CONJUGACY SEPARABILITY OF CERTAIN HNN EXTENSIONS WITH NORMAL ASSOCIATED SUBGROUPS. Communicated by Derek J. S. Robinson. 1.

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1 International Journal of Group Theory ISSN (print): , ISSN (on-line): Vol 5 No 4 (2016), pp 1-16 c 2016 University of Isfahan wwwtheoryofgroupsir wwwuiacir CONJUGACY SEPARABILITY OF CERTAIN HNN EXTENSIONS WITH NORMAL ASSOCIATED SUBGROUPS KOK BIN WONG AND PENG CHOON WONG Communicated by Derek J S Robinson Abstract In this paper, we will give necessary and sufficient conditions for certain HNN extensions of subgroup separable groups with normal associated subgroup to be conjugacy separable In fact, we will show that these HNN extensions are conjugacy separable if and only if the normalizer of one of its associated subgroup is conjugacy separable 1 Introduction Blackburn [5] first proved that the finitely generated nilpotent groups are conjugacy separable Later, Formanek [10] and Remeslennikov [29] independently extended this result by proving that polycyclicby-finite groups are conjugacy separable Dyer in [8] proved that free-by-finite groups are conjugacy separable Fuchsian groups are proved to be conjugacy separable by Fine and Rosenberger [9] At present, the conjugacy separability of HNN extensions is difficult to determine Indeed one of the simplest type of HNN extensions, the Baumslag-Solitar group, h, t; t 1 h 2 t = h 3 is not even residually finite (see [4]) However, Kim and Tang had characterized the conjugacy separability of HNN extensions of finitely generated abelian groups with cyclic associated subgroups in [18] and they gave a criterion for the conjugacy separability in HNN extensions with cyclic associated subgroups in [20] (see also [23] for a more recent result) Raptis, Talelli and Varsos [28] proved the equivalence of residual finiteness and conjugacy separability in certain HNN extensions of finitely generated abelian groups Recently, Wong and Wong [34] extended the result of Raptis, Talelli and Varsos by proving the equivalence of residual finiteness and conjugacy separability in certain HNN extensions of polycyclic-by-finite groups MSC(2010): Primary: 20E06, 20E26; Secondary: 20F10 Keywords: Residual properties, Conjugacy separability, HNN extensions Received: 11 March 2015, Accepted: 10 April 2015 Corresponding author 1

2 2 Int J Group Theory, 5 no 4 (2016) 1-16 K B Wong and P C Wong Grossman [11] showed that if all the class-preserving automorphisms of a finitely generated conjugacy separable group G are inner, then the outer automorphism group of G is residually finite This criterion has been used by many authors to show that certain outer automorphism groups are residually finite (see [2, 3, 7, 16, 22, 34, 38]) Recently, Zhou and Kim [40, 41] have studied the class-preserving automorphisms of certain groups Cyclic subgroup separability is a property that is used to show that certain generalized free products are conjugacy separable (see [17, 19, 31, 32]) Cyclic subgroup separability was introduced by Stebe [30] Kim [14, 15] gave useful criteria for certain generalized free products and HNN extensions to be cyclic subgroup separable By using Kim s criterion for HNN extensions, Wong and Wong [37] have given a characterization for certain HNN extensions with central associated subgroups to be cyclic subgroup separable (see also [1, 13, 21, 33, 35, 36] for some recent results on cyclic subgroup separability) Subgroup separability is a property stronger than cyclic subgroup separability It is well known that polycyclic groups and free groups are subgroup separable (Hall [12], Mal cev [25]) Since a finite extension of a subgroup separable group is again subgroup separable, polycyclic-by-finite groups and free-by-finite groups are subgroup separable Metaftsis and Raptis [27] gave a sufficient and necessary condition for certain HNN extensions to be subgroup separable By applying their result, Wong and Wong [39] showed that subgroup separability and conjugacy separability are equivalent in certain HNN extensions In this paper, we will give sufficient and necessary conditions for certain HNN extensions of subgroup separable groups with normal associated subgroup to be conjugacy separable In fact, we will show that these HNN extensions are conjugacy separable if and only if the normalizer of an associated subgroup is conjugacy separable This will be done in Section 3 (Theorem 37, Theorem 38 and Corollary 39) 2 Preliminaries The notation used here is standard In addition, the following will be used for any group G: (i) N f G means N is a normal subgroup of finite index in G; (ii) x G y means x and y are conjugate in G; (iii) {y} G = {g 1 yg ; g G} denotes the conjugacy class of y in G; (iv) Let X be a subset of a group G The subgroup generated by X is denoted by gp{x}; (v) G = t, A; t 1 Ht = K, φ shall denote an HNN extension where A is the base group, H, K are the associated subgroups and φ is the associated isomorphism φ : H K; (vi) g denotes the length of the element g in G = t, A; t 1 Ht = K, φ ; (vi) [X, Y ] denotes the set {[x, y] = x 1 y 1 xy : x X, y Y } Definition 21 A group G is said to be conjugacy separable if for each pair of elements x, y G such that x and y are not conjugate in G, there exists N f G, such that Nx and Ny are not conjugate in G/N Equivalently, G is conjugacy separable if for each pair x, y G and x / {y} G, there exists N f G such that x / {y} G N

3 Int J Group Theory, 5 no 4 (2016) 1-16 K B Wong and P C Wong 3 Definition 22 A group G is called H-separable for the subgroup H if for each x G\H, there exists N f G such that x / HN G is termed subgroup separable if G is H-separable for every finitely generated subgroup H G is termed residually finite if G is 1-separable G is (H, K)-double coset separable for the subgroups H, K if g 1 / Hg 2 K for some g 1, g 2 G, there exists N f G such that g 1 / Hg 2 KN Definition 23 [24, p 181] Let G = t, A; t 1 Ht = K, φ be an HNN extension A word w = w 0 t ϵ 1 w 1 t ϵ 2 w 2 t ϵ n w n, (n 0), where w i A and ϵ i = ±1 for all i, is said to be reduced if there is no j for which (a) ϵ j+1 = ϵ j = 1 and w j H, or (b) ϵ j+1 = ϵ j = 1 and w j K Definition 24 [24, p 185] Let G = t, A; t 1 Ht = K, φ be an HNN extension A word w = w 0 t ϵ 1 w 1 t ϵ 2 w 2 t ϵn, (n 0), where w i A and ϵ i = ±1 for all i, is said to be cyclically reduced if all cyclic permutations of w is reduced It is not hard to see that every element in G is conjugate to a cyclically reduced element following theorem will be used throughout this note The Theorem 25 [6] Let G = t, A; t 1 Ht = K, φ be an HNN extension Suppose x, y G and x, y are cyclically reduced If x G y, then x = y and one of the following holds; (a) x = y = 0 and there exists a finite sequence z 1, z 2,, z n of elements in H K such that x A z 1 A,t z 2 A,t A,t z n A y where u A,t v means one of: (u A v) or (u H and v = t 1 ut K) or (u K and v = tut 1 H) (b) x = y = n 1 and x = c 1 y c where y is a cyclic permutation of y and c X with X = H if ϵ n = 1 and X = K if ϵ n = 1, where ϵ n is the power of t in the last term of x expressed in cyclically reduced form, x = x 0 t ϵ1 x n 1 t ϵ n 3 The sufficient and necessary conditions Let H be a finitely generated group and S f H If S is a characteristic subgroup of H, then we set f H (S) = S Suppose S is not a characteristic subgroup of H Let [H : S] = m where m is a positive integer Since H is finitely generated, the number of subgroups of index m in H is finite Let N be the intersection of all these subgroups Then N is a characteristic subgroup of finite index in H and N S We set f H (S) = N (see [33, Lemma 31]) The following lemma is a special case of [36, Lemma 31] Lemma 31 Let A be a subgroup separable group, H, K be finitely generated normal subgroups of A, and H K = 1 If S 1 f H, S 2 f K, then there exists N f A such that N H = f H (S 1 ), N K = f K (S 2 ) and NH NK = N

4 4 Int J Group Theory, 5 no 4 (2016) 1-16 K B Wong and P C Wong Lemma 32 Let G = t, A; t 1 Ht = K, φ be an HNN extension where A is subgroup separable, H, K are finitely generated normal subgroups of A and H K = 1 Then for any N H f H there exists N f A such that φ(n H) = N K, N H N H and NH NK = N Proof Let N H f H be given Note that S 2 = φ(f H (N H )) is a characteristic subgroup of K, for φ is an isomorphism By Lemma 31, there exists N f A such that N H = f H (N H ), N K = f K (S 2 ) = S 2 and NH NK = N Note that N satisfies the conclusions of the lemma Lemma 33 Let G = t, A; t 1 Ht = K, φ be an HNN extension where H, K are non-trivial normal subgroups of A and H K = 1 Let G H be the normalizer of H in G Let g be any element in G Then the following hold: (a) If g 1 hg H for some 1 h H, then g A or g = a 0 t ϵ 1 a 1 t ϵ n a n in reduced form with n = 2m 2 and ϵ i = ( 1) i+1, (b) g G H if and only if g A or g = a 0 t ϵ 1 a 1 t ϵ n a n in reduced form with n = 2m 2 and ϵ i = ( 1) i+1, (c) G H = gp{a tat 1 } Proof (a) Suppose g 1 hg H for some 1 h H If g A, we are done Suppose g = 1 and g = a 0 t ϵ 1 a 1 in reduced form Then g 1 hg = a 1 1 t ϵ 1 a 1 0 ha 0t ϵ 1 a 1 = a 1 1 t ϵ 1 h 0 t ϵ 1 a 1 where h 0 = a 1 0 ha 0 H Since g 1 hg H, we have ϵ 1 = 1, for otherwise, g 1 hg has length 2 Thus a 1 1 t 1 h 0 ta 1 = a 1 1 k 0a 1 K where k 0 = t 1 h 0 t K But then g 1 hg H K = 1 and hence h = 1, a contradiction Thus g 2 Suppose g = 2 and g = a 0 t ϵ 1 a 1 t ϵ 2 a 2 in reduced form Then we see that g 1 hg = a 1 2 t ϵ 2 a 1 1 t ϵ 1 a 1 0 ha 0t ϵ 1 a 1 t ϵ 2 a 2 = a 1 2 t ϵ 2 a 1 1 t ϵ 1 h 0 t ϵ 1 a 1 t ϵ 2 a 2, where h 0 = a 1 0 ha 0 H As above, we have ϵ 1 = 1 Thus g 1 hg = a 1 2 t ϵ 2 a 1 1 k 0a 1 t ϵ 2 a 2 where k 0 = t 1 h 0 t K Since K is normal, we have a 1 2 t ϵ 2 k 1 t ϵ 2 a 2 H where k 1 = a 1 1 k 0a 1 K This implies that ϵ 2 = 1, for otherwise, g 1 hg has length 2 Thus g = a 0 (ta 1 t 1 )a 2 Let g = n 3 and assume that if g G, 1 g = k n 1 and g 1 hg H for some 1 h H, then g = a 0 t ϵ 1 a 1 t ϵ ka k in reduced form with k = 2m 2 and ϵ i = ( 1) i+1 Let g = a 0 t ϵ 1 a 1 t ϵ 2 a 2 t ϵ 3 a 3 t ϵ n a n in reduced form Arguing as above, we have ϵ 1 = 1 and ϵ 2 = 1 Therefore g = a 0 ta 1 t 1 a 2 t ϵ 3 a 3 t ϵ n a n Let g 1 = a 0 ta 1 t 1 a 2 and g 2 = t ϵ 3 a 3 t ϵ n a n Then g = g 1 g 2 Furthermore g 1 2 h 2g 2 H where 1 h 2 = g 1 1 hg 1 H Since 1 g 2 = n 2 n 1, by the induction hypothesis n 2 = 2m 2 and ϵ i = ( 1) i+1 for 3 i n Thus g = a 0 t ϵ 1 a 1 t ϵ n a n in reduced form with n = 2m and ϵ i = ( 1) i+1 (b) Clearly, if g A, then g G H Let g = a 0 t ϵ 1 a 1 t ϵ n a n in reduced form with n = 2m 2 and ϵ i = ( 1) i+1 Then g = a 0 ta 1 t 1 a 2 a 2m 2 ta 2m 1 t 1 a 2m can be written as g = a 0 g 1 a 2 a 2m 2 g 2m 1 a 2m where g j = ta j t 1, j = 1, 3,, 2m 1 Since tat 1, a G H for a A, we conclude that g G H If g G H then g 1 Hg = H Since H is non-trivial, there exists 1 h H such that g 1 hg H By part (a) of this lemma, the result follows

5 Int J Group Theory, 5 no 4 (2016) 1-16 K B Wong and P C Wong 5 (c) Clearly, gp{a tat 1 } G H Let g G H If g A, then g gp{a tat 1 } Suppose g 1 Then by part (b) of this lemma, g = a 0 t ϵ 1 a 1,, t ϵ n a n in reduced form with n = 2m 2 and ϵ i = ( 1) i+1 We can write g = a 0 g 1 a 2 a 2m 2 g 2m 1 a 2m where g j = ta j t 1, j = 1, 3,, 2m 1 This means that g gp{a tat 1 } The proof of this lemma is completed Lemma 34 Let G = t, A; t 1 Ht = K, φ be an HNN extension where H, K are non-trivial normal subgroups of A and H K = 1 Let G H and G K be the normalizers of H and K, respectively in G Then G K = t 1 G H t Proof Let g G H Then t 1 g 1 tkt 1 gt = t 1 g 1 Hgt = t 1 Ht = K Thus t 1 G H t G K Let g G K Then tg 1 t 1 Htgt 1 = tg 1 Kgt 1 = tkt 1 = H Hence tg K t 1 G H and G K = t 1 G H t Lemma 35 Let G = t, A; t 1 Ht = K, φ be an HNN extension where H, K are non-trivial normal subgroups of A and H K = 1 Let G H be the normalizer of H in G Then G H = A H = tkt 1 tat 1 Proof Let ψ 1 : A G H be the homomorphism defined by ψ 1 (a) = a for all a A, and let ψ 2 : tat 1 G H be the homomorphism defined by ψ 2 (tat 1 ) = tat 1 Then ψ 1 (h) = h = tφ(h)t 1 = ψ 2 (tφ(h)t 1 ) in G H Therefore ψ 1 and ψ 2 can be extended to become a homomorphism ψ : A H = tkt 1 tat 1 G H Furthermore, ψ is an epimorphism, for G H = gp{a tat 1 } by part (c) of Lemma 33 Let x A H = tkt 1 tat 1 Then x = a 1 (tb 1 t 1 )a 2 (tb 2 t 1 ) a n (tb n t 1 ) where a i / H, b i / K So ψ(x) = x Since G H is a subgroup of G = t, A; t 1 Ht = K, φ, ψ(x) G Note that ψ(x) is reduced, since it contains no consecutive subsequence t 1 ht with h H and subsequence tkt 1 with k K Hence ψ(x) 1 and ψ is an isomorphism Lemma 36 Let G = t, A; t 1 Ht = K, φ be an HNN extension where A is subgroup separable, H, K are finitely generated normal subgroups of A and H K = 1 Then G is (X, Y )-double coset separable where X, Y {H, K} Proof Let p, q G and p / XqY We may assume that p, q are in reduced forms It is sufficient to find an image G of G such that G is residually finite, XqY is finite, and p / XqY If such image can be found, then there exists N f G such that p / XqY N Let N be the preimage of N Then p / XqY N Case 1 p = q Let p = p 0 t ϵ 1 p 1 t ϵ n p n and q = q 0 t ϵ 1 q1 t ϵ m qm be reduced where n m and n, m 1 Since A is H, K-separable, there exists M A f A such that p i, q j / HM A for p i, q j / H, and p i, q j / KM A for p i, q j / K Let N H = M A H φ 1 (M A K) By Lemma 32, there exists N A f A such that N A H N H, N A H N A K = N A and φ(n A H) = N A K Let R A = N A M A Then R A f A We claim that R A H R A K = R A and φ(r A H) = R A K First we show that R A H R A K = R A Clearly, R A R A H R A K Let y R A H R A K Then y = r 1 h = r 2 k where r 1, r 2 R A and h H and k K This implies that h = r 1 1 r 2k R A K H N A K H N A, whence h N A H N H M A and h N A M A = R A Hence y R A and R A H R A K = R A Next we show that φ(r A H) = R A K Note that R A H = N A M A H = N A H M A = N A H, for N A H N H M A Also R A K = N A M A K = N A K M A = N A K, for N A K φ(n H ) M A Hence φ(r A H) = R A K

6 6 Int J Group Theory, 5 no 4 (2016) 1-16 K B Wong and P C Wong Let G = t, A; t 1 Ht = K, φ where A = A/R A, H = HR A /R A, K = KR A /R A and φ is the homomorphism induced by φ Clearly, G is a homomorphic image of G By the choice of M A, p, q are reduced, and p = p, q = q Note that xqy = q for all x X, y Y So p / XqY, for p has length n while the reduced elements in XqY have length m, and m n By [8, Theorem 13], G is conjugacy separable, and thus residually finite Since XqY is finite and p / XqY, we are done Case 2 p = q = n 0 Suppose n = 0 Then p, q A Note that q 1 p / q 1 XqY = XY, for X, Y A Since A is XY -separable, there exists M A f A such that q 1 p / XY M A It follows that p / XqY M A Let M A, N H, N A, R A and G be defined as in Case 1 Suppose p XqY Then p = xqy for some x X, y Y This implies that p XqY R A XqY M A, a contradiction Thus p / XqY As in Case 1, G is residually finite and we are done Suppose n 1 Let p = p 0 t ϵ 1 p 1 t ϵ n p n and q = q 0 t ϵ 1 q1 t ϵ n qn be in reduced forms Since A is H, K-separable, there exists M A f A such that p i, q j / HM A for p i, q j / H, and p i, q j / KM A for p i, q j / K Let M A, N H, N A, R A and G be defined as in Case 1 Then p and q are reduced and p = n = q Furthermore, H K = 1 and G is residually finite If ϵ i ϵ i for some i, then p / XqY and we are done, so assume that ϵ i = ϵ i for all i Note that p XqY if and only if the following system of equations (31) holds: p 0 = v 1 0 q 0u 1 p 1 = v 1 1 q 1u 2 (31) p n 1 = v 1 n 1 q n 1u n p n = v 1 n q n u n+1 where v 0 X, u n+1 Y, u i H, v i K if ϵ i = 1, or u i K, v i H if ϵ i = 1, and t ϵ i u i t ϵ i = v i for i = 1,, n For ease of exposition, let U i = H if ϵ i = 1 or U i = K if ϵ i = 1 and t ϵ i U i t ϵ i = V i for i = 1,, n Furthermore, let V 0 = X and U n+1 = Y By using this notation, u i U i and v i V i for all i SubCase 21 Suppose p i / V i q i U i+1 for some 0 i n Then qi 1 p i / qi 1 V i q i U i+1 = V i U i+1 Since V i U i+1 = H or K or HK, the subgroup M A above can be chosen with the additional property that q 1 i p i / V i U i+1 M A Thus p i / V i q i U i+1 M A Note that p i / V i q i U i+1 for this choice of M A Hence p / XqY and we are done SubCase 22 So, we may assume that p i V i q i U i+1 for all 0 i n We claim that V i U i+1 for some 0 i n Suppose V i = U i+1 for all i Let p 0 = v 1 0 q 0u 1 where v 0 V 0, u 1 U 1 We shall show that by matching forward, we will make equations (31) hold What

7 Int J Group Theory, 5 no 4 (2016) 1-16 K B Wong and P C Wong 7 we meant by matching forward is that given p i = vi 1 q i u i+1 for some 0 i < n, we are able to find v i+1 V i+1 and u i+2 U i+2 such that p i+1 = vi+1 1 q i+1u i+2 and t ϵ i+1 u i+1 t ϵ i+1 = v i+1 Let p 1 = v 1 1 q 1 u 2 where v 1 V 1, u 2 U 2 Suppose t ϵ 1 u 1 t ϵ 1 = v 1 Note that p 1 = v1 1 v 1v 1 1 q 1 u 2 = v1 1 q 1(q1 1 (v 1v 1 1 q 1 u 2 )) Since V 1 = U 2 = H or K, we have V 1 = U 2 A, and thus (q1 1 v 1v 1 1 q 1 u 2 ) U 2 Let u 2 = (q1 1 v 1v 1 1 q 1 u 2 ) Then we have p 0 = v0 1 q 0u 1 and p 1 = v1 1 q 1u 2 with t ϵ 1 u 1 t ϵ 1 = v 1 By continuing to match forward in this way, we have p i = vi 1 q i u i+1 and t ϵ i u i t ϵ i = v i for 0 i n So the equations (31) hold, a contradiction Hence V i U i+1 for some 0 i n Let r be the least non-negative integer such that V r U r+1 We claim that r n Suppose r = n Then V i = U i+1 for all 0 i n 1 Let p n = vn 1 q n u n+1 where v n V n, u n+1 U n+1 We shall show that by matching backward, we will make equations (31) hold What we meant by matching backward is that given p i = vi 1 q i u i+1 for some 0 < i n, we are able to find v i 1 V i 1 and u i U i such that p i 1 = vi 1 1 q i 1u i and t ϵ i u i t ϵ i = v i Let p n 1 = v 1 n 1q n 1 u n where v n 1 V n 1, u n U n Let t ϵ n v n t ϵ n = u n Note that p n 1 = v 1 n 1q n 1 u nu 1 n u n = (v 1 n 1q n 1 u nu 1 n qn 1 )q n 1u n Since V n 1 = U n = H or K, we have V n 1 = U n A, and thus (v 1 n 1q n 1 u nu 1 n qn 1 ) V n 1 Let vn 1 1 = (v 1 n 1q n 1 u nu 1 n qn 1 ) Then p n = vn 1 q n u n+1 and p n 1 = vn 1 1 q n 1u n with t ϵ n u n t ϵ n = v n By continuing to match backward in this way, we have p i = vi 1 q i u i+1 and t ϵ i u i t ϵ i = v i for 0 i n So equation (31) holds, a contradiction Hence r n By the choice of r, we have V i = U i+1 for 0 i r 1 Note that r could be 0 Let p r = v 1 r q r u r+1 where v r V r and u r+1 U r+1 Now by matching backward, we have p 0 = v 1 0 q 0u 1 (32) p r 1 = v 1 r 1 q r 1u r p r = v 1 r q r u r+1 where v i V i, u i+1 U i+1 for i = 0,, r and t ϵ i u i t ϵ i = v i for i = 1,, r Since equation (31) does not hold, there exists a s, where r + 1 s n such that p r = v 1 r q r u r+1 (33) p s 1 = v 1 s 1 q s 1u s p s = v 1 s q s u s+1 where v i V i, u i+1 U i+1 for i = r,, s and t ϵ i u i t ϵ i = v i for i = r +1,, s 1 but t ϵ s u s t ϵs v s We may also assume that s is the largest integer such that equation (33) holds

8 8 Int J Group Theory, 5 no 4 (2016) 1-16 K B Wong and P C Wong If V s = U s+1, then from p s 1 = v 1 s 1 q s 1u s, and by matching forward we would have p s = v 1 s q s u s+1 with t ϵ s u s t ϵ s = v s, but this contradicts the choice of s as the largest integer such that equation (33) holds Hence V s U s+1 Now we have V r U r+1 and V s U s+1 and (3) holds The M A above can be chosen with the additional property that v 1 s t ϵ s u s t ϵ s / M A Suppose p XqY Then the following system of equations (34) holds: (34) p 0 = z 1 0 q 0w 1 p 1 = z 1 1 q 1w 2 p n 1 = z 1 n 1 q n 1w n p n = z 1 n q n w n+1 where z i V i, w i+1 U i+1 for i = 0,, n and z i = t ϵ i w i t ϵ i for i = 1,, n From (33), we have the following system of equations: (35) p r = v 1 r q r u r+1 p s 1 = v 1 s 1 q s 1u s p s = v 1 s q s u s+1 From equations (34) and (35), we have p r = v 1 r q r u r+1 and p r = z 1 r q r w r+1 Thus z r v 1 r = q r w r+1 u 1 r+1 q 1 r V r U r+1 = H K = 1 This implies that z r = v r, w r+1 = u r+1 and z r+1 = t ϵ r+1 w r+1 t ϵ r+1 = t ϵ r+1 u r+1 t ϵ r+1 = v r+1 Again from equations (34) and (35), we have p r+1 = v 1 r+1 q r+1u r+2, p r+1 = z 1 r+1 q r+1w r+2 Since z r+1 = v r+1, we have u r+2 = w r+2 By continuing this way, we see that u j = w j for all r + 1 j s From p s = v 1 s q s u s+1 and p s = z 1 s q s w s+1, we have z s v 1 s = q s w s+1 u 1 s+1q 1 s V s U s+1 = H K = 1 This implies that z s = v s, but then t ϵ s u s t ϵ s = t ϵ s w s t ϵ s = z s = v s, which implies that v 1 s t ϵs u s t ϵs M A, contradicting the choice of M A Hence p / XqY and we are done The proof of the lemma is now complete Theorem 37 Let G = t, A; t 1 Ht = K, φ be an HNN extension where A is subgroup separable, H, K are non-trivial finitely generated normal subgroups of A and H K = 1 Let G H be the normalizer of H in G If G H is conjugacy separable then G is conjugacy separable

9 Int J Group Theory, 5 no 4 (2016) 1-16 K B Wong and P C Wong 9 Proof Let x, y G and x / {y} G We may assume that x and y are of minimal length in {x} G and {y} G, respectively It is sufficient to find an image G of G such that G is conjugacy separable and x / {y} G If such image can be found, then there exists N f G such that x / {y} G N Let N be the preimage of N Then x / {y} G N Case 1 Suppose x = y Since x, y are elements of minimal length in {x} G, {y} G respectively, x, y are cyclically reduced Let x = x 0 t ϵ 1 x 1 t ϵ2 x n 1 t ϵ n and y = y 0 t ϵ 1 y1 t ϵ 2 ym 1 t ϵ m be reduced where x i, y j A, ϵ i = ±1, ϵ j = ±1 for all i, j, and n m Then there exists M A f A such that x i, y j / HM A for x i, y j / H and x i, y j / KM A for x i, y j / K Let N H = M A H φ 1 (M A K) By Lemma 32, there exists N A f A such that N A H N H, N A H N A K = N A and φ(n A H) = N A K Let R A = N A M A Then R A H R A K = R A and φ(r A H) = R A K Let G = t, A; t 1 Ht = K, φ where A = A/R A, H = HR A /R A, K = KR A /R A and φ is the homomorphism induced by φ Note that H K = 1, and both H and K are non-trivial, for x 1 Note that x, y are cyclically reduced and x = n, y = m Furthermore, x / {y} G because x has length n while the cyclically reduced elements in {y} G have length m, and m n By [8, Theorem 13], G is conjugacy separable, we are done Case 2 Suppose x = y = 0 Then x, y A Subcase 21 Suppose x, y H and x 1 Then x / {y} G H Since G H is conjugacy separable, there exists M f G H such that x / {y} G H M and x / M Let M A = M A Let N H, N A, R A and G be as in Case 1 Again, H K = 1, and both H and K are non-trivial, for x 1 Claim 1 Let Ψ be the epimorphism of G onto G We claim that G H KerΨ M Proof of Claim 1 Now, by Lemma 35, G H = A H = tkt 1 tat 1 is a generalised free product Let GH denote the free product length of an element in G H Let w G H KerΨ and w A or w tat 1 Then Ψ(w) = 1 Since Ψ(w) = w, we have w = 1, and this implies that w R A or w = tw t 1 and w R A In either case, w M Next, we assume that if g G H KerΨ and g GH n 1, then g M Let w G H KerΨ and w GH = n 2 If n = 2r + 1, then w = a 1 (tb 1 t 1 ) a r (tb r t 1 )a r+1 or w = (tb 1 t 1 )a 2 (tb 2 t 1 ) a r+1 (tb r+1 t 1 ) If n = 2r, then w = a 1 (tb 1 t 1 ) a r (tb r t 1 ) or w = (tb 1 t 1 )a 2 (tb 2 t 1 ) (tb r t 1 )a r+1 We shall only consider the case w = a 1 (tb 1 t 1 ) a r (tb r t 1 ) (other cases are similar) Since w = 1 in G, either a i H or b i K for some i Suppose a i H Then a i = hm for some h H, m R A M Therefore w = a 1 (tb 1 t 1 ) (tb i 1 t 1 )hm(tb i t 1 ) a n (tb n t 1 ) Let g 1 = a 1 (tb 1 t 1 ) (tb i 1 t 1 ) and g 2 = (tb i t 1 ) a n (tb n t 1 ) Then g 1, g 2 G H Let w 1 = g 1 hg 2 and w 2 = g 1 2 mg 2 Then w = w 1 w 2 where w 1 G H and w 2 M, for M G H Furthermore, w 1 GH n 1 (in G H ), because the term t 1 ht is an element in K and it appears in w 1 Therefore

10 10 Int J Group Theory, 5 no 4 (2016) 1-16 K B Wong and P C Wong w 1 is of the form a 1 (tb 1 t 1 ) a i 2 (tb i 2 t 1 )a i 1 (tb i 1 kb i t 1 )a i+1 (tb i+1 t 1 ) a n (tb n t 1 ), where k = t 1 ht K Since w 2 KerΨ (m M) and w KerΨ, we have w 1 KerΨ, and thus by assumption, w 1 M because w 1 GH n 1 Now w 1, w 2 M implies that w M The proof of the case b i K is similar This establishes Claim 1 We can now continue with the proof of Subcase 21 Suppose x {y} G Then x = g 1 yg If y = 1, then x = 1 and x R A M, a contradiction Hence y 1 Since x, y H, by Lemma 33, g G H where G H is the normalizer of H in G Again, by Lemma 33, G H = gp{a, tat 1 } = G H Therefore g G H, and by Claim 1, x {y} G H KerΨ {y} G H M, a contradiction Hence x / {y} G By [8, Theorem 13], G is conjugacy separable, we are done Subcase 22 Suppose x K, y H Let x = txt 1 Then x H As in Subcase 21, one can find an image G of G such that G is conjugacy separable and x / {y} G Clearly, x / {y} G, we are done Subcase 23 Suppose x A (H K), y H Since A is H, K-separable, there exists M A f A such that x / HM A KM A Let N H, N A, R A and G be as in Case 1 Again, H K = 1, and both H and K are non-trivial, for x 1 Suppose x {y} G Let x = g 1 yg where g G If g A, we would have x H, a contradiction Let g = a 0 t ϵ 1 a 1 t ϵn a n be in reduced form where n 1 Then gxg 1 = y This implies that a n xa 1 n H K, for otherwise, gxg 1 has length 2n 2 So x H K, a contradiction Hence x / {y} G, and we are done Subcase 24 Suppose x A, y K Let y = tyt 1 Then y H and this is similar to Subcase 21, 22 or 23 Subcase 25 Suppose x A, y A (H K) If x H, then it is similar to Subcase 23 If x K, we let x = txt 1, then it is similar to Subcase 23 So we just need to consider the case x, y A (H K) Since A is H, K-separable, there exists M A f A such that x, y / HM A KM A Since x / {y}g H, there exists M f G H such that x / {y} G H M Note that x / {y} A (M A) for A G H Let M A = M A M Let N H, N A, R A and G be as in Case 1 Again, H K = 1, and both H and K are non-trivial, for x 1 Suppose x {y} G Let x = g 1 yg where g G If g A, then x {y} A and x {y} A R A {y} A (M A), a contradiction Let g = a 0 t ϵ 1 a 1 t ϵn a n be in reduced form where n 1 Then gxg 1 = y This implies that a n xa 1 n H K, for otherwise, gxg 1 has length 2n 2 So x H K, a contradiction Hence x / {y} G, and we are done Case 3 Suppose x = y = n 1 Let x = x 0 t ϵ 1 x 1 t ϵ2 x n 1 t ϵ n and y = y 0 t ϵ 1 y1 t ϵ 2 yn 1 t ϵ n Note that by Theorem 25, x {y} G if and only if x {y (j) } X for some 0 j n 1 where y (j) = y j t ϵ j+1 y j+1 t ϵ n+j 1 y n+j 1 t ϵ n+j and X = K if ϵ n = 1 Let j be fixed and the subscripts are taken modulo n and X = H if ϵ n = 1 Claim 2 We claim that if x / {y (j) } X, then there exists N j f G such that x / {y (j) } X N j

11 Int J Group Theory, 5 no 4 (2016) 1-16 K B Wong and P C Wong 11 Proof of Claim 2 For simplicity, we shall only show the case j = 0, that is y (j) = y (0) = y Suppose x / XyX By Lemma 36, there exists N f G such that x / XyXN Hence x / {y} X N, and we are done So we may assume that x XyX This implies that ϵ i = ϵ i for all 1 i n Furthermore, the following system of equations holds: x 0 = v 1 n y 0 u 1 x 1 = v 1 1 y 1u 2 (36) x n 2 = v 1 n 2 y n 2u n 1 x n 1 = v 1 n 1 y n 1u n where u i H, v i K if ϵ i = 1 or u i K, v i H if ϵ i = 1, and v i = t ϵ i u i t ϵ i for i = 1,, n 1 Note that v n, t ϵ n u n t ϵ n X and v n t ϵ n u n t ϵ n for x / {y} X For ease of exposition, let U i = H if ϵ i = 1 or U i = K if ϵ i = 1 and V i = t ϵ i U i t ϵ i for i = 1,, n By using this notation, u i U i, v i V i and X = V n So we can write x / {y} V n Subcase C1 Suppose ϵ 1 = ϵ n Then V n U 1 and V n U 1 = H K = 1 There exists M A f A such that x i, y j / HM A for x i, y j / H and x i, y j / KM A for x i, y j / K Furthermore, we may choose M A so that t ϵn u n t ϵn vn 1 / M A Let N H = M A H φ 1 (M A K) Let N A, R A and G be as in Case 1 Again, H K = 1, and both H and K are non-trivial, for x 1 Furthermore, x, y are cyclically reduced and x = n = y Suppose x {y} X Since x = x 0 t ϵ 1 x 1 t ϵ n 1 x n 1 t ϵn system of equations holds: and y = y 0 t ϵ 1 y 1 t ϵ n 1 y n 1 t ϵn, the following x 0 = z 1 n y 0 w 1 x 1 = z 1 1 y 1w 2 (37) x n 2 = z 1 n 2 y n 2w n 1 x n 1 = z 1 n 1 y n 1w n where w i U i, z i V i, and z i = t ϵ i w i t ϵ i for i = 1,, n From (36) we have the following system of equations:

12 12 Int J Group Theory, 5 no 4 (2016) 1-16 K B Wong and P C Wong x 0 = v 1 n y 0 u 1 x 1 = v 1 1 y 1u 2 (38) x n 2 = v 1 n 2 y n 2u n 1 x n 1 = v 1 n 1 y n 1u n From (37) and (38), we have z 1 n y 0 w 1 = v 1 n y 0 u 1, ie, v n z 1 n = y 0 u 1 w 1 1 y 1 0 H K = 1 Therefore v n = z n and u 1 = w 1 This implies that v 1 = t ϵ 1 u 1 t ϵ 1 = t ϵ 1 w 1 t ϵ 1 = z 1 Continuing this way, we see that u i = w i and v i = z i for i = 1,, n So v n = z n = t ϵ n w n t ϵ n t ϵ n u n t ϵ n v 1 n R A M A, a contradiction Hence x / {y}x = t ϵ n u n t ϵ n, ie, Subcase C2 Suppose ϵ i = ϵ i+1 for some i Then V i U i+1 and V i U i+1 = H K = 1 This is similar to Subcase 31 Subcase C3 Suppose ϵ 1 ϵ n and ϵ i ϵ i+1 for i = 1,, n 1 We shall show that n is even Suppose n is odd, say n = 2p + 1 Since ϵ i = 1 or 1, we have ϵ 1 = ϵ 3 = ϵ 5 = = ϵ 2p+1, but then ϵ 1 = ϵ 2p+1 = ϵ n, a contradiction, for ϵ 1 = ϵ n Let n = 2p Subcase C31 Suppose ϵ 1 = 1 Then ϵ i = ( 1) i+1 for 1 i 2p In particular, ϵ 2p = 1 and X = H In this case x = x 0 (tx 1 t 1 ) x 2p 2 (tx 2p 1 t 1 ) and y = y 0 (ty 1 t 1 ) y 2p 2 (ty 2p 1 t 1 ) By part (b) of Lemma 33, x, y G H Since G H is conjugacy separable, there exists M f G H such that x / {y} G H M So x / {y} H M There exists M A f A such that x i, y j / HM A for x i, y j / H and x i, y j / KM A for x i, y j / K Let N H = M A M H φ 1 (M A M K) Let N A, R A and G be as in Case 1 Again, H K = 1, and both H and K are non-trivial, for x 1 Furthermore, x, y are cyclically reduced and x = n = y As in Claim 1, let Ψ be the epimorphism of G onto G Then KerΨ G H M If x {y} H, we have x {y} H (KerΨ G H ) {y} H M, a contradiction Hence x / {y} H Subcase C32 Suppose ϵ 1 = 1 Then ϵ i = ( 1) i for 1 i 2p In particular, ϵ 2p = 1 and X = K Then x = x 0 (t 1 x 1 t) x 2p 2 (t 1 x 2p 1 t) and y = y 0 (t 1 y 1 t) y 2p 2 (t 1 y 2p 1 t) Let x = txt 1 = (tx 0 t 1 )x 1 (tx 2 t 1 ) (tx 2p 2 t 1 )x 2p 1 By part (b) of Lemma 33, x G H Then by Lemma 34, x G K Let y = tyt 1 Again by Lemma 33 and Lemma 34, y G K Note that x / {y} K if and only if x / {y } H As in Subcase C31, there is a epimorphic image of G, say G, for which x / {y} K Note that in each of these cases, there is a epimorphic image of G, say G, for which x / {y} X Since {y} X is finite and G is conjugacy separable by [8, Theorem 13] (and thus G is residually finite), there exists N f G such that x / {y} X N Let N be the preimage of N in G Then x / {y} X N The proof of the Claim 2 is complete

13 Int J Group Theory, 5 no 4 (2016) 1-16 K B Wong and P C Wong 13 We can now continue with the proof of Case 3 By Claim 2, for each j, there exists N j f G such that x / {y (j) } X N j for all j Let N = j N j There exists M A f A such that x i, y j / HM A for x i, y j / H and x i, y j / KM A for x i, y j / K Let N H = M A N H φ 1 (M A N K) Let N A, R A and G be as in Case 1 Again, H K = 1, and both H and K are non-trivial, for x 1 Furthermore, x, y are cyclically reduced and x = n = y Then x and y have length n and x / {y} G By [8, Theorem 13], G is conjugacy separable, we are done The proof of this theorem is now complete Theorem 38 Let G = t, A; t 1 Ht = K, φ be an HNN extension where A is subgroup separable, H, K are non-trivial finitely generated normal subgroups of A and H K = 1 Let G H be the normalizer of H in G Suppose A is conjugacy separable Then G H is conjugacy separable if and only if G is conjugacy separable Proof By Theorem 37, it is sufficient to show that if G is conjugacy separable, then G H is conjugacy separable Let x, y G H and x / {y} G H Since G is residually finite, G H is residually finite So, we may assume that x, y 1 Suppose x / {y} G Since G is conjugacy separable, there exists N G f G such that x / {y} G N G Let N = N G G H Then x / {y} G H N and we are done Suppose x {y} G Then the conclusions of Theorem 25 hold Case 1 Suppose x = y = n 1 Since x, y G H, by Lemma 33, we have x = a 0 (ta 1 t 1 ) a 2s 2 (ta 2s 1 t 1 ) and y = b 0 (tb 1 t 1 ) b 2s 2 (tb 2s 1 t 1 ) in reduced form in the HNN extension G By Lemma 35, G H = A H = tkt tat 1 is a generalised free product Let y 1 2k = b 2k, y 2k+1 = tb 2k+1 t 1, x 2k = a 2k and x 2k+1 = ta 2k+1 t 1 for 0 k s 1 Then x = x 0 x n 1 and y = y 0 y n 1 in reduced form in the generalised free product G H Note that by [26, Theorem 46 on p 212], x {y} G H if and only if x {y (j) } H for some j = 0, 1,, n 1 where y (j) = y j y j+1 y j+n 1 Now x {y} G implies that x = c 1 y c, where y is a cyclic permutation of y in G and c X with X = H if ϵ n = 1 and X = K if ϵ n = 1, where ϵ n is the power of t appearing in the last term of x In this case, ϵ n = 1 So X = H Furthermore y = (tb j t 1 )b j+1 (tb j+2s 2 t 1 )b j+2s 1 if j is odd or y = b j (tb j+1 t 1 ) b j+2s 2 (tb j+2s 1 t 1 ) if j is even Here the subscripts are taken modulo 2s (see Theorem 23) So y = y (j) and x {y (j) } H This implies that x {y} G H, a contradiction Hence Case 1 does not occur Case 2 Suppose x = y = 0 Then x, y H K, for H, K A Subcase 21 Suppose x, y H Let x = g 1 yg for some g G Then by Lemma 33, g G H and x {y} G H, a contradiction

14 14 Int J Group Theory, 5 no 4 (2016) 1-16 K B Wong and P C Wong Subcase 22 Suppose x, y K Since A is H-separable and x, y / H, there exists M A f A such that x, y / HM A Note that x / {y}a, for A G H Since A is conjugacy separable, there exists M A f A such that x / {y} A M A Let N H = M A M A H φ 1 (M A M A K) By Lemma 32, there exists N A f A such that N A H N H, N A H N A K = N A and φ(n A H) = N A K Let R A = N A M A M A Then R AH R A K = R A and φ(r A H) = R A K So t 1 (R A H)t = R A K and R A H = t(r A K)t 1 = tkt 1 tr A t 1 Therefore HR A /R A = H/(H RA ) = tkt 1 /(tkt 1 tr A t 1 ) = (tkt 1 )(tr A t 1 )/tr A t 1 Note that tr A t 1 f tat 1 and tkt 1 tat 1 So we may form an epimorphic image of G H, which is G H = A H = tkt tat 1, where A = A/R 1 A, tat 1 = tat 1 /tr A t 1, H = HR A /R A and tkt 1 = (tkt 1 )(tr A t 1 )/tr A t 1 Note that H K = 1 Now, in G H, we have x, y / H and x / {y} A By [26, Theorem 46 on p 212], x / {y} G H By [8, Theorem 4], G H is conjugacy separable So there exists N f G H such that x / {y} G H N Let N be the preimage of N in G H Then x / {y} G H N and we are done Subcase 23 Suppose x H and y K, or y H and x K We shall assume that x H and y K Since A is H-separable and y / H, there exists M A f A such that y / HM A Let N H = M A H φ 1 (M A K) Let N A, R A and G H be as in Subcase 12 Note that H K = 1 Now, in G H, we have x / H, y / H and H A So, by [26, Theorem 46 on p 212], x / {y} G H By [8, Theorem 4], G H is conjugacy separable So there exists N f G H such that x / {y} G H N Let N be the preimage of N in G H Then x / {y} G H N and we are done The proof of this theorem is now complete Corollary 39 Let G = t, A; t 1 Ht = K, φ be an HNN extension where A is a polycyclic-by-finite group, H, K are non-trivial normal subgroups of A, and H K = 1 Let G H be the normalizer of H in G Then G H is conjugacy separable if and only if G is conjugacy separable Proof Follows from Theorem 38 Acknowledgments We would like to thank the anonymous referees for the comments that helped us make several improvements to this paper This project is supported by the Frontier Science Research Cluster, University of Malaya (RG267-13AFR) References [1] R B J T Allenby, Polygonal products of polycyclic by finite groups, Bull Austral Math Soc, 54 (1996) [2] R B J T Allenby, G Kim and C Y Tang, Residual finiteness of outer automorphism groups of certain pinched 1-relator groups, J Algebra, 246 (2001)

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