Page 1

Discrete Mathematics 309 (2009) 2687–2695

Contents lists available at ScienceDirect

Discrete Mathematics

journal homepage: www.elsevier.com/locate/disc

Automorphisms of cubic Cayley graphs of order 2pq

Cui Zhang, Jin-Xin Zhou, Yan-Quan Feng∗

Department of Mathematics, Beijing Jiaotong University, Beijing 100044, PR China

a r t i c l ei n f o

Article history:

Received 19 August 2007

Received in revised form 19 May 2008

Accepted 12 June 2008

Available online 26 July 2008

Keywords:

Symmetric graph

Cayley graph

Normal Cayley graph

a b s t r a c t

In this paper the automorphism groups of connected cubic Cayley graphs of order 2pq for

distinct odd primes p and q are determined. As an application, all connected cubic non-

symmetric Cayley graphs of order 2pq are classified and this, together with classifications

of connected cubic symmetric graphs and vertex-transitive non-Cayley graphs of order

2pq given by the last two authors, completes a classification of connected cubic vertex-

transitive graphs of order 2pq.

© 2008 Elsevier B.V. All rights reserved.

1. Introduction

Let G be a permutation group on a setΩ andα ∈ Ω. Denote by Gαthe stabilizer ofα in G, that is, the subgroup of G fixing

the point α. We say that G is semiregular on Ω if Gα = 1 for every α ∈ Ω and regular if G is transitive and semiregular.

Throughout this paper a graph means a finite, simple, connected and undirected one. For a graph X, we use V(X), E(X)

and Aut(X) to denote its vertex set, edge set and full automorphism group, respectively. For u,v ∈ V(X), {u,v} is the edge

incident to u andv in X. An s-arc in a graph is an ordered(s+1)-tuple(v0,v1,...,vs−1,vs) of vertices of the graph such that

any two consecutive vertices are adjacent and any three consecutive vertices are distinct. A 1-arc is called an arc for short

and a 0-arc is a vertex. A graph X is said to be s-arc-transitive if Aut(X) is transitive on the set of s-arcs in X. In particular,

0-arc-transitive means vertex-transitive, and 1-arc-transitive means arc-transitive or symmetric. A symmetric graph X is said

to be s-regular if the automorphism group Aut(X) acts regularly on the set of s-arcs in X.

Let G be a finite group and S a subset of G such that 1 ?∈ S. The Cayley digraph Cay(G,S) on G with respect to S is

defined to have vertex set V(Cay(G,S)) = G and arc set E(Cay(G,S)) = {(g,sg) | g ∈ G,s ∈ S}. If S = S−1then

Cay(G,S), called a Cayley graph, is viewed as a graph by identifying two opposite arcs with one edge. It is known that a

Cayley digraph Cay(G,S) is connected if and only if S generates G. Furthermore, Aut(G,S) = {α ∈ Aut(G) | Sα= S} is

a subgroup of the automorphism group Aut(Cay(G,S)) of Cay(G,S). Given a g ∈ G, define the permutation R(g) on G by

x ?→ xg,x ∈ G. Then R(G) = {R(g) | g ∈ G}, called the right regular representation of G, is a permutation group isomorphic to

G.TheCayleydigraphCay(G,S)isvertex-transitivebecauseitadmitsR(G)asaregularsubgroupoftheautomorphismgroup

Aut(Cay(G,S)). A Cayley digraph Cay(G,S) is said to be normal if R(G) is normal in Aut(Cay(G,S)). Xu [30, Proposition 1.5]

proved that Cay(G,S) is normal if and only if Aut(Cay(G,S))1= Aut(G,S), where Aut(Cay(G,S))1is the stabilizer of 1 in

Aut(Cay(G,S)). A graph X is isomorphic to a Cayley graph on G if and only if Aut(X) has a subgroup isomorphic to G, acting

regularly on vertices (see [3, Lemma 16.3] or [26, Lemma 4]). For two subsets S and T of G not containing the identity 1, if

there is an α ∈ Aut(G) such that Sα= T then S and T are said to be equivalent, denoted by S ≡ T. One may easily show that

if S ≡ T then Cay(G,S)∼= Cay(G,T) (graph isomorphic) and then Cay(G,S) is normal if and only if Cay(G,T) is normal.

Itiswell-knownthateverytransitivepermutationgroupofprimedegreepiseither2-transitiveorsolvablewitharegular

normal Sylow p-subgroup (for example, see [8, Corollary 3.5B]). This implies that a Cayley graph of prime order is normal if

∗Corresponding author.

E-mail address: yqfeng@bjtu.edu.cn (Y.-Q. Feng).

0012-365X/$ – see front matter © 2008 Elsevier B.V. All rights reserved.

doi:10.1016/j.disc.2008.06.023

Page 2

2688

C. Zhang et al. / Discrete Mathematics 309 (2009) 2687–2695

the graph is neither the empty graph nor the complete graph. Du et al. [11] and Dobson et al., [9] determined the normality

of Cayley graphs on groups of order twice a prime and prime square, respectively. Wang et al. [27] obtained all disconnected

normalCayleygraphs.LetCay(G,S)beaconnectedcubicCayleygraphonanon-abeliansimplegroupG.Praeger[23]proved

that if NAut(Cay(G,S))(R(G)) is transitive on edges then the Cayley graph Cay(G,S) is normal, and Fang et al. [12] proved that

the vast majority of connected cubic Cayley graphs on non-abelian simple groups are normal. Recently, Wang and Xu [28]

determined the normality of 1-regular tetravalent Cayley graphs on dihedral groups and Feng and Xu [15] proved that

every connected tetravalent Cayley graph on a regular p-group is normal when p ?= 2,5. For more results on the normality

of Cayley graphs, we refer the reader to [13,16,19,20,30]. The normality of cubic Cayley graphs of order 2p2and 4p was

determined in [31,32] and in this paper we determine the normality of cubic Cayley graphs of order 2pq for distinct odd

primes p and q. Furthermore, all cubic non-symmetric Cayley graphs of order 2pq are classified, while the classifications of

cubic symmetric graphs and vertex-transitive non-Cayley graphs of order 2pq were given in [33].

Let Znbe the cyclic group of order n, as well as the ring of integers modulo n. Denote by Z∗

Znconsisting of numbers coprime to n and by D2nthe dihedral group of order 2n. For two groups M and N, N ≤ M means

that N is a subgroup of M and N < M means that N is a proper subgroup of M. By elementary group theory, we know that,

up to isomorphism, there are six groups of order 2pq(p > q > 2) defined as

nthe multiplicative group of

G1(2pq) = ?a?,

G2(2pq) = ?a,b | apq= b2= 1,b−1ab = a−1?,

G3(2pq) = ?a,b,c | ap= bq= c2= 1,ab = ba,cac = a−1,bc = cb?,

G4(2pq) = ?a,b,c | ap= bq= c2= 1,ab = ba,ac = ca,cbc = b−1?,

G5(2pq) = ?a,b,c | ap= bq= c2= 1,ac = ca,bc = cb,b−1ab = ar?,

G6(2pq) = ?a,b,c | ap= bq= c2= 1,cac = a−1,bc = cb,b−1ab = ar?,

where r is an element of order q in Z∗

(1)

p.

2. Preliminaries

For a subgroup H of a group G, denote by CG(H) the centralizer of H in G and by NG(H) the normalizer of H in G. Then

CG(H) is normal in NG(H).

Proposition 2.1 ([18, I. Theorem 4.5]). The quotient group NG(H)/CG(H) is isomorphic to a subgroup of the automorphism group

Aut(H) of H.

The following proposition is a basic fact in permutation group theory.

Proposition 2.2 ([29, Proposition 4.4]). Every transitive abelian group G on a set Ω is regular and the centralizer CSΩ(G) of G in

the symmetric group SΩis G.

In view of [7, pp.285, summary], one may extract the following proposition.

Proposition 2.3. Every maximal subgroup of PSL(2,7) is isomorphic to Z7? Z3or S4. Let p = 7,11 or 23. All subgroups of

PGL(2,p) of order p(p − 1) are conjugate and isomorphic to Zp? Zp−1, a Frobenius group of degree p.

The following proposition is known as Burnside’s p-q Theorem.

Proposition 2.4 ([25, Theorem 8.5.3]). Let p and q be primes and let m and n be non-negative integers. Then, any group of order

pmqnis solvable.

Let p and q be distinct odd primes. The following result gives the number of solutions of the equation x2+ x + 1 = 0 in

Zpq.

Lemma 2.5. Let p > q be odd primes and O3

?2

0

otherwise.

pqthe set of solutions of the equation x2+ x + 1 = 0 in Zpq. Then,

|O3

pq| =

3 | (p − 1) and q = 3,

3 | (p − 1) and 3 | (q − 1),

4

Proof. Since x3− 1 = (x − 1)(x2+ x + 1), a solution of the equation x2+ x + 1 = 0 must be an element of order 3 in Z∗

implying that either 3 | (p − 1) and q = 3 or 3 | (p − 1) and 3 | (q − 1). For 3 | (p − 1) and q = 3, there are two elements

pq,

Page 3

C. Zhang et al. / Discrete Mathematics 309 (2009) 2687–2695

2689

of order 3 in Z∗

follows that x1and x2are solutions of x2+ x+ 1 = 0 in Z3p. That is |O3

x2+x+1 = 0 in Zpqimplies that k is an element of order 3 in both Z∗

3 in Z∗

such that k = k1(mod p) and k = k2(mod q) and this can be easily proved by Eq. (2) in the proof of Lemma 3.1 in [21] which

claims that for any i ∈ Zpand j ∈ Zq, |(i + P) ∩ (j + Q)| = 1, where P = {sp | s ∈ Zq} and Q = {sq | s ∈ Zp}. It follows that

|O3

Let p > q be primes such that 3 | (p− 1) and 3 | (q− 1). By Lemma 2.5, there are exactly two elements of order 3, say λ

and λ2, in the ring Z3p, and exactly four elements, say λ1, λ2, λ2

Zpq. Define

SC6p= Cay(D6p,{b,ba,ba−λ}),

SC1

SC2

3p, say x1and x2= x2

1. Then, xi= 1 in Z3for each i = 1,2. Since (xi− 1)(x2

i+ xi+ 1) = x3

i− 1 = 0 in Z3p, it

3p| = 2. For 3 | (p− 1) and 3 | (q− 1), a solution k of

pand Z∗

q, there is a unique element k in Zpqsatisfying the equation x2+ x+ 1 = 0

q. Conversely, for every element, say k1, of order

pand every element, say k2, of order 3 in Z∗

pq| = 4 because there are exactly two elements of order 3 in Z∗

pand in Z∗

q, respectively.

?

1and λ2

2, of order 3 satisfying the equation x2+ x + 1 = 0 in

2pq= Cay(D2pq,{b,ba,ba−λ1}),

2pq= Cay(D2pq,{b,ba,ba−λ2}),

where D2n = ?a,b | an= b2= 1,bab = a−1? with n = 3p or pq. It is easy to show that SC6p, SC1

independent of the choices λ, λ1and λ2.

Take H1= G6(2·5·11) = G6(110) and let S1= {c,abc,(abc)−1} be a subset of H1. Take H2= G6(2·11·13) = G6(506)

and let S2= {c,ab3c,(ab3c)−1} be a subset of H2. In the groups H1and H2given in Eq. (1), set r = 3 because 3 is an element

of order 5 in Z∗

CF110= Cay(H1,S1),

SC506= Cay(H2,S2).

WiththehelpofsoftwarepackageMAGMA[4],onemayeasilycheckAut(CF110)∼= PGL(2,11)andAut(SC506)∼= PGL(2,23).

By [5], there is a unique cubic 3-regular graph of order 110 and a unique cubic 4-regular graph of order 506. It follows

that these two graphs must be CF110and SC506because |PGL(2,11)| = 1320 and |PGL(2,23)| = 12144, of which the

first is called Coxeter–Frucht graph (see [6]). Note that PGL(2,11) and PGL(2,23) have subgroups of order 110 and 506

by Proposition 2.3 and since these subgroups are Frobenius, they are isomorphic to G6(110) and G6(506), respectively. A

classification of cubic symmetric graphs of order 2pq was given in [33] and one may easily extract those which are Cayley.

2pqand SC2

2pqare

11and an element of order 11 in Z∗

23. Define

Proposition 2.6. Let X = Cay(G,S) be a connected cubic symmetric Cayley graph on a group G of order 2pq, where p > q are

odd primes. Then, X is s-regular for s = 1,3 or 4. Furthermore,

(1) X is 1-regular if and only if either q = 3 and 3 | (p− 1) or 3 | (p− 1) and 3 | (q− 1). If X is 1-regular then it is isomorphic

either to SC6pfor q = 3 and 3 | (p − 1), or to SC1

(2) X is 3-regular if and only if it is isomorphic to CF110. In this case, G = G6(110), S ≡ {c,abc,(abc)−1} (take r = 3) and

Aut(X)∼= PGL(2,11);

(3) X is 4-regular if and only if it is isomorphic to SC506. In this case, G = G6(506), S ≡ {c,ab3c,(ab3c)−1} (take r = 3) and

Aut(X)∼= PGL(2,23).

Let X = Cay(G,S) be a Cayley graph on G and A = Aut(X). It is known that Aut(G,S) = {α ∈ Aut(G) | Sα= S} is a

subgroup of A. Normal Cayley graphs are those which have the smallest possible automorphism groups.

2pqor SC2

2pqfor 3 | (p − 1) and 3 | (q − 1);

Proposition 2.7 ([30, Propositions 1.3 and 1.5]). The Cayley graph X = Cay(G,S) is normal if and only if A1= Aut(G,S) if and

only if A = R(G) ? Aut(G,S), where A1is the stabilizer of 1 in A and R(G) is the right regular representation of G.

By [10, Theorem 1 and Lemma 3.4], we have the following proposition, which can also be deduced from [14,22].

Proposition 2.8. Let D2n= ?a,b | an= b2= 1,bab = a−1? be a dihedral group of order 2n. A cubic Cayley graph Cay(D2n,S)

on D2nis 1-regular if and only if S is equivalent to{b,ba,ba−k} for n ≥ 13 and k2+k+1 ≡ 0(mod n). Further, these 1-regular

Cayley graphs are normal.

Let X and Y be two graphs. The lexicographic product X[Y] is defined as the graph with vertex set V(X[Y]) = V(X)×V(Y)

such that for any two vertices u = (x1,y1) and v = (x2,y2) in V(X[Y]), u is adjacent to v in X[Y] whenever {x1,x2} ∈ E(X)

or x1= x2and{y1,y2} ∈ E(Y). Denote by Knthe complete graph of order n, Cnthe cycle of length n, and Kn,n−nK2the graph

by deleting a one factor from the complete bipartite graph Kn,nof order 2n. The following proposition gives all non-normal

connected Cayley graphs of valency at most 4 on cyclic groups.

Proposition 2.9 ([2, Corollary 1.3]). All connected Cayley graphs with valency at most 4 on a finite cyclic group are normal,

except for G = Z4and X = K4, G = Z6and X = K3,3, G = Z5and X = K5, G = Z2mand X = Cm[2K1](m ≥ 3), or G = Z10

and X = K5,5− 5K2.

Page 4

2690

C. Zhang et al. / Discrete Mathematics 309 (2009) 2687–2695

Given a subset S of a group G with 1 ?∈ S, we call S a CI-subset of G and Cay(G,S) a CI-graph, if Cay(G,S)∼= Cay(G,T)

implies that S and T are equivalent, that is, there exists aγ ∈ Aut(G) such that Sγ= T. The following result is a well-known

criterion for CI-subset due to Babai [1].

Proposition 2.10. Let G be a finite group and S a subset of G not containing the identity element 1. Let X = Cay(G,S) and

A = Aut(X). Then S is a CI-subset of G if and only if for any σ ∈ SGwith σ−1R(G)σ ≤ A, there exists an α ∈ A such that

σ−1R(G)σ = α−1R(G)α, where SGdenotes the symmetric group on G.

Qu and Yu [24] investigated the CI-property of Cayley graphs on dihedral groups.

Proposition 2.11 ([24, Theorem 3.5]). Let G be a dihedral group of order 2n with n odd and S a subset of G not containing the

identity 1. If |S| ≤ 3 then S is a CI-subset.

3. Automorphism groups of cubic Cayley graphs of order 2pq

In this section, we shall determine the automorphism groups of cubic Cayley graphs of order 2pq for two distinct odd

primes p and q. First we prove a lemma which will be used later.

Lemma 3.1. Let G be a regular subgroup of Aut(SC6p). Then, G∼= G2(6p) or G6(6p). Furthermore, as a Cayley graph on G2(6p),

SC6pis normal and as a Cayley graph on G6(6p), SC6pis non-normal and SC6p∼= Cay(G6(6p),S) with S ≡ {c,abc,(abc)−1}.

Proof. Let X = SC6pand A = Aut(X). We first claim that A contains regular subgroups isomorphic to G6(6p). By definition

of the graph SC6p, one may assume that X = Cay(G2(6p),S), where G2(6p) = ?a,b | a3p= b2= 1,b−1ab = a−1? and

S = {b,ba,ba−k}withk2+k+1 = 0inZ3q.Clearly,khasorder3inZ∗

is normal. Thus, A = R(G2(6p)) ? ?α?, where α is an automorphism of order 3 of G2(6p) induced by aα= ak2and bα= ba.

Note that ?R(a)? ? A. Since each subgroup of ?R(a)? is characteristic in ?R(a)?, one has ?R(a3)? ? A and ?R(ap)? ? A. Thus,

?R(ap),α?∼= Z3× Z3and hence R(ap)α has order 3. Note that k2+ k + 1 = 0 (in Z3p) implies that (k,3) = 1. It follows

k2= 1(mod 3). Clearly, k2?= 1(mod p) because k2?= 1 in Z3p. Thus, a1−k2has order p and since 3 | (p − 1), ap−1also has

order p, implying that a1−p= at(1−k2)for some integer t. Now it is easy to show that R(bat)R(ap)α= R(bat). Furthermore,

R(a3)R(ap)α= R(a3)α= (R(a3))k2and R(a3)R(bat)= R(a3)−1. Thus, H = ?R(a3),R(ap)α,R(bat)?∼= G6(6p). If the stabilizer H1

of the identity 1 in H is not trivial, then H1= Aut(G2(6p),S) = ?α?, forcing A = H, a contradiction. Thus, H is regular on

V(X), that is, A contains regular subgroups isomorphic to G6(6p), as claimed.

LetM beanarbitraryregularsubgroupofA.IfM∼= G2(6p)thenProposition2.8impliesthatX,asaCayleygraphonG2(6p),

is normal. Now assume M ?∼= G2(6p). Since |A| = 18p, one has A = R(G2(6p))M, implying that |M ∩ R(G2(6p))| = 2p. Since

G2(6p) has no normal subgroups of order 2p, M is not normal in A, namely, X, as a Cayley graph on M (M ? G2(6p)), is

non-normal. Further, since |M ∩ R(G2(6p))| = 2p and ?R(a3)? is a normal Sylow p-subgroup of A, one has ?R(a3)? ≤ M.

As the centralizer CA(R(a3)) of R(a3) in A is ?R(a)?∼= Z3p, one has CM(R(a3)) = M ∩ CA(R(a3)) = ?R(a3)?. For any given

group in Eq. (1), if the centralizer of a Sylow p-subgroup of the group is the Sylow p-subgroup itself then the group must

be G6(6p). It follows that M∼= G6(6p). Without loss of generality, let M = G6(6p) = ?a,b,c | ap= b3= c2= 1,cac =

a−1,bc = cb,b−1ab = ar? with r an element of order 3 in Z∗

conjugate and are contained in ?a,c?, by the connectivity of X, one may assume S = {c,y,y−1}, where y has order 3 or 6. If

y has order 3 then there is a 3-cycle (1,y,y−1,1) passing through 1, y and y−1, but there is no 3-cycle passing through the

vertices 1,c,y, contrary to the symmetry of X. Thus, y has order 6 and one of y and y−1has form aibc, 1 ≤ i ≤ p − 1. Since

the map a ?→ ai, b ?→ b, c ?→ c induces an automorphism of G6(6p), one has S ≡ {c,abc,(abc)−1}.

The following is the main result of this section.

3p.ByProposition2.8,X is1-regularandCay(G2(6p),S)

p, and let X = Cay(G6(6p),S). Since all involutions of G6(6p) are

?

Theorem 3.2. Let p > q be odd primes and let X = Cay(G,S) be a connected cubic Cayley graph of order 2pq. Then either

Aut(X) = R(G) ? Aut(G,S) or one of the following holds:

(1) G = G6(6p) with 3 | (p − 1), S ≡ {c,abc,(abc)−1} and Aut(X)∼= G6(6p)Z3;

(2) G = G6(110), S ≡ {c,abc,(abc)−1} (take r = 3) and Aut(X)∼= PGL(2,11);

(3) G = G6(506), S ≡ {c,ab3c,(ab3c)−1} (take r = 3) and Aut(X)∼= PGL(2,23);

(4) G = G6(42), S ≡ {c,ab,(ab)−1} and Aut(X)∼= PGL(2,7).

Proof. Let A = Aut(X). Assume that Aut(X) > R(G) ? Aut(G,S), that is, R(G) is not normal in A. We deal with two cases

depending on the symmetry of X.

Case I: X is symmetric.

By Proposition 2.6, X is isomorphic to CF110, SC506, SC6p, SC1

and S ≡ {c,abc,(abc)−1}, that is the case (1) in the theorem. Assume X∼= CF110. Then Aut(X)∼= PGL(2,11), and by

Proposition 2.3, one may assume that X = Cay(G6(110),S), where G6(110) = ?a,b,c | a11= b5= c2= 1,cac =

2pqor SC2

2pq. If X∼= SC6p, then by Lemma 3.1, G∼= G6(6p)

Page 5

C. Zhang et al. / Discrete Mathematics 309 (2009) 2687–2695

2691

a−1,bc = cb,b−1ab = a3? (take r = 3). For any σ ∈ SGsuch that σ−1R(G6(110))σ ≤ Aut(X), again by Proposition 2.3,

σ−1R(G6(110))σ and R(G6(110)) are conjugate in Aut(X) because they have the same order 11·10, and by Proposition 2.10,

X is a CI-graph. This implies that S ≡ {c,cab,(cab)−1} by Proposition 2.6, which is the Case 2 in the theorem. Similarly, if

X∼= SC506then we have the Case 3 in the theorem.

Assume X∼= SC1

on the dihedral group G2(2pq) = ?a,b | apq= b2= 1,b−1ab = a−1?. The Cayley graph is normal by Proposition 2.8. By the

1-regularity of X, A∼= R(G2(2pq)) ? Z3and it is easy to show that R(G2(2pq)) is the unique regular subgroup of A because

p > q > 3, implying that X cannot be a Cayley graph on Gi(2pq) for i = 1,3,4,5 or 6.

Case II: X is non-symmetric.

In this case, the stabilizer Avof v ∈ V(X) in A is a 2-group and hence |A| = 2?· p · q with ? ≥ 2. We claim that A has no

normal 2-subgroups. Suppose to the contrary that H is a normal 2-subgroup of A. Let XHbe the quotient graph of X relative

to H, that is, the graph with vertices the orbits of H in V(X) and with two orbits adjacent if there is an edge in X between

those two orbits. Let K be the kernel of A acting on V(XH). Then, H ≤ K and A/K is transitive on V(XH). Since |V(X)| = 2pq,

every orbit of H in V(X) has length 2, implying|V(XH)| = pq. As X has valency 3 and H ≤ K, XHhas valency 2 or 3, and since

pq is odd, XHhas valency 2. By the connectivity of X, XHis a cycle of length pq, say V(XH) = {B0,B1,...,Bpq−1}, where Biis

adjacent to Bi+1for each i ∈ Zpq. If there is no edge in each Bi, then one may assume that each vertex in B1is adjacent to one

vertex in B0and two vertices in B2. By the transitivity of A/K on V(XH), the length of the cycle XHmust be even, contrary

to the fact that pq is odd. If there is an edge in some Bi0then there is an edge in each Bi, 0 ≤ i ≤ pq − 1, because of the

transitivity of A/K on V(XH). Since K fixes each orbit of H setwise, the stabilizer Kvof v ∈ V(X) in K fixes every neighbor of

v in X. The connectivity of X gives Kv= 1, forcing K = H∼= Z2. Since A/K = A/H ≤ Aut(XH)∼= D2pq, one has |A| ≤ 4pq

and hence |A : R(G)| ≤ 2, implying R(G) ? A, a contradiction. Thus, the claim is true, that is, A has no normal 2-subgroups.

In what follows we assume that N is a minimal normal subgroup of A. Then N is Zpor Zq, or a non-abelian simple group

because |N| | 2?· p · q.

Assume that N is solvable. Then N∼= Zpor Zq. By Proposition 2.4, A/N is solvable. Let C = CA(N). By Proposition 2.1,

A/C ≤ Aut(N)∼= Zp−1or Zq−1. Clearly, N ≤ C. There are two subcases: N = C and N < C, that is, N is a proper subgroup

of C.

Suppose N = C. Then A/N ≤ Aut(N)∼= Zp−1or Zq−1. Since p > q > 2, one has N∼= Zpand A/N ≤ Zp−1. Let XNbe the

quotient graph of X relative to the orbits of N, and K the kernel of A acting on V(XN). Then, N ≤ K and A/K is transitive on

V(XN). Since N is normal in A, XNhas valency at most 3, and since N∼= Zp, one has |V(XN)| = 2q > 1, implying that XNhas

valency 2 or 3. If XNhas valency 3 then K has trivial stabilizers and hence K = N. By Proposition 2.2, A/N is regular on V(XN)

because A/N ≤ Zp−1. It follows that |A| = 2pq, forcing R(G) ? A, a contradiction. If XNhas valency 2 then XNis a cycle of

length 2q because of the connectivity of X. Let V(XN) = {B0,B1,...,B2q−1} with Biadjacent to Bi+1for every i ∈ Z2q−1. If

there is an edge of X in each Bithen the induced subgraph ?Bi? of Biin X must be a cycle of length p because |Bi| = p is odd.

In this case, XNhas valency 1, a contradiction. Thus, there is no edge in each Biand one may assume that each vertex in B1

connects one vertex in B0and two vertices in B2. It follows that the induced subgraph ?B0∪ B1? of B0∪ B1in X is a perfect

matching and the induced subgraph ?B1∪ B2? of B1∪ B2in X is a cycle of length 2p because |B1| = p is odd. Thus, A/K is not

arc-transitive on XN, and hence A/K < Aut(X/N)∼= D4q, implying |A/K| = 2q by the vertex-transitivity of A/K on V(XN).

Further, K acts faithfully on B1and K < Aut(?B1∪ B2?)∼= D4p. It follows that |K| ≤ 2p and hence |A| ≤ 4pq. Thus, R(G) ? A

because |A : R(G)| ≤ 2, a contradiction.

Suppose N < C. Take a minimal normal subgroup of A/N, say M/N, in C/N. Since A/N is solvable, M/N is elementary

abelian. It follows that either M/N is a 2-group, or M/N∼= Zqor Zp. For the former, one has |M| = 2s· p or 2s· q for some

integer s ≥ 1. Since M ≤ C, a Sylow 2-subgroup of M is characteristic in M, and hence normal in A because M ? A. This

is impossible because A has no normal 2-subgroups. Thus, M/N∼= Zqor Zp, and hence M∼= Zpqbecause M ≤ C. Clearly,

M ≤ CA(M). If M = CA(M) then, by Proposition 2.1, A/M ≤ Aut(M)∼= Zp−1× Zq−1. Since M ? A, one has M ≤ R(G),

implying R(G)/M ? A/M, that is, R(G) ? A, a contradiction. If M < CA(M) then CA(M)/M must be a 2-group. It follows

that CA(M) = M × Q, where Q is a Sylow 2-subgroup of CA(M). Thus, Q is characteristic in CA(M) and normal in A because

CA(M) ? A, contrary to the fact that A has no normal 2-subgroups.

Assume that N is insolvable. Since |N| | 2?· p· q and p > q > 2, N must be non-abelian simple and by [17, pp. 12–14], N

is one of the following groups:

2pqor SC2

2pq. By Proposition 2.6, one has 3 | (p − 1) and 3 | (q − 1), and X is a 1-regular Cayley graph

A5,A6,PSL(2,7),PSL(2,8),PSL(2,17),PSL(3,3),PSU(3,3)

Since p2? |N| and q2? |N|, by checking the orders of the above groups, one has N = A5or PSL(2,7). Let C = CA(N).

Then N ∩ C = 1 because N is simple. It follows that either C is a 2-subgroup or C = 1. Thus, C = 1 because A has no

normal 2-subgroups, and by Proposition 2.1, one has A ≤ Aut(N). If N = A5then A = A5or S5. However, both S5and A5

have no subgroups of order 30, implying that X is a non-Cayley graph, a contradiction. It follows that N = PSL(2,7) and A ≤

Aut(N)∼= PGL(2,7).SinceX isaCayleygraph,Acontainsaregularsubgroupoforder42andbyProposition2.3,PSL(2,7)has

no subgroups of order 42, implying A = PGL(2,7). By Proposition 2.3, every subgroup of order 42 in PGL(2,7) is conjugate

to G6(42). Without loss of generality, let G = G6(42) = ?a,b,c | a7= b3= c2= 1,cac = a−1,bc = cb,b−1ab = a2?.

Clearly, all involutions in G are conjugate and hence one may assume c ∈ S. Note that the centralizer of c in G has order 6

and so there are seven involutions in G, of which all are contained in ?a,c?. Since S generates G, S = {c,y,y−1}, where y has

andPSU(4,2).