :, then G is said to be H-potent. A torsion-free group G is said to be potent if it is ² h:-potent for all 1 / h g G. Let P be a tree. Assign a group G¨ to each vertex ¨ and a group Ge to each edge e of P. Let a e and b e be monomorphisms which embed Ge as a subgroup of the two vertex groups at the ends of the edge e. The tree product of P is defined to be the group generated by the generators and relations of the vertex groups together with the extra relations obtained by identifying g e a e and g e b e for each g e g Ge . We use N to denote the class of finitely generated torsion-free nilpotent groups and R N to denote the class of residually N-groups. We shall make extensive use of the following results. THEOREM 1.1 w15, Theorem 4.6x. Let G s A)H B and let x g G be of minimal length in its conjugacy class. Suppose that y g G is cyclically reduced and that x ;G y.
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Ž1. If 5 x 5 s 0, then 5 y 5 F 1 and, if y g A, then there is a sequence h1 , h 2 , . . . , h r of elements in H such that y ;A h1 ;B h 2 ;A ??? ; h r s x. Ž2. If 5 x 5 s 1, then 5 y 5 s 1 and either x, y g A and x ;A y or x, y g B and x ;B y. Ž3. If 5 x 5 G 2, then 5 x 5 s 5 y 5 and y ;H xU , where xU is a cyclic permutation of x. THEOREM 1.2 w6, Theorem 4x. If A and B are conjugacy separable and H is finite, then A)H B is conjugacy separable. 2. A CRITERION In this section we prove a criterion for the conjugacy separability of generalized free products of conjugacy separable groups amalgamating a cyclic subgroup. LEMMA 2.1. Let G be ² h e : x ² k e :-separable, where h, k g G are of infinite order. If ² xy1 hx : l ² k : s 1, then there exists N 1f G such that xy1 h i x s k j only if e < i, j, where G s GrN. Proof. Note hyi 1 xkyj 1 f ² h e : x ² k e : for 0 F i1 , j1 - e except i1 s j1 s 0. Hence there exists N 1f G such that hyi 1 xkyj 1 f N² h e : x ² k e : for 0 F i1 , j1 - e except i1 s j1 s 0. Therefore, if xy1 h i x s k j, then e < i, j, where G s GrN. Note that the condition xy1 h i x s k j only if e < i, j always implies e < < h <, < k <. DEFINITION 2.2. A group G is cyclic conjugacy separable for ² h: if, for each x g G such that x 4G l ² h: s B, there exists N 1f G such that x 4G l ² h: s B, where G s GrN. If G is cyclic conjugacy separable for every cyclic subgroup ² h: of G, then G is called cyclic conjugacy separable. THEOREM 2.3. Let G s G1 )² h: G 2 . Suppose that G1 , G 2 are conjugacy separable, cyclic conjugacy separable for ² h: and ² h:-potent. If each Gk Ž k s 1, 2. satisfies: C1. if h i ;G k h j, then i s j; C2. for e¨ ery e ) 0, Gk is ² h e : x ² h e :-separable for x g Gk ; C3. for e¨ ery n ) 0, there exists M 1f Gk such that < h < s n and h i ¤G k h j for h i / h j, where Gk s GkrM, then G is conjugacy separable. Proof. Let x, y g G such that x ¤G y. Without loss of generality we can assume that x and y are of minimal lengths in their conjugacy classes in G. Since G1 , G 2 are ² h:-potent and ² h:-separable by C2, G is residually finite by w2x. Hence, we may assume x / 1 / y. To prove our result, we
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shall find M 1f G1 and L 1f G 2 such that M l ² h: s L l ² h: and x ¤G y, where G s G1rM )² h: G 2rL. Then, by Theorem 1.2, G is conjugacy separable, whence there exists K 1f G such that Kx ¤G r K Ky. Let K be the preimage of K in G. Then we have K 1f G and Kx ¤G r K Ky, as required. Case 1. 5 x 5 s 0 s 5 y 5, say, x s h t and y s h s. Clearly s / t. Ži. < t < / < s <. Since G1 , G 2 are potent, there exist M 1f G1 and L 1f G 2 such that M l ² h: s ² h st : s L l ² h:. Then, in G s G1rM )² h: G 2rL, we have < x < s < s < / < t < s < y <, hence x ¤G y. Žii. t s ys, where t ) 0. By C3, there exists M 1f G1 such that < h < s 3t and h i ¤G 1 h j for h i / h j, where G1 s G1rM. Similarly, there exists L 1f G 2 such that < h < s 3t and h i ¤G 2 h j for h i / h j, where G 2 s G 2rL. Then, in G s G1rM )² h: G 2rL, if h t ;G hyt then, by Theorem 1.1, h t s hyt . Since < h < s 3t, it follows that x ¤G y. Case 2. 5 x 5 s 0 and 5 y 5 s 1 Žor 5 y 5 s 0 and 5 x 5 s 1., say, y g G1 _ ² h: and x g ² h:. Since y is of minimal length 1 in its conjugacy class, we have y4G 1 l ² h: s B. Moreover, since G1 is cyclic conjugacy separable for ² h:, there exists M 1f G1 such that y4G 1 l ² h: s B, where G1 s G1rM. Now, by the potency of G 2 , there exists L 1f G 2 such that M l ² h: s L l ² h:. Let G s G1rM )² h: G 2rL. Then, by Theorem 1.1, y is of minimal length 1 in its conjugacy class in G. Hence x ¤G y, as required. Case 3. 5 x 5 / 5 y 5 and 5 x 5 G 2 Žor 5 y 5 G 2.. Since x is of minimal length in its conjugacy class, it is cyclically reduced. Let x s a1 b1 ??? a n bn Žsay, other cases being similar., where a i g G1 _ ² h: and bi g G 2 _ ² h:. Since G1 , G 2 are ² h:-separable Žby C2. and ² h:-potent, we can find M 1f G1 and L 1f G 2 such that M l ² h: s L l ² h: and 5 x 5 s 5 x 5, 5 y 5 s 5 y 5, where G s G1rM )² h: G 2rL. Thus x is cyclically reduced and is of minimal length 2 n in its conjugacy class. Therefore 5 x 5 / 5 y 5. Hence, by Theorem 1.1, x ¤G y, as required. Case 4. 5 x 5 s 1 s 5 y 5. Ži. x, y g G1 _ ² h: Žor x, y g G 2 _ ² h:.. Since x, y are of minimal length 1 in their conjugacy classes, x 4G l ² h: s B and y4G l ² h: s B. Now G1 is conjugacy separable and cyclic conjugacy separable for ² h:. Therefore, there exists M 1f G1 such that x ¤G 1 y, x 4G 1 l ² h: s B, and y4G 1 l ² h: s B, where G1 s G1rM. Since G 2 is potent, we can choose L 1f G 2 such that M l ² h: s L l ² h:. Thus, in G s G1rM )² h: G 2rL, x, y are of minimal length 1 in their conjugacy classes, and x ¤G 1 y. Hence, by Theorem 1.1, x ¤G y, as required. Žii. Suppose x g G1 _ ² h: and y g G 2 _ ² h: Žor x g G 2 _ ² h: and y g G1 _ ² h:.. As in Ži. above, there exist M 1f G1 and L 1f G 2 such
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that M l ² h: s L l ² h:, x 4G 1 l ² h: s B, and y4G 2 l ² h: s B, where G1 s G1rM and G 2 s G 2rL. Let G s G1rM )² h: G 2rL. Then x, y are of minimal length 1 in their conjugacy classes with x g G1 and y g G 2 . Hence, by Theorem 1.1, x ¤G y, as required. Case 5. 5 x 5 s 5 y 5 s r G 2. Let x s u1 u 2 ??? u r and y s ¨ 1¨ 2 ??? ¨ r be cyclically reduced. Since x ¤G y, by Theorem 1.1, x ¤² h: yU for any cyclic permutation yU of y. Thus, for each i, the equation I Ž i . : u1 u 2 ??? u r s hya i ¨ i ??? ¨ r¨ 1 ??? ¨ iy1 h a i
Ž 1.
has no integer solution a i . Hence, for each i, we shall find MiX 1f G1 and LXi 1f G 2 such that MiX l ² h : s LXi l ² h :. Moreover, in G s G1rMiX )² h: G 2rLXi , 5 x 5 s 5 x 5, 5 y 5 s 5 y 5, and the equation I Ž i . : u1 u 2 ??? u r s hya i¨ i ??? ¨ r¨ 1 ??? ¨ iy1 h a i
Ž 2.
r r has no integer solution a i . Let M s F is1 MiX , L s F is1 LXi , and G s G1rM )² h: G 2rL. Then x and y are cyclically reduced and x ¤² h: yU for all cyclic permutation yU of y. Therefore, by Theorem 1.1, we have x ¤G y, as required. Since all the cases are similar, we shall only consider the case i s 1. Since equation I Ž1. has no solution a 1 , we have x s u1 u 2 ??? u r ¤² h: ¨ 1¨ 2 ??? ¨ r s y. If u i and ¨ i are not in the same factor of G, then we can easily find G s G1rM1X )² h: G 2rLX1 such that 5 x 5 s 5 x 5 and 5 y 5 s 5 y 5. Clearly equation I Ž 1 . has no solution, since u i and ¨ i are not in the same factor of G. Thus we need only consider the case x s c1 d1 ??? c n d n and y s a1 b1 ??? a n bn , where a i , c i g G1 _ ² h: and bi , d i g G 2 _ ² h:. We proceed by finding M 1f G1 and L 1f G 2 such that M l ² h: s L l ² h: so that, in G s G1rM )² h: G 2rL, 5 x 5 s 2 n s 5 y 5. Moreover, equation I Ž 1 . has no integer solution a 1 in G, whence x ¤² h: y. Since G1 , G 2 are ² h:-separable and ² h:-potent, there exist M1 1f G1 and L1 1f G 2 such that M1 l ² h: s L1 l ² h:, a i , c i f M1² h:, and bi , d i f L1² h: for all i.
Ž1. There exists i such that c i f ² h: a i ² h: Žor d i f ² h: bi ² h:.. By C2, there exists M2 1f G1 such that c i f M2 ² h: a i ² h:. Now, by the potency of G 2 , we can choose L2 1f G 2 such that M2 l ² h: s L2 l ² h:. Let M s M1 l M2 and L s L1 l L2 . Then, in G s G1rM )² h: G 2rL, x and y are cyclically reduced and c1 d1 ??? c n d nf ² h:a1 b1 ??? a n bn² h:. Hence x ¤² h: y, as required. Ž2. There exists i such that c1 d1 ??? c i s hya a1 b1 ??? a i h b , d i s h l bi h d , and that c1 d1 ??? c i d i f ² h: a1 b1 ??? a i bi ² h:. This implies a1 b1 ??? a i h bq l bi f ² h: a1 b1 ??? a i bi ² h:. For convenience, we write w s a1 b1 ??? a i . Suppose C² h:Ž w . s ² h t : and C² h:Ž bi . s ² h s :, where s, t G 0.
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Ža. t / 0 / s. Then clearly h bq l f ² h t :² h s : s ² h d :, where d s j ² : gcdŽ t, s .. Now wy1 h j w f ² h: for 1 F j - t and by1 i h bi f h for 1 F j - s by C1. Since G1 , G 2 are ² h:-separable and ² h:-potent, G is ² h:-separable w2x. Thus there exists N 1f G such that wy1 h j w f N² h: for all j ² : 1 F j - t and by1 i h bi f N h for all 1 F j - s. Since G 1 , G 2 are potent, there exist M2 1f G1 and L2 1f G 2 such that M2 l ² h: s ² h d : s L2 l ² h:. Moreover h bq l f ² h d : implies h bq l f M2 ² h d :. Let M s M1 l M2 l N and L s L1 l L2 l N. Then, in G s G1rM )² h: G 2rL, x and y are cyclically reduced and a1 b1 ??? a i h bq l bif ² h:a1 b1 ??? a i bi² h:. For, if not, wh bq l bi s hy m 0 wbih m i for some m 0 , m i . Then w s hy m 0 wh l1 and h bq l bi s hyl 1 bi h m i for some l1. By the choice of N, t < m 0 , whence h m 0 s h l1 g ² h t :, and s < b q l q l1. Hence h bq l g ² h t :² h s : s ² h d :, contradicting the choice of M2 . This shows that a1 b1 ??? a i h bq l bif ² h:a1 b1 ??? a i bi² h:. Thus c1 d1 ??? c i d if ² h:a1 b1 ??? a i bi² h:. Hence x ¤² h: y, as required. Žb. t / 0 and s s 0. Then clearly h bq l f ² h t :. As in Ža., there exists N 1f G such that wy1 h j w f N² h: for 1 F j - t and there exists M2 1f G1 such that h bq l f M2 ² h t :. Because of C1, C² h:Ž bi . s 1 implies ² by1 : ² : i hbi l h s 1. Hence, by Lemma 2.1, for e s t, there exists M3 1f k < G1 such that if M3 h j s M3 by1 i h bi , then e s t k, j. As before, we can find M 1f G1 and L 1f G 2 such that M l ² h: s L l ² h:, M ; M1 l M2 l M3 l N, and L ; L1 l N. Then, in G s G1rM )² h: G 2rL, x and y are cyclically reduced, and a1 b1 ??? a i h bq l bif ² h:a1 b1 ??? a i bi² h:. For, if not, wh bq l bi s hy m 0 wbih m i for some m 0 , m i . This implies w s hy m 0 wh l1 and h bq l bi s hy l1 bi h m i for some l1. By the choice of N, we have t < m 0 , whence h m 0 s h l1 g ² h t :, and e s t < b q l q l1. Hence h bq l g ² h t :, contradicting the choice of M2 . This shows that a1 b1 ??? a i h bq l bif ² h:a1 b1 ??? a i bi² h:. Therefore, c1 d1 ??? c i d if ² h:a1 b1 ??? a i bi² h:. Hence x ¤² h: y, as required. Žc. t s 0 and s / 0. Then clearly h bq l f ² h s :. Since C² h:Ž w . s 1, : there exists a k Žor bk . with k F i such that ² h: l ² ay1 k ha k s 1 by C1. Thus, by Lemma 2.1, there exists M2 1f G1 such that if M2 h r s j < ² h s :-separable, we can find M2 ay1 k h a k , then e s s r, j. Since G 1 is bq l s: ² M3 1f G1 such that h f M3 h . Also we can find N 1f G such j ² : for 1 F j - s. Let M 1f G1 and L 1f G 2 such that by1 i h bi f N h that M l ² h: s L l ² h:, M ; M1 l M2 l M3 l N, and L ; L1 l N. Then, in G s G1rM )² h: G 2rL, x and y are cyclically reduced and a1 b1 ??? a i h bq l bif ² h:a1 b1 ??? a i bi² h:. For, if a1 b1 ??? a i h bq l bis h m 0 a1 b1 ??? a i bi h m i for some m 0 , m i , then a1 s h m 0 a1 h l1 , b1 s hyl 1 b1 h m 1 , . . . , a k s hy m ky 1 a k h l k , . . . , a i s hy m iy 1 a i h l i , and h bq l bi s hyl i bi h m i for some l j , m j . Thus, by the choice of M2 , e s s < m ky1 , l k . Since e 5 h <, e < l i . Now, by the choice of N, s < b q l q l i . Hence e s s < b q l, thus h bq l g ² h s :, a contradiction. This shows that a1 b1 ??? a i h bq l bif
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² h:a1 b1 ??? a i bi² h:. Therefore, c1 d1 ??? c i d if ² h:a1 b1 ??? a i bi² h:. Hence x ¤² h: y, as required. Žd. s s 0 and t s 0. Then clearly we have ² h: l ² by1 : i hbi s 1. Moreover, since C² h:Ž w . s 1, there exists a k Žor bk . with k F i such that ² h: l ² ay1 : < < k ha k s 1. By Lemma 2.1, for e s 2 b q l , there exist M2 1f j G1 and L2 1f G 2 such that if M2 h i s M2 ay1 h a , then e < i, j, and that if k k k < L2 h j s L2 by1 h b , then e j, k. As before, we can find M3 1f G1 such i i that h bq l f M3 ² h 2Ž bq l . :. Again, as before, we can find M 1f G1 and L 1f G 2 such that M l ² h: s L l ² h:, M ; M1 l M2 l M3 , and L ; L1 l L2 . Thus, in G s G1rM )² h: G 2rL, x and y are cyclically reduced, and a1 b1 ??? a i h bq l bif ² h:a1 b1 ??? a i bi² h:. For, if a1 b1 ??? a i h bq l bis h m 0 a1 b1 ??? a i bi h m i for some m 0 , m i , then a1 s h m 0 a1 h l1 , b1 s hyl 1 b1 h m 1 , . . . , a k s hy m ky 1 a k h l k , . . . , a i s hy m iy 1 a i h l i , and h bq l bi s hyl i bi h m i for some l j , m j . Then, by the choice of M2 , e < m ky1 , l k . Since e 5 h <, e < l i . Again, by the choice of L2 , e < b q l q l i . Hence e < b q l. This means h bq l g ² h 2Ž b q l . :, a contradiction. This shows that a1 b1 ??? a i h bq l bi f ² h: a1 b1 ??? a i bi ² h:. Therefore, c1 d1 ??? c i d i f ² h:a1 b1 ??? a i bi² h:. Hence x ¤² h: y, as required. Ž3. c1 d1 ??? c n d n s h a a1 b1 ??? a n bn h b and h aq b / 1. Choose M2 1f G1 such that h aq b f M2 . Let M1 l ² h: s ² h s : s L1 l ² h: and M2 l ² h: s ² h t :. By C3, there exist M3 1f G1 and L3 1f G 2 such that M3 l ² h: s ² h st : s L3 l ² h: and M3 h i ¤G 1 r M 3 M3 h j for M3 h i / M3 h j and L3 h i ¤G 2 r L 3 L3 h j for L3 h i / L3 h j. Let M s M1 l M2 l M3 , L s L1 l L3 , and G s G1rM )² h: G 2rL. Then x and y are cyclically reduced and x ¤² h: y. For, if x ;² h: y, then x s h a yh b s hy l yh l for some l. This im plies th at h a a 1 s h y l a 1 h l 1 , b 1 s h y l 1 b 1 h m 1 , . . . , a k s hym ky 1 a k h l k , . . . , a n s hy m ny 1 a n h l n , and bn h b s hy l n bn h l for some l j , m j . Thus, by the choice of M3 , L3 , we have h aq l s h l1 , h l1 s h m 1 , . . . , h m ky 1 s h lk , . . . , h m ny 1 s h l n , and h l n s h ly b . This means h aq l s h ly b , whence h aq b s 1, contradicting the choice of M2 . Hence x ¤² h: y, as required. This completes our proof that there exist M1X 1f G1 and LX1 1f G 2 such X ˜ s G1rM1X )² h: that equation I Ž 1 . has no solution in G ˜ G 2rL1 . Similarly, X X for each i, we can find Mi 1f G1 and L i 1f G 2 such that equation I Ž i . r ˆ s G1rMiX )² h: G2rLXi . Let M s F is1 has no solution in G MiX and L s X r F is1 L i . Then M 1f G1 , L 1f G 2 , and, in G s G1rM )² h: G 2rL, I Ž i . has no solution for each i. Hence x ¤² h: yU for any cyclic permutation yU of y. Thus it follows from Theorem 1.1 that x ¤G y, as required. It is easy to see that R N-groups are potent and they satisfy C1, C2, and C3 Žsee Lemmas 3.2 and 3.4 and w23, Corollary 2.2x.. Since, by w6x and w23x,
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free groups, finitely generated nilpotent groups and surface groups are cyclic conjugacy separable, we have: COROLLARY 2.4 w6, 23x. Let Gi Ž i s 1, 2. be a finitely generated torsionfree nilpotent or free or surface group. Then G s G1 )² c: G 2 is conjugacy separable. A generalized free product of two free groups amalgamating an infinite cyclic subgroup is called a cyclically pinched one-relator group. Hence Corollary 2.4 implies in particular that cyclically pinched one-relator groups are conjugacy separable. It was also known by w12x that generalized free products of two free or surface groups, amalgamating a maximal cyclic subgroup, satisfy C1 and C2 in Theorem 2.3. Thus, by Lemmas 4.1, 4.2, and 3.1, we have: COROLLARY 2.5 w12x. Let A i be free or surface groups and let ² u: and ² ¨ : be maximal cyclic subgroups of A1 , A 2 and A 3 , A 4 , respecti¨ ely, such that A1 l A 2 s ² u: and A 3 l A 4 s ² ¨ :. If A 2 l A 3 s ² c :, then A1 )² u: A 2 )² c: A 3 )² ¨ : A 4 and A1 )² u: A 2 )² c: A 3 are conjugacy separable.
3. DOUBLE COSET SEPARABILITY OF CERTAIN TREE PRODUCTS Let G be a tree product of a finite number of R N-groups, amalgamating cyclic subgroups. Suppose G has r vertices. Then, either G is a R N-group when G has one vertex or G s A)² c: B, where B is a vertex group, A is a subtree product of r y 1 vertices, and c is in a vertex of A. In the latter case, if h is in a vertex of G, then either h is in a vertex of A or h g B. Throughout this section, we shall use the above description for G and show that G is ² h: x ² k :-separable for x g G, where h, k are in the vertices of G. LEMMA 3.1. Let G be a tree product of a finite number of R N-groups amalgamating cyclic subgroups. If h is in a ¨ ertex group of G, then G is ² h:-potent. Proof. Clearly a R N-group is potent. Let G s A)² c: B as described above. By induction, we can assume that A is ² u:-potent for any u in a vertex of A. Then A, B are ² c :-potent. Hence, by Lemma 3.2 in w2x, G is ² h:-potent for any h in a vertex of G. LEMMA 3.2. Let G be a tree product of a finite number of R N-groups amalgamating cyclic subgroups. Let h be in a ¨ ertex group of G. If h i ;G h j, then i s j.
CONJUGACY SEPARABILITY
1061
Proof. Clearly a R N-group has the property. Hence we let G s A)² c: B as above and, by induction, we suppose that A has the property that u i ;A u j only if i s j for u in a vertex of A. Let h be in a vertex of G and h i ;G h j. Suppose h g A. If h i 4G l ² c : s B, then, by Theorem 1.1, h i ;A h j. Thus, by induction, i s j. Hence we assume h i ;G c n for some n. Then, by Theorem 1.1, h i ;A c e 1 ;B c e 2 ;A ??? ; c e r s c n for some e k . This implies e 1 s e 2 s ??? e r s n, since c is in B and is in a vertex of A. Thus h i ;A c n. Similarly h j ;A c n, since h j ;G h i ;A c n. It follows that h i ;A h j. Since h is in a vertex of A, by induction i s j. The case of h g B is similar. LEMMA 3.3. Let G be a tree product of a finite number of R N-groups amalgamating cyclic subgroups. Then G is pc . Proof. A R N-group is pc by w13, Lemma 2.6x. So let G s A)² c: B as above. By induction, we assume that A is pc . Since A, B are ² c :-potent, G is pc by Corollary 2.2 in w11x. LEMMA 3.4. separable.
Let G be a R N-group. Let h, k g G. Then G is ² h:² k :-
Proof. Let g f ² h:² k :, where g, h, k g G. Case 1. ² h: l ² k : s ² h n : / 1, where h n s k m and n ) 0. Then h gkyj f ² h n :² k m : s ² h n : for all 0 F i - n and 0 F j - < m <. Since G is pc by Lemma 3.3, there exists N 1f G such that hyi gkyj f N² h n : for all 0 F i - n and 0 F j - < m <. Then clearly g f N² h:² k :. yi
Case 2. ² h: l ² k : s 1. First, we shall find N1 1 G such that h, k f N1 and ² N1 h: l ² N1 k : s N1 with GrN1 g N. Since G is R N, there exists M1 1 G such that h, k f M1 and GrM1 g N. If ² M1 h: l ² M1 k : s M1 , then let N1 s M1. Otherwise, let M1 h s s M1 k t , where h s kyt / 1. Therefore, there exists M2 1 G such that h s kyt f M2 and GrM2 g N. Let N1 s M1 l M2 . Since GrN1 can be imbedded in a direct product GrM1 = GrM2 g N, GrN1 g N. We note that ² N1 h: l ² N1 k : s N1. For, if N1 h n s N1 k m , then N1 h n s s N1 k m s. This implies M1 k nt s M1 h n s s M1 k m s. Since GrM1 g N and < M1 k < s `, nt s ms. This means N1 h n s s N1 k m s s N1 k nt. Now GrN1 g N has the unique root property. Therefore N1 h s s N1 k t , contradicting the choice of M2 . Hence h, k f N1 and ² N1 h: l ² N1 k : s N1. Since GrN1 g N, GrN1 is ² N1 h:² N1 k :-separable. Therefore, if N1 g f ² N1 h:² N1 k :, then we can find N 1f G such that g f N² h:² k :. So we can assume N1 g s N1 h s N1 k t for some s, t. Since ² N1 h: l ² N1 k : s N1 , N1 g s N1 h s N1 k t is unique. Now choose N2 1 G such that gy1 h s k t f N2
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and GrN2 g N. Let M s N1 l N2 . Then, as before, GrM g N. Moreover Mg f ² Mh:² Mk :. Then we can find N 1f G such that g f N² h:² k :. LEMMA 3.5. Let G be a tree product of a finite number of R N-groups amalgamating cyclic subgroups. Then G is ² u:² ¨ :-separable for u, ¨ in ¨ ertices of G. Proof. By Lemma 3.4, the result holds for tree products with one vertex. Hence we assume that G s A)² c: B, where c is in a vertex of the subtree product A and B is a R N-group. By induction, we can assume that A is ² h:² k :-separable for h, k in vertices of A. Let g f ² u:² ¨ :, where g g G and u, ¨ are elements of vertices of G. We shall find M 1f A and L 1f B such that M l ² c : s L l ² c : and g f ² u:² ¨ :, where G s ArM )² c: BrL. Since G is residually finite and <² u:² ¨ :< - `, G is ² u:² ¨ :-separable. Therefore, there exists N 1f G such that g f N² u:² ¨ :. Let N be the preimage of N. Then N 1f G and g f N² u:² ¨ :. Case 1. u, ¨ g A Žor u, ¨ g B . Ž1. g g A. By the induction hypothesis, there exists M 1f A such that g f ² u:² ¨ :, where A s ArM. Since B is potent, there exists L 1f B such that M l ² c : s L l ² c :. Then, in G s ArM )² c: BrL, g f ² u:² ¨ :. Ž2. g f A. By Lemmas 3.3 and 3.1, A, B are pc and ² c :-potent. Therefore, there exist M 1f A and L 1f B such that M l ² c : s L l ² c : and 5 g 5 s 5 g 5, where G s ArM )² c: BrL. Clearly g f ² u:² ¨ :. Case 2. u g A and ¨ g B Žor u g B and ¨ g A.. If ² u: l ² c : s ² c s : / 1 Žsimilarly, if ² ¨ : l ² c : / 1., say, c s s u t , then u i g f ² u t :² ¨ : s ² c s :² ¨ : for 0 F i - < t <. Since c s, ¨ g B, by Case 1, there exists N 1f G such that u i g f N² c s :² ¨ : s N² u t :² ¨ : for 0 F i - < t <. It follows that g f N² u:² ¨ :. So we assume that ² u: l ² c : s 1 s ² ¨ : l ² c :. Ž1. g g ² c :. Let g s c a f ² u:² ¨ :. Since ² u: l ² c : s 1 s ² ¨ : l ² c :, i c f ² u:² c 2 a : and c i f ² c 2 a :² ¨ : for all 1 F i - 2 < a <. By induction, A is ² u:² c 2 a :-separable. Also, by Lemma 3.4, B is ² c 2 a :² ¨ :-separable. Therefore, there exist M 1f A and L 1f B such that c a f M² c 2 a :, c i f M² u:² c 2 a :, and c i f L² c 2 a :² ¨ : for all 1 F i - 2 < a <. Since A, B are ² c :-potent, we can also assume M l ² c : s L l ² c :. Then, in G s ArM )² c: BrL, g s c a f ² u:² ¨ :. For, if c a s u i ¨ j, then u i s c d 1 and ¨ j s c d 2 . Thus, from the choice of M, L, 2 a < d 1 , d 2 . This implies c a s c d 1 c d 2 g ² c 2 a :, contradicting the choice of M. Hence, in G s ArM )² c: BrL, g s c a f ² u:² ¨ :. Ž2. g g A _ ² c : Žor g g B _ ² c :.. If g f ² u:² c :, then, by induction, there exists M 1f A such that g f M² u:² c :. Since B is ² c :-potent, there exists L 1f B such that L l ² c : s M l ² c :. Then g f ² u:² ¨ : in G s ArM )² c: BrL. If g s u i c a , then we have c a f ² u:² ¨ :. Thus, as in Ž1.,
CONJUGACY SEPARABILITY
1063
there exist M 1f A and L 1f B such that c a f ² u:² ¨ : in G s ArM )² c: BrL. Hence g f ² u:² ¨ :. Ž3. g s ab, where a g A _ ² c : and b g B _ ² c :. If a f ² u:² c : Žor, similarly, b f ² c :² ¨ :., then there exists M 1f A and L 1f B with M l ² c : s L l ² c : such that a f ² u:² c : and b g B _ ² c : in G s ArM )² c: BrL. Then g f ² u:² ¨ :. So assume a s u i c a and b s c b ¨ j. Since g s ab f ² u:² ¨ :, c aq b f ² u:² ¨ :. As in Ž1., there exist M 1f A and L 1f B such that c aq b f ² u:² ¨ : in G s ArM )² c: BrL. Hence g f ² u:² ¨ :. Ž4. g s ba or 5 g 5 G 3, where b g B _ ² c : and a g A _ ² c :. As in Case 1, we can find M 1f A and L 1f B such that 5 g 5 s 5 g 5, where G s ArM )² c: BrL. Then g f ² u:² ¨ :. Recall that a tree product of r vertices of R N-groups Ž r ) 1., amalgamating cyclic subgroups, can be written as G s A)² c: B, where A is a subtree product of r y 1 vertices, B is a R N-group, and c is in a vertex of A. For the rest of this section, we shall show that G is ² h: x ² k :-separable for x g G and h, k in some vertices of G. Applying induction, we can assume that A is ² u: a² ¨ :-separable for a g A whenever u, ¨ are in some vertices of A. LEMMA 3.6. Let G, A, B, c be as abo¨ e and h, k be in the ¨ ertices of G. We can find M 1f A and L 1f B with M l ² c : s L l ² c : such that, in G s ArM )² c: BrL, Ž1. if xc a f ² h: x ² k :, where x, h g A, k g B, and ² k : l ² c : s 1, then xc af ² h: x ² k :; Ž2. if xc a f ² h: x ² k :, where x, h g B, k g A, and ² k : l ² c : s 1, then xc af ² h: x ² k :. Proof. Ž1. Case 1. There exists a minimal integer m ) 0 such that xy1 h m x s c s, that is, h m x s xc s. Clearly c a f ² c s :² k :. By Lemma 3.5, we can find G s ArM )² c: BrL with M l ² c : s L l ² c : such that c a f ² c s :² k :, xy1 h i xf ² c : for all 0 - i - m. If xc as h i xk j for some i, j, then k j s c d for some d . Thus xy1 h i x s c ay d . This implies m < i. It follows that c ay d g ² c s :, thus c a g ² c s :² k :, a contradiction. Therefore, xc af ² h: x ² k :. Case 2. ² xy1 hx : l ² c : s 1. Since a / 0 and ² k : l ² c : s 1, c a f ² c 2 a :² k :. By induction, A is ² h e : x ² c e :-separable. Therefore, by Lemmas 2.1 and 3.5, we can find G s ArM )² c: BrL such that xy1 h i1 xs c i 2 only if e s 2 a < i1 , i 2 and c a f ² c 2 a :² k :. If xc as h i xk j for some i, j, then k j s c d for some d . Thus xy1 h i x s c ay d . This implies e < i, a y d , whence c ay d g ² c 2 a :. This means c a g ² c 2 a :² k :, a contradiction. Therefore, xc af ² h: x ² k :. Ž2. is similar to Ž1..
1064 LEMMA 3.7.
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Let G, A, B, c be as in Lemma 3.6 and let h, k be in the
¨ ertices of G. Then G is ² h: x ² k :-separable for x g A j B.
Proof. We only consider x g A, the case x g B being similar. Let g g G and g f ² h: x ² k :, where h, k are in the vertices of G. We shall find G s ArM )² c: BrL such that g f ² h: x ² k :, where M 1f A, L 1f B, and M l ² c : s L l ² c :. Then, since G is residually finite and < h <, < k < - `, ˜ s GrN. we can easily find N 1f G such that ˜ g f ²˜ h: ˜ x² ˜ k :, where G Case 1. h, k g A. If g g A then, by induction, there exists M 1f A such that g f ² h: x ² k : in A s ArM. Since B is ² c :-potent, there exists L 1f B such that M l ² c : s L l ² c :. Then, in G s ArM )² c: BrL, we have g f ² h: x ² k :. If g g B _ ² c : or 5 g 5 G 2, then we can find G s ArM )² c: BrL such that 5 g 5 s 5 g 5. Therefore g f ² h: x ² k :. Case 2. h g A and k g B or h g B and k g A. Since g f ² h: x ² k : is equivalent to gy1 f ² k : xy1 ² h:, it suffices to consider h g A and k g B. If ² k : l ² c : s ² k t : / 1, where k t s c s, then gkyi f ² h: x ² k t : s ² h: x ² c s : for 0 F i - t. Thus, by Case 1, we can find G such that gkyi f ² h: x ² c s : for 0 F i - t. This implies g f ² h: x ² k :. Therefore we may assume ² k : l ² c : s 1. Ža. g g A. If g f ² h: x ² c :, then, by Case 1, we can find G such that g f ² h: x ² c :. Since k g B, this implies g f ² h: x ² k :. So let g s h i xc a for some i, a . Then xc a f ² h: x ² k :. By Lemma 3.6, we can find G such a that xc f ² h: x ² k :. Hence g f ² h: x ² k :. Žb. g g B _ ² c :. If x f ² h:² c :, then, by Lemma 3.5, we can find G such that x f ² h:² c : and g f ² c :. This implies g f ² h: x ² k :. So let x s h i c a for some i, a . Then g f ² h: c a ² k :. This means gcya f ² h:² c a kcy a :. Since c a kcy a g B, by Lemma 3.5, we can find G such that gcya f ² h:²c a kcy a:. Hence g f ² h: x ² k :. Žc. g s a1 b1 , where a1 g A _ ² c : and b1 g B _ ² c :. If a1 f ² h: x ² c : or b1 f ² c :² k :, then, by induction and Lemma 3.5, we can find G such that 5 g 5 s 5 g 5 and a1 f ² h: x ² c : or b1 f ² c :² k :. Since k g B, we have g f ² h: x ² k :. So let a1 s h i xc a and b1 s c b k j for some i, j, a , b . This implies xc aq b f ² h: x ² k :. Thus, by Lemma 3.6, we can find G such that xcaq b f ² h: x ² k :. Hence g f ² h: x ² k :. Žd. g s b1 a1 or 5 g 5 G 3, where a1 g A _ ² c : and b1 g B _ ² c :. We can find G such that 5 g 5 s 5 g 5. Then g f ² h: x ² k :. Case 3. h, k g B. In this case, since G is ² h:² c e kcy e :-separable by Lemma 3.5, which is equivalent to ² h: c e ² k :-separable, we may assume x g A _ ² c :. Suppose ² h: l ² c : s ² h m :, where m ) 0. Let h m s c s. Then hyi g f ² h m : x ² k : s ² c s : x ² k : for all 0 F i - m. Thus, by Case 2, we can find G such that hyi gf ²c s: x ² k : for all 0 F i - m. This means ˜g f ² h: x ² k :. Therefore, we can assume ² h: l ² c : s 1 and similarly ² k : l ² c : s 1.
CONJUGACY SEPARABILITY
1065
Ža. g g A. Suppose g f ² c : x ² c :. Then, by Case 1, we can find G such that g f ² c : x ² c :. This means g f ² h: x ² k :. So let g s c a xc b . Moreover, we can assume a / 0 Žor b / 0.. Ži. C² c:Ž x . s 1. By Lemma 3.2, ² xy1 cx : l ² c : s 1. Since, by Case 1, G is ² c e : x ² c e :-separable, we can find G such that xy1 c i1 x s c i 2 only if e s 2 a < i1 , i 2 ; x f ² c : and c af ² h:² c 2 a : by Lemmas 2.1 and 3.5. Since h,k g B and x g A _ ² c :, if g s c a xc bs h i xk j for some i, j, then h i s c d 1 b and k j s c d 2 . Hence c a xc s c d 1 xcd 2 . This implies e < a y d 1 , b y d 2 . Thus c ay d 1 g ² c e :. This means c a s ² h:² c e :, a contradiction. Therefore g s c a xc bf ² h: x ² k :. Žii. C² c:Ž x . s ² c t : / 1. If c a f ² h:² c t : Žor similarly c a f ² c t :² k :., then we can find G such that x f ² c :, xy1 c i x f ² c : for all 0 - i - t, and c af ² h:² c t : by Lemma 3.5. If g s c a xc bs h i xk j for some i, j, then h i s c d 1 and k j s c d 2 . This implies t < a y d 1. Thus c a g ² h:² c t :, a contradiction. Hence g s c a xc bf ² h: x ² k :. Therefore we can assume c a g ² h:² c t : l ² c t :² k :. Since ² h: l ² c : s 1, c a s c n1 t. Similarly, c b s c n 2 t. This means a q b / 0, since c t g C² c:Ž x .. Thus we can find G such that x f ² c :, c aq b f ² c e : for e s 2Ž a q b ., xy1 c i x f ² c : for all 0 - i - t, h i1 s c i 2 only if e < i1 , i 2 , and k j1 s c j 2 only if e < j1 , j2 by Lemmas 3.3 and 2.1. If g s c a xc bs h i xk j for some i, j, then h i s c d 1 and k j s c d 2 . Thus e < d 1 , d 2 . Since t < a y d 1 , this implies c aq b s c d 1q d 2 g ² c e :, a contradiction. Therefore g s c a xc bf ² h: x ² k :. Žb. g g B. Since g f ² h: x ² k :, we have x f ² h: g ² k :, where g, h, k g B. Therefore, as in Case 1, we can find G such that x f ² h: g ² k :. Hence g f ² h: x ² k :. Žc. g s b1 a1 or g s a1 b 2 or g s b1 a1 b 2 , where a1 g A _ ² c : and bi g B _ ² c :. We consider only the case g s b1 a1 b 2 , other cases being similar. If b1 f ² h:² c : or b 2 f ² c :² k :, then, by Lemma 3.5, we can find G such that 5 g 5 s 5 g 5, x f ² c :, and b1 f ² h:² c : or b 2 f ² c :² k :. This implies g f ² h: x ² k :. Therefore, we assume b1 s h i c a and b 2 s c b k j. Thus hyi gkyj s c a a1 c b f ² h: x ² k :. Hence, by Ža., we can find G such that c a xc bf ² h: x ² k :. Therefore g f ² h: x ² k :. Žd. g s a1 b1 a2 or 5 g 5 G 4, where a i g A _ ² c : and b1 g B _ ² c :. We choose G such that 5 g 5 s 5 g 5. Then g f ² h: x ² k :. LEMMA 3.8. Let G, A, B, c be as in Lemma 3.6. Let h, k be in ¨ ertices of G. For a i g A _ ² c : and bi g B _ ² c :, we can find M 1f A and L 1f B with M l ² c : s L l ² c : such that, in G s ArM )² c: BrL. Ž1. if a1 c a b1 ??? a n bn f ² h: a1 b1 ??? a n bn² k :, then a1 c a b1 ??? a n bnf ² h:a1 b1 ??? a n bn² k :; Ž 2 . if a 1 c a b 1 ??? b n y 1 a n f ² h : a 1 b 1 ??? b n y 1 a n² k : , then a1 c a b1 ??? bny1 a nf ² h:a1 b1 ??? bny1 a n² k :;
1066
KIM AND TANG
Ž3. if b1 c a a1 ??? bn a n f ² h: b1 a1 ??? bn a n² k :, then b1 c a a1 ??? bn a nf ² h:b1 a1 ??? bn a n² k :; Ž 4 . if b 1 c a a 1 ??? a n y 1 b n f ² h : b 1 a 1 ??? a n y 1 b n² k : , then b1 c a a1 ??? a ny1 bnf ² h:b1 a1 ??? a ny1 bn² k :. Proof. We shall only prove Ž1., since the others are similar. Let u s b1 a2 ??? bny1 a n and C² c:Ž u. s ² c t :, where t G 0. Case 1. t s 0. By Lemma 3.2, there exists br Žor a r . in u such that ² bry1 cbr : l ² c : s 1. Subcase 1. There exists a minimal integer m ) 0 such that a1y1 h m a1 s c s. Thus a1y1 h i1 a1 f ² c : for 0 - i1 - m. Clearly c a f ² c s :. Let u s b1 a2 ??? bny 1 a n . Choose M 1f A and L 1f B with M l ² c : s L l ² c : such that, in G s ArM )² c: BrL, 5a1 ubn5 s 5 a1 ubn 5, c a f ² c s :, a1y1 h i1 a1f ² c : for 0 - i1 - m, and, moreover, by Lemma 3.7, since G is ² c e : br² c e :-separable, we can also have bry1 c i 2 br s c i 3 only if e s s < i 2 , i 3 Žby Lemma 2.1.. j If a1 c a ubns h ia1 ubn k for some i, j, then a1 s h i a1 c l1 , c a b1 s cyl 1 b1 c m 1 , . . . , br s cy l r br c m r , . . . , bn s cy l n bn k j for some l j , m j . This implies m < i, whence c l1 g ² c s :. Moreover, e s s < l r , m r . Thus e < m 1 , a q l1. It follows that c aq l1 g ² c s :, which means c a g ² c s :, a contradiction. Therefore, a1 c a ubnf ² h:a1 ubn² k :. Subcase 2. ² a1y1 ha1 : l ² c : s 1. Clearly a / 0. Let e s 2 a . We can find G such that 5a1 ubn5 s 5 a1 ubn 5, c a f ² c e :, and, moreover, by Lemma 3.7, since G is ² h e : a1² c e :-separable and ² c e : br² c e :-separable, we can also have a1y1 h i1 a1s c i 2 only if e < i1 , i 2 and bry1 c i 3 br s c i 4 only if e < i 3 , i 4 Žby j Lemma 2.1.. If a1 c a ubns h ia1 ubn k for some i, j, then a1 s h ia1 c l1 , c a b1 yl 1 m1 ylr mr s c b1 c , . . . , br s c br c , . . . , bn s cy l n bn k j for some l j , m j . This implies e < i, l1 and e < l r , m r . It follows e < m 1 , a q l1. Hence e s 2 a < a , a contradiction. Therefore, a1 c a ubnf ² h:a1 ubn² k :. Case 2. C² c:Ž u. s ² c t : / 1. Subcase 1. There exists a minimal integer m ) 0 such that a1y1 hX m a1 s c sX. Ža. There exists a minimal integer mX ) 0 such that bn k m bny1 s c s . X Thus bn k i1 bny1 f ² c : for 0 - i1 - mX . Clearly c a f ² c s :Ž² c t : l ² c s :.. Let ² c t : l ² c s9 : s ² c dt :. Choose G such that 5a1 ubn5 s 5 a1 ubn 5, c a f X ² c s :² c dt :, c i1 t f ² c s : for 0 - i1 - d, uy1 c i 2 uf ² c : for 0 - i 2 - t, a1y1 h i 3 a1f ² c : for 0 - i 3 - m, and bn k i 4 bny1f ² c : for 0 - i 4 - mX . If j a1 c a ubns h ia1 ubn k for some i, j, then a1 s h ia1 c l1 , c a u s cy l1ucl n , and yln j bn s c bn k for some l1 , l n . Hence we have m < i, t < a qX l1 , and mX < j. This implies cX l1 g ² c s :, c aq l1 s c l n g ² c t :, and c l n g ² c s :. Thus c aq l1 g ² c t : l ² c s : s ² c dt :. This means c a g ² c s :² c dt :, a contradiction. Therefore, a1 c a ubnf ² h:a1 ubn² k :.
CONJUGACY SEPARABILITY
1067
Žb. ² bn kbny1 : l ² c : s 1. Then c a f ² c s :. Choose G such that 5a1 ubn5 s 5 a1 ubn 5, c a f ² c s :, uy1 c i1 uf ² c : for 0 - i1 - t, a1y1 h i 2 a1f ² c : for 0 - i 2 - m, and bn k i 3 bny1s c i 4 only if e s s < i 3 , i 4 Žby Lemmas 3.7 and 2.1.. Applying similar argument in Ža., we get a1 c a ubnf ² h:a1 ubn² k :. Subcase 2. ² a1y1 ha1 : l ² c : s 1. X X Ža. There exists a minimal integer mX ) 0 such that bn k m bny1 s c s . X X i 1 y1 a t s t Thus bX n k bn f ² c : for 0 - i1 - m . Clearly c f ² c : l ² c :. Let ² c : l ² c s : s ² c dt :. Choose G such that 5a1 ubn5 s 5 a1 ubn 5, c a f ² c dt :, c i1 t X f ² c s : for 0 - i1 - d, uy1 c i 2 uf ² c : for 0 - i 2 - t, a1y1 h i 3 a1s c i 4 only if e s dt < i 3 , i 4 , and bn k i 5 bny1f ² c : for 0 - i 5 - mX . Then, as before, we get a1 c a ubnf ² h:a1 ubn² k :. Žb. ² bn kbny1 : l ² c : s 1. Clearly c a f ² c 2 a :. Choose G such that 5a1 ubn5 s 5 a1 ubn 5, c a f ² c 2 a :, uy1 c i1 uf ² c : for 0 - i1 - t, a1y1 h i 2 a1s c i 3 only if e s 2 a < i 2 , i 3 , and bn k i 4 bny1s c i 5 only if e s 2 a < i 4 , i 5 . Again, as before, we get a1 c a ubnf ² h:a1 ubn² k :. THEOREM 3.9. Let G be a tree product of a finite number of R N-groups amalgamating cyclic subgroups. Then G is ² h: x ² k :-separable for x g G, where h, k are in the ¨ ertices of G. Proof. If G has only one vertex, then the result follows from Lemma 3.4. So suppose G has r vertices. Then G s A)² c: B, where A is a subtree product of G consisting of r y 1 vertices with B as the remaining vertex and c in a vertex of A. By induction, we can assume that A is ² u: a² ¨ :separable for a g A and u, ¨ in the vertices of A. We shall prove the theorem by induction on 5 x 5. By Lemma 3.7, G is ² h: x ² k :-separable for x g A j B. Thus we assume that G is ² h: y ² k :-separable for any y g G with 5 y 5 - 5 x 5. Let g f ² h: x ² k :, where g, x g G and h, k are in the vertices of G. We shall find M 1f A and L 1f B with M l ² c : s L l ² c : such that, in G s ArM )² c: BrL, 5 x 5 s 5 x 5, 5 g 5 s 5 g 5, and g f ² h: x ² k :. Then, since G is residually finite and < h <, < k < - `, we can easily ˜ s GrN. Throughout the find N 1f G such that ˜ g f ²˜ h: ˜ x² ˜ k :, where G following, we let a i , c j g A _ ² c : and bi , d j g B _ ² c :. Moreover, we shall only consider the case x s a1 b1 ??? a n bn , other cases being similar. If 5 g 5 ) 5 x 5 q 2, then we can find G such that 5 g 5 s 5 g 5 and 5 x 5 s 5 x 5. Then g f ² h: x ² k :. If 5 g 5 - 5 x 5, then, by induction, we can find G such that x f ² h: g ² k :. This implies g f ² h: x ² k :. Thus we need only consider g g G with 5 x 5 F 5 g 5 F 5 x 5 q 2. Case 1. h, k g A. Ž1. 5 g 5 s 5 x 5. Suppose g s d1 c1 ??? d n c n . If a1 f ² h:² c :, then, by Lemma 3.5, we can find G s ArM )² c: BrL such that 5 g 5 s 5 g 5, 5 x 5 s 5 x 5, and a1 f ² h:² c :. Thus g f ² h: x ² k :. So let a1 s h i c a . This implies
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g f ² h: c a b1 ??? a n bn² k :. Hence, by induction, there exists G such that g f ² h:c a b1 ??? a n bn² k :. It follows that g f ² h: x ² k :. So let g s c1 d1 ??? c n d n . If c1 f ² h: a1² c : or d1 c2 ??? c n d n f ² c : b1 a2 ??? a n bn² k :, then, by induction, we can find G such that 5 g 5 s 5 g 5, 5 x 5 s 5 x 5, and c1 f ² h: a1² c : or d1 c2 ??? c n d nf ² c :b1 a2 ??? a n bn² k :. This implies g f ² h: x ² k :. Therefore, let c1 s h ia1 c a and d1 c 2 ??? c n d n s c b b1 a2 ??? a n bn k j. Then a1 c aq b b1 ??? a n bn f ² h: a1 b1 ??? a n bn² k :. Hence, by Lemma 3.8, we can find G such that a1 c aq b b 1 ??? a n bnf ² h:a1 b1 ??? a n bn² k :. It follows that g f ² h: x ² k :. Ž2. 5 g 5 s 5 x 5 q 1. Suppose g s d1 c1 ??? d n c n d nq1. Then we can find G such that 5 g 5 s 5 g 5 and 5 x 5 s 5 x 5. This implies g f ² h: x ² k :. So let g s c1 d1 ??? c n d n c nq1. If c nq1 f ² c :² k :, then we can choose G such that 5 g 5 s 5 g 5, 5 x 5 s 5 x 5, and c nq 1 f ² c :² k :. Thus g f ² h: x ² k :. Therefore, we let c nq 1 s c a k j. This means gkyj s c1 d1 ??? c n d n c a f ² h: x ² k :. Hence, by Ž1., we can find G such that gkyj f ² h: x ² k :. Thus g f ² h: x ² k :. Ž3. 5 g 5 s 5 x 5 q 2. We choose G such that 5 g 5 s 5 g 5 and 5 x 5 s 5 x 5. Then g f ² h: x ² k :. Case 2. h g A and k g B. Ž1. 5 g 5 ) 5 x 5. We choose G such that 5 g 5 s 5 g 5 and 5 x 5 s 5 x 5. Then g f ² h: x ² k :. Ž2. 5 g 5 s 5 x 5. Suppose g s d1 c1 ??? d n c n . Then we can find G such that 5 g 5 s 5 g 5 and 5 x 5 s 5 x 5. This implies g f ² h: x ² k :. So let g s c1 d1 ??? c n d n . Then we can apply the method of the second part of Case 1Ž1.. Case 3. h, k g B. Ž1. 5 g 5 s 5 x 5. Suppose g s d1 c1 ??? d n c n . If bn f ² c :² k :, then, by Lemma 3.5, we can find G such that 5 g 5 s 5 g 5, 5 x 5 s 5 x 5, and bn f ² c :² k :. This implies g f ² h: x ² k :. So suppose bn s c a k j. Then g f ² h: a1 b1 ??? a n c a ² k :. Hence, by induction, we can find G such that g f ² h:a1 b1 ??? a n c a² k :. It follows that g f ² h: x ² k :. So let g s c1 d1 ??? c n d n . Then we can apply the method of the second part of Case 1Ž1.. Ž2. 5 g 5 s 5 x 5 q 1 or 5 g 5 s 5 x 5 q 2. Suppose g s d1 c1 ??? d n c n d nq1. If d1 c1 f ² h: a1² c : or d 2 c 2 ??? c n d nq1 f ² c : b1 a2 ??? a n bn² k :, then, by induction, we can find G such that 5 g 5 s 5 g 5, 5 x 5 s 5 x 5, and d1 c1f ² h: a1² c : or d 2 c 2 ??? c n d nq1f ² c :b1 a2 ??? a n bn² k :. This implies g f ² h: x ² k :. So let d1 c1 s h ia1 c a and d 2 c 2 ??? c n d nq1 s c b b1 a2 ??? a n bn k j. Then a1 c aq b b1 ??? a n bn f ² h: a1 b1 ??? a n bn² k :. Hence, by Lemma 3.8, we can find G such that a1 c aq b b1 ??? a n bnf ² h:a1 b1 ??? a n bn² k :. It follows that g f ² h: x ² k :. Now suppose g s c1 d1 ??? c n d n c nq1 or 5 g 5 s 5 x 5 q 2. Then we can choose G such that 5 g 5 s 5 g 5 and 5 x 5 s 5 x 5. This implies g f ² h: x ² k :.
CONJUGACY SEPARABILITY
1069
Case 4. h g B and k g A. Ž1. 5 g 5 s 5 x 5. Suppose g s d1 c1 ??? d n c n . Then we can choose G such that 5 g 5 s 5 g 5 and 5 x 5 s 5 x 5. This implies g f ² h: x ² k :. So let g s c1 d1 ??? c n d n . Then we can apply the method of the second part of Case 1Ž1.. Ž2. 5 g 5 s 5 x 5 q 1. Suppose g s c1 d1 ??? c n d n c nq1. Again we can apply the method of the second part of Case 1Ž1.. So let g s d1 c1 ??? d n c n d nq1. In this case, we can apply the method of Case 3Ž2.. Ž3. 5 g 5 s 5 x 5 q 2. Suppose g s c1 d1 ??? d n c nq1 d nq1. Then we can choose G such that 5 g 5 s 5 g 5 and 5 x 5 s 5 x 5. This implies g f ² h: x ² k :. So let g s d1 c1 ??? c n d nq1 c nq1. Again we can apply the method of Case 3Ž2..
4. MAIN RESULTS To prove our main result we need the following lemmas. LEMMA 4.1. Let G be a tree product of a finite number of R N-subgroups amalgamating cyclic subgroups. If h is in a ¨ ertex group of G, then, for each n, there exists N 1f G such that Ž1. < ˜ h < s n and Ž2. ˜ h i ¤G˜ ˜ h j for ˜ hi / ˜ h j, ˜ s GrN. where G Proof. Let G be a tree product of r R N-groups. We shall prove the lemma by induction on r. By Corollary 2.2 in w23x, every R N-group has properties Ž1. and Ž2.. Now let G s A)² c: B, where B g R N, A is a subtree product of G with r y 1 vertices, and c is in a vertex of A. By induction, A has the properties Ž1. and Ž2.. We shall find M 1f A and L 1f B such that M l ² c : s L l ² c : and < h < s n and h i ¤G h j for h i / h j in G s ArM )² c: BrL. Since G is conjugacy separable by Theoi j rem 1.2, we can find N 1f G such that N l ² h: s 1 and Nh ¤G r N Nh i j for h / h . Let N be the preimage of N in G. Then N 1f G and < ˜ h< s n ˜ s GrN. and ˜ h i ¤G˜ ˜ h j for ˜ hi / ˜ h j, where G Suppose h g A Žthe case h g B is similar.. Since h is in a vertex of A, by induction there exists M 1f A such that < h < s n and h i ¤A h j for h i / h j, where A s ArM. Let M l ² c : s ² c s :. Now B g R N implies that there exists N 1f B such that < c < s s and c i ¤B c j for c i / c j, where B s BrL. Therefore, in G s ArM )² c: BrL and h i ¤G h j for h i / h j as in the proof of Lemma 3.2. LEMMA 4.2. Let G be a tree product of a finite number of conjugacy separable R N-groups amalgamating cyclic subgroups. Let h be in a ¨ ertex group of G. If there is x g G such that x 4G l ² h: s B, then there exists ˜ ˜ s GrN. N 1f G such that ˜ x 4G l ² ˜ h: s B, where G
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Proof. By Theorem 3.1 in w14x, every conjugacy separable R N-group is cyclic conjugacy separable. So let G s A)² c: B as in the proof of Lemma 4.1. Thus, by induction, A is cyclic conjugacy separable for ² u:, where u is in any vertex of A. Let h be in a vertex group of G and x g G such that x 4G l ² h: s B. We may assume that x has the minimal length in its conjugacy class in G. We shall find M 1f A and L 1f B such that M l ² c : s L l ² c : and x 4G l ² h: s B, where G s ArM ) ² c: BrL. Since G is conjugacy separable and ² h: is finite, there exists N 1f G such that Nx4G r N l ²Nh: s B. Let N be the preimage of N in G. Then ˜ ˜ s GrN. Thus, without loss of generalN 1f G and ˜ x 4G l ² ˜ h: s B in G ity, we assume h is in a vertex of A. Ž1. 5 x 5 G 2. Since x is of minimal length in its conjugacy class in G, x is cyclically reduced. By Lemmas 3.1 and 3.3, A, B are ² c :-potent and ² c :-separable. Therefore, there exist M 1f A and L 1f B such that M l ² c : s L l ² c : and 5 x 5 s 5 x 5, where G s ArM )² c: BrL. Since h g A and x is cyclically reduced with length G 2 in its conjugacy class in G, by Theorem 1.1, x 4G l ² h: s B. Ž2. x g A _ ² c :. Clearly x 4 A l ² c : s B and x 4 A l ² h: s B. Since c, h are in vertices of A, by induction, there exists M 1f A such that x 4 A l ² c : s B and x 4 A l ² h: s B, where A s ArM. Since B is ² c :potent, there exists L 1f B such that M l ² c : s L l ² c :. Thus, in G s ArM )² c: BrL, x is of minimal length 1 in its conjugacy class in G. By Theorem 1.1, x ¤G h i for all i. Hence x 4G l ² h: s B. Ž3. x g B _ ² c :. Clearly x 4 B l ² c : s B. Since B is cyclic conjugacy separable by w14, Theorem 3.1x, there exists L 1f B such that x 4 B l ² c : s B, where B s BrL. By Lemma 3.1, A is ² c :-potent. So there exists M 1f A such that M l ² c : s L l ² c :. Thus, in G s ArM )² c: BrL, x g B is of minimal length 1 in its conjugacy class in G. Since h g A, by Theorem 1.1, x ¤G h i for all i. Hence x 4G l ² h: s B. Ž4. x g ² c :. Clearly x 4 A l ² h: s B. By induction, since h is in a vertex of A and x g ² c :, there exists M 1f A such that x 4 A l ² h: s B, where A s ArM. By Lemma 4.1, there exists L 1f B such that M l ² c : s L l ² c : and c i ¤B c j for c i / c j, where B s BrL. Thus, in G s ArM )² c: BrL, x ¤G h i for all i by Theorem 1.1. Hence x 4G l ² h: s B. THEOREM 4.3. Let G be a tree product of a finite number of conjugacy separable R N-groups amalgamating cyclic subgroups. Then G is conjugacy separable. Proof. where A separable that A is
Let G be a tree product of r R N-groups. Then G s A)² h: B, is a subtree product of r y 1 vertices of G, B is a conjugacy R N-group, and h is in a vertex of A. By induction, we assume conjugacy separable. Then A, B are conjugacy separable, cyclic
CONJUGACY SEPARABILITY
1071
conjugacy separable for ² h: ŽLemma 4.2., and ² h:-potent ŽLemma 3.1.. Moreover, by Lemmas 3.2, 3.9, and 4.1, A, B satisfy C1, C2, and C3 of Theorem 2.3. Hence, by Theorem 2.3, G is conjugacy separable. Applying Theorem 4.3, we immediately have the following: THEOREM 4.4. Let G be a tree product of a finite number of finitely generated torsion-free nilpotent or free or surface groups, amalgamating cyclic subgroups. Then G is conjugacy separable.
REFERENCES 1. R. B. J. T. Allenby, Conjugacy separability of a class of 1-relator products, Proc. Amer. Math. Soc. 116 Ž3. Ž1992., 621]628. 2. R. B. J. T. Allenby and C. Y. Tang, The residual finiteness of some one-relator groups with torsion, J. Algebra 71 Ž1. Ž1981., 132]140. 3. R. B. J. T. Allenby and C. Y. Tang, Conjugacy separability of certain 1-relator groups with torsion, J. Algebra 103 Ž2. Ž1986., 619]637. 4. N. Blackburn, Conjugacy in nilpotent groups, Proc. Amer. Math. Soc. 16 Ž1965., 143]148. 5. R. G. Burns, A. Karrass, and D. Solitar, A note on groups with separable finitely generated subgroups, Bull. Austral. Math. Soc. 36 Ž1987., 153]160. 6. J. L. Dyer, Separating conjugates in amalgamated free products and HNN extensions, J. Austral. Math. Soc. Ser. A 29 Ž1. Ž1980., 35]51. 7. B. Fine and G. Rosenberger, Conjugacy separability of Fuchsian groups and related questions, Contemp. Math., Amer. Math. Soc. 109 Ž1990., 11]18. 8. E. Formanek, Conjugate separability in polycyclic groups, J. Algebra 42 Ž1976., 1]10. 9. R. Gitik and E. Rips, On separability properties, I, preprint. 10. M. Hall, Jr., Coset representations in free groups, Trans. Amer. Math. Soc. 67 Ž1949., 421]432. 11. G. Kim, Cyclic subgroup separability of generalized free products, Canad. Math. Bull. 36 Ž3. Ž1993., 296]302. 12. G. Kim, J. McCarron, and C. Y. Tang, On generalized free products of conjugacy separable groups, J. Algebra 180 Ž1996., 121]135. 13. G. Kim and C. Y. Tang, Conjugacy separability of generalized free products of finite extensions of residually nilpotent groups, preprint. 14. G. Kim and C. Y. Tang, Cyclic conjugacy separability of groups, in ‘‘Groups-Korea 94,’’ pp. 173]179, de Gruyter, Berlin, 1995. 15. W. Magnus, A. Karrass, and D. Solitar, ‘‘Combinatorial group theory,’’ Pure and Appl. Math., Vol. XIII, Wiley]Interscience, New YorkrLondonrSydney, 1966. 16. A. W. Mostowski, On the decidability of some problems in special classes of groups, Fund. Math. 59 Ž1966., 123]135. 17. G. A. Niblo, Separability properties of free groups and surface groups, J. Pure Appl. Algebra 78 Ž1992., 77]84. 18. V. M. Remeslennikov, Conjugacy in polycyclic groups, Algebra Logic 8 Ž1969., 404]411. 19. P. Scott, Subgroups of surface groups are almost geometric, J. London Math. Soc. 17 Ž1978., 555]565.
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20. M. Shirvani, On conjugacy separability of fundamental groups of graphs of groups, Trans. Amer. Math. Soc. 334 Ž1. Ž1992., 229]243. 21. P. F. Stebe, Conjugacy separability of certain free products with amalgamation, Trans. Amer. Math. Soc. 156 Ž1971., 119]129. 22. P. F. Stebe, Conjugacy separability of certain Fuchsian groups, Trans. Amer. Math. Soc. 163 Ž1972., 173]188. 23. C. Y. Tang, Conjugacy separability of generalized free products of surface groups. J. Pure Appl. Algebra, to appear. 24. W. F. Thurston, Three dimensional manifolds, Kleinian groups and hyperbolic geometry, Bull. Amer. Math. Soc. 6 Ž1982., 357]381.
184, 1052]1072 Ž1996.
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A Criterion for the Conjugacy Separability of Amalgamated Free Products of Conjugacy Separable Groups Goansu KimU Department of Mathematics, Kangnung National Uni¨ ersity, Kangnung, 210-702 Korea
and C. Y. Tang† Uni¨ ersity of Waterloo, Waterloo, Ontario, N2L 3G1, Canada Communicated by Walter Feit Received August 4, 1995
We first prove a criterion for the conjugacy separability of generalized free products of two conjugacy separable groups amalgamating a cyclic subgroup. Applying this result, we show that tree products of a finite number of conjugacy separable, residually finitely generated torsion-free nilpotent groups amalgamating cyclic subgroups are conjugacy separable. From this we derive that tree products of finitely generated torsion-free nilpotent groups, free groups, or surface groups Q 1996 Academic Press, Inc. amalgamating cyclic subgroups are conjugacy separable.
1. INTRODUCTION Let S be a subset of a group G. Then G is said to be S-separable if for each g f S there exists a finite index normal subgroup N of G with g f NS. This is equivalent to S being a closed subset in the profinite topology on G. If G is S-separable for every finitely generated subgroup S of G, then G is said to be subgroup separable Žor LERF.. In the special case of S s 14 , G is S-separable means G is residually finite. If G is U
Supported by GARC-KOSEF and NDRF, Korea Research Foundation, 1995. E-mail: [email protected]. † Partially supported by the Natural Science and Engineering Research Council of Canada, Grant No. A-6064. E-mail: [email protected]. 1052 0021-8693r96 $18.00 Copyright Q 1996 by Academic Press, Inc. All rights of reproduction in any form reserved.
CONJUGACY SEPARABILITY
1053
S-separable for every conjugacy class S of G, then G is called conjugacy separable. Separability properties of groups are of great interest to both group theorists and three-manifold topologists. In particular, Mostowski w16x showed that: Ž1. finitely presented residually finite groups have solvable word problem, Ž2. finitely presented conjugacy separable groups have solvable conjugacy problem, and Ž3. finitely presented subgroup separable groups have solvable generalized word problem. It is well known that free groups and surface groups are both subgroup separable and conjugacy separable ŽHall w10x, Scott w19x, Stebe w21x.. Thurston w24x showed that the fundamental groups of Haken manifolds are residually finite. Scott asked whether these groups are subgroup separable. Burns, Karrass and Solitar w5x showed that these groups needs not be subgroup separable. The question of whether these groups are always conjugacy separable is still unknown. In particular, whether knot groups are subgroup separable or conjugacy separable is still unknown. Blackburn w4x showed that finitely generated nilpotent groups are conjugacy separable. Subsequently, Formanek w8x Žalso Remeslennikov w18x. showed that polycyclic-by-finite groups are conjugacy separable. Stebe w21, 22x first considered the conjugacy separability of generalized free products of free groups and that of Fuchsian groups. The former was completed by Dyer w6x, who showed that generalized free products of two free groups or finitely generated nilpotent groups amalgamating a cyclic subgroup are conjugacy separable. The latter was completed by Fine and Rosenberger w7x using some results of Stebe w22x and Allenby and Tang w3x. This naturally leads us to ask whether generalized free products of finitely generated Fuchsian groups amalgamating a cyclic subgroup are conjugacy separable. These groups are proved to be conjugacy separable by Kim and Tang w13x. In the same paper we asked whether generalized free products of polycyclic groups amalgamating a cyclic subgroup are conjugacy separable. L. Ribes recently informed us that together with D. Segal and P. A. Zalesskii, they have proved the conjugacy separability of these groups. Thus it is natural to ask whether tree products of conjugacy separable groups amalgamating cyclic subgroups are conjugacy separable. In this paper we show that tree products of a finite number of conjugacy separable residually nilpotent groups are conjugacy separable. The main difficulty in the proof lies in proving the double coset separability of these groups. Other results on conjugacy separability and double coset separability can be found in Allenby w1x, Shirvani w20x, Niblo w17x, and Gitik and Rips w9x. In Section 2 we prove one of our main results, a criterion for the generalized free products of two conjugacy separable groups amalgamating a cyclic subgroup to be conjugacy separable; this is Theorem 2.3. The main body of our proof is in Section 3, where we prove that tree products of
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residually nilpotent groups amalgamating cyclic subgroups are double coset separable with respect to the subgroups ² h: and ² k :, where h and k are elements of some vertex groups. This involves a detailed analysis of the various cases. Finally, in Section 4 we prove our main result, Theorem 4.3, that tree products of a finite number of conjugacy separable residually nilpotent groups, amalgamating cyclic subgroups, are conjugacy separable. From this it follows that tree products of a finite number of free groups, finitely generated nilpotent groups, and surface groups, amalgamating cyclic subgroups, are conjugacy separable. At this point it is natural to ask whether tree products of Fuchsian groups amalgamating cyclic subgroups are conjugacy separable. In view of Ribes, Segal, and Zalesskii’s result we also ask whether tree products of polycyclic groups amalgamating cyclic subgroups are conjugacy separable. Throughout this paper we use standard notations and terminology. For convenience we list the following: The letter G always denotes a group. x 4G denotes the set of all conjugates of x in G. x ;G y means x, y are conjugate in G. If x g G s A)H B, then 5 x 5 denotes the free product length of x in G. N 1f G means N is a normal subgroup of finite index in G. A group G is pc if, for every cyclic subgroup H of G and every x g G _ H, there exists N 1f G such that in G s GrN, x f H. A group G is conjugacy separable if, for each pair of nonconjugate elements x, y g G, there exists N 1f G such that in G s GrN, x ¤G y. CH Ž x . is the centralizer of x g G in the subgroup H of G. Let H s ² h: be a infinite cyclic subgroup of G. If, for every positive integer n, there exists Nn 1f G such that Nn l H s ² h n :, then G is said to be H-potent. A torsion-free group G is said to be potent if it is ² h:-potent for all 1 / h g G. Let P be a tree. Assign a group G¨ to each vertex ¨ and a group Ge to each edge e of P. Let a e and b e be monomorphisms which embed Ge as a subgroup of the two vertex groups at the ends of the edge e. The tree product of P is defined to be the group generated by the generators and relations of the vertex groups together with the extra relations obtained by identifying g e a e and g e b e for each g e g Ge . We use N to denote the class of finitely generated torsion-free nilpotent groups and R N to denote the class of residually N-groups. We shall make extensive use of the following results. THEOREM 1.1 w15, Theorem 4.6x. Let G s A)H B and let x g G be of minimal length in its conjugacy class. Suppose that y g G is cyclically reduced and that x ;G y.
CONJUGACY SEPARABILITY
1055
Ž1. If 5 x 5 s 0, then 5 y 5 F 1 and, if y g A, then there is a sequence h1 , h 2 , . . . , h r of elements in H such that y ;A h1 ;B h 2 ;A ??? ; h r s x. Ž2. If 5 x 5 s 1, then 5 y 5 s 1 and either x, y g A and x ;A y or x, y g B and x ;B y. Ž3. If 5 x 5 G 2, then 5 x 5 s 5 y 5 and y ;H xU , where xU is a cyclic permutation of x. THEOREM 1.2 w6, Theorem 4x. If A and B are conjugacy separable and H is finite, then A)H B is conjugacy separable. 2. A CRITERION In this section we prove a criterion for the conjugacy separability of generalized free products of conjugacy separable groups amalgamating a cyclic subgroup. LEMMA 2.1. Let G be ² h e : x ² k e :-separable, where h, k g G are of infinite order. If ² xy1 hx : l ² k : s 1, then there exists N 1f G such that xy1 h i x s k j only if e < i, j, where G s GrN. Proof. Note hyi 1 xkyj 1 f ² h e : x ² k e : for 0 F i1 , j1 - e except i1 s j1 s 0. Hence there exists N 1f G such that hyi 1 xkyj 1 f N² h e : x ² k e : for 0 F i1 , j1 - e except i1 s j1 s 0. Therefore, if xy1 h i x s k j, then e < i, j, where G s GrN. Note that the condition xy1 h i x s k j only if e < i, j always implies e < < h <, < k <. DEFINITION 2.2. A group G is cyclic conjugacy separable for ² h: if, for each x g G such that x 4G l ² h: s B, there exists N 1f G such that x 4G l ² h: s B, where G s GrN. If G is cyclic conjugacy separable for every cyclic subgroup ² h: of G, then G is called cyclic conjugacy separable. THEOREM 2.3. Let G s G1 )² h: G 2 . Suppose that G1 , G 2 are conjugacy separable, cyclic conjugacy separable for ² h: and ² h:-potent. If each Gk Ž k s 1, 2. satisfies: C1. if h i ;G k h j, then i s j; C2. for e¨ ery e ) 0, Gk is ² h e : x ² h e :-separable for x g Gk ; C3. for e¨ ery n ) 0, there exists M 1f Gk such that < h < s n and h i ¤G k h j for h i / h j, where Gk s GkrM, then G is conjugacy separable. Proof. Let x, y g G such that x ¤G y. Without loss of generality we can assume that x and y are of minimal lengths in their conjugacy classes in G. Since G1 , G 2 are ² h:-potent and ² h:-separable by C2, G is residually finite by w2x. Hence, we may assume x / 1 / y. To prove our result, we
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shall find M 1f G1 and L 1f G 2 such that M l ² h: s L l ² h: and x ¤G y, where G s G1rM )² h: G 2rL. Then, by Theorem 1.2, G is conjugacy separable, whence there exists K 1f G such that Kx ¤G r K Ky. Let K be the preimage of K in G. Then we have K 1f G and Kx ¤G r K Ky, as required. Case 1. 5 x 5 s 0 s 5 y 5, say, x s h t and y s h s. Clearly s / t. Ži. < t < / < s <. Since G1 , G 2 are potent, there exist M 1f G1 and L 1f G 2 such that M l ² h: s ² h st : s L l ² h:. Then, in G s G1rM )² h: G 2rL, we have < x < s < s < / < t < s < y <, hence x ¤G y. Žii. t s ys, where t ) 0. By C3, there exists M 1f G1 such that < h < s 3t and h i ¤G 1 h j for h i / h j, where G1 s G1rM. Similarly, there exists L 1f G 2 such that < h < s 3t and h i ¤G 2 h j for h i / h j, where G 2 s G 2rL. Then, in G s G1rM )² h: G 2rL, if h t ;G hyt then, by Theorem 1.1, h t s hyt . Since < h < s 3t, it follows that x ¤G y. Case 2. 5 x 5 s 0 and 5 y 5 s 1 Žor 5 y 5 s 0 and 5 x 5 s 1., say, y g G1 _ ² h: and x g ² h:. Since y is of minimal length 1 in its conjugacy class, we have y4G 1 l ² h: s B. Moreover, since G1 is cyclic conjugacy separable for ² h:, there exists M 1f G1 such that y4G 1 l ² h: s B, where G1 s G1rM. Now, by the potency of G 2 , there exists L 1f G 2 such that M l ² h: s L l ² h:. Let G s G1rM )² h: G 2rL. Then, by Theorem 1.1, y is of minimal length 1 in its conjugacy class in G. Hence x ¤G y, as required. Case 3. 5 x 5 / 5 y 5 and 5 x 5 G 2 Žor 5 y 5 G 2.. Since x is of minimal length in its conjugacy class, it is cyclically reduced. Let x s a1 b1 ??? a n bn Žsay, other cases being similar., where a i g G1 _ ² h: and bi g G 2 _ ² h:. Since G1 , G 2 are ² h:-separable Žby C2. and ² h:-potent, we can find M 1f G1 and L 1f G 2 such that M l ² h: s L l ² h: and 5 x 5 s 5 x 5, 5 y 5 s 5 y 5, where G s G1rM )² h: G 2rL. Thus x is cyclically reduced and is of minimal length 2 n in its conjugacy class. Therefore 5 x 5 / 5 y 5. Hence, by Theorem 1.1, x ¤G y, as required. Case 4. 5 x 5 s 1 s 5 y 5. Ži. x, y g G1 _ ² h: Žor x, y g G 2 _ ² h:.. Since x, y are of minimal length 1 in their conjugacy classes, x 4G l ² h: s B and y4G l ² h: s B. Now G1 is conjugacy separable and cyclic conjugacy separable for ² h:. Therefore, there exists M 1f G1 such that x ¤G 1 y, x 4G 1 l ² h: s B, and y4G 1 l ² h: s B, where G1 s G1rM. Since G 2 is potent, we can choose L 1f G 2 such that M l ² h: s L l ² h:. Thus, in G s G1rM )² h: G 2rL, x, y are of minimal length 1 in their conjugacy classes, and x ¤G 1 y. Hence, by Theorem 1.1, x ¤G y, as required. Žii. Suppose x g G1 _ ² h: and y g G 2 _ ² h: Žor x g G 2 _ ² h: and y g G1 _ ² h:.. As in Ži. above, there exist M 1f G1 and L 1f G 2 such
CONJUGACY SEPARABILITY
1057
that M l ² h: s L l ² h:, x 4G 1 l ² h: s B, and y4G 2 l ² h: s B, where G1 s G1rM and G 2 s G 2rL. Let G s G1rM )² h: G 2rL. Then x, y are of minimal length 1 in their conjugacy classes with x g G1 and y g G 2 . Hence, by Theorem 1.1, x ¤G y, as required. Case 5. 5 x 5 s 5 y 5 s r G 2. Let x s u1 u 2 ??? u r and y s ¨ 1¨ 2 ??? ¨ r be cyclically reduced. Since x ¤G y, by Theorem 1.1, x ¤² h: yU for any cyclic permutation yU of y. Thus, for each i, the equation I Ž i . : u1 u 2 ??? u r s hya i ¨ i ??? ¨ r¨ 1 ??? ¨ iy1 h a i
Ž 1.
has no integer solution a i . Hence, for each i, we shall find MiX 1f G1 and LXi 1f G 2 such that MiX l ² h : s LXi l ² h :. Moreover, in G s G1rMiX )² h: G 2rLXi , 5 x 5 s 5 x 5, 5 y 5 s 5 y 5, and the equation I Ž i . : u1 u 2 ??? u r s hya i¨ i ??? ¨ r¨ 1 ??? ¨ iy1 h a i
Ž 2.
r r has no integer solution a i . Let M s F is1 MiX , L s F is1 LXi , and G s G1rM )² h: G 2rL. Then x and y are cyclically reduced and x ¤² h: yU for all cyclic permutation yU of y. Therefore, by Theorem 1.1, we have x ¤G y, as required. Since all the cases are similar, we shall only consider the case i s 1. Since equation I Ž1. has no solution a 1 , we have x s u1 u 2 ??? u r ¤² h: ¨ 1¨ 2 ??? ¨ r s y. If u i and ¨ i are not in the same factor of G, then we can easily find G s G1rM1X )² h: G 2rLX1 such that 5 x 5 s 5 x 5 and 5 y 5 s 5 y 5. Clearly equation I Ž 1 . has no solution, since u i and ¨ i are not in the same factor of G. Thus we need only consider the case x s c1 d1 ??? c n d n and y s a1 b1 ??? a n bn , where a i , c i g G1 _ ² h: and bi , d i g G 2 _ ² h:. We proceed by finding M 1f G1 and L 1f G 2 such that M l ² h: s L l ² h: so that, in G s G1rM )² h: G 2rL, 5 x 5 s 2 n s 5 y 5. Moreover, equation I Ž 1 . has no integer solution a 1 in G, whence x ¤² h: y. Since G1 , G 2 are ² h:-separable and ² h:-potent, there exist M1 1f G1 and L1 1f G 2 such that M1 l ² h: s L1 l ² h:, a i , c i f M1² h:, and bi , d i f L1² h: for all i.
Ž1. There exists i such that c i f ² h: a i ² h: Žor d i f ² h: bi ² h:.. By C2, there exists M2 1f G1 such that c i f M2 ² h: a i ² h:. Now, by the potency of G 2 , we can choose L2 1f G 2 such that M2 l ² h: s L2 l ² h:. Let M s M1 l M2 and L s L1 l L2 . Then, in G s G1rM )² h: G 2rL, x and y are cyclically reduced and c1 d1 ??? c n d nf ² h:a1 b1 ??? a n bn² h:. Hence x ¤² h: y, as required. Ž2. There exists i such that c1 d1 ??? c i s hya a1 b1 ??? a i h b , d i s h l bi h d , and that c1 d1 ??? c i d i f ² h: a1 b1 ??? a i bi ² h:. This implies a1 b1 ??? a i h bq l bi f ² h: a1 b1 ??? a i bi ² h:. For convenience, we write w s a1 b1 ??? a i . Suppose C² h:Ž w . s ² h t : and C² h:Ž bi . s ² h s :, where s, t G 0.
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Ža. t / 0 / s. Then clearly h bq l f ² h t :² h s : s ² h d :, where d s j ² : gcdŽ t, s .. Now wy1 h j w f ² h: for 1 F j - t and by1 i h bi f h for 1 F j - s by C1. Since G1 , G 2 are ² h:-separable and ² h:-potent, G is ² h:-separable w2x. Thus there exists N 1f G such that wy1 h j w f N² h: for all j ² : 1 F j - t and by1 i h bi f N h for all 1 F j - s. Since G 1 , G 2 are potent, there exist M2 1f G1 and L2 1f G 2 such that M2 l ² h: s ² h d : s L2 l ² h:. Moreover h bq l f ² h d : implies h bq l f M2 ² h d :. Let M s M1 l M2 l N and L s L1 l L2 l N. Then, in G s G1rM )² h: G 2rL, x and y are cyclically reduced and a1 b1 ??? a i h bq l bif ² h:a1 b1 ??? a i bi² h:. For, if not, wh bq l bi s hy m 0 wbih m i for some m 0 , m i . Then w s hy m 0 wh l1 and h bq l bi s hyl 1 bi h m i for some l1. By the choice of N, t < m 0 , whence h m 0 s h l1 g ² h t :, and s < b q l q l1. Hence h bq l g ² h t :² h s : s ² h d :, contradicting the choice of M2 . This shows that a1 b1 ??? a i h bq l bif ² h:a1 b1 ??? a i bi² h:. Thus c1 d1 ??? c i d if ² h:a1 b1 ??? a i bi² h:. Hence x ¤² h: y, as required. Žb. t / 0 and s s 0. Then clearly h bq l f ² h t :. As in Ža., there exists N 1f G such that wy1 h j w f N² h: for 1 F j - t and there exists M2 1f G1 such that h bq l f M2 ² h t :. Because of C1, C² h:Ž bi . s 1 implies ² by1 : ² : i hbi l h s 1. Hence, by Lemma 2.1, for e s t, there exists M3 1f k < G1 such that if M3 h j s M3 by1 i h bi , then e s t k, j. As before, we can find M 1f G1 and L 1f G 2 such that M l ² h: s L l ² h:, M ; M1 l M2 l M3 l N, and L ; L1 l N. Then, in G s G1rM )² h: G 2rL, x and y are cyclically reduced, and a1 b1 ??? a i h bq l bif ² h:a1 b1 ??? a i bi² h:. For, if not, wh bq l bi s hy m 0 wbih m i for some m 0 , m i . This implies w s hy m 0 wh l1 and h bq l bi s hy l1 bi h m i for some l1. By the choice of N, we have t < m 0 , whence h m 0 s h l1 g ² h t :, and e s t < b q l q l1. Hence h bq l g ² h t :, contradicting the choice of M2 . This shows that a1 b1 ??? a i h bq l bif ² h:a1 b1 ??? a i bi² h:. Therefore, c1 d1 ??? c i d if ² h:a1 b1 ??? a i bi² h:. Hence x ¤² h: y, as required. Žc. t s 0 and s / 0. Then clearly h bq l f ² h s :. Since C² h:Ž w . s 1, : there exists a k Žor bk . with k F i such that ² h: l ² ay1 k ha k s 1 by C1. Thus, by Lemma 2.1, there exists M2 1f G1 such that if M2 h r s j < ² h s :-separable, we can find M2 ay1 k h a k , then e s s r, j. Since G 1 is bq l s: ² M3 1f G1 such that h f M3 h . Also we can find N 1f G such j ² : for 1 F j - s. Let M 1f G1 and L 1f G 2 such that by1 i h bi f N h that M l ² h: s L l ² h:, M ; M1 l M2 l M3 l N, and L ; L1 l N. Then, in G s G1rM )² h: G 2rL, x and y are cyclically reduced and a1 b1 ??? a i h bq l bif ² h:a1 b1 ??? a i bi² h:. For, if a1 b1 ??? a i h bq l bis h m 0 a1 b1 ??? a i bi h m i for some m 0 , m i , then a1 s h m 0 a1 h l1 , b1 s hyl 1 b1 h m 1 , . . . , a k s hy m ky 1 a k h l k , . . . , a i s hy m iy 1 a i h l i , and h bq l bi s hyl i bi h m i for some l j , m j . Thus, by the choice of M2 , e s s < m ky1 , l k . Since e 5 h <, e < l i . Now, by the choice of N, s < b q l q l i . Hence e s s < b q l, thus h bq l g ² h s :, a contradiction. This shows that a1 b1 ??? a i h bq l bif
CONJUGACY SEPARABILITY
1059
² h:a1 b1 ??? a i bi² h:. Therefore, c1 d1 ??? c i d if ² h:a1 b1 ??? a i bi² h:. Hence x ¤² h: y, as required. Žd. s s 0 and t s 0. Then clearly we have ² h: l ² by1 : i hbi s 1. Moreover, since C² h:Ž w . s 1, there exists a k Žor bk . with k F i such that ² h: l ² ay1 : < < k ha k s 1. By Lemma 2.1, for e s 2 b q l , there exist M2 1f j G1 and L2 1f G 2 such that if M2 h i s M2 ay1 h a , then e < i, j, and that if k k k < L2 h j s L2 by1 h b , then e j, k. As before, we can find M3 1f G1 such i i that h bq l f M3 ² h 2Ž bq l . :. Again, as before, we can find M 1f G1 and L 1f G 2 such that M l ² h: s L l ² h:, M ; M1 l M2 l M3 , and L ; L1 l L2 . Thus, in G s G1rM )² h: G 2rL, x and y are cyclically reduced, and a1 b1 ??? a i h bq l bif ² h:a1 b1 ??? a i bi² h:. For, if a1 b1 ??? a i h bq l bis h m 0 a1 b1 ??? a i bi h m i for some m 0 , m i , then a1 s h m 0 a1 h l1 , b1 s hyl 1 b1 h m 1 , . . . , a k s hy m ky 1 a k h l k , . . . , a i s hy m iy 1 a i h l i , and h bq l bi s hyl i bi h m i for some l j , m j . Then, by the choice of M2 , e < m ky1 , l k . Since e 5 h <, e < l i . Again, by the choice of L2 , e < b q l q l i . Hence e < b q l. This means h bq l g ² h 2Ž b q l . :, a contradiction. This shows that a1 b1 ??? a i h bq l bi f ² h: a1 b1 ??? a i bi ² h:. Therefore, c1 d1 ??? c i d i f ² h:a1 b1 ??? a i bi² h:. Hence x ¤² h: y, as required. Ž3. c1 d1 ??? c n d n s h a a1 b1 ??? a n bn h b and h aq b / 1. Choose M2 1f G1 such that h aq b f M2 . Let M1 l ² h: s ² h s : s L1 l ² h: and M2 l ² h: s ² h t :. By C3, there exist M3 1f G1 and L3 1f G 2 such that M3 l ² h: s ² h st : s L3 l ² h: and M3 h i ¤G 1 r M 3 M3 h j for M3 h i / M3 h j and L3 h i ¤G 2 r L 3 L3 h j for L3 h i / L3 h j. Let M s M1 l M2 l M3 , L s L1 l L3 , and G s G1rM )² h: G 2rL. Then x and y are cyclically reduced and x ¤² h: y. For, if x ;² h: y, then x s h a yh b s hy l yh l for some l. This im plies th at h a a 1 s h y l a 1 h l 1 , b 1 s h y l 1 b 1 h m 1 , . . . , a k s hym ky 1 a k h l k , . . . , a n s hy m ny 1 a n h l n , and bn h b s hy l n bn h l for some l j , m j . Thus, by the choice of M3 , L3 , we have h aq l s h l1 , h l1 s h m 1 , . . . , h m ky 1 s h lk , . . . , h m ny 1 s h l n , and h l n s h ly b . This means h aq l s h ly b , whence h aq b s 1, contradicting the choice of M2 . Hence x ¤² h: y, as required. This completes our proof that there exist M1X 1f G1 and LX1 1f G 2 such X ˜ s G1rM1X )² h: that equation I Ž 1 . has no solution in G ˜ G 2rL1 . Similarly, X X for each i, we can find Mi 1f G1 and L i 1f G 2 such that equation I Ž i . r ˆ s G1rMiX )² h: G2rLXi . Let M s F is1 has no solution in G MiX and L s X r F is1 L i . Then M 1f G1 , L 1f G 2 , and, in G s G1rM )² h: G 2rL, I Ž i . has no solution for each i. Hence x ¤² h: yU for any cyclic permutation yU of y. Thus it follows from Theorem 1.1 that x ¤G y, as required. It is easy to see that R N-groups are potent and they satisfy C1, C2, and C3 Žsee Lemmas 3.2 and 3.4 and w23, Corollary 2.2x.. Since, by w6x and w23x,
1060
KIM AND TANG
free groups, finitely generated nilpotent groups and surface groups are cyclic conjugacy separable, we have: COROLLARY 2.4 w6, 23x. Let Gi Ž i s 1, 2. be a finitely generated torsionfree nilpotent or free or surface group. Then G s G1 )² c: G 2 is conjugacy separable. A generalized free product of two free groups amalgamating an infinite cyclic subgroup is called a cyclically pinched one-relator group. Hence Corollary 2.4 implies in particular that cyclically pinched one-relator groups are conjugacy separable. It was also known by w12x that generalized free products of two free or surface groups, amalgamating a maximal cyclic subgroup, satisfy C1 and C2 in Theorem 2.3. Thus, by Lemmas 4.1, 4.2, and 3.1, we have: COROLLARY 2.5 w12x. Let A i be free or surface groups and let ² u: and ² ¨ : be maximal cyclic subgroups of A1 , A 2 and A 3 , A 4 , respecti¨ ely, such that A1 l A 2 s ² u: and A 3 l A 4 s ² ¨ :. If A 2 l A 3 s ² c :, then A1 )² u: A 2 )² c: A 3 )² ¨ : A 4 and A1 )² u: A 2 )² c: A 3 are conjugacy separable.
3. DOUBLE COSET SEPARABILITY OF CERTAIN TREE PRODUCTS Let G be a tree product of a finite number of R N-groups, amalgamating cyclic subgroups. Suppose G has r vertices. Then, either G is a R N-group when G has one vertex or G s A)² c: B, where B is a vertex group, A is a subtree product of r y 1 vertices, and c is in a vertex of A. In the latter case, if h is in a vertex of G, then either h is in a vertex of A or h g B. Throughout this section, we shall use the above description for G and show that G is ² h: x ² k :-separable for x g G, where h, k are in the vertices of G. LEMMA 3.1. Let G be a tree product of a finite number of R N-groups amalgamating cyclic subgroups. If h is in a ¨ ertex group of G, then G is ² h:-potent. Proof. Clearly a R N-group is potent. Let G s A)² c: B as described above. By induction, we can assume that A is ² u:-potent for any u in a vertex of A. Then A, B are ² c :-potent. Hence, by Lemma 3.2 in w2x, G is ² h:-potent for any h in a vertex of G. LEMMA 3.2. Let G be a tree product of a finite number of R N-groups amalgamating cyclic subgroups. Let h be in a ¨ ertex group of G. If h i ;G h j, then i s j.
CONJUGACY SEPARABILITY
1061
Proof. Clearly a R N-group has the property. Hence we let G s A)² c: B as above and, by induction, we suppose that A has the property that u i ;A u j only if i s j for u in a vertex of A. Let h be in a vertex of G and h i ;G h j. Suppose h g A. If h i 4G l ² c : s B, then, by Theorem 1.1, h i ;A h j. Thus, by induction, i s j. Hence we assume h i ;G c n for some n. Then, by Theorem 1.1, h i ;A c e 1 ;B c e 2 ;A ??? ; c e r s c n for some e k . This implies e 1 s e 2 s ??? e r s n, since c is in B and is in a vertex of A. Thus h i ;A c n. Similarly h j ;A c n, since h j ;G h i ;A c n. It follows that h i ;A h j. Since h is in a vertex of A, by induction i s j. The case of h g B is similar. LEMMA 3.3. Let G be a tree product of a finite number of R N-groups amalgamating cyclic subgroups. Then G is pc . Proof. A R N-group is pc by w13, Lemma 2.6x. So let G s A)² c: B as above. By induction, we assume that A is pc . Since A, B are ² c :-potent, G is pc by Corollary 2.2 in w11x. LEMMA 3.4. separable.
Let G be a R N-group. Let h, k g G. Then G is ² h:² k :-
Proof. Let g f ² h:² k :, where g, h, k g G. Case 1. ² h: l ² k : s ² h n : / 1, where h n s k m and n ) 0. Then h gkyj f ² h n :² k m : s ² h n : for all 0 F i - n and 0 F j - < m <. Since G is pc by Lemma 3.3, there exists N 1f G such that hyi gkyj f N² h n : for all 0 F i - n and 0 F j - < m <. Then clearly g f N² h:² k :. yi
Case 2. ² h: l ² k : s 1. First, we shall find N1 1 G such that h, k f N1 and ² N1 h: l ² N1 k : s N1 with GrN1 g N. Since G is R N, there exists M1 1 G such that h, k f M1 and GrM1 g N. If ² M1 h: l ² M1 k : s M1 , then let N1 s M1. Otherwise, let M1 h s s M1 k t , where h s kyt / 1. Therefore, there exists M2 1 G such that h s kyt f M2 and GrM2 g N. Let N1 s M1 l M2 . Since GrN1 can be imbedded in a direct product GrM1 = GrM2 g N, GrN1 g N. We note that ² N1 h: l ² N1 k : s N1. For, if N1 h n s N1 k m , then N1 h n s s N1 k m s. This implies M1 k nt s M1 h n s s M1 k m s. Since GrM1 g N and < M1 k < s `, nt s ms. This means N1 h n s s N1 k m s s N1 k nt. Now GrN1 g N has the unique root property. Therefore N1 h s s N1 k t , contradicting the choice of M2 . Hence h, k f N1 and ² N1 h: l ² N1 k : s N1. Since GrN1 g N, GrN1 is ² N1 h:² N1 k :-separable. Therefore, if N1 g f ² N1 h:² N1 k :, then we can find N 1f G such that g f N² h:² k :. So we can assume N1 g s N1 h s N1 k t for some s, t. Since ² N1 h: l ² N1 k : s N1 , N1 g s N1 h s N1 k t is unique. Now choose N2 1 G such that gy1 h s k t f N2
1062
KIM AND TANG
and GrN2 g N. Let M s N1 l N2 . Then, as before, GrM g N. Moreover Mg f ² Mh:² Mk :. Then we can find N 1f G such that g f N² h:² k :. LEMMA 3.5. Let G be a tree product of a finite number of R N-groups amalgamating cyclic subgroups. Then G is ² u:² ¨ :-separable for u, ¨ in ¨ ertices of G. Proof. By Lemma 3.4, the result holds for tree products with one vertex. Hence we assume that G s A)² c: B, where c is in a vertex of the subtree product A and B is a R N-group. By induction, we can assume that A is ² h:² k :-separable for h, k in vertices of A. Let g f ² u:² ¨ :, where g g G and u, ¨ are elements of vertices of G. We shall find M 1f A and L 1f B such that M l ² c : s L l ² c : and g f ² u:² ¨ :, where G s ArM )² c: BrL. Since G is residually finite and <² u:² ¨ :< - `, G is ² u:² ¨ :-separable. Therefore, there exists N 1f G such that g f N² u:² ¨ :. Let N be the preimage of N. Then N 1f G and g f N² u:² ¨ :. Case 1. u, ¨ g A Žor u, ¨ g B . Ž1. g g A. By the induction hypothesis, there exists M 1f A such that g f ² u:² ¨ :, where A s ArM. Since B is potent, there exists L 1f B such that M l ² c : s L l ² c :. Then, in G s ArM )² c: BrL, g f ² u:² ¨ :. Ž2. g f A. By Lemmas 3.3 and 3.1, A, B are pc and ² c :-potent. Therefore, there exist M 1f A and L 1f B such that M l ² c : s L l ² c : and 5 g 5 s 5 g 5, where G s ArM )² c: BrL. Clearly g f ² u:² ¨ :. Case 2. u g A and ¨ g B Žor u g B and ¨ g A.. If ² u: l ² c : s ² c s : / 1 Žsimilarly, if ² ¨ : l ² c : / 1., say, c s s u t , then u i g f ² u t :² ¨ : s ² c s :² ¨ : for 0 F i - < t <. Since c s, ¨ g B, by Case 1, there exists N 1f G such that u i g f N² c s :² ¨ : s N² u t :² ¨ : for 0 F i - < t <. It follows that g f N² u:² ¨ :. So we assume that ² u: l ² c : s 1 s ² ¨ : l ² c :. Ž1. g g ² c :. Let g s c a f ² u:² ¨ :. Since ² u: l ² c : s 1 s ² ¨ : l ² c :, i c f ² u:² c 2 a : and c i f ² c 2 a :² ¨ : for all 1 F i - 2 < a <. By induction, A is ² u:² c 2 a :-separable. Also, by Lemma 3.4, B is ² c 2 a :² ¨ :-separable. Therefore, there exist M 1f A and L 1f B such that c a f M² c 2 a :, c i f M² u:² c 2 a :, and c i f L² c 2 a :² ¨ : for all 1 F i - 2 < a <. Since A, B are ² c :-potent, we can also assume M l ² c : s L l ² c :. Then, in G s ArM )² c: BrL, g s c a f ² u:² ¨ :. For, if c a s u i ¨ j, then u i s c d 1 and ¨ j s c d 2 . Thus, from the choice of M, L, 2 a < d 1 , d 2 . This implies c a s c d 1 c d 2 g ² c 2 a :, contradicting the choice of M. Hence, in G s ArM )² c: BrL, g s c a f ² u:² ¨ :. Ž2. g g A _ ² c : Žor g g B _ ² c :.. If g f ² u:² c :, then, by induction, there exists M 1f A such that g f M² u:² c :. Since B is ² c :-potent, there exists L 1f B such that L l ² c : s M l ² c :. Then g f ² u:² ¨ : in G s ArM )² c: BrL. If g s u i c a , then we have c a f ² u:² ¨ :. Thus, as in Ž1.,
CONJUGACY SEPARABILITY
1063
there exist M 1f A and L 1f B such that c a f ² u:² ¨ : in G s ArM )² c: BrL. Hence g f ² u:² ¨ :. Ž3. g s ab, where a g A _ ² c : and b g B _ ² c :. If a f ² u:² c : Žor, similarly, b f ² c :² ¨ :., then there exists M 1f A and L 1f B with M l ² c : s L l ² c : such that a f ² u:² c : and b g B _ ² c : in G s ArM )² c: BrL. Then g f ² u:² ¨ :. So assume a s u i c a and b s c b ¨ j. Since g s ab f ² u:² ¨ :, c aq b f ² u:² ¨ :. As in Ž1., there exist M 1f A and L 1f B such that c aq b f ² u:² ¨ : in G s ArM )² c: BrL. Hence g f ² u:² ¨ :. Ž4. g s ba or 5 g 5 G 3, where b g B _ ² c : and a g A _ ² c :. As in Case 1, we can find M 1f A and L 1f B such that 5 g 5 s 5 g 5, where G s ArM )² c: BrL. Then g f ² u:² ¨ :. Recall that a tree product of r vertices of R N-groups Ž r ) 1., amalgamating cyclic subgroups, can be written as G s A)² c: B, where A is a subtree product of r y 1 vertices, B is a R N-group, and c is in a vertex of A. For the rest of this section, we shall show that G is ² h: x ² k :-separable for x g G and h, k in some vertices of G. Applying induction, we can assume that A is ² u: a² ¨ :-separable for a g A whenever u, ¨ are in some vertices of A. LEMMA 3.6. Let G, A, B, c be as abo¨ e and h, k be in the ¨ ertices of G. We can find M 1f A and L 1f B with M l ² c : s L l ² c : such that, in G s ArM )² c: BrL, Ž1. if xc a f ² h: x ² k :, where x, h g A, k g B, and ² k : l ² c : s 1, then xc af ² h: x ² k :; Ž2. if xc a f ² h: x ² k :, where x, h g B, k g A, and ² k : l ² c : s 1, then xc af ² h: x ² k :. Proof. Ž1. Case 1. There exists a minimal integer m ) 0 such that xy1 h m x s c s, that is, h m x s xc s. Clearly c a f ² c s :² k :. By Lemma 3.5, we can find G s ArM )² c: BrL with M l ² c : s L l ² c : such that c a f ² c s :² k :, xy1 h i xf ² c : for all 0 - i - m. If xc as h i xk j for some i, j, then k j s c d for some d . Thus xy1 h i x s c ay d . This implies m < i. It follows that c ay d g ² c s :, thus c a g ² c s :² k :, a contradiction. Therefore, xc af ² h: x ² k :. Case 2. ² xy1 hx : l ² c : s 1. Since a / 0 and ² k : l ² c : s 1, c a f ² c 2 a :² k :. By induction, A is ² h e : x ² c e :-separable. Therefore, by Lemmas 2.1 and 3.5, we can find G s ArM )² c: BrL such that xy1 h i1 xs c i 2 only if e s 2 a < i1 , i 2 and c a f ² c 2 a :² k :. If xc as h i xk j for some i, j, then k j s c d for some d . Thus xy1 h i x s c ay d . This implies e < i, a y d , whence c ay d g ² c 2 a :. This means c a g ² c 2 a :² k :, a contradiction. Therefore, xc af ² h: x ² k :. Ž2. is similar to Ž1..
1064 LEMMA 3.7.
KIM AND TANG
Let G, A, B, c be as in Lemma 3.6 and let h, k be in the
¨ ertices of G. Then G is ² h: x ² k :-separable for x g A j B.
Proof. We only consider x g A, the case x g B being similar. Let g g G and g f ² h: x ² k :, where h, k are in the vertices of G. We shall find G s ArM )² c: BrL such that g f ² h: x ² k :, where M 1f A, L 1f B, and M l ² c : s L l ² c :. Then, since G is residually finite and < h <, < k < - `, ˜ s GrN. we can easily find N 1f G such that ˜ g f ²˜ h: ˜ x² ˜ k :, where G Case 1. h, k g A. If g g A then, by induction, there exists M 1f A such that g f ² h: x ² k : in A s ArM. Since B is ² c :-potent, there exists L 1f B such that M l ² c : s L l ² c :. Then, in G s ArM )² c: BrL, we have g f ² h: x ² k :. If g g B _ ² c : or 5 g 5 G 2, then we can find G s ArM )² c: BrL such that 5 g 5 s 5 g 5. Therefore g f ² h: x ² k :. Case 2. h g A and k g B or h g B and k g A. Since g f ² h: x ² k : is equivalent to gy1 f ² k : xy1 ² h:, it suffices to consider h g A and k g B. If ² k : l ² c : s ² k t : / 1, where k t s c s, then gkyi f ² h: x ² k t : s ² h: x ² c s : for 0 F i - t. Thus, by Case 1, we can find G such that gkyi f ² h: x ² c s : for 0 F i - t. This implies g f ² h: x ² k :. Therefore we may assume ² k : l ² c : s 1. Ža. g g A. If g f ² h: x ² c :, then, by Case 1, we can find G such that g f ² h: x ² c :. Since k g B, this implies g f ² h: x ² k :. So let g s h i xc a for some i, a . Then xc a f ² h: x ² k :. By Lemma 3.6, we can find G such a that xc f ² h: x ² k :. Hence g f ² h: x ² k :. Žb. g g B _ ² c :. If x f ² h:² c :, then, by Lemma 3.5, we can find G such that x f ² h:² c : and g f ² c :. This implies g f ² h: x ² k :. So let x s h i c a for some i, a . Then g f ² h: c a ² k :. This means gcya f ² h:² c a kcy a :. Since c a kcy a g B, by Lemma 3.5, we can find G such that gcya f ² h:²c a kcy a:. Hence g f ² h: x ² k :. Žc. g s a1 b1 , where a1 g A _ ² c : and b1 g B _ ² c :. If a1 f ² h: x ² c : or b1 f ² c :² k :, then, by induction and Lemma 3.5, we can find G such that 5 g 5 s 5 g 5 and a1 f ² h: x ² c : or b1 f ² c :² k :. Since k g B, we have g f ² h: x ² k :. So let a1 s h i xc a and b1 s c b k j for some i, j, a , b . This implies xc aq b f ² h: x ² k :. Thus, by Lemma 3.6, we can find G such that xcaq b f ² h: x ² k :. Hence g f ² h: x ² k :. Žd. g s b1 a1 or 5 g 5 G 3, where a1 g A _ ² c : and b1 g B _ ² c :. We can find G such that 5 g 5 s 5 g 5. Then g f ² h: x ² k :. Case 3. h, k g B. In this case, since G is ² h:² c e kcy e :-separable by Lemma 3.5, which is equivalent to ² h: c e ² k :-separable, we may assume x g A _ ² c :. Suppose ² h: l ² c : s ² h m :, where m ) 0. Let h m s c s. Then hyi g f ² h m : x ² k : s ² c s : x ² k : for all 0 F i - m. Thus, by Case 2, we can find G such that hyi gf ²c s: x ² k : for all 0 F i - m. This means ˜g f ² h: x ² k :. Therefore, we can assume ² h: l ² c : s 1 and similarly ² k : l ² c : s 1.
CONJUGACY SEPARABILITY
1065
Ža. g g A. Suppose g f ² c : x ² c :. Then, by Case 1, we can find G such that g f ² c : x ² c :. This means g f ² h: x ² k :. So let g s c a xc b . Moreover, we can assume a / 0 Žor b / 0.. Ži. C² c:Ž x . s 1. By Lemma 3.2, ² xy1 cx : l ² c : s 1. Since, by Case 1, G is ² c e : x ² c e :-separable, we can find G such that xy1 c i1 x s c i 2 only if e s 2 a < i1 , i 2 ; x f ² c : and c af ² h:² c 2 a : by Lemmas 2.1 and 3.5. Since h,k g B and x g A _ ² c :, if g s c a xc bs h i xk j for some i, j, then h i s c d 1 b and k j s c d 2 . Hence c a xc s c d 1 xcd 2 . This implies e < a y d 1 , b y d 2 . Thus c ay d 1 g ² c e :. This means c a s ² h:² c e :, a contradiction. Therefore g s c a xc bf ² h: x ² k :. Žii. C² c:Ž x . s ² c t : / 1. If c a f ² h:² c t : Žor similarly c a f ² c t :² k :., then we can find G such that x f ² c :, xy1 c i x f ² c : for all 0 - i - t, and c af ² h:² c t : by Lemma 3.5. If g s c a xc bs h i xk j for some i, j, then h i s c d 1 and k j s c d 2 . This implies t < a y d 1. Thus c a g ² h:² c t :, a contradiction. Hence g s c a xc bf ² h: x ² k :. Therefore we can assume c a g ² h:² c t : l ² c t :² k :. Since ² h: l ² c : s 1, c a s c n1 t. Similarly, c b s c n 2 t. This means a q b / 0, since c t g C² c:Ž x .. Thus we can find G such that x f ² c :, c aq b f ² c e : for e s 2Ž a q b ., xy1 c i x f ² c : for all 0 - i - t, h i1 s c i 2 only if e < i1 , i 2 , and k j1 s c j 2 only if e < j1 , j2 by Lemmas 3.3 and 2.1. If g s c a xc bs h i xk j for some i, j, then h i s c d 1 and k j s c d 2 . Thus e < d 1 , d 2 . Since t < a y d 1 , this implies c aq b s c d 1q d 2 g ² c e :, a contradiction. Therefore g s c a xc bf ² h: x ² k :. Žb. g g B. Since g f ² h: x ² k :, we have x f ² h: g ² k :, where g, h, k g B. Therefore, as in Case 1, we can find G such that x f ² h: g ² k :. Hence g f ² h: x ² k :. Žc. g s b1 a1 or g s a1 b 2 or g s b1 a1 b 2 , where a1 g A _ ² c : and bi g B _ ² c :. We consider only the case g s b1 a1 b 2 , other cases being similar. If b1 f ² h:² c : or b 2 f ² c :² k :, then, by Lemma 3.5, we can find G such that 5 g 5 s 5 g 5, x f ² c :, and b1 f ² h:² c : or b 2 f ² c :² k :. This implies g f ² h: x ² k :. Therefore, we assume b1 s h i c a and b 2 s c b k j. Thus hyi gkyj s c a a1 c b f ² h: x ² k :. Hence, by Ža., we can find G such that c a xc bf ² h: x ² k :. Therefore g f ² h: x ² k :. Žd. g s a1 b1 a2 or 5 g 5 G 4, where a i g A _ ² c : and b1 g B _ ² c :. We choose G such that 5 g 5 s 5 g 5. Then g f ² h: x ² k :. LEMMA 3.8. Let G, A, B, c be as in Lemma 3.6. Let h, k be in ¨ ertices of G. For a i g A _ ² c : and bi g B _ ² c :, we can find M 1f A and L 1f B with M l ² c : s L l ² c : such that, in G s ArM )² c: BrL. Ž1. if a1 c a b1 ??? a n bn f ² h: a1 b1 ??? a n bn² k :, then a1 c a b1 ??? a n bnf ² h:a1 b1 ??? a n bn² k :; Ž 2 . if a 1 c a b 1 ??? b n y 1 a n f ² h : a 1 b 1 ??? b n y 1 a n² k : , then a1 c a b1 ??? bny1 a nf ² h:a1 b1 ??? bny1 a n² k :;
1066
KIM AND TANG
Ž3. if b1 c a a1 ??? bn a n f ² h: b1 a1 ??? bn a n² k :, then b1 c a a1 ??? bn a nf ² h:b1 a1 ??? bn a n² k :; Ž 4 . if b 1 c a a 1 ??? a n y 1 b n f ² h : b 1 a 1 ??? a n y 1 b n² k : , then b1 c a a1 ??? a ny1 bnf ² h:b1 a1 ??? a ny1 bn² k :. Proof. We shall only prove Ž1., since the others are similar. Let u s b1 a2 ??? bny1 a n and C² c:Ž u. s ² c t :, where t G 0. Case 1. t s 0. By Lemma 3.2, there exists br Žor a r . in u such that ² bry1 cbr : l ² c : s 1. Subcase 1. There exists a minimal integer m ) 0 such that a1y1 h m a1 s c s. Thus a1y1 h i1 a1 f ² c : for 0 - i1 - m. Clearly c a f ² c s :. Let u s b1 a2 ??? bny 1 a n . Choose M 1f A and L 1f B with M l ² c : s L l ² c : such that, in G s ArM )² c: BrL, 5a1 ubn5 s 5 a1 ubn 5, c a f ² c s :, a1y1 h i1 a1f ² c : for 0 - i1 - m, and, moreover, by Lemma 3.7, since G is ² c e : br² c e :-separable, we can also have bry1 c i 2 br s c i 3 only if e s s < i 2 , i 3 Žby Lemma 2.1.. j If a1 c a ubns h ia1 ubn k for some i, j, then a1 s h i a1 c l1 , c a b1 s cyl 1 b1 c m 1 , . . . , br s cy l r br c m r , . . . , bn s cy l n bn k j for some l j , m j . This implies m < i, whence c l1 g ² c s :. Moreover, e s s < l r , m r . Thus e < m 1 , a q l1. It follows that c aq l1 g ² c s :, which means c a g ² c s :, a contradiction. Therefore, a1 c a ubnf ² h:a1 ubn² k :. Subcase 2. ² a1y1 ha1 : l ² c : s 1. Clearly a / 0. Let e s 2 a . We can find G such that 5a1 ubn5 s 5 a1 ubn 5, c a f ² c e :, and, moreover, by Lemma 3.7, since G is ² h e : a1² c e :-separable and ² c e : br² c e :-separable, we can also have a1y1 h i1 a1s c i 2 only if e < i1 , i 2 and bry1 c i 3 br s c i 4 only if e < i 3 , i 4 Žby j Lemma 2.1.. If a1 c a ubns h ia1 ubn k for some i, j, then a1 s h ia1 c l1 , c a b1 yl 1 m1 ylr mr s c b1 c , . . . , br s c br c , . . . , bn s cy l n bn k j for some l j , m j . This implies e < i, l1 and e < l r , m r . It follows e < m 1 , a q l1. Hence e s 2 a < a , a contradiction. Therefore, a1 c a ubnf ² h:a1 ubn² k :. Case 2. C² c:Ž u. s ² c t : / 1. Subcase 1. There exists a minimal integer m ) 0 such that a1y1 hX m a1 s c sX. Ža. There exists a minimal integer mX ) 0 such that bn k m bny1 s c s . X Thus bn k i1 bny1 f ² c : for 0 - i1 - mX . Clearly c a f ² c s :Ž² c t : l ² c s :.. Let ² c t : l ² c s9 : s ² c dt :. Choose G such that 5a1 ubn5 s 5 a1 ubn 5, c a f X ² c s :² c dt :, c i1 t f ² c s : for 0 - i1 - d, uy1 c i 2 uf ² c : for 0 - i 2 - t, a1y1 h i 3 a1f ² c : for 0 - i 3 - m, and bn k i 4 bny1f ² c : for 0 - i 4 - mX . If j a1 c a ubns h ia1 ubn k for some i, j, then a1 s h ia1 c l1 , c a u s cy l1ucl n , and yln j bn s c bn k for some l1 , l n . Hence we have m < i, t < a qX l1 , and mX < j. This implies cX l1 g ² c s :, c aq l1 s c l n g ² c t :, and c l n g ² c s :. Thus c aq l1 g ² c t : l ² c s : s ² c dt :. This means c a g ² c s :² c dt :, a contradiction. Therefore, a1 c a ubnf ² h:a1 ubn² k :.
CONJUGACY SEPARABILITY
1067
Žb. ² bn kbny1 : l ² c : s 1. Then c a f ² c s :. Choose G such that 5a1 ubn5 s 5 a1 ubn 5, c a f ² c s :, uy1 c i1 uf ² c : for 0 - i1 - t, a1y1 h i 2 a1f ² c : for 0 - i 2 - m, and bn k i 3 bny1s c i 4 only if e s s < i 3 , i 4 Žby Lemmas 3.7 and 2.1.. Applying similar argument in Ža., we get a1 c a ubnf ² h:a1 ubn² k :. Subcase 2. ² a1y1 ha1 : l ² c : s 1. X X Ža. There exists a minimal integer mX ) 0 such that bn k m bny1 s c s . X X i 1 y1 a t s t Thus bX n k bn f ² c : for 0 - i1 - m . Clearly c f ² c : l ² c :. Let ² c : l ² c s : s ² c dt :. Choose G such that 5a1 ubn5 s 5 a1 ubn 5, c a f ² c dt :, c i1 t X f ² c s : for 0 - i1 - d, uy1 c i 2 uf ² c : for 0 - i 2 - t, a1y1 h i 3 a1s c i 4 only if e s dt < i 3 , i 4 , and bn k i 5 bny1f ² c : for 0 - i 5 - mX . Then, as before, we get a1 c a ubnf ² h:a1 ubn² k :. Žb. ² bn kbny1 : l ² c : s 1. Clearly c a f ² c 2 a :. Choose G such that 5a1 ubn5 s 5 a1 ubn 5, c a f ² c 2 a :, uy1 c i1 uf ² c : for 0 - i1 - t, a1y1 h i 2 a1s c i 3 only if e s 2 a < i 2 , i 3 , and bn k i 4 bny1s c i 5 only if e s 2 a < i 4 , i 5 . Again, as before, we get a1 c a ubnf ² h:a1 ubn² k :. THEOREM 3.9. Let G be a tree product of a finite number of R N-groups amalgamating cyclic subgroups. Then G is ² h: x ² k :-separable for x g G, where h, k are in the ¨ ertices of G. Proof. If G has only one vertex, then the result follows from Lemma 3.4. So suppose G has r vertices. Then G s A)² c: B, where A is a subtree product of G consisting of r y 1 vertices with B as the remaining vertex and c in a vertex of A. By induction, we can assume that A is ² u: a² ¨ :separable for a g A and u, ¨ in the vertices of A. We shall prove the theorem by induction on 5 x 5. By Lemma 3.7, G is ² h: x ² k :-separable for x g A j B. Thus we assume that G is ² h: y ² k :-separable for any y g G with 5 y 5 - 5 x 5. Let g f ² h: x ² k :, where g, x g G and h, k are in the vertices of G. We shall find M 1f A and L 1f B with M l ² c : s L l ² c : such that, in G s ArM )² c: BrL, 5 x 5 s 5 x 5, 5 g 5 s 5 g 5, and g f ² h: x ² k :. Then, since G is residually finite and < h <, < k < - `, we can easily ˜ s GrN. Throughout the find N 1f G such that ˜ g f ²˜ h: ˜ x² ˜ k :, where G following, we let a i , c j g A _ ² c : and bi , d j g B _ ² c :. Moreover, we shall only consider the case x s a1 b1 ??? a n bn , other cases being similar. If 5 g 5 ) 5 x 5 q 2, then we can find G such that 5 g 5 s 5 g 5 and 5 x 5 s 5 x 5. Then g f ² h: x ² k :. If 5 g 5 - 5 x 5, then, by induction, we can find G such that x f ² h: g ² k :. This implies g f ² h: x ² k :. Thus we need only consider g g G with 5 x 5 F 5 g 5 F 5 x 5 q 2. Case 1. h, k g A. Ž1. 5 g 5 s 5 x 5. Suppose g s d1 c1 ??? d n c n . If a1 f ² h:² c :, then, by Lemma 3.5, we can find G s ArM )² c: BrL such that 5 g 5 s 5 g 5, 5 x 5 s 5 x 5, and a1 f ² h:² c :. Thus g f ² h: x ² k :. So let a1 s h i c a . This implies
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g f ² h: c a b1 ??? a n bn² k :. Hence, by induction, there exists G such that g f ² h:c a b1 ??? a n bn² k :. It follows that g f ² h: x ² k :. So let g s c1 d1 ??? c n d n . If c1 f ² h: a1² c : or d1 c2 ??? c n d n f ² c : b1 a2 ??? a n bn² k :, then, by induction, we can find G such that 5 g 5 s 5 g 5, 5 x 5 s 5 x 5, and c1 f ² h: a1² c : or d1 c2 ??? c n d nf ² c :b1 a2 ??? a n bn² k :. This implies g f ² h: x ² k :. Therefore, let c1 s h ia1 c a and d1 c 2 ??? c n d n s c b b1 a2 ??? a n bn k j. Then a1 c aq b b1 ??? a n bn f ² h: a1 b1 ??? a n bn² k :. Hence, by Lemma 3.8, we can find G such that a1 c aq b b 1 ??? a n bnf ² h:a1 b1 ??? a n bn² k :. It follows that g f ² h: x ² k :. Ž2. 5 g 5 s 5 x 5 q 1. Suppose g s d1 c1 ??? d n c n d nq1. Then we can find G such that 5 g 5 s 5 g 5 and 5 x 5 s 5 x 5. This implies g f ² h: x ² k :. So let g s c1 d1 ??? c n d n c nq1. If c nq1 f ² c :² k :, then we can choose G such that 5 g 5 s 5 g 5, 5 x 5 s 5 x 5, and c nq 1 f ² c :² k :. Thus g f ² h: x ² k :. Therefore, we let c nq 1 s c a k j. This means gkyj s c1 d1 ??? c n d n c a f ² h: x ² k :. Hence, by Ž1., we can find G such that gkyj f ² h: x ² k :. Thus g f ² h: x ² k :. Ž3. 5 g 5 s 5 x 5 q 2. We choose G such that 5 g 5 s 5 g 5 and 5 x 5 s 5 x 5. Then g f ² h: x ² k :. Case 2. h g A and k g B. Ž1. 5 g 5 ) 5 x 5. We choose G such that 5 g 5 s 5 g 5 and 5 x 5 s 5 x 5. Then g f ² h: x ² k :. Ž2. 5 g 5 s 5 x 5. Suppose g s d1 c1 ??? d n c n . Then we can find G such that 5 g 5 s 5 g 5 and 5 x 5 s 5 x 5. This implies g f ² h: x ² k :. So let g s c1 d1 ??? c n d n . Then we can apply the method of the second part of Case 1Ž1.. Case 3. h, k g B. Ž1. 5 g 5 s 5 x 5. Suppose g s d1 c1 ??? d n c n . If bn f ² c :² k :, then, by Lemma 3.5, we can find G such that 5 g 5 s 5 g 5, 5 x 5 s 5 x 5, and bn f ² c :² k :. This implies g f ² h: x ² k :. So suppose bn s c a k j. Then g f ² h: a1 b1 ??? a n c a ² k :. Hence, by induction, we can find G such that g f ² h:a1 b1 ??? a n c a² k :. It follows that g f ² h: x ² k :. So let g s c1 d1 ??? c n d n . Then we can apply the method of the second part of Case 1Ž1.. Ž2. 5 g 5 s 5 x 5 q 1 or 5 g 5 s 5 x 5 q 2. Suppose g s d1 c1 ??? d n c n d nq1. If d1 c1 f ² h: a1² c : or d 2 c 2 ??? c n d nq1 f ² c : b1 a2 ??? a n bn² k :, then, by induction, we can find G such that 5 g 5 s 5 g 5, 5 x 5 s 5 x 5, and d1 c1f ² h: a1² c : or d 2 c 2 ??? c n d nq1f ² c :b1 a2 ??? a n bn² k :. This implies g f ² h: x ² k :. So let d1 c1 s h ia1 c a and d 2 c 2 ??? c n d nq1 s c b b1 a2 ??? a n bn k j. Then a1 c aq b b1 ??? a n bn f ² h: a1 b1 ??? a n bn² k :. Hence, by Lemma 3.8, we can find G such that a1 c aq b b1 ??? a n bnf ² h:a1 b1 ??? a n bn² k :. It follows that g f ² h: x ² k :. Now suppose g s c1 d1 ??? c n d n c nq1 or 5 g 5 s 5 x 5 q 2. Then we can choose G such that 5 g 5 s 5 g 5 and 5 x 5 s 5 x 5. This implies g f ² h: x ² k :.
CONJUGACY SEPARABILITY
1069
Case 4. h g B and k g A. Ž1. 5 g 5 s 5 x 5. Suppose g s d1 c1 ??? d n c n . Then we can choose G such that 5 g 5 s 5 g 5 and 5 x 5 s 5 x 5. This implies g f ² h: x ² k :. So let g s c1 d1 ??? c n d n . Then we can apply the method of the second part of Case 1Ž1.. Ž2. 5 g 5 s 5 x 5 q 1. Suppose g s c1 d1 ??? c n d n c nq1. Again we can apply the method of the second part of Case 1Ž1.. So let g s d1 c1 ??? d n c n d nq1. In this case, we can apply the method of Case 3Ž2.. Ž3. 5 g 5 s 5 x 5 q 2. Suppose g s c1 d1 ??? d n c nq1 d nq1. Then we can choose G such that 5 g 5 s 5 g 5 and 5 x 5 s 5 x 5. This implies g f ² h: x ² k :. So let g s d1 c1 ??? c n d nq1 c nq1. Again we can apply the method of Case 3Ž2..
4. MAIN RESULTS To prove our main result we need the following lemmas. LEMMA 4.1. Let G be a tree product of a finite number of R N-subgroups amalgamating cyclic subgroups. If h is in a ¨ ertex group of G, then, for each n, there exists N 1f G such that Ž1. < ˜ h < s n and Ž2. ˜ h i ¤G˜ ˜ h j for ˜ hi / ˜ h j, ˜ s GrN. where G Proof. Let G be a tree product of r R N-groups. We shall prove the lemma by induction on r. By Corollary 2.2 in w23x, every R N-group has properties Ž1. and Ž2.. Now let G s A)² c: B, where B g R N, A is a subtree product of G with r y 1 vertices, and c is in a vertex of A. By induction, A has the properties Ž1. and Ž2.. We shall find M 1f A and L 1f B such that M l ² c : s L l ² c : and < h < s n and h i ¤G h j for h i / h j in G s ArM )² c: BrL. Since G is conjugacy separable by Theoi j rem 1.2, we can find N 1f G such that N l ² h: s 1 and Nh ¤G r N Nh i j for h / h . Let N be the preimage of N in G. Then N 1f G and < ˜ h< s n ˜ s GrN. and ˜ h i ¤G˜ ˜ h j for ˜ hi / ˜ h j, where G Suppose h g A Žthe case h g B is similar.. Since h is in a vertex of A, by induction there exists M 1f A such that < h < s n and h i ¤A h j for h i / h j, where A s ArM. Let M l ² c : s ² c s :. Now B g R N implies that there exists N 1f B such that < c < s s and c i ¤B c j for c i / c j, where B s BrL. Therefore, in G s ArM )² c: BrL and h i ¤G h j for h i / h j as in the proof of Lemma 3.2. LEMMA 4.2. Let G be a tree product of a finite number of conjugacy separable R N-groups amalgamating cyclic subgroups. Let h be in a ¨ ertex group of G. If there is x g G such that x 4G l ² h: s B, then there exists ˜ ˜ s GrN. N 1f G such that ˜ x 4G l ² ˜ h: s B, where G
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Proof. By Theorem 3.1 in w14x, every conjugacy separable R N-group is cyclic conjugacy separable. So let G s A)² c: B as in the proof of Lemma 4.1. Thus, by induction, A is cyclic conjugacy separable for ² u:, where u is in any vertex of A. Let h be in a vertex group of G and x g G such that x 4G l ² h: s B. We may assume that x has the minimal length in its conjugacy class in G. We shall find M 1f A and L 1f B such that M l ² c : s L l ² c : and x 4G l ² h: s B, where G s ArM ) ² c: BrL. Since G is conjugacy separable and ² h: is finite, there exists N 1f G such that Nx4G r N l ²Nh: s B. Let N be the preimage of N in G. Then ˜ ˜ s GrN. Thus, without loss of generalN 1f G and ˜ x 4G l ² ˜ h: s B in G ity, we assume h is in a vertex of A. Ž1. 5 x 5 G 2. Since x is of minimal length in its conjugacy class in G, x is cyclically reduced. By Lemmas 3.1 and 3.3, A, B are ² c :-potent and ² c :-separable. Therefore, there exist M 1f A and L 1f B such that M l ² c : s L l ² c : and 5 x 5 s 5 x 5, where G s ArM )² c: BrL. Since h g A and x is cyclically reduced with length G 2 in its conjugacy class in G, by Theorem 1.1, x 4G l ² h: s B. Ž2. x g A _ ² c :. Clearly x 4 A l ² c : s B and x 4 A l ² h: s B. Since c, h are in vertices of A, by induction, there exists M 1f A such that x 4 A l ² c : s B and x 4 A l ² h: s B, where A s ArM. Since B is ² c :potent, there exists L 1f B such that M l ² c : s L l ² c :. Thus, in G s ArM )² c: BrL, x is of minimal length 1 in its conjugacy class in G. By Theorem 1.1, x ¤G h i for all i. Hence x 4G l ² h: s B. Ž3. x g B _ ² c :. Clearly x 4 B l ² c : s B. Since B is cyclic conjugacy separable by w14, Theorem 3.1x, there exists L 1f B such that x 4 B l ² c : s B, where B s BrL. By Lemma 3.1, A is ² c :-potent. So there exists M 1f A such that M l ² c : s L l ² c :. Thus, in G s ArM )² c: BrL, x g B is of minimal length 1 in its conjugacy class in G. Since h g A, by Theorem 1.1, x ¤G h i for all i. Hence x 4G l ² h: s B. Ž4. x g ² c :. Clearly x 4 A l ² h: s B. By induction, since h is in a vertex of A and x g ² c :, there exists M 1f A such that x 4 A l ² h: s B, where A s ArM. By Lemma 4.1, there exists L 1f B such that M l ² c : s L l ² c : and c i ¤B c j for c i / c j, where B s BrL. Thus, in G s ArM )² c: BrL, x ¤G h i for all i by Theorem 1.1. Hence x 4G l ² h: s B. THEOREM 4.3. Let G be a tree product of a finite number of conjugacy separable R N-groups amalgamating cyclic subgroups. Then G is conjugacy separable. Proof. where A separable that A is
Let G be a tree product of r R N-groups. Then G s A)² h: B, is a subtree product of r y 1 vertices of G, B is a conjugacy R N-group, and h is in a vertex of A. By induction, we assume conjugacy separable. Then A, B are conjugacy separable, cyclic
CONJUGACY SEPARABILITY
1071
conjugacy separable for ² h: ŽLemma 4.2., and ² h:-potent ŽLemma 3.1.. Moreover, by Lemmas 3.2, 3.9, and 4.1, A, B satisfy C1, C2, and C3 of Theorem 2.3. Hence, by Theorem 2.3, G is conjugacy separable. Applying Theorem 4.3, we immediately have the following: THEOREM 4.4. Let G be a tree product of a finite number of finitely generated torsion-free nilpotent or free or surface groups, amalgamating cyclic subgroups. Then G is conjugacy separable.
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20. M. Shirvani, On conjugacy separability of fundamental groups of graphs of groups, Trans. Amer. Math. Soc. 334 Ž1. Ž1992., 229]243. 21. P. F. Stebe, Conjugacy separability of certain free products with amalgamation, Trans. Amer. Math. Soc. 156 Ž1971., 119]129. 22. P. F. Stebe, Conjugacy separability of certain Fuchsian groups, Trans. Amer. Math. Soc. 163 Ž1972., 173]188. 23. C. Y. Tang, Conjugacy separability of generalized free products of surface groups. J. Pure Appl. Algebra, to appear. 24. W. F. Thurston, Three dimensional manifolds, Kleinian groups and hyperbolic geometry, Bull. Amer. Math. Soc. 6 Ž1982., 357]381.