IJCT International Journal of Combinatorics 1687-9171 1687-9163 Hindawi Publishing Corporation 125916 10.1155/2013/125916 125916 Research Article Finite 1-Regular Cayley Graphs of Valency 5 Li Jing Jian 1, 2 Lou Ben Gong 1 Zhang Xiao Jun 2, 3 Li Cai Heng 1 School of Mathematics and Statistics Yunnan University Kunming, Yunnan 650031 China ynu.edu.cn 2 School of Mathematics and Information Sciences, Guangxi University Nanning 530004 China gxu.edu.cn 3 School of Computer Science and Engineering, University of Electronic Science and Technology of China Chengdu 611731 China uestc.edu.cn 2013 27 3 2013 2013 15 11 2012 11 01 2013 28 02 2013 2013 Copyright © 2013 Jing Jian Li 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.

Let Γ=Cay(G,S) and GXAutΓ. We say Γ is (X,1)-regular Cayley graph if X acts regularly on its arcs. Γ is said to be core-free if G is core-free in some XAut(Cay(G,S)). In this paper, we prove that if an (X,1)-regular Cayley graph of valency 5 is not normal or binormal, then it is the normal cover of one of two core-free ones up to isomorphism. In particular, there are no core-free 1-regular Cayley graphs of valency 5.

1. Introduction

We assume that all graphs in this paper are finite, simple, and undirected.

Let Γ be a graph. Denote the vertex set, arc set, and full automorphism group of Γ by VΓ, AΓ, and AutΓ, respectively. A graph Γ is called X-vertex-transitive or X-arc-transitive if X acts transitively on VΓ or AΓ, where XAutΓ. Γ is simply called vertex-transitive, arc-transitive for the case where X=AutΓ. In particular, Γ is called (X,1)-regular if XAutΓ acts regularly on its arcs and then 1-regular when X=AutΓ.

Let G be a finite group with identity element 1. For a subset S of G with 1S=S-1:={x-1xS}, the Cayley graph Cay(G,S) of G (with respect to S) is defined as the graph with vertex set G such that x,yG are adjacent if and only if yx-1S. It is easy to see that a Cayley graph Cay(G,S) has valency |S|, and it is connected if and only if S=G.

Li proved in  that there are only finite number of core-free s-transitive Cayley graphs of valency k for s{2,3,4,5,7} and k3 and that, with the exceptions s=2 and (s,k)=(3,7), every s-transitive Cayley graph is a normal cover of a core-free one. It was proved in  that there are 15 core-free s-transitive cubic Cayley graphs up to isomorphism, and there are no core-free 1-regular cubic Cayley graphs. A natural problem arises. Characterize 1-transitive Cayley graphs, in particular, which graphs are 1-regular? Until now, the result about 1-regular graphs mainly focused constructing examples. For example, Frucht gave the first example of cubic 1-regular graph in . After then, Conder and Praeger constructed two infinite families of cubic 1-regular graphs in . Marušič  and Malnič et al.  constructed two infinite families of tetravalent 1-regular graphs. Classifying such graphs has aroused great interest. Motivated by above results and problem, we consider 1-regular Cayley graphs of valency 5 in this paper.

A graph Γ can be viewed as a Cayley graph of a group G if and only if AutΓ contains a subgroup that is isomorphic to G and acts regularly on the vertex set. For convenience, we denote this regular subgroup still by G. If XAutΓ contains a normal subgroup that is regular and isomorphic to G, then Γ is called an X-normal Cayley graph of G; if G is not normal in X but has a subgroup which is normal in X and semiregular on VΓ with exactly two orbits, then Γ is called an X-bi-normal Cayley graph; furthermore if X=AutΓ, Γ is called normal or bi-normal. Some characterization of normal and bi-normal Cayley graphs has given in [1, 2].

For a Cayley graph Γ=Cay(G,S), Γ is said to be core-free (with respect to G) if G is core-free in some XAutΓ; that is, CoreX(G)=xXGx=1.

The main result of this paper is the following assertion.

Theorem 1.

Let Γ= Cay(G,S) be an (X,1)-regular Cayley graph of valency 5, where GXAutΓ. Let n(G) be the number of nonisomorphic core-free (X,1)-regular Cayley graph of valency 5 with the regular subgroup equal to G. Then either

Γ  is an X-normal or X-bi-normal Cayley graph or

Γ is a nontrivial normal cover of one line of Table 1.

In particular, there are no core-free 1-regular Cayley graphs of valency 5.

Number X G n (G) 1-regular Remark
1 A 5 A 4 1 No Icosahedron
2 S 5 S 4 1 No

By Theorem 1, we can get the following remark immediately.

Remark 2.

Let Γ=Cay(G,S) be an 1-regular Cayley graph of valency 5. Then Γ is normal or bi-normal.

2. Examples

In this section we give some examples of graphs appearing in Theorem 1.

Example 3.

Let M=a11 be a cyclic group. Assume that τAut(M) is of order 10 and X=M:τ11:10. Let (1)  G=M:τ5D22. Suppose that (2)S=gτ2={g,gτ2,gτ4,gτ6,gτ8}, where gG is an involution such that gτ5. Let Γ=Cay(G,S) be the Cayley graph of the dihedral group G with respect to S. Then Γ is a connected (X,1)-regular Cayley graph of valency 5. In particular, Γ is X-normal.

Proof.

Let (3)G=a:b={1,a,a2,,a10,b,ab,a2b,,a10b}D22, where b=τ5.

Noting o(a)=11, we may assume that aτ=a9. Since the involution gG is not equal to b, we may let g=aib for some 1i<11 such that (9,i)=1. Then gτ2=(aib)τ2=a81ib=a4ib, and so gτ2g-1=a3igτ2=S. Thus the element gτ2g-1 is of order 11 as (3i,11)=1. So S=gτ2=G; that is, Γ=Cay(G,S) is connected.

Obviously, GXAutΓ and X1=τ2. However, |X|=55=|AΓ|; then Γ is an (X,1)-regular normal Cayley graph of G of valency 5.

Example 4.

Let G=a,ba5=b2=1,ab=a-1D10. Set S={b,ab,a2b,a3b,a4b} and Γ=Cay(G,S). Then ΓK5,5 and AutΓ=S5S2. Let X=(5×5):2D10×5 such that GXAutΓ. It follows that CoreX(G)5. Then Xα5 for αVΓ, and furthermore Γ is (X,1)-regular. Obviously G is not normal in X. However, CoreX(G)X is semiregular and has exactly two orbits on VΓ; then Γ is an (X,1)-regular Cayley graph of valency 5. In particular, Γ is X-bi-normal.

3. The Proof of Main Results

In this section, we will prove our main results. We first present some properties about normal Cayley graphs.

For a Cayley graph Γ=Cay(G,S), we have a subgroup of Aut(G): (4)Aut(G,S)={σAut(G)Sσ=S}.

Clearly it is a subgroup of the stabilizer in AutΓ of the vertex corresponding to the identity 1 of G. Since Γ is connected, Aut(G,S) acts faithfully on S. By Godsil [7, Lemma 2.1], the normalizer NAutΓ(G)=G:Aut(G,S). So Γ=Cay(G,S) is a normal Cayley graph if and only if Aut(G,S)=(AutΓ)1.

Let Γ=Cay(G,S) be an (X,1)-regular Cayley graph of valency 5 such that GXAutΓ. Then S contains at least one involution. Let K=CoreX(G), which is the core of G in X.

Lemma 5.

Assume that K=1. Then (X,G)=(A5,A4) or (S5,S4).

Proof.

Let H be the stabilizer in X of the vertex corresponding to the identity of G. Then H5, HG=1, and X=GH. Let [X:G] be the set of right cosets of G in X. Consider the action of X on [X:G] by the right multiplication. Then we get that X is a primitive permutation group of degree 5 and G is a stabilizer of X. Since Γ has valency 5, |G|=|VΓ|6, and so |X|=|G||H|30. Then we can show XA5 or S5, and then G=A4 or S4, respectively.

Lemma 6.

Suppose that G=A4 and X=A5. Then Γ is the icosahedron graph. Moreover, AutΓ=A5×2 and Γ is not 1-regular.

Proof.

Note that X=GH, where XA5, GA4, and H5. Since X has no nontrivial normal subgroup, Γ is not bipartite. So Γ is the icosahedron graph. Further by Magma , AutΓ=A5×2, so Γ is not 1-regular.

Lemma 7.

Suppose that G=S4 and X=S5. Then the graph Γ is not 1-regular and there is only one isomorphism class of these graphs.

Proof.

Note that G=S4, X=S5, and X=GH. Let H=σ, where σ=(1  2  3  4  5). By considering the right multiplication action of X on the right cosets of G in X, G can be viewed as a stabilizer of X acting on {1,2,3,4,5}. Without lost generality, we may assume that 1 is fixed by G. Take an involution τS. Then, by , τS5NS5(H) and we can identify S with HτHG. Note that τGS4 and NS5(H)=H:Aut(H)=(1  2  3  4  5):(2  3  5  4)5:4; then τ is one of the following: (2  5), (3  5), (2  3), (3  4), (4  5), (2  4), (2  3)(4  5), and (2  4)(3  5). Note H=(1  2  3  4  5). Assume that τ=(2  5); by calculation, we have (2  5):=h1, τ·(1  2  3  4  5)=(1  2)(3  4  5):=h2, τ·(1  3  5  2  4)=(1  3  5  4):=h3, τ·(1  4  2  5  3)=(1  4  2  3):=h4, and τ·(1  5  4  3  2)=(1  5)(2  4  3):=h5. Then H(2  5)H={Hh1,Hh2,Hh3,Hh4,Hh5}={(2  5),(1  5)(2  3  4), (1  4  5  3),(1  2)(3  5  4),(1  3  2  4),(1  5)(2  4  3),(1  4  2  3),(1  3  5  4),(1  2)(3  4  5),(2  5  3  4), (1  5  2  4),(1  4  5)(2  3),(1  3),(1  3  5  2),(1  2  5)(3  4),(2  4),(1  5  4)(2  3),(1  2  3)(4  5), (3  5),(1  5  2)(3  4),(1  4  2  5),(1  4),(1  3  2)(4  5),(1  2  5  3),(2  4  3  5)}. Thus the corresponding S is {(2  5),(2  5  3  4),(2  4),(3  5),(2  4  3  5)} since 1s=1 for each sH(2  5)H. By similar argument, for every τ, we can work out S explicitly, which is one of the following four cases: S1={(2  5),(2  5  3  4),(2  4),(3  5),(2  4  3  5)}, S2={(2  3),(3  4),(4  5),(2  3  4  5),(2  5  4  3)}, S3={(2  3)(4  5),(2  3  5),(2  5  3),(2  4  5),(2  5  4)}, and S4={(2  4)(3  5),(2  4  3),(3  5  4),(2  3  4),(3  4  5)}.

Now let A=AutΓ. We declare that XA. Assume that X=A. Note that G=NA(G)=GAut(G,S); then Aut(G,S)=1. Let σ=(2  5)(3  4). Since σ=(2  5  3  4)(2  5)·(3  5)=(3  4)(2  3)·(2  3  4  5)=(2  3)(4  5)·(2  5  3)·(2  5  4)=(2  4)(3  5)·(2  4  3)·(3  5  4), σG and Sσ=S for any possible S. Therefore σAut(G,S), which leads to a contradiction. So the assertion is right; that is, Γ is not 1-regular.

Let Gi=Si and Γi=Cay(Gi,Si) for i{1,2,3,4}. Set γ=(2  4  5  3), then S1γ=S2 and S3γ=S4. It follows that G1γ=G2 and G3γ=G4, namely, Γ1Γ2 and Γ3Γ4. Now we consider G1=S1. Note that (2  4)=(2  5)(2  4  3  5) and (3  5)=(2  5  3  4)(2  5)·(2  5  3  4)2, then G1=(2  5),(2  5  3  4). Since (2  5  4)=(2  3)(4  5)·(2  3  5), G3=S3=(2  3)(4  5),(2  3  5). On the other hand, (2  5  3  4)4=(2  5)2=((2  5)·(2  5  3  4))3=1 and (2  3  5)3=((2  3)(4  5))2=((2  3)(4  5)·(2  3  5))3=1, then G1S4 and G3A4. By the assumption, Γ3 is not the graph satisfying conditions. So far we get the result that there is only one isomorphism class of graphs when G=S4.

To finish our proof, we need to introduce some definitions and properties. Assume that Γ is an X-vertex transitive graph with X being a subgroup of AutΓ. Let N be a normal subgroup of X. Denote the set of N-orbits in VΓ by VN. The normal quotient ΓN of Γ induced by N is defined as the graph with vertex set VN, and two vertices B, CVN are adjacent if there exist uB and vC such that they are adjacent in Γ. It is easy to show that X/N acts transitively on the vertex set of ΓN. Assume further that Γ is X-edge-transitive. Then X/N acts transitively on the edge set of ΓN, and the valency val(Γ)=mval(ΓN) for some positive integer m. If m=1, then Γ is called a normal cover of ΓN.

Proof of Theorem <xref ref-type="statement" rid="thm1.1">1</xref>.

Let Γ=Cay(G,S) be an (X,1)-regular Cayley graph of valency 5, where GXAutΓ. Then it is trivial to see that Γ is connected. Let N=CoreX(G) be the core of G in X. Assume that N is not trivial. Then either G=N or |G:N|2. The former implies GX; that is, Γ is an X-normal Cayley graph with respect to G. For the case where |G:N|=2, it is easy to verify Γ is an X-bi-normal Cayley graph. Suppose that |G:N|>2; namely, N has at least three orbits on VΓ. Since val(Γ)=5 is a prime and Γ is (X,1)-regular, Γ is a cover of ΓN and G/NX/NAutΓN. We have that ΓN is a Cayley graph of G/N and ΓN is core-free with respect to G/N. Now suppose that N is trivial, then Γ is a core-free one. According to Lemmas 5, 6, and 7, there are two core-free (X,1)-regular Cayley graphs of valency 5 (up to isomorphism) as in Table 1. As far, Theorem 1 holds.

Acknowledgments

The project was sponsored by the Foundation of Guangxi University (no. XBZ110328), the Fund of Yunnan University (no. 2012CG015), NNSF (nos. 11126343, 11226141, and 11226045), and NSF of Guangxi (no. 2012GXNSFBA053010).

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