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A Criterion for the Conjugacy Separability of Certain HNN Extensions of Groups
Journal of Algebra 222, 574᎐594 Ž1999. doi:10.1006rjabr.1999.8034, available online at http:rrwww.idealibrary.com on
A Criterion for the Conjugacy Separability of Certain HNN Extensions of Groups* Goansu Kim Department of Mathematics, Yeungnam Uni¨ ersity, Kyongsan 712-749, Korea E-mail: [email protected]
and C. Y. Tang Uni¨ ersity of Waterloo, Waterloo, Ontario N2L 3G1, Canada, and California State Uni¨ ersity, Hayward, California E-mail: [email protected] Communicated by Walter Feit Received June 29, 1998
We first prove the double coset separability of certain HNN extensions with cyclic associated subgroups. Using this we prove a criterion for the conjugacy separability of HNN extensions of conjugacy separable groups with cyclic associated subgroups. Applying this result we show that certain HNN extensions of free products of cycles and certain Baumslag᎐Solitar groups with cyclic associated 䊚 2000 Academic Press subgroups are conjugacy separable. Key Words: HNN-extension, conjugacy separable, residually finite.
1. INTRODUCTION A group G is conjugacy separable if, for each x, y g G such that x and y are not conjugate in G, there exists a finite homomorphic image G of G such that the images of x and y in G are not conjugate in G. * The first author was partly supported by KOSEF through the GARC at S.N.U. and KOSEF 985-0100-002-2. The second author gratefully acknowledges the partial support by the National Science and Engineering Research Council of Canada, Grant No. A-4064. 574 0021-8693r00 $35.00 Copyright 䊚 2000 by Academic Press All rights of reproduction in any form reserved.
CONJUGACY SEPARABILITY
575
In this paper we consider the conjugacy separability of HNN extensions G s ² A, t : ty1 ht s k : of a conjugacy separable group A with cyclic associated subgroups ² h: and ² k :. Residual properties of HNN extensions are difficult to obtain. They depend very much on the choice of h, k g A. For example, the Baumslag᎐Solitar group, ² a, t: ty1 a t s a ␦ :, is an HNN extension of the cyclic group ² a:. It is residually finite w13x if and only if < < s 1 or < ␦ < s 1 or s "␦ . In w16x, Raptis, Talelli, and Varsos studied the Hopficity and the residual finiteness of HNN extensions of Baumslag᎐Solitar groups with cyclic associated subgroups of the form ² t, a, b: ty1 a t s b , by1 a b s a :. Recently, in w6, Definition 2.4x, Kim and Tang introduced the concept of a group to be quasi-regular at h, k 4 . If A is quasi-regular at h, k 4 , then the residual properties of HNN extension G of A are nicely preserved ŽTheorem 2.7.. Using this, they characterized the residual finiteness and cyclic subgroup separability of G when A is a finitely generated abelian group. They also characterized the conjugacy separability of G when A is a finitely generated abelian group w8x. In this paper we prove the double coset separability of HNN extensions of groups which are quasi-regular at the associated subgroups ŽLemma 3.4.. Using this, we derive a criterion for the conjugacy separability of HNN extensions with cyclic associated subgroups ŽTheorem 4.5.. In general it is difficult to get groups to satisfy double coset separability for associated subgroups. However, if the associated subgroups are in the center of a polycyclic-by-finite group, then we can easily apply the criterion to show that such HNN extensions are conjugacy separable ŽCorollary 4.6.. This gives us an extension of a result in w8x for HNN extensions of abelian groups. We also apply the criterion to show that HNN extensions of free products of cycles with cyclic associated subgroups are conjugacy separable ŽTheorem 5.4.. Finally, we show that certain HNN extensions of Baumslag᎐Solitar groups with cyclic associated subgroups are conjugacy separable ŽTheorem 6.7..
2. PRELIMINARIES Throughout this paper we use standard notations and terminology for this topic w12x. The letter G always denotes a group. In addition: N ef G means that N is a normal subgroup of finite index in G; x ;G y means x is conjugate to y in G. We use 5 x 5 to denote the length of x in HNN-extensions and generalized free products.
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We shall make extensive use of the following two results by D. J. Collins. THEOREM 2.1 w3, Theorem 3x. Let x and y be cyclically reduced elements of the HNN-extension G s ² B, t: ty1 Ht s K :. Suppose that x ;G y. Then 5 x 5 s 5 y 5, and one of the following holds. Ž1. 5 x 5 s 5 y 5 s 0 and there is a finite sequence z1 , z 2 , . . . , z m of elements in H j K such that x ;B z1 ;B, t* z 2 ;B, t* ⭈⭈⭈ ;B, t* z m ;B y, where u ;B, t* ¨ means one of: Ži. u ;B ¨ , Žii. u g H and ¨ s ty1 ut Žg K ., or Žiii. u g K and ¨ s tuty1 Žg H .. Ž2. 5 x 5 s 5 y 5 G 1 and y ;H j K x* where x* is a cyclic permutation of x. A group G is said to be S-separable if for each x g G _ S there exists N 1f G such that x f NS. In particular, G is residually finite Ž RF . if G is 14 -separable. A group G is said to be cyclic subgroup separable, briefly c , if G is ² x :-separable for each x g G. A group G is conjugacy separable if, for each x g G, G is x 4G -separable, where x 4G is the set of all conjugates of x in G. THEOREM 2.2 w3, Theorem 13x. If A is conjugacy separable and H, K are finite, then the HNN-extension ² A, t: ty1 Ht s K : is conjugacy separable. DEFINITION 2.3. Let A be a group and let h, k g A be of infinite order. Then A is said to be quasi-regular at h, k 4 if, for each integer ⑀ ) 0, there exist ana integer ⑀ ) 0 and N⑀ ef A, depending on ⑀ , such that N⑀ l ² h: s ² h ⑀⑀ : and N⑀ l ² k : s ² k ⑀⑀ :. In w14x, Niblo defined that a group A has regular quotients at h, k 4 if there exists a positive integer , such that for each positive integer ⑀ , there exists N ef A such that N l ² h: s ² h ⑀ : and N l ² k : s ² k ⑀ :. Thus, in the case of regular quotients, the integer is independent of ⑀ . Hence, if A has regular quotients at h, k 4 , then A is quasi-regular at h, k 4 . Remark 2.4. Let G s ² A, t: ty1 ht s k :. For N ef A such that N l ² h: s ² h s : and N l ² k : s ² k s :, there is a natural homomorphism
N : ² A, t : ty1 ht s k : ª ² A, : y1 h s k : by a N s a for a g A, where A s ArN, and t N s . Then, by Theorem 2.2, G N is conjugacy separable. LEMMA 2.5. Let G s ² A, t: ty1 ht s k :. Suppose A is ² h:-separable, ² k :-separable, and quasi-regular at h, k 4 . Then, for each M ef A, for each s ) 0 and for each reduced element x g G, there exist N ef A and ) 0 such that N ; M, N l ² h: s ² h s :, N l ² k : s ² k s : and 5 x 5 s 5 x 5 in G s G N .
CONJUGACY SEPARABILITY
577
Proof. Let x s u1 t ⑀ 1 u 2 t ⑀ 2 ⭈⭈⭈ t ⑀ n u nq1 g G be reduced where u i g A and ⑀ i s "1. Since x is reduced, we have u iq1 f ² h: if ⑀ i s y⑀ iq1 s y1 and u iq1 f ² k : if ⑀ i s y⑀ iq1 s 1. Since A is ² h:-separable and ² k :-separable, we can find Ni ef A such that if ⑀ i s ⑀ iq1 then Ni s A, or if ⑀ i s y⑀ iq1 s y1 then u iq1 f Ni ² h:, or if ⑀ i s y⑀ iq1 s 1 then u iq1 f n n Ni ² k :. Let l is1 Ni l M l ² h: s ² h s1 : and l is1 Ni l M l ² k : s ² k s 2 :. Let ⑀ s s1 s2 s. By quasi-regularity, there exist 1 and Ns ef A such that n Ns l ² h: s ² h ⑀1 : and Ns l ² k : s ² k ⑀1 :. Let N s l is1 Ni l M l Ns s s and s s1 s2 1. Then N ef A, N ; M, N l ² h: s ² h 1 2 1 s : s ² h s :, N l ² k : s ² k s1 s 2 1 s : s ² k s : and, in G s G N above, x s u1 ⑀ 1 u 2 ⑀ 2 ⭈⭈⭈ ⑀ n u nq 1 is reduced with 5 x 5 s 5 x 5. Remark 2.6. Let G s ² A, t: ty1 ht s k :. Suppose A is quasi-regular at h, k 4 and h i ;G h j. Let ⑀ s i. Then there exist and N ef A such that N l ² h: s ² h i : and N l ² k : s ² k i :. In G s G N , < h < s i s < k <. Since h i ;G h j in G s G N , we have 1 s h i ;G h j . This implies i < j . Hence i < j. Similarly j < i. Therefore, if A is quasi-regular at h, k 4 and h i ;G h j then j s "i. Similarly, if A is quasi-regular at h, k 4 and k i ;G k j or h i ;G k j then j s "i. The following theorem shows that quasi-regularity is closely related to the cyclic subgroup separability of HNN extensions. THEOREM 2.7 w6x. Let A be c and let h, k g A be of infinite order. Then G s ² A, t: ty1 ht s k : if c if and only if A is quasi-regular at h, k 4 . In this paper we consider the conjugacy separability of HNN extensions of the type G s ² A, t : ty1 ht s k : , where ² h: l ² k : s 1 and A is conjugacy separable. If < h < s < k < is finite then, by Theorem 2.2, G is conjugacy separable. Thus, we need only consider the case when h, k are of infinite order.
3. DOUBLE COSET SEPARABILITY Throughout this note, we consider the HNN extension G s ² A, t: ty1 ht s k : of a base group A with infinite cyclic associated subgroups ² h: and ² k :, where ² h: l ² k : s 1. DEFINITION 3.1. A group A is said to be double coset separable at h, k 4 if, for each u g A and for each integer ⑀ ) 0, A is ² h ⑀ : u² h ⑀ :-separable, ² h ⑀ : u² k ⑀ :-separable, and ² k ⑀ : u² k ⑀ :-separable.
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We note that if A is ² h ⑀ : u² k ⑀ :-separable for all u g A then A is ² k ⑀ : uy1 ² h ⑀ :-separable for all u g A. It follows that if A is double coset separable at h, k 4 then A is also ² k ⑀ : u² h ⑀ :-separable for all u g A. In the case of a free-by-finite or polycyclic-by-finite group G, it was known by w11, 15x that G is ² a: x ² b :-separable for all a, b, x g G. Hence G is double coset separable at h, k 4 for all h, k g G. Clearly if a group G is double coset separable at h, k 4 , then G is ² h ⑀ :-separable, ² k ⑀ :-separable, and ² h ⑀ :² k ⑀ :-separable for any ⑀ . Remark 3.2. If a group G is ² h ⑀ : u² k ⑀ :-separable for any ⑀ , then G is ² h ␣ : u² k  :-separable for any ␣ ,  . For, if g f ² h ␣ : u² k  : then l l hy␣ i gky  j f ² h l : u² k l : for each 0 F i - ␣ and 0 F j -  , where l is the least common multiple of ␣ ,  . Therefore, there exists N ef G such that l l hy␣ i gky  j f N² h l : u² k l : for all 0 F i - ␣ and 0 F j -  . This implies that g f N² h ␣ : u² k  :. Hence G is ² h ␣ : u² k  :-separable for any ␣ ,  . Similarly, if a group G is double coset separable at h, k 4 , then G is ² h ␣ : u² h  :-separable, ² k ␣ : u² k  :-separable, and ² h ␣ : u² k  :-separable for any ␣ ,  . LEMMA 3.3 w9x. Let G be ² a ⑀ : x ² b ⑀ :-separable, where x, a, b g G and a, b are of infinite order. If ² xy1 ax : l ² b : s 1, then there exists N ef G such that xy1 a i x s b j only if ⑀ < i, j, where G s GrN. Note that the condition ‘‘ xy1 a i x s b j only if ⑀ < i, j’’ always implies ⑀ < < a <, < b <. For convenience, we say N g N if N ef A such that N l ² h: s ² h ⑀ : and N l ² k : s ² k ⑀ :. In this case, we can construct G s G N as in Remark 2.4. LEMMA 3.4. Let A be quasi-regular and double coset separable at h, k 4 . Then G s ² A, t: ty1 ht s k : is double coset separable at h, k 4 . Proof. To show that, for example, G is ² h s : x ² k s :-separable for any s ) 0, we let y, x g G such that y f ² h s : x ² k s :. Then, by Lemma 2.5, we ˆ s G N1. If we can find N1 g N such that 5 ˆ y 5 s 5 y 5 and 5 ˆ x 5 s 5 x 5, in G can find N g N such that N ; N1 and y f ² h s : x ² k s : in G s G N , then there exists Lef G such that y f L² h s : x ² k s :, since G is residually finite and < h < s < k < is finite. Let L be the preimage of L in G. Then Lef G and y f L² h s : x ² k s :. This is the general sketch of our long proof. We prove the theorem by induction on 5 x 5. Suppose 5 x 5 s 0, i.e., x g A, we shall show that, for any s ) 0 and x g A, G is ² h s : x ² h s :-separable, ² h s : x ² k s :-separable, and ² k s : x ² k s :-separable. To show that G is ² h s : x ² k s :-separable for x g A, let y g G such that y f ² h s : x ² k s :. Suppose 5 y 5 s 0, that is, y g A. Then, by the double coset separability of A, there exists N2 ef A such that y f N2 ² h s : x ² k s :.
CONJUGACY SEPARABILITY
579
As in Lemma 2.5, we choose N g N such that N ; N2 . Thus y f ² h s : x ² k s : in G s G N . Suppose 5 y 5 G 1. Again, by Lemma 2.5, we choose N g N such that 5 y 5 s 5 y 5 in G s G N . Since 5 y 5 G 1 and x g A, y f ² h s : x ² k s : in G s G N . This proves that G is ² h s : x ² k s :-separable for x g A. Similarly, for any s ) 0, G is ² h s : x ² h s :-separable and ² k s : x ² k s :-separable for x g A. Let 5 x 5 s n. By induction we can assume that, for any s ) 0, G is ² h s : g ² h s :-separable, ² h s : g ² k s :-separable, and ² k s : g ² k s :-separable for all g g G such that 5 g 5 F n y 1. We need to show that G is ² h s : x ² h s :separable, ² h s : x ² k s :-separable, and ² k s : x ² k s :-separable for x g G with 5 x 5 s n. Case 1. To show G is ² h s : x ² k s :-separable for any x g G with 5 x 5 s n. Let y g G such that y f ² h s : x ² k s :. If 5 y 5 ) n or 5 y 5 - n or y s ¨ 1 t ␦ 1 ¨ 2 ⭈⭈⭈ t ␦ n ¨ nq1 and x s u1 t ⑀ 1 u 2 ⭈⭈⭈ t ⑀ n u nq1 with ␦ i / ⑀ i for some i then, by Lemma 2.5, we can find N1 g N such that, in G s G N1, 5 y 5 s 5 y 5 and 5 x 5 s 5 x 5. This implies y f ² h s : x ² k s : in G s G N1. So we consider y s ¨ 1 t ⑀ 1 ¨ 2 ⭈⭈⭈ t ⑀ n ¨ nq1 and x s u1 t ⑀ 1 u 2 ⭈⭈⭈ t ⑀ n u nq1 where y f ² h s : x ² k s :. For convenience, we let x ny 1 s u1 t ⑀ 1 u 2 ⭈⭈⭈ t ⑀ ny 1 u n and yny1 s ¨ 1 t ⑀ 1 ¨ 2 ⭈⭈⭈ t ⑀ ny 1 ¨ n . By Lemma 2.5, there exists N1 g N such that, in G s G N1, 5˜ x5 s 5 ˜ y 5 s n. We need to consider the cases ⑀ n s 1 and ⑀ n s y1. Subcase 1. Suppose ⑀ n s 1. If yny1 f ² h s : x ny1² h: or ¨ nq1 f ² k : u nq 1² k s :, then, by induction and Remark 3.2, we can find M ef G such that yny 1 f M Ž² h s : x ny1² h:. or ¨ nq1 f M² k : u nq1² k s :. By Lemma 2.5, we can find N g N such that N ; N1 l M and 5 x 5 s 5 y 5 s n in G s G N . If y s h si x k s j for some i, j, then ysi y1 xy1 yny1 s u nq1 k s j ¨ y1 ny 1 h nq1 .
Ž 3.1.
This implies ysi xy1 yny1 s h ␣ ny 1 h
and
k ␣ s u nq1 k s j ¨ y1 nq1
Ž 3.2.
for some ␣ , hence yny 1 g ² h s : x ny1² h: and ¨ nq1 g ² k : u nq1² k s :, contradicting the choice of N ; M. Thus y f ² h s : x ² k s :. Therefore we suppose yny 1 s h si1 x ny1 h i 2 and ¨ nq1 s k j1 u nq1 k s j 2 for some i1 , i 2 , j1 , j2 . Then y s yny 1 t¨ nq1 s h si1 x ny1 h i 2 tk j1 u nq1 k s j 2 f ² h s : x ny1 tu nq1² k s : . Ž 3.3. ² s : x ny1 tu nq1² k s : uy1 This implies that h i 2 tk j1 f xy1 ny 1 h nq1 . Hence, we need to find N g N such that y f ² h s : x ² k s : for each of the following four cases: : ² : ² y1 : ² : 1-a. Suppose ² xy1 ny 1 hx ny1 l h s 1 and u nq1 ku nq1 l k s 1.
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We note that i 2 q j1 / 0, since, in Ž3.3., h i 2 tk j1 s h i 2qj 1 t. Let ⑀ ) < i 2 q j1 <. By induction, G is ² h ⑀ : x ny1² h ⑀ :-separable and ² k ⑀ : u nq1² k ⑀ :-separable. Hence, by Lemma 3.3, there exist M1 ef G and M2 ef G such ˜i x ny1 s ˜h j in GrM1 then ⑀ < i, j and that if uˆy1 ˆ i ˆnq1 s ˆk j that if ˜ xy1 ny 1 h ˜ nq1 k u in GrM2 then ⑀ < i, j. Let N g N such that N ; M1 l M2 l N1 as in Lemma 2.5. In G s G N , if y g ² h s : x ² k s :, then y s h si xk s j. Thus, by the equation Ž3.2., we have yny 1 s h si x ny1 h ␣ s h si1 x ny1 h i 2 and ¨ nq1 s ky␣ u nq 1 k s j s k j1 u nq1 k s j 2 for some ␣ . This implies that sŽ iyi 1 . xy1 x ny1 s h i 2y ␣ ny 1 h
and
y ␣ yj1 uy1 u nq1 s k sŽ j 2yj. . nq1 k
Ž 3.4.
Thus ⑀ < i 2 y ␣ and ⑀ < y ␣ y j1 by the choice of N ; M1 l M2 . Hence ⑀ < i 2 q j1 , contradicting the choice of ⑀ . Hence y f ² h s : x ² k s :. : ² : ² y1 : ² : 1-b. Suppose ² xy1 ny 1 hx ny1 l h s 1 and u nq1 ku nq1 l k / 1. y1 ␣2 Let ␣ 2 ) 0 be the smallest integer such that u nq1 k u nq1 g ² k :. Then, ␣2 by Remark 2.6, we have uy1 u nq1 s k " ␣ 2 . By the equation Ž3.3., since nq 1 k i2 j1 i 2qj 1 i 2qj 1 h tk s tk , we note that k f ² k ␣ 2 : l ² k s : s ² k ␣ 2 sr d 2 :, where d 2 s Ž ␣ 2 , s .. Thus ␣ 2 srd 2 does not divide i 2 q j1. Let ⑀ s Ž ␣ 2 srd 2 . l and, i ² : as before, we can find N g N such that N ; N1 and uy1 nq1 k u nq1 f k y1 i j for all 1 F i - ␣ 2 , and if x ny1 h x ny1 s h then ⑀ < i, j. Thus, in G s G N , if y g ² h s : x ² k s : then equations of Ž3.4. hold. This implies ⑀ < i 2 y ␣ and ␣ 2 < y ␣ y j1. Now s < ⑀ and ⑀ < < h < s < k <. Hence s < < k <. This implies s < y ␣ y j1 by the second equation of Ž3.4.. Therefore Ž ␣ 2 srd 2 .< i 2 y ␣ y Žy␣ y j1 . s i 2 q j1 , a contradiction. Hence y f ² h s : x ² k s :. : ² : ² y1 : ² : 1-c. Suppose ² xy1 ny 1 hx ny1 l h / 1 and u nq1 ku nq1 l k s 1. ␣ 1 y1 Let ␣ 1 ) 0 be the smallest integer such that x ny1 h x ny1 g ² h:. Then " ␣1 by Remark 2.6 x ny 1 h ␣ 1 xy1 . Since h i 2 tk j1 s h i 2qj 1 t, from the ny1 s h i 2qj 1 Ž . ² equation 3.3 , we have h f h ␣ 1 : l ² h s : s ² h ␣ 1 sr d1 :, where d1 s Ž ␣ 1 , s .. Thus ␣ 1 srd1 does not divide i 2 q j1. Hence, as in 1-b above, for ⑀ s Ž ␣ 1 srd1 . l, we can find N g N such that N ; N1 and y f ² h s : x ² k s : in G s G N . : ² : ² y1 : ² : 1-d. Suppose ² xy1 ny 1 hx ny1 l h / 1 and u nq1 ku nq1 l k / 1. As in 1-b and 1-c above, let ␣ 1 , ␣ 2 ) 0 be the smallest integer such y1 ␣2 ² : that x ny 1 h ␣ 1 xy1 u nq1 g ² k :. By Eq. Ž3.3., h i 2qj 1 f ny1 g h and u nq1 k ␣ s ␣ s ␣ sr Ž² h 1 : l ² h :.Ž² h 2 : l ² h :. s ² h 1 d1 :² h ␣ 2 sr d 2 :, where d1 s Ž ␣ 1 , s . and d 2 s Ž ␣ 2 , s .. Hence, as before, for ⑀ s Ž ␣ 1 srd1 .Ž ␣ 2 srd 2 . l we can ² : find N g N such that N ; N1 and, in G s G N , x ny1 h i xy1 ny1 f h for all y1 j ² : 1 F i - ␣ 1 , u nq1 k u nq1 f k for all 1 F j - ␣ 2 , and, by Remark 3.2, h i 2qj1 f ² h ␣ 1 sr d1 :² h ␣ 2 sr d 2 :. In G s G N , if y g ² h s : x ² k s : then we have the equation Ž3.4.. Thus, from the first equation of Ž3.4., we have ␣ 1 < i 2 y ␣ ;
CONJUGACY SEPARABILITY
581
hence h i 2y ␣ s h " sŽ iyi1 .. Since s < ⑀ and ⑀ < < h <, s < i 2 y ␣ . Thus Ž ␣ 1 srd1 .< i 2 y ␣ . Hence h i 2y ␣ g ² h ␣ 1 sr d1 :. Similarly, from the second equation of Ž3.4., hy␣ yj1 g ² h ␣ 2 sr d 2 :. Therefore, h i 2qj1 s h i 2y ␣ h ␣qj1 g ² h ␣ 1 sr d1 :² h ␣ 2 sr d 2 :, a contradiction. Hence y f ² h s : x ² k s :. Subcase 2. Suppose ⑀ n s y1. If yny1 f ² h s : x ny1² k : or ¨ nq1 f ² h: u nq 1² k s : then by induction, as in Subcase 1 above, we can find N g N such that, in G s G N , 5 x 5 s 5 x 5, 5 y 5 s 5 y 5, and yny1 f ² h s : x ny1² k : or ¨˜nq 1 f ² h: u nq1² k s :. Moreover, if y s h si xk s j for some i, j, then ysi xy1 yny1y1 s u nq1 k s j ¨ y1 ny 1 h nq1 . This implies si ␣ xy1 ny 1 h yny1 s k
and
h ␣ s u nq1 k s j ¨ y1 nq1
Ž 3.5.
for some ␣ . This means yny 1 g ² h s : x ny1² k : and ¨ nq1 g ² h: u nq1² k s :, contradicting the choice of N. Hence y f ² h s : x ² k s :. So we can assume yny 1 s h si1 x ny1 k i 2 and ¨ nq1 s h j1 u nq1 k s j 2 for some i1 , i 2 , j1 , j2 . This implies y s yny 1 ty1 ¨ nq1 s h si1 x ny1 k i 2 ty1 h j1 u nq1 k s j 2 f ² h s : x ny1 ty1 u nq1² k s : .
Ž 3.6. ² s : x ny1 tu nq1² k s : uy1 This implies that k i 2 th j1 f xy1 ny 1 h nq1 . Hence, we need ² to find N g N such that y f h s : x ² k s : for each of the following four cases: : ² : ² y1 : ² : 2-a. Suppose ² xy1 ny 1 hx ny1 l k s1 and u nq1 hu nq1 l k s1. Equation Ž3.6. implies i 2 q j1 / 0. Let ⑀ ) < i 2 q j1 <. By induction, G is ² h ⑀ : x ny 1² k ⑀ :-separable and ² h ⑀ : u nq1² k ⑀ :-separable. Hence, by Lemma ˜i x ny1 s ˜k j in 3.3, there exist M1 ef G and M2 ef G such that if ˜ xy1 ny1 h ˜ ˆi ˆnq1 s ˆk j in GrM2 then ⑀ < i, j. Let GrM1 then ⑀ < i, j and that if u ˆy1 nq1 h u N g N such that N ; M1 l M2 l N1 as in 1-a of Subcase 1. In G s G N , if y g ² h s : x ² k s :, then y s h si xk s j. Thus, by Eq. Ž3.5., we have yny 1 s h si x ny 1 k ␣ s h si1 x ny1 k i 2 and ¨ nq1 s hy ␣ u nq1 k s j s h j1 u nq1 k s j 2 for some sŽ iyi 1 . y ␣ yj1 ␣ . This implies that xy1 x ny1 s k i 2y ␣ and uy1 u nq1 s ny 1 h nq1 h sŽ j 2 yj. k . Thus, by the choice of N ; M1 l M2 , ⑀ < i 2 y ␣ and ⑀ < y ␣ y j1. Hence ⑀ < i 2 q j1 , contradicting the choice of ⑀ . Hence y f ² h s : x ² k s :. The following three cases are also similar to Subcase 1 above. : ² : ² y1 : ² : 2-b. Suppose ² xy1 ny 1 hx ny1 l k s 1 and u nq1 hu nq1 l k / 1. y1 y1 ² : ² : ² : 2-c. Suppose x ny 1 hx ny1 l k / 1 and u nq1 hu nq1 l ² k : s 1. : ² : ² y1 : ² : 2-d. Suppose ² xy1 ny 1 hx ny1 l k / 1 and u nq1 hu nq1 l k / 1. Hence there exists N g N such that, in G s G N , y f ² h s : x ² k s :. Therefore, by induction, G is ² h s : x ² k s :-separable for any x g G.
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Case 2. G is ² h s : x ² h s :-separable for any x g G with 5 x 5 s n. Let y g G such that y f ² h s : x ² h s :. If 5 y 5 ) n or 5 y 5 - n or y s ¨ 1 t ␦ 1 ¨ 2 ⭈⭈⭈ t ␦ n ¨ nq1 and x s u1 t ⑀ 1 u 2 ⭈⭈⭈ t ⑀ n u nq1 with ␦ i / ⑀ i for some i then, as in Case 1, we can find N1 g N such that, in G s G N1, 5 y 5 s 5 y 5, 5 x 5 s 5 x 5, and y f ² h s : x ² h s :. So we consider y s ¨ 1 t ⑀ 1 ¨ 2 ⭈⭈⭈ t ⑀ n ¨ nq1 and x s u1 t ⑀ 1 u 2 ⭈⭈⭈ t ⑀ n u nq1 where y f ² h s : x ² h s :. As before, we let x ny1 s u1 t ⑀ 1 u 2 ⭈⭈⭈ t ⑀ ny 1 u n and yny1 s ¨ 1 t ⑀ 1 ¨ 2 ⭈⭈⭈ t ⑀ ny 1 ¨ n . By Lemma 2.5, there ˜ s G N1. exists N1 g N such that 5 ˜ x5 s 5 ˜ y 5 s n in G Subcase 1. Suppose ⑀ n s 1. If yny1 f ² h s : x ny1² h: or ¨ nq1 f ² k : u nq 1² h s : then by induction, as in Subcase 1 of Case 1, we can find N g N such that, in G s G N , 5 x 5 s 5 x 5, 5 y 5 s 5 y 5 and yny1 f ² h s : x ny 1² h: or ¨ nq1 f ² k : u nq1² h s :. Then, in G s G N , y f ² h s : x ² h s : as in Case 1. Hence we can assume yny 1 s h si1 x ny1 h i 2 and ¨ nq1 s k j1 u nq 1 h s j 2 for some i1 , i 2 , j1 , j2 . This implies y s yny 1 t¨ nq1 s h si1 x ny1 h i 2 tk j1 u nq1 h s j 2 f ² h s : x ny1 tu nq1² h s : . Ž 3.7. As in case 1, we have the following cases: 1-a. 1-b. 1-c. 1-d.
: ² : ² y1 : ² : Suppose ² xy1 ny 1 h x ny1 l h s 1 and u nq1 ku nq1 l h s 1. y1 y1 Suppose ² x ny 1 h x ny1 : l ² h: s 1 and ² u nq1 ku nq1 : l ² h: / 1. : ² : ² y1 : ² : Suppose ² xy1 ny 1 h x ny1 l h / 1 and u nq1 ku nq1 l h s 1. y1 y1 Suppose ² x ny 1 h x ny1 : l ² h: / 1 and ² u nq1 ku nq1 : l ² h: / 1.
In the same manner we can find N g N such that y f ² h s : x ² h s : in G s G N for each of the cases. Subcase 2. Suppose ⑀ n s y1. If yny1 f ² h s : x ny1² k : or ¨ nq1 f ² h: u nq 1² h s : then, as in Subcase 2 of Case 1, we can find N g N such that y f ² h s : x ² h s : in G s G N . Thus we can assume yny1 s h si1 x ny1 k i 2 and ¨ nq 1 s h j1 u nq1 h s j 2 for some i1 , i 2 , j1 , j2 . This implies y s yny 1 ty1 ¨ nq1 s h si1 x ny1 k i 2 ty1 h j1 u nq1 h s j 2 f ² h s : x ny1 ty1 u nq1² h s : .
Ž 3.8. Again we have the following cases: 2-a. 2-b. 2-c. 2-d.
: ² : ² y1 : ² : Suppose ² xy1 ny 1 h x ny1 l k s 1 and u nq1 hu nq1 l h s 1. y1 y1 Suppose ² x ny 1 h x ny1 : l ² k : s 1 and ² u nq1 hu nq1 : l ² h: / 1. : ² : ² y1 : ² : Suppose ² xy1 ny 1 h x ny1 l k / 1 and u nq1 hu nq1 l h s 1. y1 y1 Suppose ² x ny 1 h x ny1 : l ² k : / 1 and ² u nq1 hu nq1 : l ² h: / 1.
CONJUGACY SEPARABILITY
583
In each of the cases, we can find N g N such that y f ² h s : x ² h s : in G s G N . Hence, as in Case 1, by induction G is ² h s : x ² h s :-separable for any x g G. Similarly, G is ² k s : x ² k s :-separable for any x g G. Therefore, G is double coset separable at h, k 4 . DEFINITION 3.5. Let A be a group and let h, k g A be of infinite order. Then A is said to be s-quasi-regular at h, k 4 if ² h: l ² k : s 1 and if, for each integer ⑀ ) 0, there exist an integer ⑀ ) 0 and N⑀ ef A such that N⑀ l ² h: s ² h ⑀⑀ :, N⑀ l ² k : s ² k ⑀⑀ :, and ² h: l ² k : s 1, in A s ArN⑀ . Clearly if A is s-quasi-regular at h, k 4 , then A is quasi-regular at h, k 4 . Hence, as in Lemma 2.5, we have the following: LEMMA 3.6. Let G s ² A, t: ty1 ht s k :. Suppose A is ² h:-separable, ² k :-separable, and s-quasi-regular at h, k 4 . Then, for each M ef A, for each s ) 0 and for each reduced x g G, there exist N ef A and ) 0 such that N ; M, N l ² h: s ² h s :, N l ² k : s ² k s :, ² h: l ² k : s 1, and 5 x 5 s 5 x 5 in G s G N . THEOREM 3.7. Let A be s-quasi-regular and double coset separable at h, k 4 . Let x g G s ² A, t: ty1 ht s k : be cyclically reduced. If 5 x 5 G 1 then, for each y g G such that x ¤G y, there exists N ef G such that x ¤G y, where G s GrN. Proof. WLOG, we may assume that y is also cyclically reduced in G. Case 1 Ž5 x 5 / 5 y 5.. Since x is cyclically reduced, by Lemma 2.5 we can find N g N such that, in G s G N , 5 x 5 s 5 x 5, 5 y 5 s 5 y 5, and x, y are cyclically reduced. This implies 5 x 5 / 5 y 5. Hence, by Theorem 2.1, x ¤G y. Case 2 Ž5 x 5 s 5 y 5 G 1.. Let x s t ⑀ 1 u1 t ⑀ 2 ⭈⭈⭈ u ny1 t ⑀ n u n and y s t ¨ 1 t ␦ 2 ⭈⭈⭈ ¨ ny1 t ␦ n ¨ n be cyclically reduced, where u i , ¨ i g A. Since x ¤G y, by Theorem 2.1, x ¤ ² h: j ² k : yiU for any cyclic permutation yiU s t ␦ i ¨ i ⭈⭈⭈ t ␦ n ¨ n t ␦ 1 ¨ 1 ⭈⭈⭈ t ␦ iy1 ¨ iy1 of y. For each i, we shall find Ni g N such that, in ˆ 5 x Ni 5 s 5 x 5 s 5 yUi N i 5 s 5 y 5 and x N i , yUi N i are cyclically G N i s G, U reduced with x N i ¤ ² h: ˆ j ² ˆk : yi N i . Let N s N1 l ⭈⭈⭈ l Nn . Then, in G s G N , we have 5 x 5 s 5 x 5 s 5 y 5 s 5 y 5 with x, y cyclically reduced and x ¤ ² h: j ² k : yiU for all i. Hence, by Theorem 2.2, x ¤G y as required. To find such Ni ’s, it suffices to consider only the case i s 1, others being similar. As in Case 1, by Lemma 3.6 we can find M1 g N such that ²ˆ h: l ² k : s 1, 5 ˆ x5 s 5 x5 s 5 ˆ y 5 s 5 y 5 with ˆ x, ˆ y cyclically reduced in ˆ s G M 1. G If ⑀ i / ␦ i for some i, then we choose N1 s M1. This implies x ¤ ² h: j ² k : y in G s G N1. Hence we can assume ⑀ i s ␦ i for all i and ⑀ 1 s 1, say. ␦1
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Subcase 1. Suppose y f ² h: x ² h:. By Lemma 3.4, there exists M2 ef G such that y f M2 ² h: x ² h:. Let N1 g N such that N1 ; M1 l M2 . Then x ¤² h: y in G s G N1. Subcase 2. Suppose y s h ␣ xh  but h ␣ / hy . In this case, we need to find N1 g N such that x ¤ ² h: y in G s G N1 in each of the following cases: Ža. Suppose ² xy1 hx : l ² h: s 1. By Lemma 3.4, G is ² h: x ² h:-separable. Moreover, by Lemma 3.3, for ⑀ ) < ␣ q  <, there exists M2 ef G such that if M2 Ž xy1 h i x . s M2 h j then ⑀ < i, j. Furthermore, by Lemma 3.6, there exists N1 g N such that N1 ; M1 l M2 , ² h: l ² k : s 1, and 5 x 5 s 5 x 5 s 5 y 5 s 5 y 5 in G s G N . If y ;² h: x then y s hyi xh i s h ␣ xh  . This 1 implies xy1 hyiy ␣ x s h yi, whence ⑀ < y i y ␣ ,  y i. Therefore ⑀ < ␣ q  , contradicting the choice of ⑀ . Hence x ¤² h: y in G s G N1. Žb. Suppose there exists an integer s ) 0 such that xy1 h s x s hys. WLOG, we can assume s to be the smallest such integer. If h ␣q  s h 2 sr then y s h ␣ xh  s hy q2 sr xh  s hy qsr xh ysr , whence y ;² h: x. Hence h ␣q  f ² h 2 s :. Since G is ² h ⑀ :-separable for any ⑀ by Lemma 3.4, there exists M2 ef G such that h ␣q  f M2 ² h 2 s : and xy1 h i x f M2 ² h: for 1 F i - s. By Lemma 3.6, we can choose N1 g N such that N1 ; M1 l M2 , ² h: l ² k : s 1, h ␣q  f ² h 2 s :, and xy1 h i x f ² h: for 1 F i - s in G s G N1. If y s hyi xh i then, as before, we have xy1 hyiy ␣ x s h yi. This implies s < y i y ␣ . Let yi y ␣ s s for some , whence h yi s xy1 h s x s hys . Thus h ␣q  s h iq ␣ h yi s hy2 s g ² h 2 s :, contradicting h ␣q  f ² h2 s :. Hence x ¤² h: y in G s G N1. Žc. Suppose there exists an integer s ) 0 such that xy1 h s x s h s. WLOG, we can assume s to be the smallest such integer. Since, by Lemma 3.4, G is ² h ⑀ :-separable for any ⑀ , we can find M2 ef G such that h ␣q  f M2 and xy1 h i x f M2 ² h: for 1 F i - s. Now, by Lemma 3.6, we can choose N1 g N such that N1 ; M1 l M2 and ² h: l ² k : s 1, so that in G s G N1, if y s hyi xh i then xy1 hyiy ␣ x s h yi and s < y i y ␣ . Let yi y ␣ s s for some . Then h yi s xy1 h s x s h s . This implies h ␣q  s h iq ␣ h yi s 1, contradicting h ␣q  f M2 . Hence x ¤² h: y in G s G N1. Thus, in each case, we have found N1 such that, in G s G N1, ² h: l ² k : s 1 and x ¤² h: y. Since ⑀ 1 s 1 and ² h: l ² k : s 1, x ¤² k : y. Hence x ¤² h: j ² k : y in G N1, as required. 4. A CRITERION In this section we shall prove a criterion for the conjugacy separability of HNN extensions.
CONJUGACY SEPARABILITY
585
DEFINITION 4.1. Let A be a group and let h, k g A be of infinite order. Then A is said to be c-quasi-regular at h, k 4 if ² h: l ² k : s 1 and there exists an integer ␣ such that, for a fixed integer m ) 0 and for each integer ⑀ ) 0, there exist an integer ⑀ ) 0 and N⑀ ef A such that: Ž1. N⑀ l ² h: s ² h ⑀⑀ : and N⑀ l ² k : s ² k ⑀⑀ :, Ž2. in A s ArN⑀ , ² h: l ² k : s 1, Ž3. if h ␣ m ;A h j then h j s h " ␣ m , Ž4. if k ␣ m ;A k j then k j s k " ␣ m , and Ž5. if h i ;A k j then h i s k j s 1. For example, if A is polycyclic-by-finite and if h, k g ZŽ A. are of infinite order then A is c-quasi-regular at h, k 4 whenever ² h: l ² k : s 1. Clearly if A is c-quasi-regular at h, k 4 , then A is s-quasi-regular at h, k 4 and, hence, quasi-regular at h, k 4 . Remark 4.2. If A is c-quasi-regular at h, k 4 and h i ;G k j, then i s j s 0. For, by the above definition, for each ⑀ ) 0, there exist ⑀ and N ef A such that N l ² h: s ² h ⑀⑀ : and N l ² k : s ² k ⑀⑀ :. Then, in A s A N , h i ;A k j implies h i s k j s 1. Thus i, j must be divisible by every ⑀ ) 0. Hence i s 0 s j. LEMMA 4.3. If A is c-quasi-regular at h, k 4 then, for each M ef A, there exists an integer ␣ such that, for a fixed integer m ) 0 and for each integer s ) 0, there exist s and Ns ef A such that Ns ; M, Ns l ² h: s ² h s s :, and Ns l ² k : s ² k s s : such that, in A s ArNs , ² h: l ² k : s 1, if h ␣ m ;A h j then h j s h " ␣ m , if k ␣ m ;A k j then k j s k " ␣ m , and if h i ;A k j then h i s k j s 1. Proof. Let M l ² h: s ² h s1 : and M l ² k : s ² k s 2 :. Let ⑀ s s1 s2 s. By Definition 4.1, there exist ) 0 and N1 ef A such that N1 l ² h: s ² h ⑀ :, N1 l ² k : s ² k ⑀ :, and in A˜ s ArN1 , ² ˜ h: l ² ˜ k : s 1, if ˜ h ␣ m ;A˜ ˜ h j then ˜h j s ˜h " ␣ m , if ˜k ␣ m ;A˜ ˜k j then ˜k j s ˜k " ␣ m , and if ˜h i ;A˜ ˜k j then ˜h i s ˜k j s 1. Let Ns s N1 l M and s s s1 s2 . Then Ns ; M and Ns l ² h: s ² h s 1 s 2 s : s ² h s s : , Ns l ² k : s ² k s 1 s 2 s : s ² k s s : . Since Ns l ² h: s N1 l ² h: and Ns l ² k : s N1 l ² k :, we have, in A s ArNs , ² h: l ² k : s 1, if h ␣ m ;A h j then h j s h " ␣ m , if k ␣ m ;A k j then k j s k " ␣ m , and if h i ;A k j then h i s k j s 1. DEFINITION 4.4. A group G is said to be cyclic conjugacy separable for ² h: if x 4G l ² h: s ⭋ for x g G; then there exists N ef G such that, in G s GrN, x 4G l ² h: s ⭋. It was proved in w7x that free-by-finite groups are cyclic conjugacy separable for each ² h:. We are now ready to prove a criterion for the conjugacy separability of HNN extensions. THEOREM 4.5. Let A be c , conjugacy separable, c-quasi-regular, and double coset separable at h, k 4 . If A is cyclic conjugacy separable for ² h: and ² k :, then G s ² A, t: ty1 ht s k : is conjugacy separable.
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Proof. Let x, y g G be such that x ¤G y and that x, y are of minimal lengths in their respective conjugacy classes. By Theorem 2.7, G is c , whence residually finite. Therefore we may assume x / 1 / y. By Theorem 3.7, we need only prove the case 5 x 5 s 5 y 5 s 0. Also by Theorem 2.2, G s G N is conjugacy separable for each N g N . Therefore, throughout the proof, we shall find a suitable N g N such that, in G s G N , x ¤G y. Since we need only consider the case 5 x 5 s 5 y 5 s 0, that is x, y g A, we have the following three cases. Case 1.
x, y g ² h: Žsimilarly x, y g ² k :..
Ža. Suppose x s h m and y s h n, where m / "n. By the quasi-regularity, for ⑀ s mn, there exist and N g N such that < h < s < k < s ⑀. Then < x < s < h m < and < y < s < h n < are different; hence x ¤G y in G s G N . Žb. Suppose x s h m and y s hym . This implies h m ¤A hym and m k ¤A kym . Since A is conjugacy separable, there exists N1 ef A such that N1 h m ¤Ar N1 N1 hym and N1 k m ¤A r N1 N1 kym . Let N1 l ² h: s ² h s1 : and N1 l ² k : s ² k s 2 :. By Lemma 4.3, there exists ␣ such that, for ⑀ s 2 s1 s2 ␣ m, there exist N ef A and such that N ; N1 and, in A s ArN, ² h: l ² k : s 1 and < h < s ⑀ s < k < satisfying the conditions Ž3. ᎐ Ž5. of Definition 4.1. Now, if x ;G y then, by Theorem 2.1, there exist z1 , z 2 , . . . , z r of elements in ² h: j ² k : such that h m ;A z1 ;A , * z 2 ;A , * ⭈⭈⭈ ; A , * z r ;A hym . By the choice of N, z1 s h i g ² h:. This implies h ␣ m ;A h i ␣ . Thus h i ␣ s h " ␣ m . If h i ␣ s hy ␣ m then ⑀ s 2 s1 s2 ␣ m s < h < divides ␣ Ž i q m.. Let i q m s 2 s1 s2 for some . Since h m ;A h i s z1 , in A˜ s ArN1 , ˜ h m ;A˜ i ymq2 s s m ym m ym ␣ 1 2 ˜h s ˜h s˜ h , contradicting ˜ h ¤A˜ ˜ h . Therefore z1 s h i ␣ ␣m ␣m s h . Now z1 ;A, * z 2 . This implies h s z1␣ ;A, * z 2␣ . Therefore, either z 2␣ s h ␣ m Žas above., or z 2␣ s y1 h ␣ m s k ␣ m . Repeating this process, we have either z r␣ s h ␣ m or z r␣ s y1 h ␣ m s k ␣ m . Clearly z r␣ / k ␣ m . On the other hand, z r ;A hym implies z r␣ s h ␣ m ;A h ␣ m . Then, as above, hy␣ m s h ␣ m , contradicting < h < s 2 s1 s2 ␣ m ) 2 ␣ m. Hence x ¤G y. Case 2. Suppose x g ² h: and y g ² k : Žsimilarly x g ² k : and y g ² h:.. Let x s h m and y s k n. Since k n ;G h n, the case follows from Case 1. Case 3. Suppose x g A and x 4 A l ² h: s ⭋ Žor x 4 A l ² k : s ⭋.. By cyclic conjugacy separability of ² h:, there exists N1 ef A such that in ˜ A˜ s ArN1 , ˜ x4 A l ² ˜ h: s ⭋. Since A is conjugacy separable, there exists N2 ef A such that N2 x ¤A r N 2 N2 y. By Lemma 4.3, we can choose N g N
CONJUGACY SEPARABILITY
587
such that N ; N1 l N2 . Let G s G N . Since x 4 A l ² h: s ⭋ and x ¤A y, by Theorem 2.1, x ¤G y. The following is a generalization of Theorem 3.3 in w8x. COROLLARY 4.6. If A is polycyclic-by-finite and h, k g ZŽ A. of infinite order such that ² h: l ² k : s 1 then G s ² A, t: ty1 ht s k : is conjugacy separable. Proof. Since polycyclic groups are double coset separable w11x, this implies A is c and double coset separable at h, k 4 . It was also well known that polycyclic-by-finite groups are conjugacy separable w4x. We need to show that A is c-quasi-regular. For a given ⑀ , consider A˜ s Ar² h ⑀ :² k ⑀ :. Since A˜ is again polycyclic-by-finite, A˜ is residually ˜˜i / 1 / Nk ˜˜ j and Nh ˜˜i / Nk ˜˜ j finite. Thus there exists N˜ ef A˜ such that Nh ˜ for 1 F i, j - ⑀ . Let N be the preimage of N in A. Then N ef A. Moreover N l ² h: s ² h ⑀ :, N l ² k : s ² k ⑀ : and, in A s ArN, ² h: l ² k : s 1. Furthermore, since h, k g ZŽ A., if h m ;A h i then h i s h m , if k m ;A k i then k i s k m , if h i ;A k j then h i s k j s 1. Hence A is cquasi-regular at h, k 4 . Now h, k g ZŽ A. and A is c . This implies A is cyclic conjugacy separable for ² h: and ² k :. Therefore, by Theorem 4.5, G is conjugacy separable.
5. ON HNN EXTENSIONS OF FREE PRODUCTS OF FINITE CYCLES In this section we shall prove the conjugacy separability of certain HNN extensions of free products of cyclic groups. Let G s ² t , a1 , . . . , a m , b1 , . . . , bn : a i␣ i , bj j , ty1 ht s k : , where h g ² a1 , . . . , a m : and k g ² b1 , . . . , bn : are of infinite order. For convenience, we let C s ² a1 , . . . , a m :, D s ² b1 , . . . , bn :, and A s C ) D. LEMMA 5.1 w10, Lemma 2.8x. Let G be a finite extension of a residually finitely generated torsion-free nilpotent group B. Let h g G be of infinite order. If ² h: l B s ² h n : then, for each integer t, there exists 1 ) 0 such that for any ␣ ) 0 there exists N␣ ef G so that N␣ l ² h: s ² h ␣ n 1 : and that h nt ;G h j implies h j s h " nt , where G s GrN␣ . LEMMA 5.2. The group A s C ) D abo¨ e is c-quasi-regular at h, k 4 . Proof. Since C, D are free-by-finite, let C1 , D 1 be free subgroups of finite index in C, D, respectively. Let C1 l ² h: s ² h n1 : and D 1 l ² k : s ² k n 2 :. Let m be a given integer.
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Let t s n 2 m in Lemma 5.1; there exists 1 such that, for any s ) 0, there exists Ms ef C such that Ms l ² h: s ² h sn1 1 : and, in Cˆ s CrMs , if ˆh n1 n 2 m ;Cˆ ˆh i then ˆh i s ˆh " n1 n 2 m . Similarly, let t s n1 m in Lemma 5.1. Then there exists 2 such that, for any s ) 0, there exists L s ef D such ˆ s DrL s , if ˆk n1 n 2 m ;Dˆ ˆk i then that L s l ² k : s ² k sn 2 2 : and, in D ˆk i s ˆk " n1 n 2 m . For any ⑀ , considering s s n 2 2 ⑀ . Then there exists M1 ef C such that M1 l ² h: s ² h n1 n 2 1 2 ⑀ : and, in C˜ s CrM1 , if ˜ h n1 n 2 m ;C˜ ˜ h i then ˜ hi s " n n m 1 2 ˜h . Similarly, for s s n1 1 ⑀ , there exists L1 ef D such that ˜ s DrL1 , if ˜k n1 n 2 m ;D˜ ˜k i then ˜k i s L1 l ² k : s ² k n1 n 2 1 2 ⑀ : and, in D " n n m i ˜k 1 2 . Let A˜ s C˜) D. ˜ Then ˜h ¤A˜ ˜k j for any ˜h i / 1 / ˜k j. Since A˜ is free-by-finite, it is conjugacy separable w2x. Thus there exists N˜ ef A˜ ˜˜i ¤A˜r N˜ Nk ˜˜ j for all ˜h i / 1 / ˜k j, Nh ˜˜i ¤A˜r N˜ such that ˜ h i, ˜ k j f N˜ and Nh j i j i j i j ˜˜ ¤A˜r N˜ Nk ˜˜ for all ˜k ¤A˜ ˜k . Let N be the N˜˜ h for all ˜ h ¤A˜ ˜ h , Nk preimage of N˜ in A. Then N l ² h: s ² h n1 n 2 1 c 2 ⑀ :, N l ² k : s ² k n1 n 2 1 2 ⑀ :, and, in A s ArN, ² h: l ² k : s 1. Moreover, if h n1 n 2 m ;A h j then h j s h " n1 n 2 m , if k n1 n 2 m ;A k j then k j s k " n1 n 2 m , and if h i ;A k j then h i s k j s 1. Hence A is c-quasi-regular at h, k 4 . Since A is c-quasi-regular at h, k 4 , A is quasi-regular at h, k 4 . Hence, by Theorem 2.7, we have the following: THEOREM 5.3. The group G s ² t, a1 , . . . , a m , b1 , . . . , bn : a i␣ i , bj j , ty1 ht s k :, where h g ² a1 , . . . , a m : and k g ² b1 , . . . , bn :, is c , whence residually finite. THEOREM 5.4. The group G s ² t, a1 , . . . , a m , b1 , . . . , bn : a i␣ i , bj j , ty1 ht s k :, where h g ² a1 , . . . , a m : and k g ² b1 , . . . , bn :, is conjugacy separable. Proof. By Lemma 5.2, A s ² a1 , . . . , a m , b1 , . . . , bn : is c-quasi-regular at h, k 4 . Since A is free-by-finite and since free groups are double coset separable w15x, A is c and double coset separable at h, k 4 . Moreover, free-by-finite groups are conjugacy separable w2x and cyclic conjugacy separable w7x. The theorem follows from Theorem 4.5.
6. ON HNN EXTENSIONS OF BAUMSLAG᎐SOLITAR GROUPS In this section we consider the HNN extension G s ² t , a, b: ty1 a t s b , by1 a b s a : ,
Ž 6.1.
of the Baumslag᎐Solitar group A s ² a, b: by1 a b s a : with associated subgroups ² a : and ² b :. Hopficity and residual finiteness of this type of
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groups were studied by Raptis, Talelli, and Varsos w16x which motivates this section. Using their result on residual finiteness, we prove the cyclic subgroup separability and conjugacy separability of G. THEOREM 6.1 w16x. The group G, as gi¨ en by Ž6.1., is residually finite if and only if < < s < <. THEOREM 6.2 w12, 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. Ž1. If < x < 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 x* where x* is a cyclic permutation of x. LEMMA 6.3. The group A s ² a, b: by1 a b s a : is c-quasi-regular at a , b 4 if < < s < <. Proof. Case 1. A s ² a, b: by1 a b s a :. Clearly ² h: l ² k : s 1, where h s a and k s b . Let ⑀ be a given integer. Since A s ² a:) a sx² x, b: by1 xb s x :, we consider the natural homomorphism
⑀ : A ª ² a:r² a ⑀ : )
a ˆsxˆ
Ž ² x :r² x ⑀ : = ² b:r² b ⑀ : . ,
where Aˆ s A ⑀ . Clearly < ˆ h < s ⑀ s < ˆ k < and ² ˆ h: l ² ˆ k : s 1. Since ˆ xg ZŽ Aˆ., it follows from Theorem 6.2 that if ˆ h m ;Aˆ ˆ h i then ˆ hm s ˆ h i. Similarly, if ˆ h i ;Aˆ ˆ k j then ˆ hi s 1 s ˆ k j. m i Now suppose ˆ k ;Aˆ ˆ k . Since x g ZŽ A., this implies that if ˆ k m ;Aˆ ˆ xs m s m i for some s, then ˆ k sˆ x g ²ˆ h: l ² ˆ k : s 1. Hence ˆ k s1sˆ k . On the other hand, if ˆ k m ¤Aˆ ˆ x s for any s then, by Theorem 6.2, ˆ k m and ˆ k i are conjugate in ² ˆ x: = ²ˆ b :. Thus ˆ km s ˆ k i. ˆ We have shown that, in A s A ⑀ , < ˆ h < s ⑀ s < ˆ k < and ² ˆ h: l ² ˆ k : s 1. m i m i i j i j Moreover, if ˆ h ;Aˆ ˆ h then ˆ h sˆ h , if ˆ h ;Aˆ ˆ k then ˆ h s1sˆ k , and if ˆk m ;Aˆ ˆk i then ˆk m s ˆk i. Since Aˆ is free-by-finite, it is conjugacy separable. Thus, as in the proof of Lemma 5.2, we can find N ef A such that N l ² h: s ² h ⑀ :, N l ² k : s ² k ⑀ :, and, in A s ArN, conditions Ž2. ᎐ Ž5. of Definition 4.1 are satisfied. Hence A is c-quasi-regular at h, k 4 s a , b 4 .
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Case 2. A s ² a, b: by1 a b s ay :. Clearly ² h: l ² k : s 1, where h s a and k s b . Let ⑀ be a given integer. Since A s ² a:) a sx² x, b: by1 xb s xy1 :, we consider the natural homomorphism
⑀ : A ª ² a:r² a2 ⑀ : )
a ˆsxˆ
Ž ² x, b:r² x 2 ⑀ :² b 2 ⑀ : . ,
where Aˆ s A ⑀ . Clearly < ˆ h < s 2 ⑀ s < ˆ k < and ² ˆ h: l ² ˆ k : s 1, since ² ˆ x: l ²ˆ b : s 1. Suppose ˆ h2 m ;Aˆ ˆ h i. We note that ˆ gy1 ˆ x sˆ g sˆ x " s for any s and any 2 m s 2 m g g A. If ˆ h ;Aˆ ˆ x for some s, then ˆ h sˆ x " s and ˆ hi s ˆ x " s, thus ˆh i s ˆh " 2 m . If ˆh2 m ¤Aˆ ˆx s for any s then, by Theorem 6.2, ˆh2 m ;² a: ˆi ˆ h. i 2m ˆ ˆ This implies h s h . Now suppose ˆ k 2 m ;Aˆ ˆ k i. We note that ˆ k 2 m g ZŽ² ˆ x, ˆ b :.. If ˆ k 2 m ;Aˆ ˆ xs 2m "s ˆ ˆ ² : ² : for some s then, by Theorem 6.2, k s ˆ x g k l ˆ x s 1. Hence ˆk 2 m s 1 s ˆk i. On the other hand, if ˆk 2 m ¤Aˆ ˆx s for any s then, by Theorem 6.2, ˆ k 2 m and ˆ k i are conjugate in ² ˆ x, ˆ b :. This implies ˆ k2m s ˆ k i. i j i s ˆ ˆ ˆ Suppose h ;Aˆ k ; then, by Theorem 6.2, h ;Aˆ ˆ x for some s. Therefore, as above, ˆ hi s ˆ x " s. This implies ˆ k j ;Aˆ ˆ x " s. This means ˆ kj sˆ x" s g j ˆ ˆ ˆ ² k: l ² ˆ x : s 1. Hence h s 1 s k . Thus, as in Case 1, we can find N ef A such that conditions Ž1. ᎐ Ž5. of Definition 4.1 are satisfied. Hence A is c-quasi-regular at h, k 4 s a , b 4 . THEOREM 6.4. The group G s ² t, a, b: ty1 a t s b , by1 a b s a : is c if and only if < < s < <. Proof. Suppose < < s < <. By w17x, A s ² a, b: by1 a b s a : is c if and only if < < s < <. Therefore, by Theorem 2.7 and Lemma 6.3, G is c . Conversely, if G is c then A is c . Hence, < < s < < by w17x. LEMMA 6.5. The group A s ² a, b: by1 a b s a : is cyclic conjugacy separable for ² a : and ² b : if < < s < <. Proof. Case 1. A s ² a, b: by1 a b s a :. Let A s ² a:) a sx² x, b: by1 xb s x : and let g g A be of minimal length in its conjugate class in A. Ža. A is cyclic conjugacy separable for ² h: s ² a :. Let g 4 A l ² h: s ⭋. Suppose 5 g 5 G 2 or g g ² x, b : _ ² a:. Since a s x g ZŽ A., we can find a homomorphism ⑀ , as in Lemma 6.3, Case 1, such that, in Aˆ s A ⑀ , 5 ˆ g 5 s 5 g 5 G 2 or ˆ g g ²ˆ x, ˆ b: _ ² a ˆ:. Then, by Theorem i ˆ ˆ 6.2, ˆ g ¤Aˆ h for any i. Since A is conjugacy separable and < ˆ h < - ⬁, there
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ˆˆ ¤Aˆr Nˆ Nh ˆˆi for all i. Let N be the preimage exists Nˆ ef Aˆ such that Ng of Nˆ in A. Then, in A s ArN, g 4 A l ² h: s ⭋. Suppose g g ² a:. Since g 4 A l ² h: s ⭋, g s a s f ² a :. Then, in Aˆ s A ⑀ as above, ˆ g f ²ˆ h:. Thus, by Theorem 6.2, ˆ g ¤Aˆ ˆ h i for all i. Then, as Ž . in a , we can find N ef A such that, in A s ArN, g 4 A l ² h: s ⭋. This proves that A is cyclic conjugacy separable for ² h:. Žb. A is cyclic conjugacy separable for ² k : s ² b :. Let g 4 A l ² k : s ⭋. Suppose 5 g 5 G 2 or g g ² a: _ ² a :. As before, we can find a homomorphic ⑀ above such that, in Aˆ s A ⑀ , 5 ˆ g5 s 5 g5 G 2 : i ˆ : ² or ˆ g g ²a _ a . Then, by Theorem 6.2, g ¤ b for all i. As in Ža. we ˆ ˆ ˆ Aˆ A 4 ² : can find N ef A such that, in A s ArN, g l k s ⭋. So, suppose g g ² a , b :. Since g 4 A l ² k : s ⭋, g f ² b :. Again, we can choose Aˆ s A ⑀ such that ˆ g f ²ˆ b :. Thus, by Theorem 6.2, ˆ g ¤Aˆ ˆ bi Ž . for any i. Hence, as in a , we can find N ef A such that, A s ArN, g 4 A l ² k : s ⭋. This proves that A is cyclic conjugacy separable for ² k :. Case 2. A s ² a, b: by1 a b s ay :. Let A s ² a:) a sx² x, b: by1 xb s x and let g g A be of minimal length in its conjugate class in A. y1 :
Ža. A is cyclic conjugacy separable for ² h: s ² a :. Let g 4 A l ² h: s ⭋. Suppose 5 g 5 G 2 or g g ² a , b : _ ² a :. As in Lemma 6.3, Case 2, we can find a homomorphism ⑀ such that, in Aˆ s A ⑀ , 5 ˆ g 5 s 5 g 5 G 2 or ˆ g g ²a ˆ, ˆb : _ ² aˆ:. Then, by Theorem 6.2, i ˆg ¤Aˆ ˆh for all i. If g g ² a:, then g s a s f ² h: s ² a :. If a s f ² a :, we can choose Aˆ s A ⑀ such that ˆ g f ²a ˆ: l ² aˆ :. Then, by Theorem 6.2, i s ˆg ¤Aˆ ˆh for all i. If g s a g ² a : we can choose Aˆ s A ⑀ such that ˆg f ² aˆ :. Then, again by Theorem 6.2, ˆg ¤Aˆ ˆh i for all i. Hence, as in Case 1, A is cyclic conjugacy separable for ² h:. Žb. A is cyclic conjugacy separable for ² k : s ² b :. Let g 4 A l ² k : s ⭋. Suppose 5 g 5 G 2 or g g ² a: _ ² a :. As in Ža., we can find a homomorphism ⑀ above such that, in Aˆ s A ⑀ , 5 ˆ g5 s 5 g5 G 2 or ˆ g g ²a ˆ: _ ² aˆ:. Then, by Theorem 6.2, ˆg ¤Aˆ ˆk i for any i. Let B s ² a , b :. If g g B, then g 4 B l ² k : s ⭋. Let g s b s x r where x s a . We note that b s x r ;B b i s k i if and only if b s s b i and that r is even if s is odd, and r s 0 if s is even. Thus, if b s f ² b : then, in Bˆ s B⑀ , we have ˆ b sˆ x r ¤Bˆ ˆ b i for any i. So we can assume that b s s b j r for some j. This means x / 1. If s is even, then we choose Bˆ s B⑀ such that ˆ x r / 1. This implies ˆ b sˆ x r ¤Bˆ ˆ b i for any i. If s is odd, then r is not even. Hence we choose Bˆ s B⑀ such that ˆ xr f ²ˆ x 2 :. Then ˆ b sˆ x r ¤Bˆ ˆ bi ˆ B for any i. In this way we can find Aˆ s A ⑀ such that ˆ g4 l ²ˆ k : s ⭋. Since any nontrivial element ˆ k i has the minimal length 1 in its conjugate ˆ by Theorem 6.2, ˆg ¤Aˆ ˆk i for any i. Therefore, as in Case 1, A class in A, is cyclic conjugacy separable for ² k :.
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LEMMA 6.6. The group A s ² a, b: by1 a b s a : is double coset separable at a , b 4 if < < s < <. Proof. Case 1. A s ² a, b: by1 a b s a .. Let A s ² a:) a sx B, where B s ² x, b: by1 xb s x :. Ža. A is ² h s : u² k s :-separable for h s a , k s b . We first show that A is ² a :² b ␦ :-separable for any , ␦ . Let ¨ g A and ¨ f ² a :² b ␦ :. If ¨ f B then we choose Aˆ s A ⑀ , as in Lemma 6.5, Case 1, such that ¨ˆ f Bˆ by considering the length of ¨ in A s ² a:) a sx B. ˆ If ¨ g B, we let ⑀ s ␦ in Aˆ s A ⑀ . Then Then ¨ˆ f ² a ˆ :² ˆb ␦ : ; B. ␦: ˆ ˆ :² ¨ˆ f ² a b . Since A is free-by-finite, whence residually finite, there ˆ exists Nˆ ef Aˆ such that ¨ˆ f Nˆ² a ˆ :² ˆb ␦ :. Let N be the preimage of Nˆ in A. Then N ef G and ¨ f N² a :² b ␦ :. Thus A is ² a :² b ␦ :-separable for any , ␦ . To show that A is ² a s : u² b s :-separable for any s, let ¨ f ² a s : u² b s :, that is, uy1 ¨ f ² uy1 a s u:² b s :. Let d s Ž s, .. Since a g ZŽ A., r d.y1 y1 si ² Ž sr d. : ² uy1 a s u: s D Žis0 u a u a . Thus, for each 0 F i - d , uy1 ay si u ⭈ uy1 ¨ f ² aŽ sr d. :² b s :. By the above note, there exists N ef A such that uy1 ay si u ⭈ uy1 ¨ f N² aŽ sr d. :² b s : for all 0 F i - d . It follows that uy1 ¨ f N² uy1 a s u:² b s :; hence ¨ f N² a s : u² b s :. Žb. A is ² h s : u² h s :-separable for h s a . This follows easily by using similar argument as in Ža.. Žc. A is ² k s : u² k s :-separable for k s b .
By Theorem 6.4, A is c . Therefore it is sufficient to show that A is ² k s : u² k s :-separable for u s a r 1 b s1 ⭈⭈⭈ b s ny 1 a r n where n G 1, 0 r 1 , . . . , rny1 - and rn / 0. Let ¨ f ² k s : u² k s :. Suppose 5 ¨ 5 ) 5 u 5 q 2 or ¨ s a e1 b c1 ⭈⭈⭈ b c n a e nq 1 with 5 ¨ 5 s 5 u 5 q 2 or ¨ s b c1 a e1 ⭈⭈⭈ a e ny 1 b c n with 5 ¨ 5 s 5 u 5 or 5 ¨ 5 - 5 u 5. In this case, we can choose Aˆ s A ⑀ such that 5 ¨ˆ5 s 5 ¨ 5 and 5 u ˆ5 s 5 u 5. It follows that s: ²ˆs: ˆ ² ¨ˆ f k u ˆ k . So, suppose ¨ s b c1 a e1 ⭈⭈⭈ a e n b c nq 1 . Since ² a , b : s ² a : = ² b :, we note that ¨ s b c1 a e1 ⭈⭈⭈ a e n b e nq 1 g ² k s : u² k s :, where k s b , if and only if b c1 s b s ␣ , a e1 s a r 1 a ␦ 1 , b c 2 s ay ␦ 1 b s1 a ␦ 1 , . . . , b c n s ay ␦ ny 1 b s ny 1 a ␦ ny 1 , a e n s ay ␦ ny 1 a r n , and b c nq 1 s b s  for some ␣ ,  , ␦ i . Thus b c1 g ² b s :, a e iyr i g ² a :, b c iq 1 s b s i , b c nq 1 g ² b s :, and a e1q ⭈⭈⭈ qe n s a r 1q ⭈⭈⭈ qr n . Hence ¨ f ² b s : u² b s : if and only if b c1 f ² b s :, or a e iyr i f ² a :, or b c iq 1 / b s i , or b c nq 1 f ² b s :, or a e1q ⭈⭈⭈ qe n / a r 1q ⭈⭈⭈ qr n . As usual, we can find Aˆ s A ⑀ such that 5 ¨ˆ5 s 5 ¨ 5 and 5 u ˆ5 s 5 u 5, and ˆb c1 f ² ˆb s :, or aˆe iyr i f ² aˆ:, or c s c iq 1 i nq 1 ˆb / ˆb , or ˆb f ² ˆb s :, or aˆe1q ⭈⭈⭈ qe n / aˆr1 q ⭈⭈⭈ qrn . This implies
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¨ˆ f ² ˆ b s:u ˆ² ˆb s :. Thus, as in Ža., we can find N ef A such that ¨ f
N² b s : u² b s :.
Case 2. A s ² a, b: by1 a b s ay :. Let A s ² a:) a s x B, where B s ² x, b: by 1 xb s xy 1 :. Since ² h s : u² h s :-separability and ² h s : u² k s :-separability are very similar to Case 1, we shall only show that A is ² k s : u² k s :-separable for k s b . Let ¨ g A such that ¨ f ² k s : u² k s : s ² b s : u² b s :. As in Žc. of Case 1, we need only consider u s a r 1 b s1 ⭈⭈⭈ b s ny 1 a r n and ¨ s b c1 a e1 ⭈⭈⭈ a e n b c nq 1 where n G 1. Since ² a : l ² b : s 1, we note that ¨ g ² b s : u² k s : iff b c1 s b s ␣ , a e1 s a r 1 a ␦ 1 , b c 2 s ay ␦ 1 b s1 a 1 , a e 2 s ay 1 a r 2 a ␦ 2 , . . . , b c n s ay ␦ ny 1 b s ny 1 a ny 1 , a e n s ay ny 1 a r n , and b c nq 1 s b s  for some ␣ ,  , ␦ i , i , where b c iq 1 s b s i . If si is even, then ␦ i s i , and if si is odd, then ␦ i s y i . Thus ¨ f ² b s : u² b s : if and only if b c1 f ² b s :, or a e iyr i f ² a :, or b c iq 1 / b s i , or b c nq 1 f ² b s :, or b c1 a e1 ⭈⭈⭈ b c n s b s ␣ a r 1 b s1 ⭈⭈⭈ b s ny 1 a ny 1 and a e n s a r n a n but a ny 1 / ay n . Then, as usual, we can find Aˆ s A ⑀ such that 5 ¨ˆ5 s 5 ¨ 5 and 5 u ˆ5 s 5 u 5, and ˆb c1 f ² ˆb s :, or aˆe iyr i f ² aˆ:, or ˆb c iq 1 / ˆb s i , or ˆb c nq 1 f ² ˆb s :, or aˆ ny 1 / a ˆy n if b c1 a e1 ⭈⭈⭈ b c n s b s ␣ a r1 b s1 ⭈⭈⭈ b s ny 1 a ny 1 and a e n s a r n a n . This implies ¨ˆ f ² ˆ b s:u ˆ² ˆb s :. As in Case 1, we can find N ef A such that s s ¨ f N² b : u² b :. THEOREM 6.7. The group G s ² t, a, b: ty1 a t s b , by1 a b s a : is conjugacy separable if and only if < < s < <. Proof. If < < s < <, then A s ² a, b: by1 a b s a : is conjugacy separable by w5x. Applying Lemmas 6.3᎐6.6 and Theorem 4.5, G is conjugacy separable. Conversely, if G is conjugacy separable then, G is residually finite. Hence, by Theorem 6.1, < < s < <. Brunner w1x has studied the epimorphisms of the class of groups G Ž l, m; k . s ² a, t : ty1 ayk ta l ty1 a k t s a m : , where l, m, k are integers and < l < ) m and obtained various results concerning the Hopficity and the automorphism groups of certain groups in GŽ l, m; k .. Clearly GŽ l, m; k . is a special case of G above when s 1. Thus we have the following: THEOREM 6.8. The following statements on GŽ l, m; k . are equi¨ alent: Ž1. Ž2. Ž3.
GŽ l, m; k . is conjugacy separable. < l < s < m <. GŽ l, m; k . is cyclic subgroup separable.
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ACKNOWLEDGMENTS The authors thank the referee for several helpful suggestions in the exposition of the paper. The second author also thanks the Department of Mathematics of the California State University at Hayward for the hospitality and the use of facilities while he was visiting there.
REFERENCES 1. A. M. Brunner, On a class of one-relator groups, Canad. J. Math. 32 Ž1980., 414᎐420. 2. J. L. Dyer, Separating conjugates in free-by-finite groups, J. London Math. Soc. Ž 2 . 20, No. 2 Ž1979., 215᎐221. 3. J. L. Dyer, Separating conjugates in amalgamated free products and HNN extensions, J. Austral. Math. Soc. Ser. A 29, No. 1 Ž1980., 35᎐51. 4. E. Formanek, Conjugate separability in polycyclic groups, J. Algebra 42 Ž1976., 1᎐10. 5. G. Kim, J. McCarron, and C. Y. Tang, Adjoining roots to conjugacy separable groups, J. Algebra 176, No. 2 Ž1995., 327᎐345. 6. G. Kim and C. Y. Tang, Cyclic subgroup separability of HNN-extensions with cyclic associated subgroups, Canad. Math. Bull. 42, No. 3 Ž1999., 335᎐343. 7. G. Kim and C. Y. Tang, Cyclic conjugacy separability of groups, in ‘‘Groups Korea 94,’’ pp. 173᎐179, de Gruyter, BerlinrNew York, 1995. 8. G. Kim and C. Y. Tang, Conjugacy separability of HNN-extensions of abelian groups, Arch. Math. 67 Ž1996., 353᎐359. 9. G. Kim and C. Y. Tang, A criterion for the conjugacy separability of amalgamated free products of conjugacy separable groups, J. Algebra 184 Ž1996., 1052᎐1072. 10. G. Kim and C. Y. Tang, Conjugacy separability of generalized free products of finite extensions of residually nilpotent groups, in ‘‘Groups Theory ŽProc. of the ’96 Beijing Int’l Symposium.,’’ pp. 10᎐24, Springer-Verlag, BerlinrNew York, 1998. 11. J. C. Lennox and J. S. Wilson, On products of subgroups in polycyclic groups, Arch. Math. 33 Ž1979., 305᎐309. 12. W. Magnus, A. Karrass, and D. Solitar, ‘‘Combinatorial Group Theory,’’ Pure and Applied Math. Vol. XIII, Wiley-Interscience, New YorkrLondonrSydney, 1966. 13. S. Meskin, Nonresidually finite one-relator groups, Trans. Amer. Math. Soc. 164 Ž1972., 105᎐114. 14. G. A. Niblo, H.N.N. extensions of a free group by Z which are subgroup separable, Proc. London Math. Soc. 61 Ž1990., 18᎐32. 15. G. A. Niblo, Separability properties of free groups and surface groups, J. Pure Appl. Algebra 78 Ž1992., 77᎐84. 16. E. Raptis, O. Talelli, and D. Varsos, On the Hopficity of certain HNN-extensions with base a Baumslag᎐Solitar group, preprint. 17. P. F. Stebe, Residual finiteness of a class of knot groups, Comm. Pure. Appl. Math. 21 Ž1968., 563᎐583.
A Criterion for the Conjugacy Separability of Certain HNN Extensions of Groups* Goansu Kim Department of Mathematics, Yeungnam Uni¨ ersity, Kyongsan 712-749, Korea E-mail: [email protected]
and C. Y. Tang Uni¨ ersity of Waterloo, Waterloo, Ontario N2L 3G1, Canada, and California State Uni¨ ersity, Hayward, California E-mail: [email protected] Communicated by Walter Feit Received June 29, 1998
We first prove the double coset separability of certain HNN extensions with cyclic associated subgroups. Using this we prove a criterion for the conjugacy separability of HNN extensions of conjugacy separable groups with cyclic associated subgroups. Applying this result we show that certain HNN extensions of free products of cycles and certain Baumslag᎐Solitar groups with cyclic associated 䊚 2000 Academic Press subgroups are conjugacy separable. Key Words: HNN-extension, conjugacy separable, residually finite.
1. INTRODUCTION A group G is conjugacy separable if, for each x, y g G such that x and y are not conjugate in G, there exists a finite homomorphic image G of G such that the images of x and y in G are not conjugate in G. * The first author was partly supported by KOSEF through the GARC at S.N.U. and KOSEF 985-0100-002-2. The second author gratefully acknowledges the partial support by the National Science and Engineering Research Council of Canada, Grant No. A-4064. 574 0021-8693r00 $35.00 Copyright 䊚 2000 by Academic Press All rights of reproduction in any form reserved.
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In this paper we consider the conjugacy separability of HNN extensions G s ² A, t : ty1 ht s k : of a conjugacy separable group A with cyclic associated subgroups ² h: and ² k :. Residual properties of HNN extensions are difficult to obtain. They depend very much on the choice of h, k g A. For example, the Baumslag᎐Solitar group, ² a, t: ty1 a t s a ␦ :, is an HNN extension of the cyclic group ² a:. It is residually finite w13x if and only if < < s 1 or < ␦ < s 1 or s "␦ . In w16x, Raptis, Talelli, and Varsos studied the Hopficity and the residual finiteness of HNN extensions of Baumslag᎐Solitar groups with cyclic associated subgroups of the form ² t, a, b: ty1 a t s b , by1 a b s a :. Recently, in w6, Definition 2.4x, Kim and Tang introduced the concept of a group to be quasi-regular at h, k 4 . If A is quasi-regular at h, k 4 , then the residual properties of HNN extension G of A are nicely preserved ŽTheorem 2.7.. Using this, they characterized the residual finiteness and cyclic subgroup separability of G when A is a finitely generated abelian group. They also characterized the conjugacy separability of G when A is a finitely generated abelian group w8x. In this paper we prove the double coset separability of HNN extensions of groups which are quasi-regular at the associated subgroups ŽLemma 3.4.. Using this, we derive a criterion for the conjugacy separability of HNN extensions with cyclic associated subgroups ŽTheorem 4.5.. In general it is difficult to get groups to satisfy double coset separability for associated subgroups. However, if the associated subgroups are in the center of a polycyclic-by-finite group, then we can easily apply the criterion to show that such HNN extensions are conjugacy separable ŽCorollary 4.6.. This gives us an extension of a result in w8x for HNN extensions of abelian groups. We also apply the criterion to show that HNN extensions of free products of cycles with cyclic associated subgroups are conjugacy separable ŽTheorem 5.4.. Finally, we show that certain HNN extensions of Baumslag᎐Solitar groups with cyclic associated subgroups are conjugacy separable ŽTheorem 6.7..
2. PRELIMINARIES Throughout this paper we use standard notations and terminology for this topic w12x. The letter G always denotes a group. In addition: N ef G means that N is a normal subgroup of finite index in G; x ;G y means x is conjugate to y in G. We use 5 x 5 to denote the length of x in HNN-extensions and generalized free products.
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We shall make extensive use of the following two results by D. J. Collins. THEOREM 2.1 w3, Theorem 3x. Let x and y be cyclically reduced elements of the HNN-extension G s ² B, t: ty1 Ht s K :. Suppose that x ;G y. Then 5 x 5 s 5 y 5, and one of the following holds. Ž1. 5 x 5 s 5 y 5 s 0 and there is a finite sequence z1 , z 2 , . . . , z m of elements in H j K such that x ;B z1 ;B, t* z 2 ;B, t* ⭈⭈⭈ ;B, t* z m ;B y, where u ;B, t* ¨ means one of: Ži. u ;B ¨ , Žii. u g H and ¨ s ty1 ut Žg K ., or Žiii. u g K and ¨ s tuty1 Žg H .. Ž2. 5 x 5 s 5 y 5 G 1 and y ;H j K x* where x* is a cyclic permutation of x. A group G is said to be S-separable if for each x g G _ S there exists N 1f G such that x f NS. In particular, G is residually finite Ž RF . if G is 14 -separable. A group G is said to be cyclic subgroup separable, briefly c , if G is ² x :-separable for each x g G. A group G is conjugacy separable if, for each x g G, G is x 4G -separable, where x 4G is the set of all conjugates of x in G. THEOREM 2.2 w3, Theorem 13x. If A is conjugacy separable and H, K are finite, then the HNN-extension ² A, t: ty1 Ht s K : is conjugacy separable. DEFINITION 2.3. Let A be a group and let h, k g A be of infinite order. Then A is said to be quasi-regular at h, k 4 if, for each integer ⑀ ) 0, there exist ana integer ⑀ ) 0 and N⑀ ef A, depending on ⑀ , such that N⑀ l ² h: s ² h ⑀⑀ : and N⑀ l ² k : s ² k ⑀⑀ :. In w14x, Niblo defined that a group A has regular quotients at h, k 4 if there exists a positive integer , such that for each positive integer ⑀ , there exists N ef A such that N l ² h: s ² h ⑀ : and N l ² k : s ² k ⑀ :. Thus, in the case of regular quotients, the integer is independent of ⑀ . Hence, if A has regular quotients at h, k 4 , then A is quasi-regular at h, k 4 . Remark 2.4. Let G s ² A, t: ty1 ht s k :. For N ef A such that N l ² h: s ² h s : and N l ² k : s ² k s :, there is a natural homomorphism
N : ² A, t : ty1 ht s k : ª ² A, : y1 h s k : by a N s a for a g A, where A s ArN, and t N s . Then, by Theorem 2.2, G N is conjugacy separable. LEMMA 2.5. Let G s ² A, t: ty1 ht s k :. Suppose A is ² h:-separable, ² k :-separable, and quasi-regular at h, k 4 . Then, for each M ef A, for each s ) 0 and for each reduced element x g G, there exist N ef A and ) 0 such that N ; M, N l ² h: s ² h s :, N l ² k : s ² k s : and 5 x 5 s 5 x 5 in G s G N .
CONJUGACY SEPARABILITY
577
Proof. Let x s u1 t ⑀ 1 u 2 t ⑀ 2 ⭈⭈⭈ t ⑀ n u nq1 g G be reduced where u i g A and ⑀ i s "1. Since x is reduced, we have u iq1 f ² h: if ⑀ i s y⑀ iq1 s y1 and u iq1 f ² k : if ⑀ i s y⑀ iq1 s 1. Since A is ² h:-separable and ² k :-separable, we can find Ni ef A such that if ⑀ i s ⑀ iq1 then Ni s A, or if ⑀ i s y⑀ iq1 s y1 then u iq1 f Ni ² h:, or if ⑀ i s y⑀ iq1 s 1 then u iq1 f n n Ni ² k :. Let l is1 Ni l M l ² h: s ² h s1 : and l is1 Ni l M l ² k : s ² k s 2 :. Let ⑀ s s1 s2 s. By quasi-regularity, there exist 1 and Ns ef A such that n Ns l ² h: s ² h ⑀1 : and Ns l ² k : s ² k ⑀1 :. Let N s l is1 Ni l M l Ns s s and s s1 s2 1. Then N ef A, N ; M, N l ² h: s ² h 1 2 1 s : s ² h s :, N l ² k : s ² k s1 s 2 1 s : s ² k s : and, in G s G N above, x s u1 ⑀ 1 u 2 ⑀ 2 ⭈⭈⭈ ⑀ n u nq 1 is reduced with 5 x 5 s 5 x 5. Remark 2.6. Let G s ² A, t: ty1 ht s k :. Suppose A is quasi-regular at h, k 4 and h i ;G h j. Let ⑀ s i. Then there exist and N ef A such that N l ² h: s ² h i : and N l ² k : s ² k i :. In G s G N , < h < s i s < k <. Since h i ;G h j in G s G N , we have 1 s h i ;G h j . This implies i < j . Hence i < j. Similarly j < i. Therefore, if A is quasi-regular at h, k 4 and h i ;G h j then j s "i. Similarly, if A is quasi-regular at h, k 4 and k i ;G k j or h i ;G k j then j s "i. The following theorem shows that quasi-regularity is closely related to the cyclic subgroup separability of HNN extensions. THEOREM 2.7 w6x. Let A be c and let h, k g A be of infinite order. Then G s ² A, t: ty1 ht s k : if c if and only if A is quasi-regular at h, k 4 . In this paper we consider the conjugacy separability of HNN extensions of the type G s ² A, t : ty1 ht s k : , where ² h: l ² k : s 1 and A is conjugacy separable. If < h < s < k < is finite then, by Theorem 2.2, G is conjugacy separable. Thus, we need only consider the case when h, k are of infinite order.
3. DOUBLE COSET SEPARABILITY Throughout this note, we consider the HNN extension G s ² A, t: ty1 ht s k : of a base group A with infinite cyclic associated subgroups ² h: and ² k :, where ² h: l ² k : s 1. DEFINITION 3.1. A group A is said to be double coset separable at h, k 4 if, for each u g A and for each integer ⑀ ) 0, A is ² h ⑀ : u² h ⑀ :-separable, ² h ⑀ : u² k ⑀ :-separable, and ² k ⑀ : u² k ⑀ :-separable.
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We note that if A is ² h ⑀ : u² k ⑀ :-separable for all u g A then A is ² k ⑀ : uy1 ² h ⑀ :-separable for all u g A. It follows that if A is double coset separable at h, k 4 then A is also ² k ⑀ : u² h ⑀ :-separable for all u g A. In the case of a free-by-finite or polycyclic-by-finite group G, it was known by w11, 15x that G is ² a: x ² b :-separable for all a, b, x g G. Hence G is double coset separable at h, k 4 for all h, k g G. Clearly if a group G is double coset separable at h, k 4 , then G is ² h ⑀ :-separable, ² k ⑀ :-separable, and ² h ⑀ :² k ⑀ :-separable for any ⑀ . Remark 3.2. If a group G is ² h ⑀ : u² k ⑀ :-separable for any ⑀ , then G is ² h ␣ : u² k  :-separable for any ␣ ,  . For, if g f ² h ␣ : u² k  : then l l hy␣ i gky  j f ² h l : u² k l : for each 0 F i - ␣ and 0 F j -  , where l is the least common multiple of ␣ ,  . Therefore, there exists N ef G such that l l hy␣ i gky  j f N² h l : u² k l : for all 0 F i - ␣ and 0 F j -  . This implies that g f N² h ␣ : u² k  :. Hence G is ² h ␣ : u² k  :-separable for any ␣ ,  . Similarly, if a group G is double coset separable at h, k 4 , then G is ² h ␣ : u² h  :-separable, ² k ␣ : u² k  :-separable, and ² h ␣ : u² k  :-separable for any ␣ ,  . LEMMA 3.3 w9x. Let G be ² a ⑀ : x ² b ⑀ :-separable, where x, a, b g G and a, b are of infinite order. If ² xy1 ax : l ² b : s 1, then there exists N ef G such that xy1 a i x s b j only if ⑀ < i, j, where G s GrN. Note that the condition ‘‘ xy1 a i x s b j only if ⑀ < i, j’’ always implies ⑀ < < a <, < b <. For convenience, we say N g N if N ef A such that N l ² h: s ² h ⑀ : and N l ² k : s ² k ⑀ :. In this case, we can construct G s G N as in Remark 2.4. LEMMA 3.4. Let A be quasi-regular and double coset separable at h, k 4 . Then G s ² A, t: ty1 ht s k : is double coset separable at h, k 4 . Proof. To show that, for example, G is ² h s : x ² k s :-separable for any s ) 0, we let y, x g G such that y f ² h s : x ² k s :. Then, by Lemma 2.5, we ˆ s G N1. If we can find N1 g N such that 5 ˆ y 5 s 5 y 5 and 5 ˆ x 5 s 5 x 5, in G can find N g N such that N ; N1 and y f ² h s : x ² k s : in G s G N , then there exists Lef G such that y f L² h s : x ² k s :, since G is residually finite and < h < s < k < is finite. Let L be the preimage of L in G. Then Lef G and y f L² h s : x ² k s :. This is the general sketch of our long proof. We prove the theorem by induction on 5 x 5. Suppose 5 x 5 s 0, i.e., x g A, we shall show that, for any s ) 0 and x g A, G is ² h s : x ² h s :-separable, ² h s : x ² k s :-separable, and ² k s : x ² k s :-separable. To show that G is ² h s : x ² k s :-separable for x g A, let y g G such that y f ² h s : x ² k s :. Suppose 5 y 5 s 0, that is, y g A. Then, by the double coset separability of A, there exists N2 ef A such that y f N2 ² h s : x ² k s :.
CONJUGACY SEPARABILITY
579
As in Lemma 2.5, we choose N g N such that N ; N2 . Thus y f ² h s : x ² k s : in G s G N . Suppose 5 y 5 G 1. Again, by Lemma 2.5, we choose N g N such that 5 y 5 s 5 y 5 in G s G N . Since 5 y 5 G 1 and x g A, y f ² h s : x ² k s : in G s G N . This proves that G is ² h s : x ² k s :-separable for x g A. Similarly, for any s ) 0, G is ² h s : x ² h s :-separable and ² k s : x ² k s :-separable for x g A. Let 5 x 5 s n. By induction we can assume that, for any s ) 0, G is ² h s : g ² h s :-separable, ² h s : g ² k s :-separable, and ² k s : g ² k s :-separable for all g g G such that 5 g 5 F n y 1. We need to show that G is ² h s : x ² h s :separable, ² h s : x ² k s :-separable, and ² k s : x ² k s :-separable for x g G with 5 x 5 s n. Case 1. To show G is ² h s : x ² k s :-separable for any x g G with 5 x 5 s n. Let y g G such that y f ² h s : x ² k s :. If 5 y 5 ) n or 5 y 5 - n or y s ¨ 1 t ␦ 1 ¨ 2 ⭈⭈⭈ t ␦ n ¨ nq1 and x s u1 t ⑀ 1 u 2 ⭈⭈⭈ t ⑀ n u nq1 with ␦ i / ⑀ i for some i then, by Lemma 2.5, we can find N1 g N such that, in G s G N1, 5 y 5 s 5 y 5 and 5 x 5 s 5 x 5. This implies y f ² h s : x ² k s : in G s G N1. So we consider y s ¨ 1 t ⑀ 1 ¨ 2 ⭈⭈⭈ t ⑀ n ¨ nq1 and x s u1 t ⑀ 1 u 2 ⭈⭈⭈ t ⑀ n u nq1 where y f ² h s : x ² k s :. For convenience, we let x ny 1 s u1 t ⑀ 1 u 2 ⭈⭈⭈ t ⑀ ny 1 u n and yny1 s ¨ 1 t ⑀ 1 ¨ 2 ⭈⭈⭈ t ⑀ ny 1 ¨ n . By Lemma 2.5, there exists N1 g N such that, in G s G N1, 5˜ x5 s 5 ˜ y 5 s n. We need to consider the cases ⑀ n s 1 and ⑀ n s y1. Subcase 1. Suppose ⑀ n s 1. If yny1 f ² h s : x ny1² h: or ¨ nq1 f ² k : u nq 1² k s :, then, by induction and Remark 3.2, we can find M ef G such that yny 1 f M Ž² h s : x ny1² h:. or ¨ nq1 f M² k : u nq1² k s :. By Lemma 2.5, we can find N g N such that N ; N1 l M and 5 x 5 s 5 y 5 s n in G s G N . If y s h si x k s j for some i, j, then ysi y1 xy1 yny1 s u nq1 k s j ¨ y1 ny 1 h nq1 .
Ž 3.1.
This implies ysi xy1 yny1 s h ␣ ny 1 h
and
k ␣ s u nq1 k s j ¨ y1 nq1
Ž 3.2.
for some ␣ , hence yny 1 g ² h s : x ny1² h: and ¨ nq1 g ² k : u nq1² k s :, contradicting the choice of N ; M. Thus y f ² h s : x ² k s :. Therefore we suppose yny 1 s h si1 x ny1 h i 2 and ¨ nq1 s k j1 u nq1 k s j 2 for some i1 , i 2 , j1 , j2 . Then y s yny 1 t¨ nq1 s h si1 x ny1 h i 2 tk j1 u nq1 k s j 2 f ² h s : x ny1 tu nq1² k s : . Ž 3.3. ² s : x ny1 tu nq1² k s : uy1 This implies that h i 2 tk j1 f xy1 ny 1 h nq1 . Hence, we need to find N g N such that y f ² h s : x ² k s : for each of the following four cases: : ² : ² y1 : ² : 1-a. Suppose ² xy1 ny 1 hx ny1 l h s 1 and u nq1 ku nq1 l k s 1.
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We note that i 2 q j1 / 0, since, in Ž3.3., h i 2 tk j1 s h i 2qj 1 t. Let ⑀ ) < i 2 q j1 <. By induction, G is ² h ⑀ : x ny1² h ⑀ :-separable and ² k ⑀ : u nq1² k ⑀ :-separable. Hence, by Lemma 3.3, there exist M1 ef G and M2 ef G such ˜i x ny1 s ˜h j in GrM1 then ⑀ < i, j and that if uˆy1 ˆ i ˆnq1 s ˆk j that if ˜ xy1 ny 1 h ˜ nq1 k u in GrM2 then ⑀ < i, j. Let N g N such that N ; M1 l M2 l N1 as in Lemma 2.5. In G s G N , if y g ² h s : x ² k s :, then y s h si xk s j. Thus, by the equation Ž3.2., we have yny 1 s h si x ny1 h ␣ s h si1 x ny1 h i 2 and ¨ nq1 s ky␣ u nq 1 k s j s k j1 u nq1 k s j 2 for some ␣ . This implies that sŽ iyi 1 . xy1 x ny1 s h i 2y ␣ ny 1 h
and
y ␣ yj1 uy1 u nq1 s k sŽ j 2yj. . nq1 k
Ž 3.4.
Thus ⑀ < i 2 y ␣ and ⑀ < y ␣ y j1 by the choice of N ; M1 l M2 . Hence ⑀ < i 2 q j1 , contradicting the choice of ⑀ . Hence y f ² h s : x ² k s :. : ² : ² y1 : ² : 1-b. Suppose ² xy1 ny 1 hx ny1 l h s 1 and u nq1 ku nq1 l k / 1. y1 ␣2 Let ␣ 2 ) 0 be the smallest integer such that u nq1 k u nq1 g ² k :. Then, ␣2 by Remark 2.6, we have uy1 u nq1 s k " ␣ 2 . By the equation Ž3.3., since nq 1 k i2 j1 i 2qj 1 i 2qj 1 h tk s tk , we note that k f ² k ␣ 2 : l ² k s : s ² k ␣ 2 sr d 2 :, where d 2 s Ž ␣ 2 , s .. Thus ␣ 2 srd 2 does not divide i 2 q j1. Let ⑀ s Ž ␣ 2 srd 2 . l and, i ² : as before, we can find N g N such that N ; N1 and uy1 nq1 k u nq1 f k y1 i j for all 1 F i - ␣ 2 , and if x ny1 h x ny1 s h then ⑀ < i, j. Thus, in G s G N , if y g ² h s : x ² k s : then equations of Ž3.4. hold. This implies ⑀ < i 2 y ␣ and ␣ 2 < y ␣ y j1. Now s < ⑀ and ⑀ < < h < s < k <. Hence s < < k <. This implies s < y ␣ y j1 by the second equation of Ž3.4.. Therefore Ž ␣ 2 srd 2 .< i 2 y ␣ y Žy␣ y j1 . s i 2 q j1 , a contradiction. Hence y f ² h s : x ² k s :. : ² : ² y1 : ² : 1-c. Suppose ² xy1 ny 1 hx ny1 l h / 1 and u nq1 ku nq1 l k s 1. ␣ 1 y1 Let ␣ 1 ) 0 be the smallest integer such that x ny1 h x ny1 g ² h:. Then " ␣1 by Remark 2.6 x ny 1 h ␣ 1 xy1 . Since h i 2 tk j1 s h i 2qj 1 t, from the ny1 s h i 2qj 1 Ž . ² equation 3.3 , we have h f h ␣ 1 : l ² h s : s ² h ␣ 1 sr d1 :, where d1 s Ž ␣ 1 , s .. Thus ␣ 1 srd1 does not divide i 2 q j1. Hence, as in 1-b above, for ⑀ s Ž ␣ 1 srd1 . l, we can find N g N such that N ; N1 and y f ² h s : x ² k s : in G s G N . : ² : ² y1 : ² : 1-d. Suppose ² xy1 ny 1 hx ny1 l h / 1 and u nq1 ku nq1 l k / 1. As in 1-b and 1-c above, let ␣ 1 , ␣ 2 ) 0 be the smallest integer such y1 ␣2 ² : that x ny 1 h ␣ 1 xy1 u nq1 g ² k :. By Eq. Ž3.3., h i 2qj 1 f ny1 g h and u nq1 k ␣ s ␣ s ␣ sr Ž² h 1 : l ² h :.Ž² h 2 : l ² h :. s ² h 1 d1 :² h ␣ 2 sr d 2 :, where d1 s Ž ␣ 1 , s . and d 2 s Ž ␣ 2 , s .. Hence, as before, for ⑀ s Ž ␣ 1 srd1 .Ž ␣ 2 srd 2 . l we can ² : find N g N such that N ; N1 and, in G s G N , x ny1 h i xy1 ny1 f h for all y1 j ² : 1 F i - ␣ 1 , u nq1 k u nq1 f k for all 1 F j - ␣ 2 , and, by Remark 3.2, h i 2qj1 f ² h ␣ 1 sr d1 :² h ␣ 2 sr d 2 :. In G s G N , if y g ² h s : x ² k s : then we have the equation Ž3.4.. Thus, from the first equation of Ž3.4., we have ␣ 1 < i 2 y ␣ ;
CONJUGACY SEPARABILITY
581
hence h i 2y ␣ s h " sŽ iyi1 .. Since s < ⑀ and ⑀ < < h <, s < i 2 y ␣ . Thus Ž ␣ 1 srd1 .< i 2 y ␣ . Hence h i 2y ␣ g ² h ␣ 1 sr d1 :. Similarly, from the second equation of Ž3.4., hy␣ yj1 g ² h ␣ 2 sr d 2 :. Therefore, h i 2qj1 s h i 2y ␣ h ␣qj1 g ² h ␣ 1 sr d1 :² h ␣ 2 sr d 2 :, a contradiction. Hence y f ² h s : x ² k s :. Subcase 2. Suppose ⑀ n s y1. If yny1 f ² h s : x ny1² k : or ¨ nq1 f ² h: u nq 1² k s : then by induction, as in Subcase 1 above, we can find N g N such that, in G s G N , 5 x 5 s 5 x 5, 5 y 5 s 5 y 5, and yny1 f ² h s : x ny1² k : or ¨˜nq 1 f ² h: u nq1² k s :. Moreover, if y s h si xk s j for some i, j, then ysi xy1 yny1y1 s u nq1 k s j ¨ y1 ny 1 h nq1 . This implies si ␣ xy1 ny 1 h yny1 s k
and
h ␣ s u nq1 k s j ¨ y1 nq1
Ž 3.5.
for some ␣ . This means yny 1 g ² h s : x ny1² k : and ¨ nq1 g ² h: u nq1² k s :, contradicting the choice of N. Hence y f ² h s : x ² k s :. So we can assume yny 1 s h si1 x ny1 k i 2 and ¨ nq1 s h j1 u nq1 k s j 2 for some i1 , i 2 , j1 , j2 . This implies y s yny 1 ty1 ¨ nq1 s h si1 x ny1 k i 2 ty1 h j1 u nq1 k s j 2 f ² h s : x ny1 ty1 u nq1² k s : .
Ž 3.6. ² s : x ny1 tu nq1² k s : uy1 This implies that k i 2 th j1 f xy1 ny 1 h nq1 . Hence, we need ² to find N g N such that y f h s : x ² k s : for each of the following four cases: : ² : ² y1 : ² : 2-a. Suppose ² xy1 ny 1 hx ny1 l k s1 and u nq1 hu nq1 l k s1. Equation Ž3.6. implies i 2 q j1 / 0. Let ⑀ ) < i 2 q j1 <. By induction, G is ² h ⑀ : x ny 1² k ⑀ :-separable and ² h ⑀ : u nq1² k ⑀ :-separable. Hence, by Lemma ˜i x ny1 s ˜k j in 3.3, there exist M1 ef G and M2 ef G such that if ˜ xy1 ny1 h ˜ ˆi ˆnq1 s ˆk j in GrM2 then ⑀ < i, j. Let GrM1 then ⑀ < i, j and that if u ˆy1 nq1 h u N g N such that N ; M1 l M2 l N1 as in 1-a of Subcase 1. In G s G N , if y g ² h s : x ² k s :, then y s h si xk s j. Thus, by Eq. Ž3.5., we have yny 1 s h si x ny 1 k ␣ s h si1 x ny1 k i 2 and ¨ nq1 s hy ␣ u nq1 k s j s h j1 u nq1 k s j 2 for some sŽ iyi 1 . y ␣ yj1 ␣ . This implies that xy1 x ny1 s k i 2y ␣ and uy1 u nq1 s ny 1 h nq1 h sŽ j 2 yj. k . Thus, by the choice of N ; M1 l M2 , ⑀ < i 2 y ␣ and ⑀ < y ␣ y j1. Hence ⑀ < i 2 q j1 , contradicting the choice of ⑀ . Hence y f ² h s : x ² k s :. The following three cases are also similar to Subcase 1 above. : ² : ² y1 : ² : 2-b. Suppose ² xy1 ny 1 hx ny1 l k s 1 and u nq1 hu nq1 l k / 1. y1 y1 ² : ² : ² : 2-c. Suppose x ny 1 hx ny1 l k / 1 and u nq1 hu nq1 l ² k : s 1. : ² : ² y1 : ² : 2-d. Suppose ² xy1 ny 1 hx ny1 l k / 1 and u nq1 hu nq1 l k / 1. Hence there exists N g N such that, in G s G N , y f ² h s : x ² k s :. Therefore, by induction, G is ² h s : x ² k s :-separable for any x g G.
582
KIM AND TANG
Case 2. G is ² h s : x ² h s :-separable for any x g G with 5 x 5 s n. Let y g G such that y f ² h s : x ² h s :. If 5 y 5 ) n or 5 y 5 - n or y s ¨ 1 t ␦ 1 ¨ 2 ⭈⭈⭈ t ␦ n ¨ nq1 and x s u1 t ⑀ 1 u 2 ⭈⭈⭈ t ⑀ n u nq1 with ␦ i / ⑀ i for some i then, as in Case 1, we can find N1 g N such that, in G s G N1, 5 y 5 s 5 y 5, 5 x 5 s 5 x 5, and y f ² h s : x ² h s :. So we consider y s ¨ 1 t ⑀ 1 ¨ 2 ⭈⭈⭈ t ⑀ n ¨ nq1 and x s u1 t ⑀ 1 u 2 ⭈⭈⭈ t ⑀ n u nq1 where y f ² h s : x ² h s :. As before, we let x ny1 s u1 t ⑀ 1 u 2 ⭈⭈⭈ t ⑀ ny 1 u n and yny1 s ¨ 1 t ⑀ 1 ¨ 2 ⭈⭈⭈ t ⑀ ny 1 ¨ n . By Lemma 2.5, there ˜ s G N1. exists N1 g N such that 5 ˜ x5 s 5 ˜ y 5 s n in G Subcase 1. Suppose ⑀ n s 1. If yny1 f ² h s : x ny1² h: or ¨ nq1 f ² k : u nq 1² h s : then by induction, as in Subcase 1 of Case 1, we can find N g N such that, in G s G N , 5 x 5 s 5 x 5, 5 y 5 s 5 y 5 and yny1 f ² h s : x ny 1² h: or ¨ nq1 f ² k : u nq1² h s :. Then, in G s G N , y f ² h s : x ² h s : as in Case 1. Hence we can assume yny 1 s h si1 x ny1 h i 2 and ¨ nq1 s k j1 u nq 1 h s j 2 for some i1 , i 2 , j1 , j2 . This implies y s yny 1 t¨ nq1 s h si1 x ny1 h i 2 tk j1 u nq1 h s j 2 f ² h s : x ny1 tu nq1² h s : . Ž 3.7. As in case 1, we have the following cases: 1-a. 1-b. 1-c. 1-d.
: ² : ² y1 : ² : Suppose ² xy1 ny 1 h x ny1 l h s 1 and u nq1 ku nq1 l h s 1. y1 y1 Suppose ² x ny 1 h x ny1 : l ² h: s 1 and ² u nq1 ku nq1 : l ² h: / 1. : ² : ² y1 : ² : Suppose ² xy1 ny 1 h x ny1 l h / 1 and u nq1 ku nq1 l h s 1. y1 y1 Suppose ² x ny 1 h x ny1 : l ² h: / 1 and ² u nq1 ku nq1 : l ² h: / 1.
In the same manner we can find N g N such that y f ² h s : x ² h s : in G s G N for each of the cases. Subcase 2. Suppose ⑀ n s y1. If yny1 f ² h s : x ny1² k : or ¨ nq1 f ² h: u nq 1² h s : then, as in Subcase 2 of Case 1, we can find N g N such that y f ² h s : x ² h s : in G s G N . Thus we can assume yny1 s h si1 x ny1 k i 2 and ¨ nq 1 s h j1 u nq1 h s j 2 for some i1 , i 2 , j1 , j2 . This implies y s yny 1 ty1 ¨ nq1 s h si1 x ny1 k i 2 ty1 h j1 u nq1 h s j 2 f ² h s : x ny1 ty1 u nq1² h s : .
Ž 3.8. Again we have the following cases: 2-a. 2-b. 2-c. 2-d.
: ² : ² y1 : ² : Suppose ² xy1 ny 1 h x ny1 l k s 1 and u nq1 hu nq1 l h s 1. y1 y1 Suppose ² x ny 1 h x ny1 : l ² k : s 1 and ² u nq1 hu nq1 : l ² h: / 1. : ² : ² y1 : ² : Suppose ² xy1 ny 1 h x ny1 l k / 1 and u nq1 hu nq1 l h s 1. y1 y1 Suppose ² x ny 1 h x ny1 : l ² k : / 1 and ² u nq1 hu nq1 : l ² h: / 1.
CONJUGACY SEPARABILITY
583
In each of the cases, we can find N g N such that y f ² h s : x ² h s : in G s G N . Hence, as in Case 1, by induction G is ² h s : x ² h s :-separable for any x g G. Similarly, G is ² k s : x ² k s :-separable for any x g G. Therefore, G is double coset separable at h, k 4 . DEFINITION 3.5. Let A be a group and let h, k g A be of infinite order. Then A is said to be s-quasi-regular at h, k 4 if ² h: l ² k : s 1 and if, for each integer ⑀ ) 0, there exist an integer ⑀ ) 0 and N⑀ ef A such that N⑀ l ² h: s ² h ⑀⑀ :, N⑀ l ² k : s ² k ⑀⑀ :, and ² h: l ² k : s 1, in A s ArN⑀ . Clearly if A is s-quasi-regular at h, k 4 , then A is quasi-regular at h, k 4 . Hence, as in Lemma 2.5, we have the following: LEMMA 3.6. Let G s ² A, t: ty1 ht s k :. Suppose A is ² h:-separable, ² k :-separable, and s-quasi-regular at h, k 4 . Then, for each M ef A, for each s ) 0 and for each reduced x g G, there exist N ef A and ) 0 such that N ; M, N l ² h: s ² h s :, N l ² k : s ² k s :, ² h: l ² k : s 1, and 5 x 5 s 5 x 5 in G s G N . THEOREM 3.7. Let A be s-quasi-regular and double coset separable at h, k 4 . Let x g G s ² A, t: ty1 ht s k : be cyclically reduced. If 5 x 5 G 1 then, for each y g G such that x ¤G y, there exists N ef G such that x ¤G y, where G s GrN. Proof. WLOG, we may assume that y is also cyclically reduced in G. Case 1 Ž5 x 5 / 5 y 5.. Since x is cyclically reduced, by Lemma 2.5 we can find N g N such that, in G s G N , 5 x 5 s 5 x 5, 5 y 5 s 5 y 5, and x, y are cyclically reduced. This implies 5 x 5 / 5 y 5. Hence, by Theorem 2.1, x ¤G y. Case 2 Ž5 x 5 s 5 y 5 G 1.. Let x s t ⑀ 1 u1 t ⑀ 2 ⭈⭈⭈ u ny1 t ⑀ n u n and y s t ¨ 1 t ␦ 2 ⭈⭈⭈ ¨ ny1 t ␦ n ¨ n be cyclically reduced, where u i , ¨ i g A. Since x ¤G y, by Theorem 2.1, x ¤ ² h: j ² k : yiU for any cyclic permutation yiU s t ␦ i ¨ i ⭈⭈⭈ t ␦ n ¨ n t ␦ 1 ¨ 1 ⭈⭈⭈ t ␦ iy1 ¨ iy1 of y. For each i, we shall find Ni g N such that, in ˆ 5 x Ni 5 s 5 x 5 s 5 yUi N i 5 s 5 y 5 and x N i , yUi N i are cyclically G N i s G, U reduced with x N i ¤ ² h: ˆ j ² ˆk : yi N i . Let N s N1 l ⭈⭈⭈ l Nn . Then, in G s G N , we have 5 x 5 s 5 x 5 s 5 y 5 s 5 y 5 with x, y cyclically reduced and x ¤ ² h: j ² k : yiU for all i. Hence, by Theorem 2.2, x ¤G y as required. To find such Ni ’s, it suffices to consider only the case i s 1, others being similar. As in Case 1, by Lemma 3.6 we can find M1 g N such that ²ˆ h: l ² k : s 1, 5 ˆ x5 s 5 x5 s 5 ˆ y 5 s 5 y 5 with ˆ x, ˆ y cyclically reduced in ˆ s G M 1. G If ⑀ i / ␦ i for some i, then we choose N1 s M1. This implies x ¤ ² h: j ² k : y in G s G N1. Hence we can assume ⑀ i s ␦ i for all i and ⑀ 1 s 1, say. ␦1
584
KIM AND TANG
Subcase 1. Suppose y f ² h: x ² h:. By Lemma 3.4, there exists M2 ef G such that y f M2 ² h: x ² h:. Let N1 g N such that N1 ; M1 l M2 . Then x ¤² h: y in G s G N1. Subcase 2. Suppose y s h ␣ xh  but h ␣ / hy . In this case, we need to find N1 g N such that x ¤ ² h: y in G s G N1 in each of the following cases: Ža. Suppose ² xy1 hx : l ² h: s 1. By Lemma 3.4, G is ² h: x ² h:-separable. Moreover, by Lemma 3.3, for ⑀ ) < ␣ q  <, there exists M2 ef G such that if M2 Ž xy1 h i x . s M2 h j then ⑀ < i, j. Furthermore, by Lemma 3.6, there exists N1 g N such that N1 ; M1 l M2 , ² h: l ² k : s 1, and 5 x 5 s 5 x 5 s 5 y 5 s 5 y 5 in G s G N . If y ;² h: x then y s hyi xh i s h ␣ xh  . This 1 implies xy1 hyiy ␣ x s h yi, whence ⑀ < y i y ␣ ,  y i. Therefore ⑀ < ␣ q  , contradicting the choice of ⑀ . Hence x ¤² h: y in G s G N1. Žb. Suppose there exists an integer s ) 0 such that xy1 h s x s hys. WLOG, we can assume s to be the smallest such integer. If h ␣q  s h 2 sr then y s h ␣ xh  s hy q2 sr xh  s hy qsr xh ysr , whence y ;² h: x. Hence h ␣q  f ² h 2 s :. Since G is ² h ⑀ :-separable for any ⑀ by Lemma 3.4, there exists M2 ef G such that h ␣q  f M2 ² h 2 s : and xy1 h i x f M2 ² h: for 1 F i - s. By Lemma 3.6, we can choose N1 g N such that N1 ; M1 l M2 , ² h: l ² k : s 1, h ␣q  f ² h 2 s :, and xy1 h i x f ² h: for 1 F i - s in G s G N1. If y s hyi xh i then, as before, we have xy1 hyiy ␣ x s h yi. This implies s < y i y ␣ . Let yi y ␣ s s for some , whence h yi s xy1 h s x s hys . Thus h ␣q  s h iq ␣ h yi s hy2 s g ² h 2 s :, contradicting h ␣q  f ² h2 s :. Hence x ¤² h: y in G s G N1. Žc. Suppose there exists an integer s ) 0 such that xy1 h s x s h s. WLOG, we can assume s to be the smallest such integer. Since, by Lemma 3.4, G is ² h ⑀ :-separable for any ⑀ , we can find M2 ef G such that h ␣q  f M2 and xy1 h i x f M2 ² h: for 1 F i - s. Now, by Lemma 3.6, we can choose N1 g N such that N1 ; M1 l M2 and ² h: l ² k : s 1, so that in G s G N1, if y s hyi xh i then xy1 hyiy ␣ x s h yi and s < y i y ␣ . Let yi y ␣ s s for some . Then h yi s xy1 h s x s h s . This implies h ␣q  s h iq ␣ h yi s 1, contradicting h ␣q  f M2 . Hence x ¤² h: y in G s G N1. Thus, in each case, we have found N1 such that, in G s G N1, ² h: l ² k : s 1 and x ¤² h: y. Since ⑀ 1 s 1 and ² h: l ² k : s 1, x ¤² k : y. Hence x ¤² h: j ² k : y in G N1, as required. 4. A CRITERION In this section we shall prove a criterion for the conjugacy separability of HNN extensions.
CONJUGACY SEPARABILITY
585
DEFINITION 4.1. Let A be a group and let h, k g A be of infinite order. Then A is said to be c-quasi-regular at h, k 4 if ² h: l ² k : s 1 and there exists an integer ␣ such that, for a fixed integer m ) 0 and for each integer ⑀ ) 0, there exist an integer ⑀ ) 0 and N⑀ ef A such that: Ž1. N⑀ l ² h: s ² h ⑀⑀ : and N⑀ l ² k : s ² k ⑀⑀ :, Ž2. in A s ArN⑀ , ² h: l ² k : s 1, Ž3. if h ␣ m ;A h j then h j s h " ␣ m , Ž4. if k ␣ m ;A k j then k j s k " ␣ m , and Ž5. if h i ;A k j then h i s k j s 1. For example, if A is polycyclic-by-finite and if h, k g ZŽ A. are of infinite order then A is c-quasi-regular at h, k 4 whenever ² h: l ² k : s 1. Clearly if A is c-quasi-regular at h, k 4 , then A is s-quasi-regular at h, k 4 and, hence, quasi-regular at h, k 4 . Remark 4.2. If A is c-quasi-regular at h, k 4 and h i ;G k j, then i s j s 0. For, by the above definition, for each ⑀ ) 0, there exist ⑀ and N ef A such that N l ² h: s ² h ⑀⑀ : and N l ² k : s ² k ⑀⑀ :. Then, in A s A N , h i ;A k j implies h i s k j s 1. Thus i, j must be divisible by every ⑀ ) 0. Hence i s 0 s j. LEMMA 4.3. If A is c-quasi-regular at h, k 4 then, for each M ef A, there exists an integer ␣ such that, for a fixed integer m ) 0 and for each integer s ) 0, there exist s and Ns ef A such that Ns ; M, Ns l ² h: s ² h s s :, and Ns l ² k : s ² k s s : such that, in A s ArNs , ² h: l ² k : s 1, if h ␣ m ;A h j then h j s h " ␣ m , if k ␣ m ;A k j then k j s k " ␣ m , and if h i ;A k j then h i s k j s 1. Proof. Let M l ² h: s ² h s1 : and M l ² k : s ² k s 2 :. Let ⑀ s s1 s2 s. By Definition 4.1, there exist ) 0 and N1 ef A such that N1 l ² h: s ² h ⑀ :, N1 l ² k : s ² k ⑀ :, and in A˜ s ArN1 , ² ˜ h: l ² ˜ k : s 1, if ˜ h ␣ m ;A˜ ˜ h j then ˜h j s ˜h " ␣ m , if ˜k ␣ m ;A˜ ˜k j then ˜k j s ˜k " ␣ m , and if ˜h i ;A˜ ˜k j then ˜h i s ˜k j s 1. Let Ns s N1 l M and s s s1 s2 . Then Ns ; M and Ns l ² h: s ² h s 1 s 2 s : s ² h s s : , Ns l ² k : s ² k s 1 s 2 s : s ² k s s : . Since Ns l ² h: s N1 l ² h: and Ns l ² k : s N1 l ² k :, we have, in A s ArNs , ² h: l ² k : s 1, if h ␣ m ;A h j then h j s h " ␣ m , if k ␣ m ;A k j then k j s k " ␣ m , and if h i ;A k j then h i s k j s 1. DEFINITION 4.4. A group G is said to be cyclic conjugacy separable for ² h: if x 4G l ² h: s ⭋ for x g G; then there exists N ef G such that, in G s GrN, x 4G l ² h: s ⭋. It was proved in w7x that free-by-finite groups are cyclic conjugacy separable for each ² h:. We are now ready to prove a criterion for the conjugacy separability of HNN extensions. THEOREM 4.5. Let A be c , conjugacy separable, c-quasi-regular, and double coset separable at h, k 4 . If A is cyclic conjugacy separable for ² h: and ² k :, then G s ² A, t: ty1 ht s k : is conjugacy separable.
586
KIM AND TANG
Proof. Let x, y g G be such that x ¤G y and that x, y are of minimal lengths in their respective conjugacy classes. By Theorem 2.7, G is c , whence residually finite. Therefore we may assume x / 1 / y. By Theorem 3.7, we need only prove the case 5 x 5 s 5 y 5 s 0. Also by Theorem 2.2, G s G N is conjugacy separable for each N g N . Therefore, throughout the proof, we shall find a suitable N g N such that, in G s G N , x ¤G y. Since we need only consider the case 5 x 5 s 5 y 5 s 0, that is x, y g A, we have the following three cases. Case 1.
x, y g ² h: Žsimilarly x, y g ² k :..
Ža. Suppose x s h m and y s h n, where m / "n. By the quasi-regularity, for ⑀ s mn, there exist and N g N such that < h < s < k < s ⑀. Then < x < s < h m < and < y < s < h n < are different; hence x ¤G y in G s G N . Žb. Suppose x s h m and y s hym . This implies h m ¤A hym and m k ¤A kym . Since A is conjugacy separable, there exists N1 ef A such that N1 h m ¤Ar N1 N1 hym and N1 k m ¤A r N1 N1 kym . Let N1 l ² h: s ² h s1 : and N1 l ² k : s ² k s 2 :. By Lemma 4.3, there exists ␣ such that, for ⑀ s 2 s1 s2 ␣ m, there exist N ef A and such that N ; N1 and, in A s ArN, ² h: l ² k : s 1 and < h < s ⑀ s < k < satisfying the conditions Ž3. ᎐ Ž5. of Definition 4.1. Now, if x ;G y then, by Theorem 2.1, there exist z1 , z 2 , . . . , z r of elements in ² h: j ² k : such that h m ;A z1 ;A , * z 2 ;A , * ⭈⭈⭈ ; A , * z r ;A hym . By the choice of N, z1 s h i g ² h:. This implies h ␣ m ;A h i ␣ . Thus h i ␣ s h " ␣ m . If h i ␣ s hy ␣ m then ⑀ s 2 s1 s2 ␣ m s < h < divides ␣ Ž i q m.. Let i q m s 2 s1 s2 for some . Since h m ;A h i s z1 , in A˜ s ArN1 , ˜ h m ;A˜ i ymq2 s s m ym m ym ␣ 1 2 ˜h s ˜h s˜ h , contradicting ˜ h ¤A˜ ˜ h . Therefore z1 s h i ␣ ␣m ␣m s h . Now z1 ;A, * z 2 . This implies h s z1␣ ;A, * z 2␣ . Therefore, either z 2␣ s h ␣ m Žas above., or z 2␣ s y1 h ␣ m s k ␣ m . Repeating this process, we have either z r␣ s h ␣ m or z r␣ s y1 h ␣ m s k ␣ m . Clearly z r␣ / k ␣ m . On the other hand, z r ;A hym implies z r␣ s h ␣ m ;A h ␣ m . Then, as above, hy␣ m s h ␣ m , contradicting < h < s 2 s1 s2 ␣ m ) 2 ␣ m. Hence x ¤G y. Case 2. Suppose x g ² h: and y g ² k : Žsimilarly x g ² k : and y g ² h:.. Let x s h m and y s k n. Since k n ;G h n, the case follows from Case 1. Case 3. Suppose x g A and x 4 A l ² h: s ⭋ Žor x 4 A l ² k : s ⭋.. By cyclic conjugacy separability of ² h:, there exists N1 ef A such that in ˜ A˜ s ArN1 , ˜ x4 A l ² ˜ h: s ⭋. Since A is conjugacy separable, there exists N2 ef A such that N2 x ¤A r N 2 N2 y. By Lemma 4.3, we can choose N g N
CONJUGACY SEPARABILITY
587
such that N ; N1 l N2 . Let G s G N . Since x 4 A l ² h: s ⭋ and x ¤A y, by Theorem 2.1, x ¤G y. The following is a generalization of Theorem 3.3 in w8x. COROLLARY 4.6. If A is polycyclic-by-finite and h, k g ZŽ A. of infinite order such that ² h: l ² k : s 1 then G s ² A, t: ty1 ht s k : is conjugacy separable. Proof. Since polycyclic groups are double coset separable w11x, this implies A is c and double coset separable at h, k 4 . It was also well known that polycyclic-by-finite groups are conjugacy separable w4x. We need to show that A is c-quasi-regular. For a given ⑀ , consider A˜ s Ar² h ⑀ :² k ⑀ :. Since A˜ is again polycyclic-by-finite, A˜ is residually ˜˜i / 1 / Nk ˜˜ j and Nh ˜˜i / Nk ˜˜ j finite. Thus there exists N˜ ef A˜ such that Nh ˜ for 1 F i, j - ⑀ . Let N be the preimage of N in A. Then N ef A. Moreover N l ² h: s ² h ⑀ :, N l ² k : s ² k ⑀ : and, in A s ArN, ² h: l ² k : s 1. Furthermore, since h, k g ZŽ A., if h m ;A h i then h i s h m , if k m ;A k i then k i s k m , if h i ;A k j then h i s k j s 1. Hence A is cquasi-regular at h, k 4 . Now h, k g ZŽ A. and A is c . This implies A is cyclic conjugacy separable for ² h: and ² k :. Therefore, by Theorem 4.5, G is conjugacy separable.
5. ON HNN EXTENSIONS OF FREE PRODUCTS OF FINITE CYCLES In this section we shall prove the conjugacy separability of certain HNN extensions of free products of cyclic groups. Let G s ² t , a1 , . . . , a m , b1 , . . . , bn : a i␣ i , bj j , ty1 ht s k : , where h g ² a1 , . . . , a m : and k g ² b1 , . . . , bn : are of infinite order. For convenience, we let C s ² a1 , . . . , a m :, D s ² b1 , . . . , bn :, and A s C ) D. LEMMA 5.1 w10, Lemma 2.8x. Let G be a finite extension of a residually finitely generated torsion-free nilpotent group B. Let h g G be of infinite order. If ² h: l B s ² h n : then, for each integer t, there exists 1 ) 0 such that for any ␣ ) 0 there exists N␣ ef G so that N␣ l ² h: s ² h ␣ n 1 : and that h nt ;G h j implies h j s h " nt , where G s GrN␣ . LEMMA 5.2. The group A s C ) D abo¨ e is c-quasi-regular at h, k 4 . Proof. Since C, D are free-by-finite, let C1 , D 1 be free subgroups of finite index in C, D, respectively. Let C1 l ² h: s ² h n1 : and D 1 l ² k : s ² k n 2 :. Let m be a given integer.
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Let t s n 2 m in Lemma 5.1; there exists 1 such that, for any s ) 0, there exists Ms ef C such that Ms l ² h: s ² h sn1 1 : and, in Cˆ s CrMs , if ˆh n1 n 2 m ;Cˆ ˆh i then ˆh i s ˆh " n1 n 2 m . Similarly, let t s n1 m in Lemma 5.1. Then there exists 2 such that, for any s ) 0, there exists L s ef D such ˆ s DrL s , if ˆk n1 n 2 m ;Dˆ ˆk i then that L s l ² k : s ² k sn 2 2 : and, in D ˆk i s ˆk " n1 n 2 m . For any ⑀ , considering s s n 2 2 ⑀ . Then there exists M1 ef C such that M1 l ² h: s ² h n1 n 2 1 2 ⑀ : and, in C˜ s CrM1 , if ˜ h n1 n 2 m ;C˜ ˜ h i then ˜ hi s " n n m 1 2 ˜h . Similarly, for s s n1 1 ⑀ , there exists L1 ef D such that ˜ s DrL1 , if ˜k n1 n 2 m ;D˜ ˜k i then ˜k i s L1 l ² k : s ² k n1 n 2 1 2 ⑀ : and, in D " n n m i ˜k 1 2 . Let A˜ s C˜) D. ˜ Then ˜h ¤A˜ ˜k j for any ˜h i / 1 / ˜k j. Since A˜ is free-by-finite, it is conjugacy separable w2x. Thus there exists N˜ ef A˜ ˜˜i ¤A˜r N˜ Nk ˜˜ j for all ˜h i / 1 / ˜k j, Nh ˜˜i ¤A˜r N˜ such that ˜ h i, ˜ k j f N˜ and Nh j i j i j i j ˜˜ ¤A˜r N˜ Nk ˜˜ for all ˜k ¤A˜ ˜k . Let N be the N˜˜ h for all ˜ h ¤A˜ ˜ h , Nk preimage of N˜ in A. Then N l ² h: s ² h n1 n 2 1 c 2 ⑀ :, N l ² k : s ² k n1 n 2 1 2 ⑀ :, and, in A s ArN, ² h: l ² k : s 1. Moreover, if h n1 n 2 m ;A h j then h j s h " n1 n 2 m , if k n1 n 2 m ;A k j then k j s k " n1 n 2 m , and if h i ;A k j then h i s k j s 1. Hence A is c-quasi-regular at h, k 4 . Since A is c-quasi-regular at h, k 4 , A is quasi-regular at h, k 4 . Hence, by Theorem 2.7, we have the following: THEOREM 5.3. The group G s ² t, a1 , . . . , a m , b1 , . . . , bn : a i␣ i , bj j , ty1 ht s k :, where h g ² a1 , . . . , a m : and k g ² b1 , . . . , bn :, is c , whence residually finite. THEOREM 5.4. The group G s ² t, a1 , . . . , a m , b1 , . . . , bn : a i␣ i , bj j , ty1 ht s k :, where h g ² a1 , . . . , a m : and k g ² b1 , . . . , bn :, is conjugacy separable. Proof. By Lemma 5.2, A s ² a1 , . . . , a m , b1 , . . . , bn : is c-quasi-regular at h, k 4 . Since A is free-by-finite and since free groups are double coset separable w15x, A is c and double coset separable at h, k 4 . Moreover, free-by-finite groups are conjugacy separable w2x and cyclic conjugacy separable w7x. The theorem follows from Theorem 4.5.
6. ON HNN EXTENSIONS OF BAUMSLAG᎐SOLITAR GROUPS In this section we consider the HNN extension G s ² t , a, b: ty1 a t s b , by1 a b s a : ,
Ž 6.1.
of the Baumslag᎐Solitar group A s ² a, b: by1 a b s a : with associated subgroups ² a : and ² b :. Hopficity and residual finiteness of this type of
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groups were studied by Raptis, Talelli, and Varsos w16x which motivates this section. Using their result on residual finiteness, we prove the cyclic subgroup separability and conjugacy separability of G. THEOREM 6.1 w16x. The group G, as gi¨ en by Ž6.1., is residually finite if and only if < < s < <. THEOREM 6.2 w12, 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. Ž1. If < x < 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 x* where x* is a cyclic permutation of x. LEMMA 6.3. The group A s ² a, b: by1 a b s a : is c-quasi-regular at a , b 4 if < < s < <. Proof. Case 1. A s ² a, b: by1 a b s a :. Clearly ² h: l ² k : s 1, where h s a and k s b . Let ⑀ be a given integer. Since A s ² a:) a sx² x, b: by1 xb s x :, we consider the natural homomorphism
⑀ : A ª ² a:r² a ⑀ : )
a ˆsxˆ
Ž ² x :r² x ⑀ : = ² b:r² b ⑀ : . ,
where Aˆ s A ⑀ . Clearly < ˆ h < s ⑀ s < ˆ k < and ² ˆ h: l ² ˆ k : s 1. Since ˆ xg ZŽ Aˆ., it follows from Theorem 6.2 that if ˆ h m ;Aˆ ˆ h i then ˆ hm s ˆ h i. Similarly, if ˆ h i ;Aˆ ˆ k j then ˆ hi s 1 s ˆ k j. m i Now suppose ˆ k ;Aˆ ˆ k . Since x g ZŽ A., this implies that if ˆ k m ;Aˆ ˆ xs m s m i for some s, then ˆ k sˆ x g ²ˆ h: l ² ˆ k : s 1. Hence ˆ k s1sˆ k . On the other hand, if ˆ k m ¤Aˆ ˆ x s for any s then, by Theorem 6.2, ˆ k m and ˆ k i are conjugate in ² ˆ x: = ²ˆ b :. Thus ˆ km s ˆ k i. ˆ We have shown that, in A s A ⑀ , < ˆ h < s ⑀ s < ˆ k < and ² ˆ h: l ² ˆ k : s 1. m i m i i j i j Moreover, if ˆ h ;Aˆ ˆ h then ˆ h sˆ h , if ˆ h ;Aˆ ˆ k then ˆ h s1sˆ k , and if ˆk m ;Aˆ ˆk i then ˆk m s ˆk i. Since Aˆ is free-by-finite, it is conjugacy separable. Thus, as in the proof of Lemma 5.2, we can find N ef A such that N l ² h: s ² h ⑀ :, N l ² k : s ² k ⑀ :, and, in A s ArN, conditions Ž2. ᎐ Ž5. of Definition 4.1 are satisfied. Hence A is c-quasi-regular at h, k 4 s a , b 4 .
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Case 2. A s ² a, b: by1 a b s ay :. Clearly ² h: l ² k : s 1, where h s a and k s b . Let ⑀ be a given integer. Since A s ² a:) a sx² x, b: by1 xb s xy1 :, we consider the natural homomorphism
⑀ : A ª ² a:r² a2 ⑀ : )
a ˆsxˆ
Ž ² x, b:r² x 2 ⑀ :² b 2 ⑀ : . ,
where Aˆ s A ⑀ . Clearly < ˆ h < s 2 ⑀ s < ˆ k < and ² ˆ h: l ² ˆ k : s 1, since ² ˆ x: l ²ˆ b : s 1. Suppose ˆ h2 m ;Aˆ ˆ h i. We note that ˆ gy1 ˆ x sˆ g sˆ x " s for any s and any 2 m s 2 m g g A. If ˆ h ;Aˆ ˆ x for some s, then ˆ h sˆ x " s and ˆ hi s ˆ x " s, thus ˆh i s ˆh " 2 m . If ˆh2 m ¤Aˆ ˆx s for any s then, by Theorem 6.2, ˆh2 m ;² a: ˆi ˆ h. i 2m ˆ ˆ This implies h s h . Now suppose ˆ k 2 m ;Aˆ ˆ k i. We note that ˆ k 2 m g ZŽ² ˆ x, ˆ b :.. If ˆ k 2 m ;Aˆ ˆ xs 2m "s ˆ ˆ ² : ² : for some s then, by Theorem 6.2, k s ˆ x g k l ˆ x s 1. Hence ˆk 2 m s 1 s ˆk i. On the other hand, if ˆk 2 m ¤Aˆ ˆx s for any s then, by Theorem 6.2, ˆ k 2 m and ˆ k i are conjugate in ² ˆ x, ˆ b :. This implies ˆ k2m s ˆ k i. i j i s ˆ ˆ ˆ Suppose h ;Aˆ k ; then, by Theorem 6.2, h ;Aˆ ˆ x for some s. Therefore, as above, ˆ hi s ˆ x " s. This implies ˆ k j ;Aˆ ˆ x " s. This means ˆ kj sˆ x" s g j ˆ ˆ ˆ ² k: l ² ˆ x : s 1. Hence h s 1 s k . Thus, as in Case 1, we can find N ef A such that conditions Ž1. ᎐ Ž5. of Definition 4.1 are satisfied. Hence A is c-quasi-regular at h, k 4 s a , b 4 . THEOREM 6.4. The group G s ² t, a, b: ty1 a t s b , by1 a b s a : is c if and only if < < s < <. Proof. Suppose < < s < <. By w17x, A s ² a, b: by1 a b s a : is c if and only if < < s < <. Therefore, by Theorem 2.7 and Lemma 6.3, G is c . Conversely, if G is c then A is c . Hence, < < s < < by w17x. LEMMA 6.5. The group A s ² a, b: by1 a b s a : is cyclic conjugacy separable for ² a : and ² b : if < < s < <. Proof. Case 1. A s ² a, b: by1 a b s a :. Let A s ² a:) a sx² x, b: by1 xb s x : and let g g A be of minimal length in its conjugate class in A. Ža. A is cyclic conjugacy separable for ² h: s ² a :. Let g 4 A l ² h: s ⭋. Suppose 5 g 5 G 2 or g g ² x, b : _ ² a:. Since a s x g ZŽ A., we can find a homomorphism ⑀ , as in Lemma 6.3, Case 1, such that, in Aˆ s A ⑀ , 5 ˆ g 5 s 5 g 5 G 2 or ˆ g g ²ˆ x, ˆ b: _ ² a ˆ:. Then, by Theorem i ˆ ˆ 6.2, ˆ g ¤Aˆ h for any i. Since A is conjugacy separable and < ˆ h < - ⬁, there
CONJUGACY SEPARABILITY
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ˆˆ ¤Aˆr Nˆ Nh ˆˆi for all i. Let N be the preimage exists Nˆ ef Aˆ such that Ng of Nˆ in A. Then, in A s ArN, g 4 A l ² h: s ⭋. Suppose g g ² a:. Since g 4 A l ² h: s ⭋, g s a s f ² a :. Then, in Aˆ s A ⑀ as above, ˆ g f ²ˆ h:. Thus, by Theorem 6.2, ˆ g ¤Aˆ ˆ h i for all i. Then, as Ž . in a , we can find N ef A such that, in A s ArN, g 4 A l ² h: s ⭋. This proves that A is cyclic conjugacy separable for ² h:. Žb. A is cyclic conjugacy separable for ² k : s ² b :. Let g 4 A l ² k : s ⭋. Suppose 5 g 5 G 2 or g g ² a: _ ² a :. As before, we can find a homomorphic ⑀ above such that, in Aˆ s A ⑀ , 5 ˆ g5 s 5 g5 G 2 : i ˆ : ² or ˆ g g ²a _ a . Then, by Theorem 6.2, g ¤ b for all i. As in Ža. we ˆ ˆ ˆ Aˆ A 4 ² : can find N ef A such that, in A s ArN, g l k s ⭋. So, suppose g g ² a , b :. Since g 4 A l ² k : s ⭋, g f ² b :. Again, we can choose Aˆ s A ⑀ such that ˆ g f ²ˆ b :. Thus, by Theorem 6.2, ˆ g ¤Aˆ ˆ bi Ž . for any i. Hence, as in a , we can find N ef A such that, A s ArN, g 4 A l ² k : s ⭋. This proves that A is cyclic conjugacy separable for ² k :. Case 2. A s ² a, b: by1 a b s ay :. Let A s ² a:) a sx² x, b: by1 xb s x and let g g A be of minimal length in its conjugate class in A. y1 :
Ža. A is cyclic conjugacy separable for ² h: s ² a :. Let g 4 A l ² h: s ⭋. Suppose 5 g 5 G 2 or g g ² a , b : _ ² a :. As in Lemma 6.3, Case 2, we can find a homomorphism ⑀ such that, in Aˆ s A ⑀ , 5 ˆ g 5 s 5 g 5 G 2 or ˆ g g ²a ˆ, ˆb : _ ² aˆ:. Then, by Theorem 6.2, i ˆg ¤Aˆ ˆh for all i. If g g ² a:, then g s a s f ² h: s ² a :. If a s f ² a :, we can choose Aˆ s A ⑀ such that ˆ g f ²a ˆ: l ² aˆ :. Then, by Theorem 6.2, i s ˆg ¤Aˆ ˆh for all i. If g s a g ² a : we can choose Aˆ s A ⑀ such that ˆg f ² aˆ :. Then, again by Theorem 6.2, ˆg ¤Aˆ ˆh i for all i. Hence, as in Case 1, A is cyclic conjugacy separable for ² h:. Žb. A is cyclic conjugacy separable for ² k : s ² b :. Let g 4 A l ² k : s ⭋. Suppose 5 g 5 G 2 or g g ² a: _ ² a :. As in Ža., we can find a homomorphism ⑀ above such that, in Aˆ s A ⑀ , 5 ˆ g5 s 5 g5 G 2 or ˆ g g ²a ˆ: _ ² aˆ:. Then, by Theorem 6.2, ˆg ¤Aˆ ˆk i for any i. Let B s ² a , b :. If g g B, then g 4 B l ² k : s ⭋. Let g s b s x r where x s a . We note that b s x r ;B b i s k i if and only if b s s b i and that r is even if s is odd, and r s 0 if s is even. Thus, if b s f ² b : then, in Bˆ s B⑀ , we have ˆ b sˆ x r ¤Bˆ ˆ b i for any i. So we can assume that b s s b j r for some j. This means x / 1. If s is even, then we choose Bˆ s B⑀ such that ˆ x r / 1. This implies ˆ b sˆ x r ¤Bˆ ˆ b i for any i. If s is odd, then r is not even. Hence we choose Bˆ s B⑀ such that ˆ xr f ²ˆ x 2 :. Then ˆ b sˆ x r ¤Bˆ ˆ bi ˆ B for any i. In this way we can find Aˆ s A ⑀ such that ˆ g4 l ²ˆ k : s ⭋. Since any nontrivial element ˆ k i has the minimal length 1 in its conjugate ˆ by Theorem 6.2, ˆg ¤Aˆ ˆk i for any i. Therefore, as in Case 1, A class in A, is cyclic conjugacy separable for ² k :.
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LEMMA 6.6. The group A s ² a, b: by1 a b s a : is double coset separable at a , b 4 if < < s < <. Proof. Case 1. A s ² a, b: by1 a b s a .. Let A s ² a:) a sx B, where B s ² x, b: by1 xb s x :. Ža. A is ² h s : u² k s :-separable for h s a , k s b . We first show that A is ² a :² b ␦ :-separable for any , ␦ . Let ¨ g A and ¨ f ² a :² b ␦ :. If ¨ f B then we choose Aˆ s A ⑀ , as in Lemma 6.5, Case 1, such that ¨ˆ f Bˆ by considering the length of ¨ in A s ² a:) a sx B. ˆ If ¨ g B, we let ⑀ s ␦ in Aˆ s A ⑀ . Then Then ¨ˆ f ² a ˆ :² ˆb ␦ : ; B. ␦: ˆ ˆ :² ¨ˆ f ² a b . Since A is free-by-finite, whence residually finite, there ˆ exists Nˆ ef Aˆ such that ¨ˆ f Nˆ² a ˆ :² ˆb ␦ :. Let N be the preimage of Nˆ in A. Then N ef G and ¨ f N² a :² b ␦ :. Thus A is ² a :² b ␦ :-separable for any , ␦ . To show that A is ² a s : u² b s :-separable for any s, let ¨ f ² a s : u² b s :, that is, uy1 ¨ f ² uy1 a s u:² b s :. Let d s Ž s, .. Since a g ZŽ A., r d.y1 y1 si ² Ž sr d. : ² uy1 a s u: s D Žis0 u a u a . Thus, for each 0 F i - d , uy1 ay si u ⭈ uy1 ¨ f ² aŽ sr d. :² b s :. By the above note, there exists N ef A such that uy1 ay si u ⭈ uy1 ¨ f N² aŽ sr d. :² b s : for all 0 F i - d . It follows that uy1 ¨ f N² uy1 a s u:² b s :; hence ¨ f N² a s : u² b s :. Žb. A is ² h s : u² h s :-separable for h s a . This follows easily by using similar argument as in Ža.. Žc. A is ² k s : u² k s :-separable for k s b .
By Theorem 6.4, A is c . Therefore it is sufficient to show that A is ² k s : u² k s :-separable for u s a r 1 b s1 ⭈⭈⭈ b s ny 1 a r n where n G 1, 0 r 1 , . . . , rny1 - and rn / 0. Let ¨ f ² k s : u² k s :. Suppose 5 ¨ 5 ) 5 u 5 q 2 or ¨ s a e1 b c1 ⭈⭈⭈ b c n a e nq 1 with 5 ¨ 5 s 5 u 5 q 2 or ¨ s b c1 a e1 ⭈⭈⭈ a e ny 1 b c n with 5 ¨ 5 s 5 u 5 or 5 ¨ 5 - 5 u 5. In this case, we can choose Aˆ s A ⑀ such that 5 ¨ˆ5 s 5 ¨ 5 and 5 u ˆ5 s 5 u 5. It follows that s: ²ˆs: ˆ ² ¨ˆ f k u ˆ k . So, suppose ¨ s b c1 a e1 ⭈⭈⭈ a e n b c nq 1 . Since ² a , b : s ² a : = ² b :, we note that ¨ s b c1 a e1 ⭈⭈⭈ a e n b e nq 1 g ² k s : u² k s :, where k s b , if and only if b c1 s b s ␣ , a e1 s a r 1 a ␦ 1 , b c 2 s ay ␦ 1 b s1 a ␦ 1 , . . . , b c n s ay ␦ ny 1 b s ny 1 a ␦ ny 1 , a e n s ay ␦ ny 1 a r n , and b c nq 1 s b s  for some ␣ ,  , ␦ i . Thus b c1 g ² b s :, a e iyr i g ² a :, b c iq 1 s b s i , b c nq 1 g ² b s :, and a e1q ⭈⭈⭈ qe n s a r 1q ⭈⭈⭈ qr n . Hence ¨ f ² b s : u² b s : if and only if b c1 f ² b s :, or a e iyr i f ² a :, or b c iq 1 / b s i , or b c nq 1 f ² b s :, or a e1q ⭈⭈⭈ qe n / a r 1q ⭈⭈⭈ qr n . As usual, we can find Aˆ s A ⑀ such that 5 ¨ˆ5 s 5 ¨ 5 and 5 u ˆ5 s 5 u 5, and ˆb c1 f ² ˆb s :, or aˆe iyr i f ² aˆ:, or c s c iq 1 i nq 1 ˆb / ˆb , or ˆb f ² ˆb s :, or aˆe1q ⭈⭈⭈ qe n / aˆr1 q ⭈⭈⭈ qrn . This implies
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¨ˆ f ² ˆ b s:u ˆ² ˆb s :. Thus, as in Ža., we can find N ef A such that ¨ f
N² b s : u² b s :.
Case 2. A s ² a, b: by1 a b s ay :. Let A s ² a:) a s x B, where B s ² x, b: by 1 xb s xy 1 :. Since ² h s : u² h s :-separability and ² h s : u² k s :-separability are very similar to Case 1, we shall only show that A is ² k s : u² k s :-separable for k s b . Let ¨ g A such that ¨ f ² k s : u² k s : s ² b s : u² b s :. As in Žc. of Case 1, we need only consider u s a r 1 b s1 ⭈⭈⭈ b s ny 1 a r n and ¨ s b c1 a e1 ⭈⭈⭈ a e n b c nq 1 where n G 1. Since ² a : l ² b : s 1, we note that ¨ g ² b s : u² k s : iff b c1 s b s ␣ , a e1 s a r 1 a ␦ 1 , b c 2 s ay ␦ 1 b s1 a 1 , a e 2 s ay 1 a r 2 a ␦ 2 , . . . , b c n s ay ␦ ny 1 b s ny 1 a ny 1 , a e n s ay ny 1 a r n , and b c nq 1 s b s  for some ␣ ,  , ␦ i , i , where b c iq 1 s b s i . If si is even, then ␦ i s i , and if si is odd, then ␦ i s y i . Thus ¨ f ² b s : u² b s : if and only if b c1 f ² b s :, or a e iyr i f ² a :, or b c iq 1 / b s i , or b c nq 1 f ² b s :, or b c1 a e1 ⭈⭈⭈ b c n s b s ␣ a r 1 b s1 ⭈⭈⭈ b s ny 1 a ny 1 and a e n s a r n a n but a ny 1 / ay n . Then, as usual, we can find Aˆ s A ⑀ such that 5 ¨ˆ5 s 5 ¨ 5 and 5 u ˆ5 s 5 u 5, and ˆb c1 f ² ˆb s :, or aˆe iyr i f ² aˆ:, or ˆb c iq 1 / ˆb s i , or ˆb c nq 1 f ² ˆb s :, or aˆ ny 1 / a ˆy n if b c1 a e1 ⭈⭈⭈ b c n s b s ␣ a r1 b s1 ⭈⭈⭈ b s ny 1 a ny 1 and a e n s a r n a n . This implies ¨ˆ f ² ˆ b s:u ˆ² ˆb s :. As in Case 1, we can find N ef A such that s s ¨ f N² b : u² b :. THEOREM 6.7. The group G s ² t, a, b: ty1 a t s b , by1 a b s a : is conjugacy separable if and only if < < s < <. Proof. If < < s < <, then A s ² a, b: by1 a b s a : is conjugacy separable by w5x. Applying Lemmas 6.3᎐6.6 and Theorem 4.5, G is conjugacy separable. Conversely, if G is conjugacy separable then, G is residually finite. Hence, by Theorem 6.1, < < s < <. Brunner w1x has studied the epimorphisms of the class of groups G Ž l, m; k . s ² a, t : ty1 ayk ta l ty1 a k t s a m : , where l, m, k are integers and < l < ) m and obtained various results concerning the Hopficity and the automorphism groups of certain groups in GŽ l, m; k .. Clearly GŽ l, m; k . is a special case of G above when s 1. Thus we have the following: THEOREM 6.8. The following statements on GŽ l, m; k . are equi¨ alent: Ž1. Ž2. Ž3.
GŽ l, m; k . is conjugacy separable. < l < s < m <. GŽ l, m; k . is cyclic subgroup separable.
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ACKNOWLEDGMENTS The authors thank the referee for several helpful suggestions in the exposition of the paper. The second author also thanks the Department of Mathematics of the California State University at Hayward for the hospitality and the use of facilities while he was visiting there.
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