Abstract

A Cayley graph of a finite group 𝐺 is called normal edge transitive if its automorphism group has a subgroup which both normalizes 𝐺 and acts transitively on edges. In this paper we determine all cubic, connected, and undirected edge-transitive Cayley graphs of dihedral groups, which are not normal edge transitive. This is a partial answer to the question of Praeger (1999).

1. Introduction

Let 𝐺 be a finite group, and let 𝑆 be a subset of 𝐺 such that 1𝐺𝑆. The Cayley graph 𝑋= Cay(𝐺,𝑆) of 𝐺 on 𝑆 is defined as the graph with a vertex set 𝑉(𝑋)=𝐺 and edge set 𝐸(𝑋)={{𝑔,𝑠𝑔}𝑔𝐺,𝑠𝑆}. Immediately from the definition there are three obvious facts (1) Aut(𝑋), the automorphism group of 𝑋, contains the right regular representation 𝑅(𝐺) of 𝐺; (2) 𝑋 is connected if and only if 𝐺=𝑆; (3) 𝑋 is undirected if and only if 𝑆=𝑆1.

A part of Aut(𝑋) may be described in terms of automorphisms of 𝐺, that is, the normalizer 𝑁Aut(𝑋)(𝐺)=𝐺Aut(𝐺,𝑆), a semidirect product of 𝐺 by Aut(𝐺,𝑆), where Aut(𝐺,𝑆)={𝜎Aut(𝐺)𝑆𝜎=𝑆}.

We simply use 𝐴(𝑋) to denote the arc set of 𝑋. A Cayley graph 𝑋= Cay(𝐺,𝑆) is said to be vertex transitive, edge transitive, and arc transitive if its automorphism group Aut(𝑋) is transitive on the vertex set 𝑉(𝑋), edge set 𝐸(𝑋), and arc set 𝐴(𝑋), respectively. For 𝑠1, an 𝑠-arc in a graph 𝑋 is an ordered (𝑠+1)-tuple (𝑣0,𝑣1,,𝑣𝑠) of vertices of 𝑋 such that 𝑣𝑖1 is adjacent to 𝑣𝑖 for 1𝑖𝑠 and 𝑣𝑖1𝑣𝑖+1 for 1𝑖<𝑠 in other words, a directed walk of length 𝑠 which never includes a backtracking. 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 subgroup of the automorphism group of a graph 𝑋 is said to be 𝑠-regular if it acts regularly on the set of 𝑠-arcs of 𝑋.

It is difficult to find the full automorphism group of a graph in general, and so this makes it difficult to decide whether it is edge-transitive, even for a Cayley graph. As an accessible kind of edge transitive graphs, Praeger [1] focuses attention on those graphs for which 𝑁Aut(𝑋)(𝐺) is transitive on edges, and such a graph is said to be normal edge transitive. By the definition, every normal edge-transitive Cayley graph is edge-transitive, but not every edge-transitive Cayley graph is normal edge-transitive.

Independently for our investigation, and as another attempt to study the structure of finite Cayley graphs, Xu [2] defined a Cayley graph 𝑋= Cay(𝐺,𝑆) to be normal if 𝑅(𝐺) is normal subgroup of the full automorphism group Aut(𝑋). Xu's concept of normality for a Cayley graph is a very strong condition. For example, 𝐾𝑛 is normal if and only if 𝑛<4. However any edge-transitive Cayley graph which is normal, in the sense of Xu's definition, is automatically normal edge transitive.

Praeger posed the following question in [1]: what can be said about the structure of Cayley graphs which are edge transitive but not normal edge transitive? In [3], Alaeiyan et al. have given partial answer to this question for abelian groups of valency at most 5, and also Sim and Kim [4] determined normal edge-transitive circulant graphs. In the next theorem we will identify all cubic edge transitive Cayley graphs of dihedral group which are not normal edge-transitive. This is a partial answer to Question 5 of [1]. Throughout of this paper, we suppose that 𝐷2𝑛=𝑎,𝑏𝑎𝑛=𝑏2=1,𝑏𝑎𝑏1=𝑎1, and 𝑋= Cay(𝐷2𝑛,𝑆) is connected and undirected cubic Cayley graph. The main result of this paper is the following theorem.

Theorem 1.1. Let 𝐺=𝐷2𝑛 be a dihedral group, and let 𝑋= Cay(G,S) be a connected cubic Cayley graph. If 𝑋 is an edge-transitive Cayley graph but is not normal edge transitive, then 𝑋, 𝐺 satisfy one of the following: (1)𝑛=4, 𝑆={𝑏,𝑎𝑏,𝑎2𝑏}, 𝑋𝐾4,44𝐾2;(2)𝑛=8, 𝑆={𝑏,𝑎𝑏,𝑎3𝑏}, 𝑋𝑃(8,3), the generalized Peterson graph.

2. Basic Facts

In this section we give some facts on Cayley graphs, which will be useful for our purpose. First we make some comments about the normalizer 𝑁Aut(𝑋)(𝐺) of the regular subgroup 𝐺. As before, the normalizer of the regular subgroup 𝐺 in the symmetric group Sym(𝐺) is the holomorph of 𝐺, that is, the semidirect product 𝐺Aut(𝐺). Thus,𝑁(Aut(𝑋))(𝐺)=(𝐺Aut(𝐺))Aut(𝑋)=𝐺(Aut(𝐺)Aut(𝑋))=𝐺Aut(𝐺,𝑆).(2.1)

The following lemmas are basic for our purpose. Now we have the first lemma from [1].

Lemma 2.1 (see [1, Proposition 1]). Let 𝑋= Cay(G,S) be a Cayley graph for a finite group 𝐺. Then 𝑋 is normal edge transitive if and only if Aut(𝐺,𝑆) is either transitive on 𝑆 or has two orbits in 𝑆 which are inverse of each other.

Lemma 2.2 (see [2, Proposition 1.5]). Let 𝑋= Cay(G,S), and 𝐴=Aut(𝑋). Then 𝑋 is normal if and only if 𝐴1=Aut(𝐺,𝑆), where 𝐴1 is the stabilizer of 1 in 𝐴.

Lemma 2.3 (see [5, Lemma 4.4]). All 1-regular cubic Cayley graphs on the dihedral group 𝐷2𝑛 are normal.

Lemma 2.4 (see [6, Lemma 3.2]). Let Γ be a connected cubic graph on dihedral group 𝐷2𝑛, and let 𝐵1 and 𝐵2 be two orbits of 𝐶=𝑎. Also let 𝐺 be the subgroup of 𝐺 fixing setwise 𝐵1 and 𝐵2, respectively. If 𝐺 acts unfaithfully on one of 𝐵1 and 𝐵2, then Γ𝐾3,3.

Let 𝐶𝐺 be the core of 𝐶=𝑎 in Aut(𝑋). By assuming the hypothesis in the above lemma, we have the following results

Lemma 2.5 (see [6, Lemma 3.5]). If 𝐶𝐺 is a proper subgroup of 𝐶, then 𝑋 is isomorphic to 𝐶𝑎𝑦(𝐷14,{𝑏,𝑎𝑏,𝑎3𝑏}) or 𝐶𝑎𝑦(𝐷16,{𝑏,𝑎𝑏,𝑎3𝑏}).

Lemma 2.6 (see [6, Lemma 3.6]). If 𝐶𝐺=𝐶, then 𝑋 is isomorphic to 𝐶𝑎𝑦(𝐷2𝑛,{𝑏,𝑎𝑏,𝑎𝑘𝑏}), where 𝑘2𝑘+1=0 (mod 𝑛), and 𝑛13.

Let 𝐺=𝐷2𝑛. Then the elements of 𝐺 are 𝑎𝑖 and 𝑎𝑖𝑏, where 𝑖=0,1,,𝑛1. All 𝑎𝑖𝑏 are involutions, and 𝑎𝑖 is an involution if and only if 𝑛 is even and 𝑖=𝑛/2. Finally in this section we obtain a preliminary result restricting 𝑆 for cubic Cayley graphs of Cay(𝐷2𝑛,𝑆). We can easily prove the following lemma

Lemma 2.7. Let Γ= Cay (𝐺,𝑆) be Cayley graphs of 𝐺=𝐷2𝑛. Then Γ is cubic, connected and, undirected if and only if one of the following conditions holds (1)When 𝑛 is odd, one has𝑆𝑜1=𝑎𝑖𝑏,𝑎𝑗𝑏,𝑎𝑘𝑏𝑆,0𝑖<𝑗<𝑘<𝑛,𝑜2=𝑎𝑖,𝑎𝑖,𝑎𝑗𝑏𝑛,0<𝑖<2,0𝑗<𝑛,(2.2)(2)When 𝑛 is even, one has𝑆𝑒1=𝑎𝑖𝑏,𝑎𝑗𝑏,𝑎𝑘𝑏𝑆,0𝑖<𝑗<𝑘<𝑛,𝑒2=𝑎𝑖,𝑎𝑖,𝑎𝑗𝑏𝑛,0<𝑖<2𝑆,0𝑗<𝑛,𝑒3=𝑎𝑛2,𝑎𝑖,𝑎𝑖𝑛,0<𝑖<2,𝑆𝑒4=𝑎𝑛/2,𝑎𝑖𝑏,𝑎𝑗𝑏,0𝑖<𝑗<𝑛.(2.3)

Let 𝑋 and 𝑌 be two graphs. The direct 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,𝑣] is an edge in 𝑋×𝑌 whenever 𝑥1=𝑥2 and [𝑦1,𝑦2]𝐸(𝑌) or 𝑦1=𝑦2 and [𝑥1,𝑥2]𝐸(𝑋). Two graphs are called relatively prime if they have no nontrivial common direct factor. 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,𝑣] is an edge in 𝑋[𝑌] whenever [𝑥1,𝑥2]𝐸(𝑋) or 𝑥1=𝑥2 and [𝑦1,𝑦2]𝐸(𝑌). Let 𝑉(𝑌)={𝑦1,𝑦2,,𝑦𝑛}. Then there is a natural embedding 𝑛𝑋 in 𝑋[𝑌], where for 1𝑖𝑛, the 𝑖th copy of 𝑋 is the subgraph induced on the vertex subset {(𝑥,𝑦𝑖)𝑥𝑉(𝑋)} in 𝑋[𝑌]. The deleted lexicographic product 𝑋[𝑌]𝑛𝑋 is the graph obtained by deleting all the edges (natural embedding) of 𝑛𝑋 from 𝑋[𝑌].

3. Proof of Theorem 1.1

As we have seen in Section 1, each edge transitive Cayley graph which is normal is automatically normal edge transitive. Hence for the proof of Theorem 1.1, we must determine all nonnormal connected undirected cubic Cayley graphs for dihedral group 𝐷2𝑛. If 𝑛=2, then dihedral group 𝐷4 is isomorphic to 2×2, and so it is easy to show that the cubic Cayley graph Cay(𝐷4,𝑆) is normal. So from now we assume that 𝑛3. Also, since Cay(𝐷2𝑛,𝑆) when 𝑆=𝑆𝑒3 is disconnected, thus we do not consider this case for the proof of the main theorem. First we prove the following lemma.

Lemma 3.1. Let G be the dihedral group D2n with n3, and let Γ = Cay (G,S) be a cubic Cayley graph. Then (a)if 𝑆 is 𝑆𝑒4 and Γ is connected, then 𝑆(𝑆2{1})= holds;(b)if 𝑆 is 𝑆𝑜2 or 𝑆𝑒2 and Γ is connected, then 𝑆(𝑆2{1})= if 𝑛>3. For 𝑛=3, and 𝑆 is 𝑆𝑜2 or 𝑆𝑒2, one has; 𝑆(𝑆2{1}) and Γ=𝐶𝑎𝑦(𝐷6,𝑆) is connected and normal;(c)if 𝑆 is 𝑆𝑜1 or 𝑆𝑒1, then 𝑆(𝑆2{1})= always holds.

Proof. (a) Suppose first that 𝑆=𝑆𝑒4={𝑎𝑛/2,𝑎𝑖𝑏,𝑎𝑗𝑏}. Then 𝑆2𝑎{1}=n/2+𝑖𝑏,𝑎n/2+𝑗𝑏,𝑎𝑖𝑗,𝑎𝑗𝑖.(3.1) We show that 𝑆(𝑆2{1})=. Suppose to the contrary that 𝑆(𝑆2{1}). We may suppose that 𝑎𝑖𝑗=𝑎𝑗𝑖=𝑎𝑛/2. Now we have Γ=𝑛𝐾1[𝑌], where 𝑌=𝐾4. Hence Γ is not connected, which is a contradiction.
(b) Now suppose that 𝑆=𝑆𝑜2 or 𝑆=𝑆𝑒2, that is, 𝑆={𝑎𝑖,𝑎𝑖,𝑎𝑗𝑏}. For 𝑛>3, we have 𝑆2{1}={𝑎2𝑖,𝑎2𝑖,𝑎𝑖+𝑗𝑏,𝑎𝑗𝑖𝑏}. We claim that 𝑆(𝑆2{1})=. Suppose to the contrary that 𝑆(𝑆2{1}). We may suppose that 𝑎2𝑖=𝑎𝑖. Then Γ=𝑚𝐾1[𝑌], where 𝑌=Cay(𝑆,𝑆) and |𝐷2𝑛𝑆|=𝑚. So Γ is not connected, which is a contradiction. Now let 𝑛=3. Then 𝑆={𝑎,𝑎1,𝑏}, {𝑎,𝑎1,𝑎𝑏}, or {𝑎,𝑎1,𝑎2𝑏}, respectively. Therefore 𝑆21={𝑎2,𝑎𝑏,𝑎2𝑏,𝑎}, {𝑎2,𝑎2𝑏,𝑎,𝑏}, or {𝑎2,𝑏,𝑎,𝑎𝑏}, respectively. Obviously Γ is connected, and 𝐺𝐷6. Also we have 𝑆(𝑆2{1}), and Cay(𝐷6,{𝑎,𝑎1,𝑏}) Cay(𝐷6,{𝑎,𝑎1,𝑎𝑏}) Cay(𝐷6,{𝑎,𝑎1,𝑎2𝑏}). Let 𝜎 be an automorphism of Γ=Cay(𝐷6,{𝑎,𝑎1,𝑏}), which fixes 1 and all elements of 𝑆. Since 𝑎𝜎=𝑎, and (𝑎2)𝜎=𝑎2, we have {1,𝑎2,𝑎2𝑏}𝜎={1,𝑎2,𝑎2𝑏} and {1,𝑎,𝑎𝑏}𝜎={1,𝑎,𝑎𝑏}. Therefore (𝑎𝑏)𝜎=𝑎𝑏, and (𝑎2𝑏)𝜎=𝑎2𝑏, and hence 𝜎 fixes all elements of 𝑆2. Thus 𝜎=1, and 𝐴1 acts faithfully on 𝑆. So we may view 𝐴1 as a permutation group on 𝑆. Now let 𝛼 be an arbitrary element of 𝐴1. Since 1𝛼=1, we have {𝑎,𝑎2,𝑏}𝛼={𝑎,𝑎2,𝑏}. If 𝑏𝛼=𝑎 or 𝑏𝛼=𝑎2, then {1,𝑎𝑏,𝑎2𝑏}𝛼={1,𝑎2𝑏,𝑎2} or {1,𝑎𝑏,𝑎2𝑏}𝛼={1,𝑎𝑏,𝑎}, which is a contradiction. Thus 𝑏𝛼=𝑏, and 𝐴1 is generated by the permutation (𝑎,𝑎2). So |𝐴1|=2. On the other hand, 𝛽𝑎𝑡𝑏𝑙𝑎2𝑡𝑏𝑙 is an element of Aut(𝐺,𝑆). Therefore |𝐴1|=|Aut(𝐺,𝑆)|=2, and hence by Lemma 2.2, Γ is normal.
(c) Finally, suppose that 𝑆=𝑆𝑜1 or 𝑆=𝑆𝑒1, that is, 𝑆={𝑎𝑖𝑏,𝑎𝑗𝑏,𝑎𝑘𝑏}. Then 𝑆2{1}={𝑎𝑖𝑗,𝑎𝑗𝑖,𝑎𝑖𝑘,𝑎𝑘𝑖,𝑎𝑗𝑘,𝑎𝑘𝑗}. Clearly 𝑆(𝑆2{1})=. The results now follow.
By considering this lemma, we prove the following proposition. This result will be used in the proof of Theorem 1.1.

Proposition 3.2. Let 𝐺 be the dihedral group 𝐷2𝑛(𝑛3), and let 𝑋=Cay(𝐺,𝑆) be a connected and undirected cubic Cayley graph. Then 𝑋 is normal except y one of the following cases happens: (1)𝑛=4,𝑆={𝑏,𝑎𝑏,𝑎2𝑏},𝑋𝐾4,44𝐾2;(2)𝑛=8,𝑆={𝑏,𝑎𝑏,𝑎3𝑏},𝑋𝑃(8,3), (the generalized Peterson graph);(3)𝑛=3,𝑆={𝑏,𝑎𝑏,𝑎2𝑏},𝑋𝐾3,3;(4)𝑛=7,𝑆={𝑏,𝑎𝑏,𝑎3𝑏},𝑋𝑆(7), (Heawood's graph).

Proof. First assume that 𝑆=𝑆𝑒4. Since Γ is connected, by Lemma 3.1(a), 𝑆(𝑆2{1})=. Now consider the graph Γ2(1), and let 𝜎 be an automorphism of Γ=Cay(𝐷2𝑛,{𝑎𝑛/2,𝑎𝑖𝑏,𝑎𝑗𝑏}, which fixes 1 and all elements of 𝑆. Since (𝑎𝑛/2)𝜎=𝑎𝑛/2, (𝑎𝑖𝑏)𝜎=𝑎𝑖𝑏, and (𝑎𝑗𝑏)𝜎=𝑎𝑗𝑏, we have {1,𝑎𝑛/2+𝑖𝑏,𝑎𝑛/2+𝑗𝑏}𝜎={1,𝑎𝑛/2+𝑖𝑏,𝑎𝑛/2+𝑗𝑏},{1,𝑎𝑛/2+𝑖𝑏,𝑎𝑗𝑖}𝜎={1,𝑎𝑛/2+𝑖𝑏,𝑎𝑗𝑖}, and {1,𝑎𝑛/2+𝑗𝑏,𝑎𝑖𝑗}𝜎={1,𝑎𝑛/2+𝑗𝑏,𝑎𝑖𝑗}, respectively. Therefore (𝑎𝑛/2+𝑖𝑏)𝜎=𝑎𝑛/2+𝑖𝑏, (𝑎𝑛/2+𝑗𝑏)𝜎=𝑎𝑛/2+𝑗𝑏,(𝑎𝑗𝑖)𝜎=𝑎𝑗𝑖, and (𝑎𝑖𝑗)𝜎=𝑎𝑖𝑗, and hence 𝜎 fixes all elements of 𝑆2. Because of the connectivity of Γ, this automorphism is the identity in Aut(Γ). Therefore 𝐴1 acts faithfully on 𝑆. So we may view 𝐴1 as a permutation group on 𝑆. Now let 𝛼 be an arbitrary element of 𝐴1. Since 1𝛼=1, we have {𝑎𝑛/2,𝑎𝑖𝑏,𝑎𝑗𝑏}𝛼={𝑎𝑛/2,𝑎𝑖𝑏,𝑎𝑗𝑏}. If (𝑎𝑛/2)𝛼=𝑎𝑖𝑏, or (𝑎𝑛/2)𝛼=𝑎𝑗𝑏, then {1,𝑎𝑛/2+𝑖𝑏,𝑎𝑛/2+𝑗𝑏}𝛼={1,𝑎𝑛/2+𝑖𝑏,𝑎𝑗𝑖}, or {1,𝑎𝑛/2+𝑖𝑏,𝑎𝑛/2+𝑗𝑏}𝛼={1,𝑎𝑛/2+𝑗𝑏,𝑎𝑖𝑗}, respectively. Now again we consider Γ2(1). In this subgraph, 𝑎𝑛/2+𝑖𝑏 and 𝑎𝑛/2+𝑗𝑏 have valency 2, and 𝑎𝑖𝑗, 𝑎𝑗𝑖 have valency 1. This implies a contradiction. Thus (𝑎𝑛/2)𝛼=𝑎𝑛/2, and 𝐴1 is generated by the permutation (𝑎𝑖𝑏,𝑎𝑗𝑏). So |𝐴1|=2. On the other hand, 𝛽𝑎𝑡𝑏𝑙𝑎𝑡(𝑎𝑖+𝑗𝑏)𝑙 is an element of Aut(𝐺,𝑆). Therefore |𝐴1|=|Aut(𝐺,𝑆)|=2, and hence by Lemma 2.2, Γ is normal.
Now assume that 𝑆=𝑆𝑒2={𝑎𝑖,𝑎𝑖,𝑎𝑗𝑏}, or 𝑆=𝑆𝑜2={𝑎𝑖,𝑎𝑖,𝑎𝑗𝑏}. If 𝑛=3, then by Lemma 3.1 (b), Γ=Cay(𝐷6,𝑆), and Γ is normal. Now if 𝑛>3, then again by Lemma 3.1(b), 𝑆(𝑆2{1})=. Considering the graph Γ2(1), with the same reason as before if an automorphism of Γ fixes 1 and all elements of 𝑆, then it also fixes all elements of 𝑆2. Because of the connectivity of Γ, this automorphism is the identity in Aut(Γ). Therefore 𝐴1 acts faithfully on 𝑆. So we may view 𝐴1 as a permutation group on 𝑆. We can easily see that 𝐴1 is generated by the permutation (𝑎𝑖,𝑎𝑖). So |𝐴1|=2. On the other hand, 𝜎𝑎𝑡𝑏𝑙𝑎𝑡(𝑎2𝑗𝑏)𝑙 is an element of Aut(𝐺,𝑆). Therefore |𝐴1|=|Aut(𝐺,𝑆)|=2, and hence by Lemma 2.2, Γ is normal.
Finally assume that 𝑆=𝑆𝑒1={𝑎𝑖𝑏,𝑎𝑗𝑏,𝑎𝑘𝑏}, or 𝑆=𝑆𝑜1={𝑎𝑖𝑏,𝑎𝑗𝑏,𝑎𝑘𝑏}. Up to graph isomorphism, 𝑆={𝑏,𝑎𝑗𝑏,𝑎𝑘𝑏}, where <𝑗,𝑘𝑍𝑛. In this case, Γ is a bipartite graph with the partition 𝐵=𝐵1𝐵2, where 𝐵1 and 𝐵2 are just two orbits of 𝐶=𝑎, and we assume the block 𝐵1 contains 1. Let 𝐺 be the subgroup of 𝐺 fixing setwise 𝐵1 and 𝐵2, respectively. If 𝐺 acts unfaithfully on one of 𝐵1 and 𝐵2, then by Lemma 2.4, Γ𝐾3,3, and 𝜎=(𝑏,𝑎𝑏) is not in Aut(𝐺,𝑆) but in 𝐴1, and so Γ is not normal. Let 𝐺 act faithfully on 𝐵1 and 𝐵2. Then 𝑛3. If 𝑛=4, then Γ is isomorphic to 𝐾4,44𝐾2, and 𝜎=(𝑏,𝑎𝑏)(𝑎2,𝑎3) is not in Aut(𝐺,𝑆) but in 𝐴1, and so Γ is not normal. From now on we assume 𝑛5. Now suppose that 𝐶𝐺, the core of 𝐶 in 𝐺, is a proper subgroup of 𝐶. Then by Lemma 2.5, Γ Cay(𝐷14,{𝑏,𝑎𝑏,𝑎3𝑏}) or Γ Cay(𝐷16,{𝑏,𝑎𝑏,𝑎3𝑏}). For the first case, 𝜎=(𝑎,𝑎2,𝑎3,𝑎6)(𝑎4,𝑎5)(𝑎2𝑏,𝑎6𝑏,𝑎5𝑏,𝑎4𝑏)(𝑎𝑏,𝑏) is not in Aut(𝐺,𝑆) but in 𝐴1, and so Γ is not normal. For the second case, 𝜎=(𝑎,𝑎7,𝑎6)(𝑎2,𝑎5,𝑎3)(𝑏,𝑎𝑏,𝑎3𝑏)(𝑎4𝑏,𝑎5𝑏,𝑎7𝑏) is not in Aut(𝐺,𝑆) but in 𝐴1, and so Γ is not normal. Finally we suppose that 𝐶𝐺=𝐶. Then by Lemma 2.6, Γ is isomorphic to the Cay(𝐷2𝑛,{𝑏,𝑎𝑏,𝑎𝑘𝑏}), where 𝑘2𝑘+10 (mod 𝑛) and 𝑛13. The Cayley graph Γ is 1-regular, and by Lemma 2.3, Γ is normal. The result now follows.
Now we complete the proof of Theorem 1.1. We remind that any edge transitive Cayley graph which is normal, in the sense of Xu's definition, is also normal edge transitive. Thus this implies that we must consider nonnormal Cayley graphs which were obtained in Proposition 3.2. So we consider four cases in Proposition 3.2. For case (1), we claim that there is no automorphism of 𝐺 such that 𝑏 maps to 𝑎𝑏. Suppose to the contrary that there is an automorphism 𝜎 such that 𝑏 maps to 𝑎𝑏. Then 𝑎 must be mapped to 𝑎𝑖, where (𝑖,4)=1, and so with the simple check it is easy to see that this is a contradiction. Also in case (2), with the same reason as above there is a contradiction. Hence Aut(𝐺,𝑆) does not act transitively on 𝑆 also does not have two orbits in 𝑆 which are inverse of each other. Now by using Lemma 2.1 these graphs are not normal edge transitive. For the last two cases it is easy to show that Aut(𝐺,𝑆) acts transitively on 𝑆, and hence by Lemma 2.1, these graphs are normal edge transitive. Now the proof is complete as claimed.