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ISRN Algebra
VolumeΒ 2011Β (2011), Article IDΒ 428959, 6 pages
http://dx.doi.org/10.5402/2011/428959
Research Article

Nonnormal Edge-Transitive Cubic Cayley Graphs of Dihedral Groups

1Department of Mathematics, Iran University of Science and Technology, Narmak, Tehran 16844, Iran
2Payame Noor University, Babol, Iran
3Department of Mathematics, Urmia University, Urmia 57135, Iran

Received 22 June 2011; Accepted 11 July 2011

Academic Editors: B.Β Bakalov and M.Β Goze

Copyright Β© 2011 Mehdi Alaeiyan et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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 1πΊβˆ‰π‘†. 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}βˆ£π‘”βˆˆπΊ,π‘ βˆˆπ‘†}. 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=βŸ¨π‘†βŸ©; (3) 𝑋 is undirected if and only if 𝑆=π‘†βˆ’1.

A part of Aut(𝑋) may be described in terms of automorphisms of 𝐺, that is, the normalizer 𝑁Aut(𝑋)(𝐺)=πΊβ‹ŠAut(𝐺,𝑆), a semidirect product of 𝐺 by Aut(𝐺,𝑆), where Aut(𝐺,𝑆)={𝜎∈Aut(𝐺)βˆ£π‘†πœŽ=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 𝑠β‰₯1, an 𝑠-arc in a graph 𝑋 is an ordered (𝑠+1)-tuple (𝑣0,𝑣1,…,𝑣𝑠) of vertices of 𝑋 such that π‘£π‘–βˆ’1 is adjacent to 𝑣𝑖 for 1≀𝑖≀𝑠 and π‘£π‘–βˆ’1≠𝑣𝑖+1 for 1≀𝑖<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=βŸ¨π‘Ž,π‘βˆ£π‘Žπ‘›=𝑏2=1,π‘π‘Žπ‘βˆ’1=π‘Žβˆ’1⟩, 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,π‘Žπ‘,π‘Ž2𝑏}, 𝑋≅𝐾4,4βˆ’4𝐾2;(2)𝑛=8, 𝑆={𝑏,π‘Žπ‘,π‘Ž3𝑏}, 𝑋≅𝑃(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 πΊβ‹ŠAut(𝐺). Thus,𝑁(Aut(𝑋))(𝐺)=(πΊβ‹ŠAut(𝐺))∩Aut(𝑋)=πΊβ‹Š(Aut(𝐺)∩Aut(𝑋))=πΊβ‹ŠAut(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 Ξ“ be a connected cubic graph on dihedral group 𝐷2𝑛, and let 𝐡1 and 𝐡2 be two orbits of 𝐢=βŸ¨π‘ŽβŸ©. Also let πΊβˆ— be the subgroup of 𝐺 fixing setwise 𝐡1 and 𝐡2, respectively. If πΊβˆ— acts unfaithfully on one of 𝐡1 and 𝐡2, then Γ≅𝐾3,3.

Let 𝐢𝐺 be the core of 𝐢=βŸ¨π‘ŽβŸ© 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 𝐢𝐺 is a proper subgroup of 𝐢, then 𝑋 is isomorphic to πΆπ‘Žπ‘¦(𝐷14,{𝑏,π‘Žπ‘,π‘Ž3𝑏}) or πΆπ‘Žπ‘¦(𝐷16,{𝑏,π‘Žπ‘,π‘Ž3𝑏}).

Lemma 2.6 (see [6, Lemma 3.6]). If 𝐢𝐺=𝐢, then 𝑋 is isomorphic to πΆπ‘Žπ‘¦(𝐷2𝑛,{𝑏,π‘Žπ‘,π‘Žπ‘˜π‘}), where π‘˜2βˆ’π‘˜+1=0 (mod 𝑛), and 𝑛β‰₯13.

Let 𝐺=𝐷2𝑛. Then the elements of 𝐺 are π‘Žπ‘– and π‘Žπ‘–π‘, where 𝑖=0,1,…,π‘›βˆ’1. All π‘Žπ‘–π‘ are involutions, and π‘Žπ‘– 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 Ξ“= Cay (𝐺,𝑆) be Cayley graphs of 𝐺=𝐷2𝑛. Then Ξ“ is cubic, connected and, undirected if and only if one of the following conditions holds (1)When 𝑛 is odd, one hasπ‘†π‘œ1=ξ€½π‘Žπ‘–π‘,π‘Žπ‘—π‘,π‘Žπ‘˜π‘ξ€Ύπ‘†,0≀𝑖<𝑗<π‘˜<𝑛,π‘œ2=ξ€½π‘Žπ‘–,π‘Žβˆ’π‘–,π‘Žπ‘—π‘ξ€Ύπ‘›,0<𝑖<2,0≀𝑗<𝑛,(2.2)(2)When 𝑛 is even, one has𝑆𝑒1=ξ€½π‘Žπ‘–π‘,π‘Žπ‘—π‘,π‘Žπ‘˜π‘ξ€Ύπ‘†,0≀𝑖<𝑗<π‘˜<𝑛,𝑒2=ξ€½π‘Žπ‘–,π‘Žβˆ’π‘–,π‘Žπ‘—π‘ξ€Ύπ‘›,0<𝑖<2𝑆,0≀𝑗<𝑛,𝑒3=ξƒ―π‘Žπ‘›2,π‘Žπ‘–,π‘Žβˆ’π‘–ξƒ°π‘›,0<𝑖<2,𝑆𝑒4=ξ€½π‘Žπ‘›/2,π‘Žπ‘–π‘,π‘Žπ‘—π‘ξ€Ύ,0≀𝑖<𝑗<𝑛.(2.3)

Let 𝑋 and π‘Œ be two graphs. The direct product π‘‹Γ—π‘Œ is defined as the graph with vertex set 𝑉(π‘‹Γ—π‘Œ)=𝑉(𝑋)×𝑉(π‘Œ) such that for any two vertices 𝑒=[π‘₯1,𝑦1] and 𝑣=[π‘₯2,𝑦2] in 𝑉(π‘‹Γ—π‘Œ), [𝑒,𝑣] is an edge in π‘‹Γ—π‘Œ whenever π‘₯1=π‘₯2 and [𝑦1,𝑦2]∈𝐸(π‘Œ) or 𝑦1=𝑦2 and [π‘₯1,π‘₯2]∈𝐸(𝑋). Two graphs are called relatively prime if they have no nontrivial common direct factor. The lexicographic product 𝑋[π‘Œ] is defined as the graph with vertex set 𝑉(𝑋[π‘Œ])=𝑉(𝑋)×𝑉(π‘Œ) such that for any two vertices 𝑒=[π‘₯1,𝑦1] and 𝑣=[π‘₯2,𝑦2] in 𝑉(𝑋[π‘Œ]), [𝑒,𝑣] is an edge in 𝑋[π‘Œ] whenever [π‘₯1,π‘₯2]∈𝐸(𝑋) or π‘₯1=π‘₯2 and [𝑦1,𝑦2]∈𝐸(π‘Œ). Let 𝑉(π‘Œ)={𝑦1,𝑦2,…,𝑦𝑛}. Then there is a natural embedding 𝑛𝑋 in 𝑋[π‘Œ], where for 1≀𝑖≀𝑛, the 𝑖th copy of 𝑋 is the subgraph induced on the vertex subset {(π‘₯,𝑦𝑖)∣π‘₯βˆˆπ‘‰(𝑋)} in 𝑋[π‘Œ]. The deleted lexicographic product 𝑋[π‘Œ]βˆ’π‘›π‘‹ is the graph obtained by deleting all the edges (natural embedding) of 𝑛𝑋 from 𝑋[π‘Œ].

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 β„€2Γ—β„€2, and so it is easy to show that the cubic Cayley graph Cay(𝐷4,𝑆) is normal. So from now we assume that 𝑛β‰₯3. 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 nβ‰₯3, and let Ξ“ = Cay (G,S) be a cubic Cayley graph. Then (a)if 𝑆 is 𝑆𝑒4 and Ξ“ is connected, then π‘†βˆ©(𝑆2βˆ’{1})=βˆ… holds;(b)if 𝑆 is π‘†π‘œ2 or 𝑆𝑒2 and Ξ“ is connected, then π‘†βˆ©(𝑆2βˆ’{1})=βˆ… if 𝑛>3. For 𝑛=3, and 𝑆 is π‘†π‘œ2 or 𝑆𝑒2, one has; π‘†βˆ©(𝑆2βˆ’{1})β‰ βˆ… and Ξ“=πΆπ‘Žπ‘¦(𝐷6,𝑆) is connected and normal;(c)if 𝑆 is π‘†π‘œ1 or 𝑆𝑒1, then π‘†βˆ©(𝑆2βˆ’{1})=βˆ… always holds.

Proof. (a) Suppose first that 𝑆=𝑆𝑒4={π‘Žπ‘›/2,π‘Žπ‘–π‘,π‘Žπ‘—π‘}. Then 𝑆2βˆ’ξ€½π‘Ž{1}=n/2+𝑖𝑏,π‘Žn/2+𝑗𝑏,π‘Žπ‘–βˆ’π‘—,π‘Žπ‘—βˆ’π‘–ξ€Ύ.(3.1) We show that π‘†βˆ©(𝑆2βˆ’{1})=βˆ…. Suppose to the contrary that π‘†βˆ©(𝑆2βˆ’{1})β‰ βˆ…. We may suppose that π‘Žπ‘–βˆ’π‘—=π‘Žπ‘—βˆ’π‘–=π‘Žπ‘›/2. Now we have Ξ“=𝑛𝐾1[π‘Œ], where π‘Œ=𝐾4. Hence Ξ“ is not connected, which is a contradiction.
(b) Now suppose that 𝑆=π‘†π‘œ2 or 𝑆=𝑆𝑒2, that is, 𝑆={π‘Žπ‘–,π‘Žβˆ’π‘–,π‘Žπ‘—π‘}. For 𝑛>3, we have 𝑆2βˆ’{1}={π‘Ž2𝑖,π‘Žβˆ’2𝑖,π‘Žπ‘–+𝑗𝑏,π‘Žπ‘—βˆ’π‘–π‘}. We claim that π‘†βˆ©(𝑆2βˆ’{1})=βˆ…. Suppose to the contrary that π‘†βˆ©(𝑆2βˆ’{1})β‰ βˆ…. We may suppose that π‘Ž2𝑖=π‘Žβˆ’π‘–. Then Ξ“=π‘šπΎ1[π‘Œ], where π‘Œ=Cay(𝑆,βŸ¨π‘†βŸ©) and |𝐷2π‘›βˆΆβŸ¨π‘†βŸ©|=π‘š. So Ξ“ is not connected, which is a contradiction. Now let 𝑛=3. Then 𝑆={π‘Ž,π‘Žβˆ’1,𝑏}, {π‘Ž,π‘Žβˆ’1,π‘Žπ‘}, or {π‘Ž,π‘Žβˆ’1,π‘Ž2𝑏}, respectively. Therefore 𝑆2βˆ’1={π‘Ž2,π‘Žπ‘,π‘Ž2𝑏,π‘Ž}, {π‘Ž2,π‘Ž2𝑏,π‘Ž,𝑏}, or {π‘Ž2,𝑏,π‘Ž,π‘Žπ‘}, respectively. Obviously Ξ“ is connected, and 𝐺≅𝐷6. Also we have π‘†βˆ©(𝑆2βˆ’{1})β‰ βˆ…, and Cay(𝐷6,{π‘Ž,π‘Žβˆ’1,𝑏})β‰… Cay(𝐷6,{π‘Ž,π‘Žβˆ’1,π‘Žπ‘})β‰… Cay(𝐷6,{π‘Ž,π‘Žβˆ’1,π‘Ž2𝑏}). Let 𝜎 be an automorphism of Ξ“=Cay(𝐷6,{π‘Ž,π‘Žβˆ’1,𝑏}), which fixes 1 and all elements of 𝑆. Since π‘ŽπœŽ=π‘Ž, and (π‘Ž2)𝜎=π‘Ž2, we have {1,π‘Ž2,π‘Ž2𝑏}𝜎={1,π‘Ž2,π‘Ž2𝑏} and {1,π‘Ž,π‘Žπ‘}𝜎={1,π‘Ž,π‘Žπ‘}. Therefore (π‘Žπ‘)𝜎=π‘Žπ‘, and (π‘Ž2𝑏)𝜎=π‘Ž2𝑏, 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 {π‘Ž,π‘Ž2,𝑏}𝛼={π‘Ž,π‘Ž2,𝑏}. If 𝑏𝛼=π‘Ž or 𝑏𝛼=π‘Ž2, then {1,π‘Žπ‘,π‘Ž2𝑏}𝛼={1,π‘Ž2𝑏,π‘Ž2} or {1,π‘Žπ‘,π‘Ž2𝑏}𝛼={1,π‘Žπ‘,π‘Ž}, which is a contradiction. Thus 𝑏𝛼=𝑏, and 𝐴1 is generated by the permutation (π‘Ž,π‘Ž2). So |𝐴1|=2. On the other hand, π›½βˆΆπ‘Žπ‘‘π‘π‘™β†’π‘Ž2𝑑𝑏𝑙 is an element of Aut(𝐺,𝑆). Therefore |𝐴1|=|Aut(𝐺,𝑆)|=2, and hence by Lemma 2.2, Ξ“ is normal.
(c) Finally, suppose that 𝑆=π‘†π‘œ1 or 𝑆=𝑆𝑒1, that is, 𝑆={π‘Žπ‘–π‘,π‘Žπ‘—π‘,π‘Žπ‘˜π‘}. Then 𝑆2βˆ’{1}={π‘Žπ‘–βˆ’π‘—,π‘Žπ‘—βˆ’π‘–,π‘Žπ‘–βˆ’π‘˜,π‘Žπ‘˜βˆ’π‘–,π‘Žπ‘—βˆ’π‘˜,π‘Žπ‘˜βˆ’π‘—}. Clearly π‘†βˆ©(𝑆2βˆ’{1})=βˆ…. 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(𝑛β‰₯3), 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,π‘Žπ‘,π‘Ž2𝑏},𝑋≅𝐾4,4βˆ’4𝐾2;(2)𝑛=8,𝑆={𝑏,π‘Žπ‘,π‘Ž3𝑏},𝑋≅𝑃(8,3), (the generalized Peterson graph);(3)𝑛=3,𝑆={𝑏,π‘Žπ‘,π‘Ž2𝑏},𝑋≅𝐾3,3;(4)𝑛=7,𝑆={𝑏,π‘Žπ‘,π‘Ž3𝑏},𝑋≅𝑆(7), (Heawood's graph).

Proof. First assume that 𝑆=𝑆𝑒4. Since Ξ“ is connected, by Lemma 3.1(a), π‘†βˆ©(𝑆2βˆ’{1})=βˆ…. Now consider the graph Ξ“2(1), and let 𝜎 be an automorphism of Ξ“=Cay(𝐷2𝑛,{π‘Žπ‘›/2,π‘Žπ‘–π‘,π‘Žπ‘—π‘}, which fixes 1 and all elements of 𝑆. Since (π‘Žπ‘›/2)𝜎=π‘Žπ‘›/2, (π‘Žπ‘–π‘)𝜎=π‘Žπ‘–π‘, and (π‘Žπ‘—π‘)𝜎=π‘Žπ‘—π‘, we have {1,π‘Žπ‘›/2+𝑖𝑏,π‘Žπ‘›/2+𝑗𝑏}𝜎={1,π‘Žπ‘›/2+𝑖𝑏,π‘Žπ‘›/2+𝑗𝑏},{1,π‘Žπ‘›/2+𝑖𝑏,π‘Žπ‘—βˆ’π‘–}𝜎={1,π‘Žπ‘›/2+𝑖𝑏,π‘Žπ‘—βˆ’π‘–}, and {1,π‘Žπ‘›/2+𝑗𝑏,π‘Žπ‘–βˆ’π‘—}𝜎={1,π‘Žπ‘›/2+𝑗𝑏,π‘Žπ‘–βˆ’π‘—}, respectively. Therefore (π‘Žπ‘›/2+𝑖𝑏)𝜎=π‘Žπ‘›/2+𝑖𝑏, (π‘Žπ‘›/2+𝑗𝑏)𝜎=π‘Žπ‘›/2+𝑗𝑏,(π‘Žπ‘—βˆ’π‘–)𝜎=π‘Žπ‘—βˆ’π‘–, and (π‘Žπ‘–βˆ’π‘—)𝜎=π‘Žπ‘–βˆ’π‘—, and hence 𝜎 fixes all elements of 𝑆2. Because of the connectivity of Ξ“, this automorphism is the identity in Aut(Ξ“). 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 {π‘Žπ‘›/2,π‘Žπ‘–π‘,π‘Žπ‘—π‘}𝛼={π‘Žπ‘›/2,π‘Žπ‘–π‘,π‘Žπ‘—π‘}. If (π‘Žπ‘›/2)𝛼=π‘Žπ‘–π‘, or (π‘Žπ‘›/2)𝛼=π‘Žπ‘—π‘, then {1,π‘Žπ‘›/2+𝑖𝑏,π‘Žπ‘›/2+𝑗𝑏}𝛼={1,π‘Žπ‘›/2+𝑖𝑏,π‘Žπ‘—βˆ’π‘–}, or {1,π‘Žπ‘›/2+𝑖𝑏,π‘Žπ‘›/2+𝑗𝑏}𝛼={1,π‘Žπ‘›/2+𝑗𝑏,π‘Žπ‘–βˆ’π‘—}, respectively. Now again we consider Ξ“2(1). In this subgraph, π‘Žπ‘›/2+𝑖𝑏 and π‘Žπ‘›/2+𝑗𝑏 have valency 2, and π‘Žπ‘–βˆ’π‘—, π‘Žπ‘—βˆ’π‘– have valency 1. This implies a contradiction. Thus (π‘Žπ‘›/2)𝛼=π‘Žπ‘›/2, and 𝐴1 is generated by the permutation (π‘Žπ‘–π‘,π‘Žπ‘—π‘). So |𝐴1|=2. On the other hand, π›½βˆΆπ‘Žπ‘‘π‘π‘™β†’π‘Žβˆ’π‘‘(π‘Žπ‘–+𝑗𝑏)𝑙 is an element of Aut(𝐺,𝑆). Therefore |𝐴1|=|Aut(𝐺,𝑆)|=2, and hence by Lemma 2.2, Ξ“ is normal.
Now assume that 𝑆=𝑆𝑒2={π‘Žπ‘–,π‘Žβˆ’π‘–,π‘Žπ‘—π‘}, or 𝑆=π‘†π‘œ2={π‘Žπ‘–,π‘Žβˆ’π‘–,π‘Žπ‘—π‘}. If 𝑛=3, then by Lemma 3.1 (b), Ξ“=Cay(𝐷6,𝑆), and Ξ“ is normal. Now if 𝑛>3, then again by Lemma 3.1(b), π‘†βˆ©(𝑆2βˆ’{1})=βˆ…. Considering the graph Ξ“2(1), with the same reason as before if an automorphism of Ξ“ fixes 1 and all elements of 𝑆, then it also fixes all elements of 𝑆2. Because of the connectivity of Ξ“, this automorphism is the identity in Aut(Ξ“). 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 (π‘Žπ‘–,π‘Žβˆ’π‘–). So |𝐴1|=2. On the other hand, πœŽβˆΆπ‘Žπ‘‘π‘π‘™β†’π‘Žβˆ’π‘‘(π‘Ž2𝑗𝑏)𝑙 is an element of Aut(𝐺,𝑆). Therefore |𝐴1|=|Aut(𝐺,𝑆)|=2, and hence by Lemma 2.2, Ξ“ is normal.
Finally assume that 𝑆=𝑆𝑒1={π‘Žπ‘–π‘,π‘Žπ‘—π‘,π‘Žπ‘˜π‘}, or 𝑆=π‘†π‘œ1={π‘Žπ‘–π‘,π‘Žπ‘—π‘,π‘Žπ‘˜π‘}. Up to graph isomorphism, 𝑆={𝑏,π‘Žπ‘—π‘,π‘Žπ‘˜π‘}, where <𝑗,π‘˜β‰₯π‘βˆ—π‘›. In this case, Ξ“ is a bipartite graph with the partition 𝐡=𝐡1βˆͺ𝐡2, where 𝐡1 and 𝐡2 are just two orbits of 𝐢=βŸ¨π‘ŽβŸ©, and we assume the block 𝐡1 contains 1. Let πΊβˆ— be the subgroup of 𝐺 fixing setwise 𝐡1 and 𝐡2, respectively. If πΊβˆ— acts unfaithfully on one of 𝐡1 and 𝐡2, then by Lemma 2.4, Γ≅𝐾3,3, and 𝜎=(𝑏,π‘Žπ‘) is not in Aut(𝐺,𝑆) but in 𝐴1, and so Ξ“ is not normal. Let πΊβˆ— act faithfully on 𝐡1 and 𝐡2. Then 𝑛≠3. If 𝑛=4, then Ξ“ is isomorphic to 𝐾4,4βˆ’4𝐾2, and 𝜎=(𝑏,π‘Žπ‘)(π‘Ž2,π‘Ž3) is not in Aut(𝐺,𝑆) but in 𝐴1, and so Ξ“ is not normal. From now on we assume 𝑛β‰₯5. Now suppose that 𝐢𝐺, the core of 𝐢 in 𝐺, is a proper subgroup of 𝐢. Then by Lemma 2.5, Ξ“β‰… Cay(𝐷14,{𝑏,π‘Žπ‘,π‘Ž3𝑏}) or Ξ“β‰… Cay(𝐷16,{𝑏,π‘Žπ‘,π‘Ž3𝑏}). For the first case, 𝜎=(π‘Ž,π‘Ž2,π‘Ž3,π‘Ž6)(π‘Ž4,π‘Ž5)(π‘Ž2𝑏,π‘Ž6𝑏,π‘Ž5𝑏,π‘Ž4𝑏)(π‘Žπ‘,𝑏) is not in Aut(𝐺,𝑆) but in 𝐴1, and so Ξ“ is not normal. For the second case, 𝜎=(π‘Ž,π‘Ž7,π‘Ž6)(π‘Ž2,π‘Ž5,π‘Ž3)(𝑏,π‘Žπ‘,π‘Ž3𝑏)(π‘Ž4𝑏,π‘Ž5𝑏,π‘Ž7𝑏) is not in Aut(𝐺,𝑆) but in 𝐴1, and so Ξ“ is not normal. Finally we suppose that 𝐢𝐺=𝐢. Then by Lemma 2.6, Ξ“ is isomorphic to the Cay(𝐷2𝑛,{𝑏,π‘Žπ‘,π‘Žπ‘˜π‘}), where π‘˜2βˆ’π‘˜+1≑0 (mod 𝑛) and 𝑛β‰₯13. The Cayley graph Ξ“ is 1-regular, and by Lemma 2.3, Ξ“ 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 π‘Žπ‘. Suppose to the contrary that there is an automorphism 𝜎 such that 𝑏 maps to π‘Žπ‘. Then π‘Ž must be mapped to π‘Žπ‘–, 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.

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