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ARS MATHEMATICA CONTEMPORANEA 3 (2010) 151–163
Half-arc-transitive graphs of order 4p of valency
twice a prime
Xiuyun Wang , Yan-Quan Feng ∗
Department of Mathematics, Beijing Jiaotong University
Beijing 100044, P. R. China
Received 9 November 2009, accepted 1 May 2010, published online 25 October 2010
Abstract
A graph is half-arc-transitive if its automorphism group acts transitively on vertices and
edges, but not on arcs. Let p be a prime. Cheng and Oxley [On weakly symmetric graphs
of order twice a prime, J. Combin. Theory B 42(1987) 196-211] proved that there is no
half-arc-transitive graph of order 2p, and Alspach and Xu [ 12 -transitive graphs of order 3p,
J. Algebraic Combin. 3(1994) 347-355] classified half-arc-transitive graphs of order 3p. In
this paper we classify half-arc-transitive graphs of order 4p of valency 2q for each prime
q ≥ 5. It is shown that such graphs exist if and only if p− 1 is divisible by 4q. Moreover,
for such p and q a unique half-arc-transitive graph of order 4p and valency 2q exists and
this graph is a Cayley graph.
Keywords: Cayley graph, half-arc-transitive graph, transitive graph.
Math. Subj. Class.: 05C25, 20B25
1 Introduction
Throughout this paper we denote by Zn the cyclic group of order n as well as the ring
of residue classes modulo n, and by Z∗n the multiplicative group of the ring Zn. Let D2n
be the dihedral group of order 2n, and let An and Sn be the alternating and symmetric
group of degree n, respectively. All graphs are assumed to be finite, simple and undirected,
but with an implicit orientation of the edges when appropriate. For a graph X , let V (X),
E(X), A(X) and Aut(X) be the vertex set, the edge set, the arc set and the automorphism
group of X , respectively. A graph X is said to be vertex-transitive, edge-transitive or arc-
transitive (symmetric) if Aut(X) acts transitively on V (X), E(X), or A(X), respectively,
and half-arc-transitive if X is vertex-transitive and edge-transitive, but not arc-transitive.
∗Corresponding author.
E-mail addresses: wxy830815@gmail.com (Xiuyun Wang), yqfeng@bjtu.edu.cn (Yan-Quan Feng)
Copyright c© 2010 DMFA Slovenije
152 Ars Math. Contemp. 3 (2010) 151–163
More generally, by a half-arc-transitive action of a subgroup G of Aut(X) on a graph X
we shall mean a vertex-transitive and edge-transitive, but not arc-transitive action of G on
X . In this case, we shall say that the graph X is G-half-arc-transitive.
The investigation of half-arc-transitive graphs was initiated by Tutte [34] and he proved
that a vertex- and edge-transitive graph with odd valency must be arc-transitive. In 1970
Bouwer [4] constructed a 2k-valent half-arc-transitive graph for every k ≥ 2 and later
more such graphs were constructed (see [10, 15, 17, 18, 33]). Let p, q be odd primes. It
is well-known that there are no half-arc-transitive graphs of order p or p2, and by Cheng
and Oxley [6], there are no half-arc-transitive graphs of order 2p. Alspach and Xu [2]
classified half-arc-transitive graphs of order 3p and Wang [35] classified half-arc-transitive
graphs of order a product of two distinct primes. Despite all of these efforts, however,
more classifications of half-arc-transitive graphs with general valencies seem to be very
difficult. For example, classification of half-arc-transitive graphs of order 4p has been
considered for more than 10 years by many authors, but it has still not been completed.
Recently, classifications of tetravalent and hexavalent half-arc-transitive graphs of order
4p were given in [13] and [37], respectively. In fact, investigation of half-arc-transitive
graphs of small valencies is currently an active topic in algebraic graph theory. For more
information, see [1, 7, 11, 12, 14, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 32, 38, 39]. In this
paper we classify 2q-valent half-arc-transitive graphs of order 4p for each prime q ≥ 5.
It is shown that such graphs are Cayley and exist if and only if p − 1 is divisible by 4q.
Moreover, for a given order such a graph is unique.
To end this section, we introduce the so called quotient graph of a graph X . Let Σ =
{B0, B1, . . . , Bn−1} be a partition of V (X). The quotient graph XΣ of X relative to the
partition Σ is defined to have vertex set and edge set as follows:
V (XΣ) = Σ,
E(XΣ) = {{Bi, Bj} | there exist vi ∈ Bi, vj ∈ Bj such that {vi, vj} ∈ E(X)}.
In particular, if N ≤ Aut(X) then the set of orbits of N on V (X) is a partition of V (X).
In this case, the quotient graph of X relative to the orbits of N is also called the quotient
graph of X relative to N , denoted by XN . It is easy to see that if N E G ≤ Aut(X) and
G is transitive on edges of X then the valency of XN is a divisor of the valency of X .
2 Preliminary results
For a finite group G and a subset S of G such that 1 6∈ S and S = S−1, the Cayley
graph Cay(G,S) on G with respect to S is defined to have vertex set G and edge set
{{g, sg} | g ∈ G, s ∈ S}. The following facts about Cayley graph are well known (see [3]).
Given g ∈ G, define the permutationR(g) onG by x 7→ xg, x ∈ G. Then the right regular
representation R(G) = {R(g) | g ∈ G} is a regular subgroup of Aut(Cay(G,S)), and
Aut(G,S) = {α ∈ Aut(G) | Sα = S} is a subgroup of the stabilizer Aut(Cay(G,S))1 of
the vertex 1 in Aut(Cay(G,S)). Furthermore, A graph X is isomorphic to a Cayley graph
onG if and only if its automorphism group Aut(X) has a subgroup isomorphic toG, acting
regularly on vertices.
A Cayley graph Cay(G,S) is said to be normal if Aut(Cay(G,S)) contains R(G) as a
normal subgroup. The following proposition is fundamental for normal Cayley graphs.
Proposition 2.1. [38, Proposition 1.5] Let X = Cay(G,S) be a Cayley graph on a finite
group G with respect to S. Let A = Aut(X) and let A1 be the stabilizer of 1 in A. Then X
X. Wang and Y.-Q. Feng: Half-arc-transitive graphs of order 4p of valency twice a prime 153
is normal if and only if A1 = Aut(G,S).
Cheng and Oxley [6] classified the connected symmetric graphs of order 2p for a prime
p. To extract a classification of connected q-, 2q- and 4q-valent symmetric graphs of order
2p for a prime q ≥ 5, we need to define some graphs. Let V and V ′ be two disjoint
copies of Zp, say V = {i | i ∈ Zp} and V ′ = {i′ | i ∈ Zp}. Let r be a positive
integer dividing p− 1 and H(p, r) the unique subgroup of Z∗p of order r. Define the graph
G(2p, r) to have vertex set V ∪V ′ and edge set {xy′ | x, y ∈ Zp, y−x ∈ H(p, r)}. Clearly,
G(2p, p − 1) ∼= Kp,p − pK2, the complete bipartite graph of order 2p minus a 1-factor.
Furthermore, assume that r is an even integer dividing p − 1. Then the graph G(2, p, r)
is defined to have vertex set V ∪ V ′ and edge set {xy, x′y, xy′, x′y′ | x, y ∈ Zp, y − x ∈
H(p, r)}. The lexicographic productX[Y ] of graphX by graph Y is the graph with vertex
set V (X[Y ]) = V (X) × V (Y ) and with two vertices u = (x1, y1) and v = (x2, y2)
adjacent whenever x1 is adjacent to x2, or x1=x2 and y1 is adjacent to y2. Clearly, if X is
symmetric and Y is a graph with no edge, then X[Y ] is symmetric. Moreover, G(2, p, r)
is in fact the lexicographic product of a circulant Cay(Zp, H(p, r)) by 2K1.
Proposition 2.2. [6, Theorem 2.4 and Table 1] Let p, q be odd primes with q ≥ 5 and let
X be a connected edge-transitive graph of order 2p. Then X is symmetric. Furthermore,
if X has valency q then one of the following holds:
(1) X ∼= K2p, the complete graph of order 2p, and 2p− 1 = q;
(2) X ∼= Kp,p, the complete bipartite graph of order 2p, and p = q;
(3) X ∼= G(2p, q) with q | (p− 1) and (p, q) 6= (11, 5), and Aut(X) ∼= (ZpoZq)oZ2;
(4) X ∼= G(2 · 11, 5) and Aut(X) ∼= PSL(2, 11) o Z2.
If X has valency 2q then X is bipartite and one of the following holds:
(5) For 2q < p− 1, X ∼= G(2p, 2q) with 2q | (p− 1) and Aut(X) ∼= (Zp oZ2q) oZ2;
(6) For 2q = p− 1, X ∼= Kp,p − pK2 and Aut(X) ∼= Sp o Z2.
If X has valency 4q then one of the following holds:
(7) X is non-bipartite, X ∼= G(2, p, 2q) with 2q | p − 1; for 2q < p − 1, Aut(X) ∼=
Zp2 o (Zp o Z2q) and for 2q = p− 1, Aut(X) ∼= Z
p
2 o Sp;
(8) X is bipartite and X ∼= G(2p, 4q) with 4q | (p − 1); for 4q < p − 1, Aut(X) ∼=
(Zp o Z4q) o Z2 and for 4q = p− 1, X ∼= Kp,p − pK2 and Aut(X) ∼= Sp o Z2.
Let G act transitively on a set Ω. Then G induces a natural action on Ω×Ω defined by
(x, y)g = (xg, yg) for (x, y) ∈ Ω × Ω and g ∈ G. The orbits of G on Ω × Ω are called
orbitals of G. The orbital ∆ = {(x, x) | x ∈ Ω} of G is trivial and all other orbitals of
G are nontrivial. Let O be a nontrivial orbital of G. The pair (Ω, O), denoted by O, is a
directed graph with vertex set Ω and directed edge set O, called the orbital digraph of G
relative to O. For any orbital O of G, it is easy to show that O∗ = {(α, β) | (β, α) ∈ O}
is also an orbital of G, called the paired orbital of O, and O is said to be self-paired if
O∗ = O. Clearly, if O is a non-self-paired orbital then the underlying graph of O is G-
half-arc-transitive. Conversely, if X is a half-arc-transitive graph then X is the underlying
graph of an orbital digraph (V (X), O) of Aut(X) relative to a non-self-paired orbital O.
In this case, Aut(X) coincides with the automorphism group of the digraph (V (X), O).
This implies the following proposition.
154 Ars Math. Contemp. 3 (2010) 151–163
Proposition 2.3. Let X be a connected half-arc-transitive graph of valency 2n. Let A =
Aut(X) and let Au be the stabilizer of u ∈ V (X) in A. Then each prime divisor of |Au| is
a divisor of n!.
By [10, Lemma 2.2], we have the following proposition (also see [1, 19, 31, 36]).
Proposition 2.4. The smallest half-arc-transitive graph has order 27. The smallest vertex-
primitive half-arc-transitive graph of order kp, with p a prime and k < p, has order 253.
The following proposition is straightforward (also see [13, Propositions 2.1 and 2.2]).
Proposition 2.5. Let X = Cay(G,S) be a half-arc-transitive graph. Then, there is no
involution in S, and no α ∈ Aut(G,S) such that sα = s−1 for some s ∈ S. In particular,
there are no half-arc-transitive Cayley graphs on abelian groups.
LetG be a transitive permutation group on a set Ω. A nonempty subset ∆ of Ω is called
a block for G if for each g ∈ G, either ∆g = ∆ or ∆g ∩ ∆ = φ. Clearly, the set Ω, the
empty set φ, and the sets {α} consisting of only one point are blocks of G on Ω. We call
these trivial blocks. A transitive group G is said to be primitive if G has only trivial
blocks, and imprimitive if there is at least one non-trivial block. The socle of a finite
group G, denoted by soc(G), is the product of all minimal normal subgroups of G. One
may extract the following results from [20, Table 3].
Proposition 2.6. Let p be a prime and G a primitive group of degree n.
(1) For n = p, G is either solvable with a normal Sylow p-subgroup or non-solvable
with the following table, where d and k denote the degree and transitive multiplicity,
respectively.
soc(G) d k comment
Ap p p− 2 G = Ap
Ap p p G = Sp
PSL(2, 22
s
) p = 22
s
+ 1 3 s > 0
PSL(n, q) p = q
n−1
q−1 2 n ≥ 3, n odd
PSL(2, 11) 11 2
M11 11 4
M23 23 4
(2) For n = 2p, either G is 2-transitive or p = 5.
(3) For n = 4p, either G is 2-transitive or p = 7, 13, or 17.
One may deduce Proposition 2.6(1) and 2.6(2) also from [9, Corollary 3.5B] and [6,
Theorem 1.1], respectively. Moreover, if G is primitive, but not 2-transitive of degree 2p
then p = 5 and G ∼= A5 or S5. If G is primitive, but not 2-transitive of degree 4p then
G ∼= A8, S8, PSL(2, 8) or PGL(2, 7) for p = 7, G ∼= Aut(PSL(3, 3)) for p = 13, or G is
isomorphic to a subgroup between PSL(2, 16) and PΓL(2, 16) for p = 17.
X. Wang and Y.-Q. Feng: Half-arc-transitive graphs of order 4p of valency twice a prime 155
3 Main result
Let p, q be odd primes with q ≥ 5. In this section we classify the 2q-valent half-arc-
transitive graphs of order 4p. Following the notation of Kwak et al. [16], for any integer `
and any positive integer t, define
`[t] = `t−1 + `t−2 + · · ·+ 1.
LetG = 〈a, b | ap = b4 = 1, b−1ab = ar〉, where r2 ≡ −1 (mod p). It is easy to see that r
is an element of order 4 in Z∗p and the groupG is independent of the choice of r. In particu-
lar, p− 1 is divisible by 4. Furthermore, assume that p− 1 is divisible by q. Then there is a
unique subgroup of order q, say 〈s〉, in Z∗p. Set T = {b, as[1]b, as[2]b, as[3]b, . . . , as[q−1]b}
and S = T ∪ T−1. Define
C2q4p = Cay(G,S). (3.1)
Clearly, s[i] 6= 0 (mod p) for each 1 ≤ i ≤ q − 1, that is s[i] ∈ Z∗p, and one may easily
show that C2q4p is a connected graph of order 4p and of valency 2q.
Clearly, si (1 ≤ i ≤ q − 1) are all the elements of order q in Z∗p. Set Si = Ti ∪ T−1i
with Ti = {b, as
i[1]b, as
i[2]b, as
i[3]b, . . . , as
i[q−1]b}. Then T1 = T and S1 = S. For any
given 1 ≤ i ≤ q − 1, the map a 7→ as[i]−1 , b 7→ b, induces an automorphism of G, say βi.
For each 1 ≤ j ≤ q − 1, there is 1 ≤ k ≤ q − 1 such that sj = (si)k. It follows that
s[j]s[i]−1 =
1− sj
1− s
1− s
1− si
=
1− (si)k
1− si
= si[k],
which means T βi1 = Ti. Thus, βi is an isomorphism from Cay(G,S) to Cay(G,Si).
Therefore, C2q4p is independent of the choice of the element s of order q in Z∗p.
Denote by α the automorphism of G induced by a 7→ as, b 7→ ab. Note that s[q] =
sq−1 + sq−2 + · · · + s + 1 = 0 (mod p). Then bα = ab, (as[i]b)α = as[i+1]b for each
1 ≤ i ≤ q − 2, and (as[q−1]b)α = b. It follows that α ∈ Aut(G,S), implying that
Aut(C2q4p) has an edge-transitive subgroup of order 4pq, that is R(G) o 〈α〉. It will be
shown in Lemma 3.2 that Aut(C2q4p) = R(G) o 〈α〉 and hence C
2q
4p is half-arc-transitive.
The following is the main result of this paper.
Theorem 3.1. Let p, q be odd primes with q ≥ 5. ThenX is a 2q-valent half-arc-transitive
graph of order 4p if and only if 4q | (p− 1) and X ∼= C2q4p with Aut(C
2q
4p) = R(G) o 〈α〉.
The sufficiency is proved as a part of Lemma 3.2 and the necessity is proved in Lemma
3.3. For a graph X and u ∈ V (X), let Nd(u) denote the set of vertices having distance d
from u in X . If X is a directed graph, let N+(u) denote the set of out-neighbors of u and
N−(u) the set of in-neighbors of u.
Lemma 3.2. Let p, q be odd primes with q ≥ 5. Let X = Cay(G,S) be a 2q-valent
half-arc-transitive Cayley graph on a group G of order 4p. Then G ∼= 〈a, b | ap =
b4 = 1, b−1ab = ar〉 with r2 = −1 (mod p). Moreover, for each prime p satisfying
4q | (p−1), the Cayley graph C2q4p (defined in Eq (3.1)) is half-arc-transitive and Aut(C
2q
4p) =
R(G) o 〈α〉.
156 Ars Math. Contemp. 3 (2010) 151–163
Proof. Since there are no half-arc-transitive graphs of order p or 2p (see [5, 6]), X is
connected. Thus, |S| = 2q, S−1 = S and 〈S〉 = G. By Proposition 2.5, G is non-abelian
and by Proposition 2.4, p ≥ 7. From elementary group theory, all non-abelian groups of
order 4p for every odd prime p ≥ 7, up to isomorphism, can be written as follows:
G1(p) = 〈a, b | a2p = b2 = 1, b−1ab = a−1〉,
G2(p) = 〈a, b | a2p = 1, b2 = ap, b−1ab = a−1〉,
G3(p) = 〈a, b | ap = b4 = 1, b−1ab = ar〉, r2 ≡ −1 (mod p).
Clearly, G3(p) exists if and only if p ≡ 1 (mod 4), and G1(p) is the dihedral group D4p.
By Proposition 2.5, there is no involution in S, and since G = 〈S〉, one has G 6= G1(p).
Suppose G = G2(p). Then S contains at least one element of order 4 and its inverse. Each
element of order 4 is of the form aib or aib−1. The automorphism of G2(p) induced by
b 7→ b−1, a 7→ a, maps ajb to (ajb)−1 for any integer j and fixes 〈a〉 pointwise. This is
impossible by Proposition 2.5. Thus, G = G3(p) and 4 | p− 1.
Let 4q | (p− 1), and let s be an element of order q in Z∗p. Recall that C
2q
4p = Cay(G,S)
and R(G) o 〈α〉 ≤ Aut(C2q4p), where S = T ∪ T−1 with T = {b, as[1]b, as[2]b, as[3]b, . . . ,
as[q−1]b}, and α is the automorphism of G of order q induced by a 7→ as, b 7→ ab. To
finish the proof, it suffices to show that Aut(C2q4p) = R(G) o 〈α〉.
Let G∗ = 〈a, b2〉. Then G∗ is isomorphic to D2p. It is easy to see that C2q4p is bipartite
with partite sets U = G∗ and U ′ = bG∗. Set X = C2q4p and A = Aut(X). Let A∗ be the
subgroup of A fixing the partite sets U and U ′ of X setwise. Then A∗ is transitive on both
U and U ′ and |A : A∗| = 2, implyingA∗CA. SinceR(G) ≤ A, we haveA = 〈A∗, R(b)〉.
Set R(G∗) = {R(g) | g ∈ G∗}. Then R(G∗) ≤ R(G), and R(G∗) fixes U and U ′ setwise.
Hence, R(G∗) ≤ A∗. Since X has valency 2q and 4q | (p−1), one has |A| = 4pqn, where
each prime divisor of n is less than p.
Claim 1: A is solvable.
Suppose that A∗ is 2-transitive in its action on U . Then U \ {1} = N2(1), the set
of vertices having distance 2 from 1. Every vertex in U \ {1} has the same number of
neighbors in N1(1), say k. Then (2p− 1)k = 2q(2q − 1), and since (2p− 1, 2q) = 1, one
has k = 2q, implying p = q, contrary to the fact that 4q | p−1. Since q ≥ 5 and 4q | p−1,
we have p ≥ 29, and by Proposition 2.6, A∗ is imprimitive on U .
LetB be a non-trivial block ofA∗ on U containing 1. ThenB is also a block ofR(G∗),
forcing thatB is a subgroup ofG∗. Thus, Σ1 = {Bg | g ∈ G∗} is a complete block system
of A∗ on U . Furthermore, Σ2 = Σ
R(b)
1 = {Bg | g ∈ bG∗} is a complete block system of
A∗ on U ′. Recall that A = 〈A∗, R(b)〉. It follows that Σ = Σ1 ∪ Σ2 = {Bg | g ∈ G} is
a complete block system of A on V (X). Then the quotient graph XΣ is bipartite and the
edge-transitivity of X implies that XΣ is edge-transitive. Let K be the kernel of A acting
on Σ. Since |U | = 2p, one has |B| = p or 2.
Assume |B| = p. Then |Σ| = 4 and since XΣ is bipartite, XΣ is a 4-cycle, say XΣ =
(B0, B1, B2, B3) with B0 = B. The induced subgraph 〈Bi, Bi+1〉 of Bi ∪Bi+1 in X has
order 2p and valency q, which cannot be isomorphic to Kp,p because 4q | (p − 1). Since
p ≥ 29, Proposition 2.2 implies that 〈Bi, Bi+1〉 ∼= G(2p, q) and |Aut(〈Bi∪Bi+1〉)| = 2pq.
Any Sylow p-subgroup of A is a subgroup of K, and since |Bi| = p, K is primitive on
Bi. Suppose that K is unfaithful on Bi. Then the kernel of K on Bi is transitive on Bi+1,
forcing 〈Bi, Bi+1〉 ∼= Kp,p, a contradiction. Thus, K acts faithfully on Bi, implying that
X. Wang and Y.-Q. Feng: Half-arc-transitive graphs of order 4p of valency twice a prime 157
|K| is a divisor of |Aut(〈Bi ∪ Bi+1〉)| = 2pq. This means that K is solvable. Also, A/K
is solvable because A/K ≤ Aut(XΣ) ∼= D8. It follows that A is solvable.
Assume |B| = 2. ThenK is a 2-group and hence solvable. By Proposition 2.2, XΣ is a
connected symmetric graph of order 2p. Let XΣ be of valency d and let k be the number of
edges betweenB andBb inX (note thatB is indeed adjacent toBb inXΣ). Clearly, k ≤ 4
and 2 · 2q = k · d. Since q ≥ 5, k 6= 3 and d = q, 2q, or 4q. Recall that 4q | (p − 1) and
p ≥ 29. If 4q < p− 1, or 4q = p− 1 and XΣ has valency q or 2q then, by Proposition 2.2,
XΣ ∼= G(2p, q), G(2p, 2q) orG(2p, 4q) and Aut(XΣ) ∼= (ZpoZq)oZ2, (ZpoZ2q)oZ2
or (Zp o Z4q) o Z2. In all these cases, Aut(XΣ) is solvable, and since A/K ≤ Aut(XΣ),
A/K is solvable. Thus, A is solvable. Now one may assume 4q = p − 1 and XΣ has
valency 4q. In this case, XΣ ∼= Kp,p − pK2 and there is exactly one edge in X between B
and Bb, which forces that K = 1. It follows that A∗ is faithful on Σ1. Since |Σ1| = p, by
Proposition 2.6 either soc(A∗) is non-solvable and 2-transitive on Σ1 or A∗ is solvable.
Suppose that soc(A∗) is non-solvable and 2-transitive on Σ1. If soc(A∗) is not 3-
transitive on Σ1 then, by Proposition 2.6, soc(A∗) ∼= PSL(m, r) because p ≥ 29. In this
case, m ≥ 3 is an odd number, r is a prime-power and p = 1 + r + r2 + · · · + rm−1.
Since m is odd, r(1 + r) | (p − 1), which is impossible because p − 1 = 4q and q ≥ 5.
Thus, soc(A∗) is 3-transitive and henceA∗ is 3-transitive on Σ1. SinceXΣ ∼= Kp,p−pK2,
one may let Σ1 = {Bi | i ∈ Zp} and Σ2 = {B′i | i ∈ Zp} such that for k, ` ∈ Zp, Bk
is adjacent to B′` in XΣ if and only if k 6= `. Note that for i, j ∈ Zp and i 6= j, there is
exactly one edge in X between Bi and B′j . This implies that for any α ∈ A, we have: if α
fixes Bi and B′j setwise then it fixes every vertex in Bi and B
′
j because |Bi| = |B′j | = 2.
Let H be the subgroup of A∗ fixing B0 and B1. Clearly, H fixes B′0 and B
′
1, and hence H
fixes every vertex in B0 ∪B1 ∪B′0 ∪B′1. By the 3-transitivity of A∗ on Σ1, H is transitive
on Σ1\{B0, B1}, forcing that X has valency p − 1 = 4q, contrary to the fact that X has
valency 2q. Thus, A∗ is solvable, and since |A : A∗| = 2, A is solvable. This completes
the proof of Claim 1.
Let P be a Sylow p-subgroup ofA. Then |P | = p because |A| = 4pqn with each prime
divisor of n less than p.
Claim 2: P EA.
Let N be a minimal normal subgroup of A. By Claim 1, N is elementary abelian, and
since |V (X)| = 4p, N is a p-group or a 2-group. IfN is a p-group then P = NEA. Thus,
one may assume that N ∼= Zr2 for some positive integer r. Since q ≥ 5, the quotient graph
XN of X relative to N has valency q or 2q. Let H be the kernel of A acting on V (XN ).
Note that XN is edge-transitive and the orbits of N are of length 2 or 4.
Suppose first that the orbits of N have length 2. Then H is a 2-group and |XN | = 2p.
If XN has valency q, by Proposition 2.2, XN ∼= G(2p, q) is a q-valent symmetric graph
of order 2p and A/H ≤ Aut(G(2p, q)) = (Zp o Zq) o Z2. Note that G = G3(p) and
G has no non-trivial normal 2-subgroup. This implies that R(G) ∩ H = 1 and R(G) =
R(G)/(R(G) ∩H) ∼= R(G)H/H ≤ A/H . Since 4 | |R(G)|, one has 4 | |Aut(G(2p, q))|,
a contradiction. Thus, XN has valency 2q. In this case, H1 = 1 and H = N ∼= Z2.
By Proposition 2.2, XN ∼= G(2p, 2q) and Aut(XN ) has a normal Sylow p-subgroup. It
follows that PH/H E A/H , that is PH E A. Since |H| = 2, P is characteristic in PH
and hence P EA.
Suppose now that the orbits of N have length 4. Then |XN | = p and XN cannot have
valency q. Thus, XN has valency 2q. In this case, H1 = 1 and H = N ∼= Z22. Since
158 Ars Math. Contemp. 3 (2010) 151–163
4q | (p− 1), XN cannot be a complete graph, implying that A/H cannot be 2-transitive on
V (XN ). By Proposition 2.6,A/H has a normal Sylow p-subgroup, that is PH/HEA/H .
Since |H| = 4, P is characteristic in PH and hence P E A. This completes the proof of
Claim 2.
Now we are ready to finish the proof by showing Aut(C2q4p) = R(G)o〈α〉. By Claim 2,
P E A. Since X is bipartite, the quotient graph XP of X relative to P is a 4-cycle, say
XP = (O0, O1, O2, O3). Recall that α is the automorphism of G induced by a 7→ as and
b 7→ ab. Since α has order q, α fixes Oi setwise for each i ∈ Z4. The induced subgraph
〈Oi, Oi+1〉 of Oi ∪ Oi+1 in X , i ∈ Z4, is a q-valent edge-transitive graph of order 2p. By
Proposition 2.2, 〈Oi, Oi+1〉 ∼= G(2p, q) and Aut(G(2p, q)) ∼= (Zp o Zq) o Z2. Let L be
the kernel of A acting on V (XP ). Then P ≤ L and α ∈ L, implying that pq | |L|. Since
|Oi| = p (i ∈ Z4), L is primitive on Oi. Suppose that L is unfaithful on Oi. Then the
kernel of L on Oi is transitive on Oi+1 because L is primitive on Oi+1, which implies that
the induced subgraph 〈Oi, Oi+1〉 is isomorphic to Kp,p. It follows that 2p = 2q, contrary
to the fact that 4q | (p−1). Thus, L acts faithfully onOi and L . Aut(G(2p, q)), implying
|L| | 2pq. It follows that |L| = pq because L is intransitive on Oi ∪ Oi+1. Note that
A/L ≤ Aut(XP ) ∼= D8. Since R(G) o 〈α〉 ≤ A, one has |A| = 4pq or 8pq.
Suppose |A| = 8pq. Then A/L = Aut(XP ) ∼= D8 and consequently X is sym-
metric. Furthermore, R(G) o 〈α〉 E A and |A1| = 2q. Noting that R(G) is charac-
teristic in R(G) o 〈α〉, we have R(G) E A. By Proposition 2.1, A1 = Aut(G,S) and
hence Aut(G,S) = 〈α, β〉, where β is an involution in Aut(G,S). Recall that T =
{b, as[1]b, as[2]b, as[3]b, . . . , as[q−1]b}, S = T ∪T−1, X = Cay(G,S), and α permutes the
elements of T cyclically. The arc-transitivity of A implies that β interchanges T and T−1.
Thus, there is an i such that bβα
i
= b−1, and since 〈a〉 is characteristic in G, aβαi = at
for some t ∈ Z∗p. Note that G has an automorphism mapping at to a and b to b. It fol-
lows that G has an automorphism γ such that aγ = a and bγ = b−1. Since b−1ab = ar,
one has b−1ab = bab−1, that is b2a = ab2, a contradiction. Thus, |A| = 4pq and hence
A = R(G) o 〈α〉, as required.
To finish the proof of Theorem 3.1, we only need to prove the following lemma.
Lemma 3.3. Let p, q be odd primes with q ≥ 5 and letX be a 2q-valent half-arc-transitive
graph of order 4p. Then 4q | (p− 1) and X ∼= C2q4p .
Proof. Since q ≥ 5, X has valency at least 10, and since there are no half-arc-transitive
graphs of order p or 2p (see [5, 6]), X is connected. Let A = Aut(X). Recall that X is
an underlying graph of an orbital digraph O := (V (X), O) of A for some non-self-paired
orbital O. Thus, A = Aut(O) and O is a directed graph with out- and in-valency equal
to q. Furthermore, A is transitive on the directed edges of O. Since V (X) = V (O) and
A = Aut(O), in what follows we change the graph X to O when it is necessary. Let
u ∈ V (X) and denote by Au the stabilizer of u in A. Since |N+(u)| = |N−(u)| = q is a
prime, Au acts on N+(u) and N−(u) primitively, and there exits α ∈ Au such that α has
order q and permutes the elements in N+(u) cyclically. This implies that 4pq | |A|, and by
Proposition 2.4, p ≥ 7. By Proposition 2.3, |A| = 4pqn, where each prime divisor of n is
a divisor of q!. Let P be a Sylow p-subgroup of A.
Assume that A has a non-trivial normal 2-subgroup, say N . Then the orbits of N have
length 2 or 4 and since q ≥ 5, the quotient graph XN of X relative to N has valency
q or 2q. Let L be the kernel of A acting on V (XN ). Then L E A and N ≤ L. Since
X. Wang and Y.-Q. Feng: Half-arc-transitive graphs of order 4p of valency twice a prime 159
|V (X)| = 4p and N is a 2-group, by Propostion 2.2, XN is a symmetric graph of order p
or 2p.
Suppose first that the orbits of N have length 4. Then |XN | = p, XN cannot have
valency q. Thus, XN has valency 2q and the stabilizer Lu of u in L fixes the neighborhood
of u in X pointwise because X and XN have the same valency. Thus, Lu = 1 and L acts
regularly on each orbit of N , forcing |L| = 4. It follows that PL is regular on V (X) and
hence X is a Cayley graph on PL. However, Lemma 3.2 implies that L cannot be normal
in PL, a contradiction.
Suppose now that the orbits of N are of length 2. Then |XN | = 2p. By Proposition
2.2, XN is symmetric. If XN has valency q then X ∼= XN [2K1], which is symmetric, a
contradiction. Thus, XN has valency 2q. In this case, Lu = 1 and L acts regularly on
each orbit of N . It follows that L = N ∼= Z2 and the quotient graph XN of X relative
to N is A/N -half-arc-transitive. Note that PN/N is a Sylow p-subgroup of A/N . By
Proposition 2.2, PN E A or XN ∼= Kp,p − pK2 with 2q = p − 1. For the former,
P E A because |N | = 2. For the latter, XN is bipartite. Let (A/N)∗ be the subgroup of
A/N fixing the bipartite sets, say Σ and Σ′, of XN setwise. Since XN ∼= Kp,p − pK2,
(A/N)∗ acts faithfully on Σ and Σ′, respectively. Since A/N is vertex-transitive on XN ,
|A/N : (A/N)∗| = 2. Assume ∆ ∈ Σ and ∆′ ∈ Σ′ such that ∆ is not adjacent to ∆′ in
XN , and let (A/N)∆ and (A/N)∆′ be the stabilizers of ∆ and ∆′ in A/N , respectively.
Then (A/N)∆ = (A/N)∆′ ≤ (A/N)∗. By the half-arc-transitivity of A/N on XN ,
the stabilizer (A/N)∆ cannot be transitive on Σ′\{∆′}. It follows that (A/N)∗ is not 2-
transitive on Σ and Σ′. Since |Σ| = |Σ′| = p, by Proposition 2.6, (A/N)∗ has a normal
Sylow p-subgroup, which is also a normal Sylow p-subgroup of A/N because |A/N :
(A/N)∗| = 2. It follows that PN EA and hence P EA.
Now assume thatA has no non-trivial normal 2-subgroups. Again we prove that PEA.
Suppose that A is primitive on V (X). By Proposition 2.6, A is 2-transitive on V (X)
provided p 6= 7, 13 or 17. Since X is half-arc-transitive, A is not 2-transitive on V (X). It
follows that p = 7, 13 or 17, which is impossible by Proposition 2.4. Thus,A is imprimitive
on V (X). Let B be a non-trivial block of A on V (X). Since |V (X)| = 4p, we have
|B| = 2, 4, p or 2p. It follows that B = {Ba | a ∈ A} is a complete block system of A
on V (X). Consider the quotient graph XB relative to B and let K be the kernel of A on
B. Then A/K ≤ Aut(XB). Since X is A-edge-transitive, XB is A/K-edge-transitive. Let
B′ ∈ B be adjacent to B in XB and let k be the number of edges in X between B and
B′. Then XB has valency
2q|B|
k . Assume u ∈ B and choose B
′ ∈ B such that B′ contains
an out-neighbor of u in O. Recall that α ∈ Au and α permutes the elements in N+(u)
cyclically. Then α fixes B setwise. Since |N+(u)| = q is a prime, either B′ contains
exactly one out-neighbor of u in O or N+(u) ⊆ B′. In particular, if |B| = p or 2p then
N+(u) ⊆ B′ and α ∈ K. If |B| = 2 or 4 then B′ contains exactly one out-neighbor of u
inO. It follows Ku = 1 because Ku fixes each out-neighbor of u inO. Thus, |K| ≤ 4 and
since |V (X)| = 4p and A has no non-trivial normal 2-group, one has K = 1, implying
A ≤ Aut(XB). Consider the four cases |B| = 4, 2, p or 2p, respectively.
Case I: |B| = 4.
In this case, K = 1 and A ≤ Aut(XB). Since |B| = p, Proposition 2.6 implies that
either P E A or A is 2-transitive on B. First suppose that A is 2-transitive on B. Then
XB ∼= Kp and (p − 1)k = 4 · 2q = 8q, where k is the number of edges in X between
B and B′. It follows that k = 1, 2 or 4. The 2-transitivity of A on B implies that the
number of directed edges in O with direction from B to B′ is equal to that of directed
160 Ars Math. Contemp. 3 (2010) 151–163
edges with direction from B′ to B. Thus, half-arc-transitivity of X implies that k 6= 1,
and hence k = 2 or 4 and p − 1 = 2q or 4q. Again by Proposition 2.6, soc(A) ∼= Ap,
PSL(2, 22
s
) with p = 22
s
+ 1, PSL(m, r) with p = r
m−1
r−1 , PSL(2, 11), M11 or M23. If
soc(A) ∼= Ap then |A| is divisible by 12p!. From elementary number theory it is well known
that there exists a prime t between q and 2q. Since p − 1 = 2q or 4q, one has t | |A|,
which is impossible because |A| = 4pqn where each prime divisor of n is a divisor of q!.
If soc(A) ∼= PSL(2, 22
s
) then p − 1 = 22s 6= 2q or 4q, which is clearly impossible. If
soc(A) ∼= PSL(m, r) then p = r
m−1
r−1 = r
m−1 + · · ·+ r + 1 and m ≥ 3 is odd. It follows
that r(1 + r) | (p − 1), which is also impossible because p − 1 = 2q or 4q and q ≥ 5
(note that 4 · 5 + 1 = 21 is not a prime). Thus, soc(A) ∼= PSL(2, 11), M11 or M23. It
follows that A ∼= PSL(2, 11), PGL(2, 11), M11 or M23 because |Out(PSL(2, 11))| = 2,
and |Out(M11)| = |Out(M23)| = 1 (see [8]). Note that |PSL(2, 11)| = 22 · 3 · 5 · 11,
|M11| = 24 · 32 · 5 · 11 and |M23| = 27 · 32 · 5 · 7 · 11 · 23. Since |V (X)| = 4p, if
A ∼= PSL(2, 11) or PGL(2, 11) then Au is a subgroup of A of order 15 or 30, respectively,
which is not true by [8]. Similarly, A 6∼= M11 or M23 because M11 and M23 have no
subgroups of order 22 · 32 · 5 and 25 · 32 · 5 · 7 · 11, respectively. Thus, P EA.
Case II: |B| = 2.
In this case K = 1 and A ≤ Aut(XB). By Proposition 2.2, XB is symmetric. Let
B = {u, v} and B′ = {u′, v′}, and assume that (u, u′) is a directed edge in O. Suppose
that u has two neighbors inX inB′. SinceB′ contains exactly one out-neighbor of u inO,
(v′, u) is a directed edge in O. Since A is transitive on the directed edges in O, A contains
an element mapping (u, u′) to (v′, u), forcing that k ≥ 3. Recall thatXB has valency 2q|B|k .
It follows that k = 4 and hence X = XB[2K1], which is symmetric, a contradiction. Thus,
there are exactly 2 or 1 edges in X between B and B′, meaning that XB has valency 2q
or 4q. Recall that A has no non-trivial normal 2-subgroup and then, by Proposition 2.2,
P EA, or XB ∼= Kp,p−pK2 with p−1 = 2q or 4q, or XB ∼= G(2, p, 2q) with p−1 = 2q.
Assume that XB ∼= Kp,p − pK2. Let B1 and B2 be the bipartite sets of XB such that
B ∈ B1 and B′ ∈ B2, and let A∗ be the subgroup of A fixing B1 and B2 setwise. Then
|A : A∗| = 2 and A∗ E A. There is a unique block C ∈ B2 which is not adjacent to B
in XB. Let AB and AC be the block stabilizers of B and C in A, that is the subgroups of
A fixing B and C, respectively. Then AB = AC ≤ A∗ and A∗ is faithful on B1 and B2.
Clearly, P ≤ A∗. By Proposition 2.6, P EA∗ or A∗ is 2-transitive on B1 and B2.
Suppose that A∗ is 2-transitive on B1 and B2. Then AB = AC is transitive on B1\{B}
and B2\{C}, respectively. Note that p− 1 = 4q or 2q. If p− 1 = 4q then there is exactly
one directed edge in O between B and B′. Since X is half-arc-transitive, AB has two
orbits on B2\{C}, a contradiction. Thus, p − 1 = 2q and there are exactly two directed
edges in O between B and B′. These two directed edges have different direction, that is
one from B to B′ and the other from B′ to B, because AB is transitive on B2\{C}. This
means that the permutation β on V (X) interchanging the two vertices in each block of B
cannot be an automorphism ofO. On the other hand, since the induced subgraph 〈B ∪B′〉
of B ∪B′ in X is a matching, one has β ∈ A, contrary to the fact that A ≤ Aut(O). Thus,
P EA∗ and hence P EA.
Assume that XB ∼= G(2, p, 2q) with p− 1 = 2q. By Proposition 2.2, Aut(XB) ∼= Zp2 o
Sp. Thus, Aut(XB) has a normal subgroupN such thatN ∼= Zp2. Since |V (XB)| = 2p, the
orbits of N have size 2, which are blocks of Aut(XB) (in fact XB ∼= Kp[2K1]). It follows
that A has blocks of size 4 on V (X). By Case I, P EA.
X. Wang and Y.-Q. Feng: Half-arc-transitive graphs of order 4p of valency twice a prime 161
Case III: |B| = p.
In this case, N+(u) ⊆ B′ and α ∈ K, where α ∈ Au permutes the elements in N+(u)
cyclically. Since |V (XB)| = 4, any element of order p in A fixes each vertex of XB. Thus,
pq | |K| andXB is a 4-cycle, say V (XB) = (B0, B1, B2, B3), whereBi is adjacent toBi+1
for each i ∈ Z4. Let Y = 〈B0 ∪ B1〉 be the subgraph induced by B0 ∪ B1 in X . Then Y
is a q-valent edge-transitive graph of order 2p, and all edges in Y have the same direction
either from B0 to B1 or from B1 to B0 in O, forcing A/K ∼= Z4. If Y ∼= Kp,p then
X ∼= C4[pK1], which is symmetric, a contradiction. One may thus assume that Y 6∼= Kp,p.
If K is unfaithful on B0 then the kernel of K on B0 is transitive on B1 because |B1| = p,
forcing that Y ∼= Kp,p, a contradiction. Thus, K ≤ Aut(Y ). By Proposition 2.2, either
Y ∼= G(2p, q) with q | (p − 1) and (p, q) 6= (11, 5), and Aut(Y ) ∼= (Zp o Zq) o Z2, or
Y ∼= G(2 · 11, 5) and Aut(Y ) ∼= PSL(2, 11) o Z2. For the former, since K fixes B0 and
B1 setwise and pq | |K|, one has |K| = pq. For the latter, Y ∼= G(2 · 11, 5) and Aut(Y ) ∼=
PSL(2, 11) o Z2. Since K fixes B0 and B1 setwise, K . PSL(2, 11). Suppose that K ∼=
PSL(2, 11). Let CA(K) be the centralizer of K in A. Then CA(K)∩K = 1 because K is
non-abelian simple. Since CA(K) ∼= CA(K)/(K ∩ CA(K)) ∼= KCA(K)/K ≤ A/K ∼=
Z4 and A has no non-trivial normal 2-subgroup, one has CA(K) = 1. It follows that
A = A/CA(K) . Aut(PSL(2, 11)) ∼= PGL(2, 11), contrary to the fact that A/K ∼= Z4.
This implies that K is isomorphic to a proper subgroup of PSL(2, 11) of order divisible by
55. By [8], |K| = 55. Thus, we always have |K| = pq. Since |A| = 4pq, P E K and
hence P EA.
Case IV: |B| = 2p.
In this case, X is a bipartite graph with bipartite sets B and B′. If p = q, then X ∼=
K2p,2p is symmetric, a contradiction. Thus q < p. Let A∗ be the subgroup of A fixing B
andB′ setwise. Then |A : A∗| = 2 andA∗EA. Suppose thatA∗ is 2-transitive onB. Then
every vertex in B \ {u} has the same number of in-neighbors and the same number of out-
neighbors in N+(u) in O, say m1 and m2, respectively. It follows that q2 = (2p − 1)m1
and q(q − 1) = (2p − 1)m2, forcing m1 −m2 = 1 and q = 2p − 1, contrary to the fact
that q < p. Since |B| = 2p, by Proposition 2.6 A∗ is imprimitive on B. Note that A∗ is the
block stabilizer of B in A. Thus, every non-trivial block of A∗ on B is also a block of A,
which has size 2 or p. By Case II and Case III, P EA.
Now we are ready to prove the lemma. Since P E A, the orbits of P are blocks of A
of length p. By the proof in Case III, |A| = 4pq and hence A is solvable. Clearly, the
Hall {2, p}-subgroup of A acts regularly on V (X), implying that X is a Cayley graph. By
Lemma 3.2, one may let X = Cay(G,S), where G = 〈a, b | ap = b4 = 1, b−1ab = ar〉
with r2 ≡ −1 (mod p). Thus, 4q | (p− 1).
Since P E A, one has P ≤ R(G). Set C = CA(P ). Since R(G) is a Hall {2, p}-
subgroup of A and there is no involution in G commuting with a, C is a p-group or a
{q, p}-group. If q | |C| then |C| = pq and C has a normal Sylow q-subgroup, which is
normal in A because C EA. This implies that AuEA, forcing Au = 1, a contradiction. It
follows that P = C, and sinceA/C = A/P . Aut(P ) ∼= Zp−1, one hasR(G)/P EA/P ,
that is R(G) E A. Thus X = Cay(G,S) is a normal Cayley graph. By Proposition 2.1,
A = R(G)oAut(G,S) and since |A| = 4pq, there is an element δ of order q in Aut(G,S).
The connectivity of X implies 〈S〉 = G. Since G has a unique normal subgroup of
order 2p, S contains elements of order 4. Each element of order 4 inG is of the form aib or
aib−1 for some integer i. It is easy to see that Aut(G) is transitive on the set {aib | 0 ≤ i ≤
162 Ars Math. Contemp. 3 (2010) 151–163
p − 1}. Thus, one may assume b ∈ S. Since there is no automorphism of G mapping b to
b−1, one may further assume aδ = as and bδ = akb (s 6= 0 (mod p)), and since Sδ = S, we
have that E = {b, bδ, . . . , bδq−1} = {b, aks[1]b, aks[2]b, aks[3]b, . . . , aks[q−1]b} is a subset
of S, where s[i] = si−1 + · · ·+ s+ 1 for 1 ≤ i ≤ q − 1 and s[q] = sq−1 + · · ·+ s+ 1 =
0 (mod p). It follows that S = E ∪E−1 because S = S−1. Clearly, k 6= 0 (mod p). Since
the map a 7→ ak, b 7→ b, induces an automorphism of G, we have X ∼= C2q4p .
4 Acknowledgements
This work was supported by the National Natural Science Foundation of China (10871021,
10911140266), the Fundamental Research Funds for the Central Universities (2009YJS036)
and the Doctorate Foundation of Beijing Jiaotong University (141109522).
References
[1] B. Alspach, D. Marušič, L. Nowitz, Constructing graphs which are 1/2-transitive, J. Austral.
Math. Soc. A 56 (1994), 391–402.
[2] B. Alspach, M. Y. Xu, 1/2-transitive graphs of order 3p, J. Algebraic Combin. 3 (1994), 347–
355.
[3] N. Biggs, Algebraic Graph Theory, second edn., Cambridge University Press, Cambridge,
1993.
[4] I. Z. Bouwer, Vertex and edge-transitive but not 1-transitive graphs, Canad. Math. Bull. 13
(1970), 231–237.
[5] C. Y. Chao, On the classification of symmetric graphs with a prime number of vertices, Trans.
Amer. Math. Soc. 158 (1971), 247–256.
[6] Y. Cheng, J. Oxley, On weakly symmetric graphs of order twice a prime, J. Combin. Theory B
42 (1987), 196–211.
[7] M. D. E. Conder, D. Marušič, A tetravalent half-arc-transitive graph with non-abelian vertex
stabilizer, J. Combin. Theory B 88 (2003), 67–76.
[8] H. J. Conway, R. T. Curtis, S. P. Norton, R. A. Parker, R. A Wilson, Atlas of Finite Groups,
Oxford University Press, Oxford, 1985.
[9] J. D. Dixon, B. Mortimer, Permutation Groups, Springer-Verlag, New York, 1996.
[10] S. F. Du, M. Y. Xu, Vertex-primitive 1/2-arc-transitive graphs of smallest order, Comm. Alge-
bra 27 (1999), 163–171.
[11] X. G. Fang, C. H. Li, M. Y. Xu, On edge-transitive Cayley graphs of valency 4, European J.
Combin. 25 (2004), 1107–1116.
[12] Y.-Q. Feng, J. H. Kwak, M. Y. Xu, J.-X. Zhou, Tetravalent half-arc-transitive graphs of order
p4, European J. Combin. 29 (2008), 555–567.
[13] Y.-Q. Feng, K. S. Wang, C. X. Zhou, Tetravalent half-transitive graphs of order 4p, European
J. Combin. 28 (2007), 726–733.
[14] Y.-Q. Feng, J. H. Kwak, C. X. Zhou, Constructing even radius tightly attached half-arc-
transitive graphs of valency four, J. Algebraic Combin. 26 (2007), 431–451.
[15] D. F. Holt, A graph which is edge-transitive but not arc transitive, J. Graph Theory 5 (1981),
201–204.
[16] J. H. Kwak, Y. S. Kwon, J. M. Oh, Infinitely many one-regular Cayley graphs on dihedral
groups of any prescribed valency, J. Combin. Theory B 98 (2008), 585–598.
X. Wang and Y.-Q. Feng: Half-arc-transitive graphs of order 4p of valency twice a prime 163
[17] C. H. Li, Z. P. Lu, D. Marušič, On primitive permutation groups with small suborbits and their
orbital graphs, J. Algebra 279 (2004), 749–770.
[18] C. H. Li, H. S. Sim, On half-transitive metacirculant graphs of prime-power order, J. Combin.
Theory B 81 (2001), 45–57.
[19] H. L. Li, J. Wang, L. Y. Wang, M. Y. Xu, Vertex primitive graphs of order containing a large
prime factor, Comm. in Algebra 22 (1994) 3449–3477.
[20] M. W. Liebeck, J. Saxl, Primitive permutation groups containing an element of large prime
order, J. London. Math. Soc. 31 (1985), 237–249.
[21] A. Malnič, D. Marušič, Constructing 4-valent 1/2-transitive graphs with a nonsolvable auto-
morphism group, J. Combin. Theory B 75 (1999), 46–55.
[22] A. Malnič, D. Marušič, Constructing 1/2-arc-transitive graphs of valency 4 and vertex stabi-
lizer Z2 × Z2, Discrete Math. 245 (2002), 203–216.
[23] D. Marušič, Half-transitive group actions on finite graphs of valency 4, J. Combin. Theory B
73 (1998), 41–76.
[24] D. Marušič, Quartic half-transitive graphs with large vertex stabilizers, Discrete Math. 299
(2005), 180–193.
[25] D. Marušič, R. Nedela, Finite graphs of valency 4 and girth 4 admitting half-transitive group
actions, J. Austral. Math. Soc. 73(2002) 155-170.
[26] D. Marušič, R. Nedela, Partial line graph operator and half-arc-transitive group actions, Math.
Slovaca 51 (2001), 241–257.
[27] D. Marušič, C. E. Praeger, Tetravalent graphs admitting half-transitive group action: alternating
cycles, J. Combin. Theory B 75 (1999), 188–205.
[28] D. Marušič, P. Šparl, On quartic half-arc-transitive metacirculants, J. Algebraic Combin. 28
(2008), 365–395.
[29] D. Marušič, A. Waller, Half-transitive graphs of valency 4 with prescribed attachment numbers,
J. Graph Theory 34 (2000), 89–99.
[30] D. Marušič, M. Y. Xu, A 1
2
-transitive graph of valency 4 with a nonsolvable group of automor-
phisms, J. Graph Theory 25 (1997), 133–138.
[31] C. E. Praeger, M. Y. Xu, Vertex primitive graphs of order a product of two distinct primes, J.
Combin. Theory B 59 (1993), 245–266.
[32] P. Šparl, A classification of tightly attached half-arc-transitive graphs of valency 4, J. Combin.
Theory B 98 (2008), 1076–1108.
[33] D. E. Taylor, M.Y. Xu, Vertex-primitive 1/2-transitive graphs, J. Austral. Math. Soc. A 57
(1994), 113–124.
[34] W. T. Tutte, Connectivity in Graphs, University of Toronto Press, Toronto, 1966.
[35] R. J. Wang, Half-transitive graphs of order a product of two distinct primes, Comm. Algebra 22
(1994), 915–927.
[36] R. J. Wang, M. Y. Xu, A classification of symmetric graphs of order 3p, J. Combin. Theory B
58 (1993), 197–216.
[37] X. Y. Wang, Y.-Q. Feng, Hexavalent half-arc-transitive graphs of order 4p, European J. combin.
30 (2009), 1263–1270.
[38] M. Y. Xu, Half-transitive graphs of prime-cube order, J. Algebraic Combin. 1 (1992), 275–282.
[39] C. X. Zhou, Y.-Q. Feng, An infinite family of tetravalent half-arc-transitive graphs, Discrete
Math. 306 (2006), 2205–2211.