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ISSN (print): 2251-7650, ISSN (on-line): 2251-7669 Vol. 5 No. 4 (2016), pp. 1-16.
c
⃝2016 University of Isfahan
www.ui.ac.ir
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
polycyclic-by-finite groups are conjugacy separable. Dyer in [8] proved that free-polycyclic-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−1h2t=h3⟩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.
Grossman [11] showed that if all the class-preserving automorphisms of a finitely generated conjugacy
separable groupGare inner, then the outer automorphism group ofGis 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 3.7, Theorem 3.8 and Corollary 3.9).
2. Preliminaries.
The notation used here is standard. In addition, the following will be used for any group G:
(i) N ◁f G meansN is a normal subgroup of finite index inG;
(ii) x∼G y meansx andy are conjugate in G;
(iii) {y}G={g−1yg ; g∈G} denotes the conjugacy class of y inG;
(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−1Ht=K, φ⟩ shall denote an HNN extension where A is the base group, H, K are the associated subgroups andφis the associated isomorphism φ:H−→K;
(vi) ∥g∥ denotes the length of the element gin G=⟨t, A;t−1Ht=K, φ⟩; (vi) [X, Y] denotes the set {[x, y] =x−1y−1xy : x∈X, y∈Y}.
Definition 2.1. 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 inG, there exists N ◁fG, such thatN x and N y are not conjugate
in G/N. Equivalently, G is conjugacy separable if for each pair x, y ∈ G and x /∈ {y}G, there exists
Definition 2.2. A groupGis called H-separable for the subgroupH if for eachx∈G\H, there exists
N ◁f G such thatx /∈HN.
G is termed subgroup separable if Gis H-separable for every finitely generated subgroup H.
G is termed residually finite if Gis 1-separable.
G is (H, K)-double coset separable for the subgroups H, K if g1 ∈/ Hg2K for some g1, g2 ∈ G, there
exists N ◁f Gsuch that g1∈/ Hg2KN.
Definition 2.3. [24, p. 181] Let G = ⟨t, A;t−1Ht = K, φ⟩ be an HNN extension. A word w =
w0tϵ1w1tϵ2w2· · ·tϵnwn, (n≥0), wherewi∈A andϵi=±1 for all i, is said to be reducedif there is no
j for which
(a) ϵj+1 =−ϵj = 1 and wj ∈H, or (b) ϵj+1 =−ϵj =−1 andwj ∈K.
Definition 2.4. [24, p. 185] Let G = ⟨t, A;t−1Ht = K, φ⟩ be an HNN extension. A word w =
w0tϵ1w1tϵ2w2· · ·tϵn, (n≥0), where w
i ∈A and ϵi =±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. The
following theorem will be used throughout this note.
Theorem 2.5. [6] Let G=⟨t, A;t−1Ht=K, φ⟩ be an HNN extension. Suppose x, y∈Gand 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 z1, z2, . . . , zn of elements in H∪K such that
x∼A z1 ∼A,t∗ z2 ∼A,t∗∼A,t∗ zn ∼A y where u∼A,t∗ v means one of: (u∼A v) or (u ∈H and
v=t−1ut∈K) or (u∈K and v=tut−1 ∈H).
(b) ∥x∥ = ∥y∥ = n ≥ 1 and x = c−1y∗c where y∗ is a cyclic permutation of y and c ∈ X with
X = H if ϵn = −1 and X = K if ϵn = 1, where ϵn is the power of t in the last term of x
expressed in cyclically reduced form,x=x0tϵ1· · ·x
n−1tϵn.
3. The sufficient and necessary conditions.
LetH be a finitely generated group andS ◁f H. IfS is a characteristic subgroup ofH, then we set
fH(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 fH(S) =N (see [33, Lemma 3.1]).
The following lemma is a special case of [36, Lemma 3.1]
Lemma 3.1. Let A be a subgroup separable group, H, K be finitely generated normal subgroups of A, andH∩K= 1. IfS1◁fH, S2◁fK, then there existsN ◁fAsuch thatN∩H =fH(S1),N∩K=fK(S2)
Lemma 3.2. Let G=⟨t, A;t−1Ht=K, φ⟩ be an HNN extension whereA is subgroup separable, H, K
are finitely generated normal subgroups ofAandH∩K = 1. Then for anyNH◁fH there existsN ◁fA
such that φ(N ∩H) =N ∩K, N ∩H ⊆NH and N H∩N K =N.
Proof. LetNH◁fH be given. Note thatS2=φ(fH(NH)) is a characteristic subgroup of K, forφis an
isomorphism. By Lemma 3.1, there exists N ◁fA such that N∩H=fH(NH),N ∩K =fK(S2) =S2
and N H∩N K =N. Note that N satisfies the conclusions of the lemma. □
Lemma 3.3. Let G =⟨t, A;t−1Ht= K, φ⟩ be an HNN extension where H, K are non-trivial normal subgroups ofA andH∩K = 1. Let GH be the normalizer ofH in G. Letg be any element in G. Then
the following hold:
(a) If g−1hg ∈ H for some 1 ̸= h ∈ H, then g ∈ A or g = a0tϵ1a1· · ·tϵnan in reduced form with
n= 2m≥2 and ϵi = (−1)i+1,
(b) g ∈ GH if and only if g ∈ A or g = a0tϵ1a1· · ·tϵnan in reduced form with n = 2m ≥ 2 and
ϵi = (−1)i+1,
(c) GH = gp{A∪tAt−1}.
Proof. (a) Suppose g−1hg ∈H for some 1̸=h∈H. If g∈A, we are done. Suppose ∥g∥= 1 and g=
a0tϵ1a1 in reduced form. Theng−1hg =a−1
1 t−ϵ1a−01ha0tϵ1a1 =a−11t−ϵ1h0tϵ1a1 whereh0 =a−01ha0 ∈H.
Sinceg−1hg∈H, we haveϵ1 = 1, for otherwise,g−1hg has length 2. Thusa−11t−1h0ta1=a−11k0a1 ∈K
wherek0 =t−1h0t∈K. But theng−1hg∈H∩K = 1 and henceh= 1, a contradiction. Thus∥g∥ ≥2.
Suppose∥g∥= 2 andg=a0tϵ1a1tϵ2a2 in reduced form. Then we see that g−1hg =a−21t−ϵ2a−1
1 t−ϵ1a−01ha0tϵ1a1tϵ2a2 =a2−1t−ϵ2a−11t−ϵ1h0tϵ1a1tϵ2a2,
where h0 = a−01ha0 ∈ H. As above, we have ϵ1 = 1. Thus g−1hg = a−21t−ϵ2a−1
1 k0a1tϵ2a2 where k0 = t−1h0t ∈ K. Since K is normal, we have a−21t−ϵ2k1tϵ2a2 ∈ H where k1 = a−1
1 k0a1 ∈ K. This
implies thatϵ2 =−1, for otherwise,g−1hg has length 2. Thusg=a0(ta1t−1)a2.
Let ∥g∥ = n ≥ 3 and assume that if g ∈ G, 1 ≤ ∥g∥ = k ≤ n−1 and g−1hg ∈ H for some 1 ̸= h ∈ H, then g = a0tϵ1a1· · ·tϵka
k in reduced form with k = 2m ≥ 2 and ϵi = (−1)i+1. Let
g = a0tϵ1a1tϵ2a2tϵ3a3· · ·tϵna
n in reduced form. Arguing as above, we have ϵ1 = 1 and ϵ2 = −1. Therefore g =a0ta1t−1a2tϵ3a3· · ·tϵnan. Let g1 =a0ta1t−1a2 and g2 =tϵ3a3· · ·tϵnan. Then g =g1g2.
Furthermore g2−1h2g2 ∈ H where 1 ̸= h2 = g1−1hg1 ∈ H. Since 1 ≤ ∥g2∥ = n−2 ≤ n−1, by the
induction hypothesis n−2 = 2m ≥ 2 and ϵi = (−1)i+1 for 3 ≤ i≤ n. Thus g = a0tϵ1a1· · ·tϵnan in reduced form with n= 2m+ 2≥2 and ϵi = (−1)i+1.
(b) Clearly, ifg∈A, theng∈GH. Letg=a0tϵ1a1· · ·tϵnanin reduced form withn= 2m≥2 andϵi = (−1)i+1. Then g=a0ta1t−1a2· · ·a2m−2ta2m−1t−1a2m can be written as g=a0g1a2· · ·a2m−2g2m−1a2m where gj =tajt−1,j= 1,3, . . . ,2m−1. Since tat−1, a∈GH fora∈A, we conclude that g∈GH.
(c) Clearly, gp{A∪ tAt−1} ⊆ GH. Let g ∈ GH. If g ∈ A, then g ∈ gp{A ∪tAt−1}. Suppose ∥g∥ ≥ 1. Then by part (b) of this lemma, g = a0tϵ1a1, . . . , tϵnan in reduced form with n = 2m ≥ 2 and ϵi= (−1)i+1. We can write g=a0g1a2· · ·a2m−2g2m−1a2m where gj =tajt−1,j= 1,3, . . . ,2m−1. This means that g∈gp{A∪tAt−1}. The proof of this lemma is completed. □
Lemma 3.4. Let G =⟨t, A;t−1Ht= K, φ⟩ be an HNN extension where H, K are non-trivial normal subgroups of A and H∩K = 1. Let GH and GK be the normalizers of H and K, respectively in G.
Then GK =t−1GHt.
Proof. Let g∈ GH. Thent−1g−1tKt−1gt = t−1g−1Hgt =t−1Ht=K. Thus t−1GHt ⊆GK. Let g ∈
GK. Then tg−1t−1Htgt−1 =tg−1Kgt−1 =tKt−1 =H. HencetGKt−1 ⊆GH andGK =t−1GHt. □
Lemma 3.5. Let G =⟨t, A;t−1Ht= K, φ⟩ be an HNN extension where H, K are non-trivial normal subgroups of A and H∩K = 1. Let GH be the normalizer of H in G. Then GH ∼=AH=∗ tKt−1tAt−1. Proof. Letψ1:A→GH be the homomorphism defined byψ1(a) =afor alla∈A, and letψ2 :tAt−1→
GH be the homomorphism defined byψ2(tat−1) =tat−1. Then ψ1(h) =h =tφ(h)t−1 =ψ2(tφ(h)t−1)
inGH. Thereforeψ1 andψ2 can be extended to become a homomorphismψ:AH=∗ tKt−1tAt−1→GH. Furthermore, ψ is an epimorphism, for GH = gp{A∪tAt−1} by part (c) of Lemma 3.3. Let x ∈
AH=∗ tKt−1tAt−1. Then x = a1(tb1t−1)a2(tb2t−1)· · ·an(tbnt−1) where ai ∈/ H, bi ∈/ K. So ψ(x) = x. Since GH is a subgroup of G = ⟨t, A;t−1Ht = K, φ⟩, ψ(x) ∈ G. Note that ψ(x) is reduced, since it contains no consecutive subsequence t−1ht with h ∈ H and subsequence tkt−1 with k ∈ K. Hence
ψ(x)̸= 1 and ψ is an isomorphism. □
Lemma 3.6. Let G=⟨t, A;t−1Ht=K, φ⟩ be an HNN extension whereA 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 imageGofGsuch thatGis residually finite,XqY is finite, andp /∈XqY. If such image can be
found, then there exists N ◁fGsuch thatp /∈XqY N. LetN be the preimage ofN. Thenp /∈XqY N.
Case 1. ∥p∥ ̸= ∥q∥. Let p =p0tϵ1p1· · ·tϵnp
n and q =q0tϵ
′
1q1· · ·tϵ′mqm be reduced where n̸= m and
n, m≥1. SinceA is H,K-separable, there existsMA◁fA such that pi, qj ∈/ HMA forpi, qj ∈/H, and
pi, qj ∈/ KMAforpi, qj ∈/ K. LetNH =MA∩H∩φ−1(MA∩K). By Lemma 3.2, there exists NA◁fA such that NA∩H ⊆NH,NAH∩NAK =NA and φ(NA∩H) =NA∩K. Let RA=NA∩MA. Then
RA◁f A. We claim thatRAH∩RAK =RAand φ(RA∩H) =RA∩K.
First we show that RAH∩RAK =RA. Clearly, RA ⊆RAH∩RAK. Let y ∈RAH∩RAK. Then
y =r1h=r2k where r1, r2 ∈RA and h ∈H and k∈K. This implies that h=r−11r2k∈RAK∩H ⊆
NAK∩H ⊆ NA, whence h ∈ NA∩H ⊆ NH ⊆ MA and h ∈ NA∩MA = RA. Hence y ∈ RA and
RAH∩RAK=RA.
Next we show that φ(RA∩H) = RA∩K. Note that RA∩H =NA∩MA∩H =NA∩H∩MA =
NA∩H, for NA∩H ⊆NH ⊆ MA. Also RA∩K =NA∩MA∩K = NA∩K∩MA = NA∩K, for
Let G = ⟨t, A;t−1Ht = K, φ⟩ where A = A/RA, H = HRA/RA, K = KRA/RA and φ is the homomorphism induced by φ. Clearly,Gis a homomorphic image ofG. By the choice ofMA,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 lengthn while the reduced elements in XqY have length m, and m̸=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 thatq−1p /∈ q−1XqY =XY, for
X, Y ◁ A. Since A is XY-separable, there exists MA◁f A such that q−1p /∈ XY MA. It follows that
p /∈XqY MA. LetMA, NH, NA, RA and G be defined as in Case 1. Suppose p∈XqY. Then p=xqy
for some x∈X, y∈Y. This implies thatp∈XqY RA⊆XqY MA, a contradiction. Thusp /∈XqY. As
in Case 1, Gis residually finite and we are done.
Suppose n ≥ 1. Let p = p0tϵ1p1· · ·tϵnpn and q = q0tϵ ′
1q1· · ·tϵ′nqn be in reduced forms. Since A is H, K-separable, there exists MA◁f A such that pi, qj ∈/ HMA for pi, qj ∈/ H, and pi, qj ∈/ KMA
for pi, qj ∈/ K. Let MA, NH, NA, RA and G be defined as in Case 1. Then p and q are reduced and ∥p∥=n=∥q∥. Furthermore,H∩K = 1 andG is residually finite.
If ϵi ̸= ϵ′i for some i, then p /∈ XqY and we are done, so assume that ϵi = ϵ′i for all i. Note that
p∈XqY if and only if the following system of equations (3.1) holds:
p0 =v0−1q0u1 p1 =v1−1q1u2
.. . (3.1)
pn−1 =vn−−11qn−1un
pn=vn−1qnun+1.
where v0∈X,un+1 ∈Y,ui ∈H,vi ∈K ifϵi = 1, or ui ∈K,vi∈H ifϵi =−1, and t−ϵiuitϵi =vi for
i= 1, . . . , n.
For ease of exposition, let Ui =H ifϵi = 1 or Ui =K ifϵi =−1 and t−ϵiUitϵi =Vi for i= 1, . . . , n.
Furthermore, letV0 =X and Un+1=Y. By using this notation,ui ∈Ui and vi∈Vi for all i.
SubCase 2.1. Suppose pi ∈/ ViqiUi+1 for some 0≤i≤n. Then qi−1pi ∈/ qi−1ViqiUi+1 =ViUi+1. Since ViUi+1 = H or K or HK, the subgroup MA above can be chosen with the additional property that
q−i 1pi ∈/ ViUi+1MA. Thus pi ∈/ ViqiUi+1MA. Note that pi ∈/ ViqiUi+1 for this choice of MA. Hence
p /∈XqY and we are done.
SubCase 2.2. So, we may assume that pi ∈ViqiUi+1 for all 0≤i≤n.
we meant by matching forward is that given pi = v−i 1qiui+1 for some 0 ≤i < n, we are able to find vi+1∈Vi+1 and ui+2 ∈Ui+2 such that pi+1 =vi−+11qi+1ui+2 and t−ϵi+1ui+1tϵi+1 =vi+1.
Letp1 =v1′− 1q
1u′2 wherev1′ ∈V1,u′2∈U2. Supposet−ϵ1u1tϵ1 =v1. Note thatp1 =v−11v1v1′− 1q
1u′2 = v1−1q1(q−11(v1v′1−1q1u′2)). SinceV1=U2=HorK, we haveV1 =U2◁A, and thus (q1−1v1v1′−1q1u′2)∈U2. Let u2 = (q−11v1v1′−1q1u′2). Then we have p0 =v0−1q0u1 and p1=v−11q1u2 with t−ϵ1u1tϵ1 =v1.
By continuing to match forward in this way, we havepi =vi−1qiui+1 andt−ϵiuitϵi =vi for 0≤i≤n. So the equations (3.1) hold, a contradiction. Hence Vi ̸=Ui+1 for some 0≤i≤n.
Let r be the least non-negative integer such that Vr ̸=Ur+1. We claim that r ̸=n. Suppose r=n.
Then Vi=Ui+1 for all 0≤i≤n−1.
Let pn=vn−1qnun+1 where vn∈Vn,un+1 ∈Un+1. We shall show that by “matching backward”, we
will make equations (3.1) hold. What we meant by matching backward is that given pi = vi−1qiui+1
for some 0 < i ≤ n, we are able to find vi−1 ∈ Vi−1 and ui ∈ Ui such that pi−1 = vi−−11qi−1ui and
t−ϵiuitϵi =vi.
Let pn−1 = v′−n−11qn−1u′n where vn′−1 ∈ Vn−1, u′n ∈ Un. Let tϵnvnt−ϵn = un. Note that pn−1 = v′−n−11qn−1u′nu−n1un= (v′−n−11qn−1u′nun−1q−n−11)qn−1un. SinceVn−1 =Un=HorK, we haveVn−1=Un◁A, and thus (v′−n−11qn−1u′nu−n1qn−−11)∈Vn−1. Letvn−−11= (v′−
1
n−1qn−1u′nu−n1qn−−11). Then pn=vn−1qnun+1 and pn−1 =vn−−11qn−1un witht−ϵnuntϵn =vn.
By continuing to match backward in this way, we havepi =v−i 1qiui+1andt−ϵiuitϵi =vifor 0≤i≤n. So equation (3.1) holds, a contradiction. Hencer ̸=n.
By the choice ofr, we haveVi =Ui+1 for 0≤i≤r−1. Note thatr could be 0. Let pr=vr−1qrur+1
where vr ∈Vr and ur+1 ∈Ur+1. Now by matching backward, we have
p0=v−01q0u1
.. . (3.2)
pr−1=v−r−11qr−1ur
pr=v−r1qrur+1.
where vi ∈Vi,ui+1∈Ui+1 fori= 0, . . . , r and t−ϵiuitϵi =vi fori= 1, . . . , r. Since equation (3.1) does not hold, there exists as, wherer+ 1≤s≤nsuch that
pr=vr−1qrur+1
.. . (3.3)
ps−1 =vs−−11qs−1us
ps=v′−
1
s qsu′s+1.
IfVs=Us+1, then fromps−1 =v−s−11qs−1us, and by matching forward we would haveps =v−s1qsus+1
with t−ϵsustϵs =vs, but this contradicts the choice of sas the largest integer such that equation (3.3)
holds. Hence Vs̸=Us+1.
Now we have Vr ̸= Ur+1 and Vs ̸= Us+1 and (3) holds. The MA above can be chosen with the additional property that v′−s1t−ϵsustϵs ∈/ MA.
Supposep∈XqY. Then the following system of equations (3.4) holds:
p0 =z−01q0w1 p1 =z−11q1w2
.. . (3.4)
pn−1 =zn−−11qn−1wn
pn=z−n1qnwn+1.
where zi ∈Vi,wi+1∈Ui+1 fori= 0, . . . , n and zi =t−ϵiwitϵi fori= 1, . . . , n. From (3.3), we have the following system of equations:
pr =v−r1qrur+1
.. . (3.5)
ps−1 =vs−−11qs−1us
ps =v′−s1qsu′s+1.
From equations (3.4) and (3.5), we have pr = v−r1qrur+1 and pr = z−r1qrwr+1. Thus zrv−r1 =
qrwr+1u−r+11 qr−1 ∈ Vr ∩Ur+1 = H ∩K = 1. This implies that zr = vr, wr+1 = ur+1 and zr+1 = t−ϵr+1w
r+1tϵr+1 =t−ϵr+1ur+1tϵr+1 =vr+1.
Again from equations (3.4) and (3.5), we have pr+1 = v−r+11 qr+1ur+2, pr+1 = z−r+11 qr+1wr+2. Since zr+1=vr+1, we have ur+2 =wr+2. By continuing this way, we see that uj =wj for all r+ 1≤j ≤s. From ps = v′−s1qsu′s+1 and ps = z−s1qsws+1, we have zsv′−
1
s = qsws+1u′− 1
s+1q−s1 ∈ Vs∩Us+1 = H ∩ K = 1. This implies that zs = v′s, but then t−ϵsustϵs = t−ϵswstϵs = zs = v′s, which implies that
v′−s1t−ϵsustϵs ∈MA, contradicting the choice ofMA. Hencep /∈XqY and we are done.
The proof of the lemma is now complete. □
Theorem 3.7. LetG=⟨t, A;t−1Ht=K, φ⟩be an HNN extension whereAis subgroup separable,H, K
are non-trivial finitely generated normal subgroups of A and H∩K = 1. Let GH be the normalizer of
Proof. Let x, y ∈G and x /∈ {y}G. We may assume thatx 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 Gsuch that x /∈ {y}GN. LetN be the preimage of N. Thenx /∈ {y}GN.
Case 1. Suppose∥x∥ ̸=∥y∥. Sincex, yare elements of minimal length in{x}G,{y}Grespectively,x, y
are cyclically reduced. Let x =x0tϵ1x1tϵ2· · ·x
n−1tϵn and y =y0tϵ′1y1tϵ′2· · ·ym−1tϵ′m be reduced where
xi, yj ∈A,ϵi =±1,ϵ′j =±1 for alli, j, andn̸=m. Then there existsMA◁fAsuch thatxi, yj ∈/ HMA
forxi, yj ∈/ H and xi, yj ∈/ KMA forxi, yj ∈/ K.
Let NH =MA∩H∩φ−1(MA∩K). By Lemma 3.2, there existsNA◁f A such thatNA∩H⊆NH,
NAH∩NAK = NA and φ(NA∩H) =NA∩K. Let RA =NA∩MA. ThenRAH∩RAK = RA and
φ(RA∩H) =RA∩K.
Let G = ⟨t, A;t−1Ht = K, φ⟩ where A = A/R
A, H = HRA/RA, K = KRA/RA and φ is the homomorphism induced byφ. Note that H∩K = 1, and both H and K are non-trivial, for x̸= 1.
Note that x, y are cyclically reduced and ∥x∥ =n, ∥y∥ =m. Furthermore,x /∈ {y}G because x has
lengthnwhile the cyclically reduced elements in{y}G have lengthm, andm̸=n. By [8, Theorem 13],
G is conjugacy separable, we are done.
Case 2. Suppose∥x∥=∥y∥= 0. Thenx, y∈A.
Subcase 2.1. Supposex, y∈H and x̸= 1. Thenx /∈ {y}GH. Since G
H is conjugacy separable, there
exists M ◁f GH such that x /∈ {y}GHM and x /∈M. Let MA=M ∩A. Let NH, NA, RA and Gbe as
in Case 1. Again, H∩K = 1, and bothH and K are non-trivial, for x̸= 1.
Claim 1. Let Ψ be the epimorphism ofG ontoG. We claim thatGH ∩KerΨ⊆M.
Proof of Claim 1. Now, by Lemma 3.5, GH =AH=∗ tKt−1tAt−1 is a generalised free product. Let
∥ ∥GH denote the free product length of an element inGH.
Let w∈ GH ∩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∈RAorw=tw′t−1 and w′ ∈RA. In either case, w∈M.
Next, we assume that if g∈GH ∩KerΨ and∥g∥GH ≤n−1, then g∈M.
Let w ∈ GH ∩KerΨ and ∥w∥GH = n ≥ 2. If n = 2r+ 1, then w = a1(tb1t−1)· · ·ar(tbrt−1)ar+1
or w = (tb1t−1)a2(tb2t−1)· · ·ar+1(tbr+1t−1). If n = 2r, then w = a1(tb1t−1)· · ·ar(tbrt−1) or w = (tb1t−1)a2(tb2t−1)· · ·(tbrt−1)ar+1.
We shall only consider the case w=a1(tb1t−1)· · ·ar(tbrt−1) (other cases are similar). Since w= 1 inG, either ai ∈H orbi∈K for somei.
Supposeai ∈H. Then ai =hmfor someh∈H, m∈RA⊆M. Therefore
w=a1(tb1t−1)· · ·(tbi−1t−1)hm(tbit−1)· · ·an(tbnt−1).
w1 is of the form
a1(tb1t−1)· · ·ai−2(tbi−2t−1)ai−1(tbi−1kbit−1)ai+1(tbi+1t−1)· · ·an(tbnt−1),
where k=t−1ht∈K. Since w2 ∈KerΨ (m ∈M) andw∈KerΨ, we havew1 ∈KerΨ, and thus by
assumption,w1∈M because∥w1∥GH ≤n−1. Noww1, w2 ∈M implies that w∈M. The proof of the
case bi ∈K is similar. This establishes Claim 1. □
We can now continue with the proof of Subcase 2.1.
Supposex∈ {y}G. Then x=g−1yg. Ify= 1, then x= 1 and x∈RA⊆M, a contradiction. Hence
y̸= 1. Sincex, y∈H, by Lemma 3.3,g∈GH whereGH is the normalizer ofHinG. Again, by Lemma
3.3, GH =gp{A, tAt−1} =GH. Therefore g ∈ GH, and by Claim 1, x ∈ {y}GHKerΨ⊆ {y}GHM, a contradiction. Hence x /∈ {y}G. By [8, Theorem 13],Gis conjugacy separable, we are done.
Subcase 2.2. Supposex ∈K, y ∈H. Letx′ =txt−1. Then x′ ∈H. As in Subcase 2.1, one can find an imageG ofG such thatGis conjugacy separable and x′ ∈ {/ y}G. Clearly, x /∈ {y}G, we are done.
Subcase 2.3. Supposex∈A−(H∪K), y∈H. SinceA isH, K-separable, there existsMA◁fAsuch that x /∈HMA∪KMA. Let NH, NA, RA and Gbe as in Case 1. Again, H∩K = 1, and both H and
K are non-trivial, forx̸= 1.
Suppose x ∈ {y}G. Let x = g−1yg where g ∈G. If g ∈A, we would have x ∈H, a contradiction. Let g = a0tϵ1a1· · ·tϵnan be in reduced form where n ≥ 1. Then gxg−1 = y. This implies that
anxa−n1 ∈ 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 2.4. Supposex∈A, y∈K. Lety′ =tyt−1. Theny′ ∈H and this is similar to Subcase 2.1, 2.2 or 2.3.
Subcase 2.5. Supposex∈A, y∈A−(H∪K). Ifx∈H, then it is similar to Subcase 2.3. Ifx∈K, we letx′ =txt−1, then it is similar to Subcase 2.3. So we just need to consider the casex, y∈A−(H∪K). Since A is H, K-separable, there exists MA′ ◁f A such that x, y /∈HMA′ ∪KMA′ . Sincex /∈ {y}GH,
there existsM ◁f GH such thatx /∈ {y}GHM. Note that x /∈ {y}A(M∩A) for A⊆GH.
Let MA=MA′ ∩M. Let NH, NA, RA and G be as in Case 1. Again,H∩K = 1, and both H and
K are non-trivial, forx̸= 1.
Suppose x ∈ {y}G. Let x = g−1yg where g ∈ G. If g ∈ A, then x ∈ {y}A and x ∈ {y}ARA ⊆ {y}A(M ∩A), a contradiction. Let g = a0tϵ1
a1· · ·tϵnan be in reduced form where n ≥ 1. Then
gxg−1 =y. This implies thatanxa−n1∈H∪K, for otherwise, gxg−1 has length 2n≥2. Sox∈H∪K, a contradiction. Hencex /∈ {y}G, and we are done.
Case 3. Suppose∥x∥=∥y∥=n≥1. Let x=x0tϵ1x1tϵ2· · ·xn−1tϵn and y=y0tϵ ′
1y1tϵ′2· · ·yn−1tϵ′n. Note that by Theorem 2.5, x ∈ {y}G if and only if x ∈ {y(j)}X for some 0 ≤ j ≤ n−1 where
y(j)=yjtϵj+1yj+1· · ·tϵn+j−1yn+j−1tϵn+j and the subscripts are taken modulo n and X=H ifϵn=−1 and X=K ifϵn= 1. Letj be fixed.
Claim 2. We claim that if x /∈ {y(j)}X, then there exists N
Proof of Claim 2. For simplicity, we shall only show the case j= 0, that isy(j) =y(0)=y.
Suppose x /∈XyX. By Lemma 3.6, there exists N ◁f G such that x /∈XyXN. Hence x /∈ {y}XN,
and we are done.
So we may assume that x ∈ XyX. This implies that ϵi =ϵ′i for all 1 ≤ i ≤ n. Furthermore, the
following system of equations holds:
x0 =vn−1y0u1 x1 =v1−1y1u2
.. . (3.6)
xn−2 =vn−−12yn−2un−1 xn−1 =vn−−11yn−1un
where ui ∈ H, vi ∈ K if ϵi = 1 or ui ∈ K, vi ∈H if ϵi = −1, and vi = t−ϵiuitϵi for i= 1, . . . , n−1.
Note thatvn, t−ϵnuntϵn ∈X and vn̸=t−ϵnuntϵn forx /∈ {y}X.
For ease of exposition, let Ui =H ifϵi = 1 or Ui =K ifϵi =−1 and Vi =t−ϵiUitϵi for i= 1, . . . , n.
By using this notation, ui ∈Ui,vi ∈Vi and X =Vn. So we can writex /∈ {y}Vn.
Subcase C1. Suppose ϵ1 =ϵn. Then Vn̸=U1 and Vn∩U1=H∩K= 1.
There exists MA′ ◁f A such that xi, yj ∈/ HMA′ for xi, yj ∈/ H and xi, yj ∈/ KMA′ for xi, yj ∈/ K.
Furthermore, we may choose MA′ so that t−ϵnuntϵnvn−1 ∈/ MA′ . Let NH = MA′ ∩H∩φ−1(MA′ ∩K). Let NA, RA and G be as in Case 1. Again, H∩K = 1, and both H and K are non-trivial, forx̸= 1.
Furthermore, x, yare cyclically reduced and∥x∥=n=∥y∥.
Suppose x∈ {y}X. Sincex =x0tϵ1x1· · ·tϵn−1xn−1tϵn and y=y0tϵ1y1· · ·tϵn−1yn−1tϵn, the following
system of equations holds:
x0 =z−n1y0w1 x1 =z−11y1w2
.. . (3.7)
xn−2 =zn−−12yn−2wn−1 xn−1 =zn−−11yn−1wn
where wi ∈Ui,zi ∈Vi, and zi =t−ϵiwitϵi fori= 1, . . . , n.
x0=v−n1y0u1 x1=v−11 y1u2
.. . (3.8)
xn−2=vn−−12yn−2un−1 xn−1=vn−−11yn−1un.
From (3.7) and (3.8), we have z−n1y0w1 = v−n1y0u1, i.e., vnz−n1 = y0u1w−11y− 1
0 ∈ H ∩K = 1.
Therefore vn =zn and u1 =w1. This implies that v1 =t−ϵ1u1tϵ1 =t−ϵ1w1tϵ1 = z1. Continuing this
way, we see that ui = wi and vi = zi for i = 1, . . . , n. So vn = zn = t−ϵnwntϵn = t−ϵnuntϵn, i.e.,
t−ϵnuntϵnv−n1 ∈RA⊆MA′ , a contradiction. Hencex /∈ {y}X.
Subcase C2. Suppose ϵi = ϵi+1 for some i. Then Vi ̸= Ui+1 and Vi∩Ui+1 = H ∩K = 1. This is
similar to Subcase 3.1.
Subcase C3. Suppose ϵ1 ̸= ϵn and ϵi ≠ ϵi+1 for i = 1, . . . , n−1. We shall show that n is even.
Suppose n is odd, say n= 2p+ 1. Since ϵi = 1 or −1, we have ϵ1 = ϵ3 =ϵ5 =· · · =ϵ2p+1, but then ϵ1 =ϵ2p+1 =ϵn, a contradiction, for ϵ1̸=ϵn.
Letn= 2p.
Subcase C3.1. Suppose ϵ1 = 1. Then ϵi = (−1)i+1 for 1 ≤ i ≤ 2p. In particular, ϵ2p = −1 and
X = H. In this case x = x0(tx1t−1)· · ·x2p−2(tx2p−1t−1) and y = y0(ty1t−1)· · ·y2p−2(ty2p−1t−1). By
part (b) of Lemma 3.3, x, y ∈GH. Since GH is conjugacy separable, there existsM ◁f GH such that
x /∈ {y}GHM. So x /∈ {y}HM.
There exists MA′ ◁f A such that xi, yj ∈/ HMA′ forxi, yj ∈/ H and xi, yj ∈/ KMA′ forxi, yj ∈/ K. Let
NH =MA′ ∩M ∩H∩φ−1(MA′ ∩M∩K). LetNA, RA and Gbe as in Case 1. Again,H∩K = 1, and both H and K are non-trivial, forx̸= 1. Furthermore, x, yare cyclically reduced and∥x∥=n=∥y∥.
As in Claim 1, let Ψ be the epimorphism of G onto G. Then KerΨ∩GH ⊆ M. If x ∈ {y}H, we
have x∈ {y}H(KerΨ∩GH)⊆ {y}HM, a contradiction. Hencex /∈ {y}H.
Subcase C3.2. Suppose ϵ1 = −1. Then ϵi = (−1)i for 1 ≤ i ≤ 2p. In particular, ϵ2p = 1 and
X = K. Then x = x0(t−1x1t)· · ·x2p−2(t−1x2p−1t) and y = y0(t−1y1t)· · · y2p−2(t−1y2p−1t). Let x′ = txt−1 = (tx0t−1)x1(tx2t−1)· · ·(tx2p−2t−1)x2p−1. By part (b) of Lemma 3.3, x′ ∈ GH. Then by Lemma 3.4, x ∈ GK. Let y′ = tyt−1. Again by Lemma 3.3 and Lemma 3.4, y ∈ GK. Note that
x /∈ {y}K if and only if x′ ∈ {/ y′}H. As in Subcase C3.1, 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 ofG, say G, for whichx /∈ {y}X. Since {y}X is finite andG is conjugacy separable by [8, Theorem 13] (and thusGis residually finite), there exists N ◁f G such thatx /∈ {y}XN. Let N be the preimage of N in G. Then x /∈ {y}XN. The proof
We can now continue with the proof of Case 3.
By Claim 2, for each j, there exists Nj ◁f G such that x /∈ {y(j)}XNj for all j. Let N =
∩
j
Nj.
There exists MA′ ◁f A such that xi, yj ∈/ HMA′ for xi, yj ∈/ H and xi, yj ∈/ KMA′ for xi, yj ∈/ K. Let
NH =MA′ ∩N ∩H∩φ−1(MA′ ∩N ∩K). Let NA, RA and G be as in Case 1. Again,H∩K= 1, and both H and K are non-trivial, forx̸= 1. Furthermore, x, yare cyclically reduced and∥x∥=n=∥y∥.
Thenx andy have lengthnandx /∈ {y}G. By [8, Theorem 13],Gis conjugacy separable, we are done.
The proof of this theorem is now complete. □
Theorem 3.8. LetG=⟨t, A;t−1Ht=K, φ⟩be an HNN extension whereAis subgroup separable,H, K
are non-trivial finitely generated normal subgroups ofAandH∩K= 1. LetGH be the normalizer of H
in G. SupposeA is conjugacy separable. Then GH is conjugacy separable if and only ifG is conjugacy
separable.
Proof. By Theorem 3.7, it is sufficient to show that ifGis conjugacy separable, then GH is conjugacy
separable.
Let x, y ∈ GH and x /∈ {y}GH. Since G is residually finite, GH is residually finite. So, we may
assume thatx, y̸= 1.
Supposex /∈ {y}G. SinceGis conjugacy separable, there existsNG◁fGsuch thatx /∈ {y}GNG. Let
N =NG∩GH. Thenx /∈ {y}GHN and we are done.
Supposex∈ {y}G. Then the conclusions of Theorem 2.5 hold.
Case 1. Suppose∥x∥=∥y∥=n≥1. Since x, y∈GH, by Lemma 3.3, we have
x=a0(ta1t−1)· · ·a2s−2(ta2s−1t−1) and y=b0(tb1t−1)· · ·b2s−2(tb2s−1t−1) in reduced form in the HNN
extension G.
By Lemma 3.5, GH = AH=∗tKt−1tAt−1 is a generalised free product. Let y2k = b2k, y2k+1 = tb2k+1t−1,x2k=a2k andx2k+1 =ta2k+1t−1 for 0≤k≤s−1. Thenx=x0· · ·xn−1 andy=y0· · ·yn−1
in reduced form in the generalised free product GH.
Note that by [26, Theorem 4.6 on p. 212], x ∈ {y}GH if and only if x ∈ {y(j)}H for some j = 0,1, . . . , n−1 where y(j) =yjyj+1· · ·yj+n−1.
Now x ∈ {y}G implies that x= c−1y′c, where y′ is a cyclic permutation of y in G and c ∈X with
X =H if ϵn = −1 and X = K ifϵn = 1, where ϵn is the power of t appearing in the last term of x.
In this case, ϵn = −1. So X =H. Furthermore y′ = (tbjt−1)bj+1· · ·(tbj+2s−2t−1)bj+2s−1 if j is odd
or y′ =bj(tbj+1t−1)· · ·bj+2s−2(tbj+2s−1t−1) ifj is even. Here the subscripts are taken modulo 2s(see Theorem 2.3). Soy′ =y(j)andx∈ {y(j)}H. This implies thatx∈ {y}GH, a contradiction. Hence Case 1 does not occur.
Case 2. Suppose∥x∥=∥y∥= 0. Thenx, y∈H∪K, forH, K ◁ A.
Subcase 2.1. Suppose x, y∈ H. Let x =g−1yg for some g ∈G. Then by Lemma 3.3, g ∈GH and
Subcase 2.2. Supposex, y∈K. SinceA isH-separable andx, y /∈H, there existsMA′ ◁fAsuch that
x, y /∈HMA′ . Note thatx /∈ {y}A, forA ⊆GH. Since A is conjugacy separable, there exists MA◁f A
such that x /∈ {y}AM A.
LetNH =MA∩MA′ ∩H∩φ−1(MA∩MA′ ∩K). By Lemma 3.2, there existsNA◁fAsuch thatNA∩H ⊆
NH,NAH∩NAK=NAandφ(NA∩H) =NA∩K. LetRA=NA∩MA∩MA′. ThenRAH∩RAK =RA
andφ(RA∩H) =RA∩K. Sot−1(RA∩H)t=RA∩K andRA∩H =t(RA∩K)t−1 =tKt−1∩tRAt−1. Therefore HRA/RA ∼= H/(H∩RA) ∼= tKt−1/(tKt−1 ∩tRAt−1) ∼= (tKt−1)(tRAt−1)/tRAt−1. Note
that tRAt−1 ◁f tAt−1 and tKt−1 ◁ tAt−1. So we may form an epimorphic image of GH, which is
GH = AH=∗ tKt−1tAt−1, where A = A/RA, tAt−1 = tAt−1/tRAt−1, H = HRA/RA and tKt−1 = (tKt−1)(tRAt−1)/tRAt−1. Note thatH∩K = 1. Now, inGH, we have x, y /∈H andx /∈ {y}A. By [26, Theorem 4.6 on p. 212], x /∈ {y}GH. By [8, Theorem 4], G
H is conjugacy separable. So there exists
N ◁f GH such that x /∈ {y}GHN. Let N be the preimage of N inGH. Then x /∈ {y}GHN and we are
done.
Subcase 2.3. Suppose x ∈ H and y ∈ K, or y ∈ H and x ∈ K. We shall assume that x ∈ H and
y∈K.
Since A isH-separable andy /∈H, there existsMA′ ◁f Asuch that y /∈HMA′ . LetNH =MA′ ∩H∩
φ−1(MA′ ∩K). LetNA, RA andGH be as in Subcase 1.2. Note thatH∩K = 1. Now, inGH, we have
x /∈H,y /∈H and H ◁ A. So, by [26, Theorem 4.6 on p. 212], x /∈ {y}GH. By [8, Theorem 4], GH is
conjugacy separable. So there exists N ◁fGH such thatx /∈ {y}GHN. LetN be the preimage of N in
GH. Then x /∈ {y}GHN and we are done.
The proof of this theorem is now complete. □
Corollary 3.9. Let G =⟨t, A;t−1Ht =K, φ⟩ be an HNN extension where A is a polycyclic-by-finite group, H, K are non-trivial normal subgroups ofA, and H∩K= 1. LetGH be the normalizer ofH in
G. Then GH is conjugacy separable if and only if Gis conjugacy separable.
Proof. Follows from Theorem 3.8. □
Acknowledgments
We would like to thank the anonymous referees for the comments that helped us make several
improve-ments to this paper.
This project is supported by the Frontier Science Research Cluster, University of Malaya
(RG267-13AFR).
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Kok Bin Wong
Institute of Mathematical Sciences, University of Malaya, 50603 Kuala Lumpur, Malaysia
Email: [email protected]
Peng Choon Wong
Institute of Mathematical Sciences, University of Malaya, 50603 Kuala Lumpur, Malaysia
