A complete classification of cubic symmetric graphs of girth 6

Journal of Combinatorial Theory, Series B 99 (2009) 162–184

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Journal of Combinatorial Theory, Series B www.elsevier.com/locate/jctb

A complete classification of cubic symmetric graphs of girth 6 ✩ Klavdija Kutnar a , Dragan Marušiˇc a,b a b

University of Primorska, FAMNIT, Glagoljaška 8, 6000 Koper, Slovenia University of Ljubljana, IMFM, Jadranska 19, 1000 Ljubljana, Slovenia

a r t i c l e

i n f o

a b s t r a c t

Article history: Received 31 October 2007 Available online 2 July 2008

A complete classification of cubic symmetric graphs of girth 6 is given. It is shown that with the exception of the Heawood graph, the Moebius–Kantor graph, the Pappus graph, and the Desargues graph, a cubic symmetric graph X of girth 6 is a normal Cayley graph of a generalized dihedral group; in particular,

Keywords: Cubic graph Symmetric graph s-Regular graph Girth Consistent cycle Automorphism group

(i) X is 2-regular if and only if it is isomorphic to a so-called I kn (t )-path, a graph of order either n2 /2 or n2 /6, which is characterized by the fact that its quotient relative to a certain semiregular automorphism is a path. (ii) X is 1-regular if and only if there exists an integer r with e e prime decomposition r = 3s p 11 . . . pt t > 3, where s ∈ {0, 1}, t  1, and p i ≡ 1 (mod 3), such that X is isomorphic either to a Cayley graph of a dihedral group D 2r of order 2r or X is isomorphic to a certain Zr -cover of one of the following graphs: the cube Q 3 , the Pappus graph or an I kn (t )-path of order n2 /2.

© 2008 Elsevier Inc. All rights reserved.

1. Introductory remarks A graph X is said to be arc-transitive, or symmetric, the term that will be used in this paper, if its automorphism group Aut X acts transitively on the arcs of X . In particular, it is said to be s-regular if its automorphism group Aut X acts regularly on the set of s-arcs of X . Symmetric graphs, and cubic

✩

Supported in part by “Agencija za raziskovalno dejavnost Republike Slovenije,” research program P1-0285. E-mail address: [email protected] (D. Marušiˇc).

0095-8956/$ – see front matter doi:10.1016/j.jctb.2008.06.001

© 2008

Elsevier Inc. All rights reserved.

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symmetric graphs in particular, have received considerable attention over the years (see for example [5–8,13,15,17,25,27,42,43,50]). For example, it is known that there are only five connected symmetric cubic graphs with girth (the length of shortest cycle) less than 6: the complete graph K 4 , the complete bipartite graph K 3,3 , the cube Q 3 , the Petersen graph and the dodecahedron, but there exist infinitely many cubic symmetric graphs of girth 6. With respect to the latter, in 1971 Miller [38] characterized their automorphism groups as two generators groups satisfying certain relations (see [38, Theorem 2.2] for more details). In 1985 Negami [40] showed that a cubic symmetric graph which is s-regular, s  2, of girth 6 is the dual of a triangulation of the torus or the Klein bottle. In 2006 Feng and Nedela [16] showed that, with the exception of the complete graph K 3,3 , the Moebius–Kantor graph GP(8, 3), the Desargues graph GP(10, 3) and the well-known Coxeter graph of order 28, all cubic symmetric graphs of girth g  7 are one-skeletons of trivalent regular maps with face-size g. Most recently, Conder and Nedela [7] proved that a symmetric cubic graph of girth at most 9 is either 1-regular, 2-regular, or belongs to a small family of exceptional graphs. In particular, they proved that with the exception of the Heawood graph, the Pappus graph, and the Desargues graph a cubic symmetric graph of girth 6 is either 1-regular or 2-regular. Certain properties of these exceptional graphs, needed throughout this paper, are gathered in Section 3. It is the object of this paper to complete the analysis of cubic symmetric graphs of girth 6 by proving the following two theorems which give, respectively, a full classification of cubic 2-regular graphs of girth 6 and cubic 1-regular graphs of girth 6. Theorem 1.1. A cubic symmetric graph of girth 6 different from the Moebius–Kantor graph GP(8, 3) is 2-regular if and only if it is isomorphic to an I k2m (t )-path (defined in Section 2.4), a normal Cayley graph Cay(G , {x, y , z}) of a generalized dihedral group G = (Zm × Zk )  Z2 = x, y , z | x2 = y 2 = z2 = (xyz)2 = (xy )m = ( yz)m = (zx)m = 1, (xy )k = ( yz)k = (zx)k  where m > 3, either k = m or k = m/3, and t = m if k is odd and t = m + 1 if k is even. Note that the Moebius–Kantor graph GP(8, 3) is also a Cayley graph of a generalized dihedral group, namely of the group (Z4 × Z2 )  Z2 . Theorem 1.1 implies that cubic 2-regular graphs of girth 6 are of order twice times a square or of order six times a square. This fact was already observed in [16, p. 50]. Theorem 1.2. A cubic symmetric graph X of girth 6 is 1-regular if and only if there exist an integer r with e e prime decomposition r = 3s p 11 . . . pt t > 3, where s ∈ {0, 1}, t  1, and p i ≡ 1 (mod 3), and an integer m such that X is isomorphic to a normal Cayley graph of a generalized dihedral group (Zrm × Zm )  Z2 ; in particular: (i) if m = 1 then X is isomorphic to a Cayley graph Cay( D 2r , {τ , τρ , τρ k+1 }) of a dihedral group D 2r = τ , ρ | τ 2 = ρ r = 1, ρ τ = ρ −1  where r  11 is odd; (ii) if m = 2 then X is a Zr -cover of the cube Q 3 , isomorphic to C Q (k, r ) (defined in the paragraph preceding Proposition 5.6); (iii) if m = 3 then X is a Zr -cover of the Pappus graph, isomorphic to C P (k, r ) (defined in the paragraph preceding Proposition 5.7); 2m (t )-path, where t = m if m is odd and t = m + 1 if m even, (iv) if m  4 then X is a Zr -cover of the I m isomorphic to C I (m, k, r ) (defined in Section 5, in the paragraph preceding the proof of this theorem); where k ∈ Zr∗ satisfies k2 + k + 1 = 0. Group-theoretic and combinatorial techniques are used in the proofs of Theorems 1.1 and 1.2, carried out in Sections 4 and 5, respectively. An essential part of our proofs strategy is the concept of consistent cycles and a beautiful result of Conway [9] which says that given a symmetric graph X of valency d, an arc-transitive subgroup G of Aut X has d − 1 orbits on G-consistent cycles (see Proposition 2.2). (A cycle in a graph is called consistent provided there exists a so-called shunt automorphism

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acting as a one-step rotation of the cycle; for the formal definition see Section 2.3.) It transpires that in a cubic 1-regular graph of girth 6 all consistent cycles are of length 6, and that in a cubic 2-regular graph of girth 6, different from the Moebius–Kantor graph GP(8, 3), consistent cycles are of lengths 6 and 2m  8, see Proposition 4.2. (Note that the lengths of consistent cycles in GP(8, 3) are 8 and 12, see Example 3.3.) For each of these consistent 2m-cycles the corresponding shunt automorphism is semiregular. (It is known that every cubic vertex-transitive graph has a semiregular automorphism [35]. Recently this result have been extended to quartic vertex-transitive graphs [12], but the problem of existence of semiregular automorphisms in vertex-transitive graphs [32] is still open for larger valencies.) In the next section we gather various concepts that are needed in the proofs of our main results in Sections 4 and 5. 2. Preliminaries 2.1. Notation and terminology Throughout this paper graphs are simple and, unless otherwise specified, finite, undirected and connected. For group-theoretic terms not defined here we refer the reader to [41,45,48]. Given a graph X we 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. For adjacent vertices u and v in X , we write u ∼ v and denote the corresponding edge by uv. If u ∈ V ( X ) then N (u ) denotes the set of neighbors of u and N i (u ) denotes the set of vertices at distance i > 1 from u. A graph X is cubic if all of its vertices have valency 3, that is, | N (u )| = 3 for each u ∈ V ( X ). A sequence (u 0 , u 1 , u 2 , . . . , uk ) of distinct vertices in X is called a k-arc if u i is adjacent to u i +1 for every i ∈ {0, 1, . . . , k − 1}. By an n-cycle we shall always mean a cycle with n vertices. A subgroup G  Aut X is said to be vertex-transitive, edge-transitive and arc-transitive provided it acts transitively on the sets of vertices, edges and arcs of X , respectively. The graph X is said to be vertex-transitive, edge-transitive, and arc-transitive if its automorphism group is vertex-transitive, edgetransitive and arc-transitive, respectively. An arc-transitive graph is also called symmetric. A subgroup G  Aut X is said to be k-arc-transitive if it acts transitively on the set of k-arcs, and it is said to be k-regular if it is k-arc-transitive and the stabilizer of a k-arc in G is trivial. Given a transitive group G acting on a set V , we say that a partition B of V is G-invariant if the elements of G permute the parts, that is, blocks of B , setwise. If the trivial partitions { V } and {{ v }: v ∈ V } are the only G-invariant partitions of V , then G is said to be primitive, and is said to be imprimitive otherwise. In the latter case we shall refer to a corresponding G-invariant partition as to an imprimitivity block system of G (see also [3]). If the set V above is the vertex set of a vertextransitive graph X , and B is an imprimitivity block system of G, then clearly any two blocks B , B  ∈ B induce isomorphic vertex-transitive subgraphs. We will use the symbol Zr , both for cyclic group of order r and the ring of integers modulo r. In the latter case, Zr∗ will denote the multiplicative group of units of Zr . By D 2n we denote the dihedral group of order 2n. 2.2. Cubic symmetric graphs In [46] Tutte proved that every finite cubic symmetric graph is s-regular for some s  5. Later on Djokovic´ and Miller [11] proved that a vertex stabilizer in an s-regular subgroup of automorphisms of a cubic symmetric graph is isomorphic to Z3 , S 3 , S 3 × Z2 , S 4 , or S 4 × Z2 depending on whether s = 1, 2, 3, 4 or 5, respectively. It is well known that there are only five connected symmetric cubic graphs with girth less than 6, namely F004A (the complete graph K 4 ), F006A (the complete bipartite graph K 3,3 ), F008A (the cube Q 3 ), F010A (the Petersen graph) and F020A (the dodecahedron), see for example [20], while there exist infinitely many cubic symmetric graphs of girth 6. (Hereafter the notation FnA, FnB, etc. will refer to the corresponding graphs in the Foster census of all cubic symmetric graphs [2,4].)

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The next result on cubic symmetric graphs of girth 6 may be deduced from [7, Theorem 2.3] and [16, Corollary 6.3]. Proposition 2.1. (See [7,16].) Let X be a cubic symmetric graph of girth 6. Then X is bipartite and one of the following holds: (i) (ii) (iii) (iv)

X X X X

is isomorphic to the Heawood graph which is 4-regular; is isomorphic either to the Pappus graph or to the Desargues graph which are both 3-regular; is 2-regular; is 1-regular.

Let n  3 be a positive integer, and let r ∈ {1, . . . , n − 1} \ {n/2}. The generalized Petersen graph GP(n, r ) is defined to have the vertex set V (GP(n, r )) = {u i | i ∈ Zn } ∪ { v i | i ∈ Zn } and the edge set E (GP(n, r )) = {u i u i +1 | i ∈ Zn } ∪ { v i v i +r | i ∈ Zn } ∪ {u i v i | i ∈ Zn }. Note that GP(n, r ) is cubic, and it is easy to see that GP(n, r ) ∼ = GP(n, n − r ). In 1971 Frucht, Graver and Watkins [19] proved that there are only seven generalized Petersen graphs that are symmetric: GP(4, 1), GP(5, 2), GP(8, 3), GP(10, 2), GP(10, 3), GP(12, 5), and GP(24, 5) which are, respectively, 2-, 3-, 2-, 2-, 3-, 2- and 2-regular. Furthermore, GP(8, 3), GP(10, 3) and GP(12, 5) are the only symmetric generalized Petersen graphs of girth 6. 2.3. Consistent cycles

= (u 0 , . . . , u r ) in X is called G-consistent (or just Let X be a graph and G  Aut X . A walk D g consistent if G = Aut X ) if there exists g ∈ G such that u i = u i +1 for i ∈ {0, 1, . . . , r − 1}. The automor-

If D

is a simple closed walk then we say that D

is a phism g is called a shunt automorphism for D.

is called a G-consistent cycle and is G-consistent oriented cycle. The underlying nonoriented cycle of D denoted by D. The following result of Conway [9] implies that an arc-transitive group G of automorphisms of a cubic graph has exactly two orbits in its action on the set of all G-consistent oriented cycles. A written proof of this result is given by Biggs in [1] (see also [36]). Proposition 2.2. (See [1,9].) Let G be a group of automorphisms of a d-valent graph X (d  2). Assume that G is arc-transitive. Let Ω be the set of all G-consistent oriented cycles in X . Then G has exactly d − 1 orbits in its action on Ω . 2.4. Semiregular automorphisms and I kn (t )-paths

For a graph X and a partition W of V ( X ), we let X W be the associated quotient graph of X relative to W , that is, the graph with vertex set W and edge set induced naturally by the edge set E ( X ). The subgraph of X induced by W ∈ W will be denoted by  W . Similarly, we let [ W , W  ], W , W  ∈ W denote the bipartite subgraph of X induced by the edges having one endvertex in W and the other endvertex in W  . Given integers k  1 and n  2 we say that an automorphism of a graph is (k, n)-semiregular if it has k orbits of length n and no other orbit. In the case when W corresponds to the set of orbits of a semiregular automorphism ρ ∈ Aut X , the symbol X W will be replaced by X ρ . Moreover, for W , W  ∈ W we let d( W ) and d( W , W  ) denote the valency of  W  and [ W , W  ], respectively. Let X be a connected graph admitting a (k, n)-semiregular automorphism











−1 ρ = u 00 u 10 · · · un0−1 u 01 u 11 · · · un1−1 · · · uk0−1 uk1−1 · · · unk− , 1

(1)

and let W = { W i | i ∈ Zk } be the set of orbits W i = | s ∈ Zn } of ρ . Using Frucht’s notation [18] X may be represented in the following way. Each orbit of ρ is represented by a circle. Inside a circle corresponding to the orbit W i the symbol n/ T , where T = T −1 ⊆ Zn \ {0}, indicates that for each s ∈ Zn , the vertex u is is adjacent to all the vertices u is+t where t ∈ T . When | T |  2 we use a simplified

{u is

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Fig. 1. The cubic symmetric graph F050A of order 50 given in Frucht’s notation relative to a (5, 10)-semiregular automorphism.

notation n/t, n/(n/2) and n, respectively, when T = {t , −t }, T = {n/2} and T = ∅. Finally, an arrow pointing from the circle representing the orbit W i to the circle representing the orbit W j , j = i, s+ y

labeled by y ∈ Zn means that for each s ∈ Zn , the vertex u is ∈ W i is adjacent to the vertex u j . When the label is 0, the arrow on the line may be omitted. An example illustrating this notation is given in Fig. 1. We end this section with the definition of I kn (t )-paths. Let X be a cubic graph admitting a (k, n)semiregular automorphism ρ given by (1) with k  2 and n  2, and let W i = {u is | s ∈ Zn }, i ∈ Zk , be orbits of ρ , as given in the paragraph following (1). We say that X is an I kn (t )-path if the corresponding quotient graph X ρ is a path W 0 W 1 . . . W k−1 , and, in particular, for each s ∈ Zn we have N (u 0s ) = +2 −2 s {u 0s±1 , u 1s }, N (u is ) = {u is−1 , u is+1 , u is+ } if i is odd and N (u is ) = {u is−1 , u is− , u i +1 } if i is even, for i ∈ Zk \ 1 1 s−2 s+t t s s {0, k − 1}, and N (uk−1 ) = {uk−2 , uk−2 , uk−1 }, where t = n/2, if k is odd, and N (uks −1 ) = {uks −2 , uks± −1 }, where 2t ≡ 2 (mod n) (and so t = 1, or n is even and t = n2 + 1), if k is even. For example, the graph F050A, shown in Fig. 1, is the I 510 (5)-path. Observe that every I kn (t )-path is of girth 6, and that the cycle induced by the vertices in W 0 is ρ -consistent.

2.5. Normal Cayley graphs Given a group G and a generating set S of G such that S = S −1 and 1 ∈ / S, the Cayley graph Cay(G , S ) of G relative to S has vertex set G and edge set { g ∼ gs | g ∈ G , s ∈ S }. Then X = Cay(G , S ) admits a left regular action of G on V ( X ) = G, here after identified with G (this should cost no confusion). Let Aut(G , S ) = {α ∈ Aut G | S α = S } be a group of all those automorphisms of G fixing S set-wise. Note that, by Godsil [21, Lemma 2.1], for a connected Cayley graph X = Cay(G , S ) we have N Aut X (G ) = G  Aut(G , S ). Following Xu [49], X = Cay(G , S ) is called a normal Cayley graph if G is normal in Aut X , that is, if Aut(G , S ) coincides with the vertex stabilizer 1 ∈ G. It follows that for a cubic symmetric normal Cayley graph X = Cay(G , S ), the set S consists of three involutions, say x, y and z, and there exists α ∈ Aut(G , S ) cyclically permuting x, y and z: xα = y, y α = z and zα = x. 2.6. Graph covers In this subsection we give a formal definition of various concepts related to graph covers. Let X be a graph, let r be a positive integer, and let ζ : A ( X ) → S r be a permutation voltage assignment, that is, a function from the set of arcs of X into the symmetric group S r where reverse arcs carry inverse voltages. We thus have a labeling of the arcs of X by permutations in S r such that ζu , v ζ v ,u = id for all pairs of adjacent vertices u , v ∈ V ( X ), where ζu , v denotes the permutation assigned to the arc (u , v ). The voltage assignment ζ extends to walks in X in a natural way. In particular, for any walk

= u 1 u 2 · · · ut of X we let ζ denote the voltage ζu 1 ,u 2 ζu 2 ,u 3 · · · ζut −1 ,ut of D,

that is, the ζ -voltage D D

The covering graph X˜ = Cov( X , ζ ) of X with respect to ζ has vertex set V ( X ) × Zr , and edges of of D. the form (u , s)( v , s ), where uv ∈ E ( X ), s ∈ Zr and s = sζu , v . The graph X is said to be the base graph of X˜ , and the latter is referred to as an r-fold cover of X . The set of vertices (u , 0), (u , 1), . . . , (u , r − 1) is called the fibre of u. The subgroup K of all those automorphisms of X˜ which fix each of the fibres setwise is called the group of covering transformations. The graph X˜ is also called a K -cover of X . It is a simple observation that the group of covering transformations of a connected covering graph acts semiregularly on each of the fibres. In particular, if the group of covering transformations is regular on the fibres of X˜ , we say that X˜ is a regular K -cover (in short, a regular cover). In this case, the voltage group Im(ζ ) is a regular group of degree r isomorphic to the group of covering transformations K . Given a spanning tree T of the graph X , a voltage assignment ζ is said to be T -reduced if the voltages on the tree arcs equal the identity element. In [22] it is shown that every regular cover X˜ of

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a graph X can be derived from a T -reduced voltage assignment ζ with respect to an arbitrary fixed spanning tree T of X . If ζ is reduced then the corresponding covering graph Cov( X , ζ ) is connected if and only if the voltages on the cotree arcs generate the voltage group K . Another viewpoint of covering graphs is as follows. Let Y be a graph and K a subgroup of Aut Y with the set W of orbits. If the quotient projection ℘ : Y → Y W relative to the orbits of K is valency preserving then ℘ is called a regular covering projection with the group of covering transformation K , further X˜ = Y is called the covering graph and X = Y W is called the base graph. We say that α ∈ Aut X ˜ ∈ Aut X˜ , called a lift of α , such that lifts to an automorphism of X˜ if there exists an automorphism α α˜ ℘ = ℘ α . The problem whether an automorphism α of X lifts or not is expressed in terms of voltages ¯ from the set of voltages of fundamental closed as follows. Given α ∈ Aut X , we define a function α walks based at a fixed vertex v ∈ V ( X ) to the voltage group K by the rule ζCα¯ = ζC α where C ranges over all fundamental closed walks at v, and ζC and ζC α are the voltages of C and C α , respectively. ¯ does not depend on the choice of the base vertex, and the fundamental Note that if K is abelian, α closed walks at v can be substituted by the fundamental cycles generated by the cotree arcs of X . The next two propositions provide information about the relationship between automorphisms of graph covers and their base graphs. The first one is taken from [29, Theorem 4.2, Corollary 4.3], whereas the second one may be deduced from [31, Corollaries 9.4, 9.7, 9.8]. Proposition 2.3. (See [29].) Let X be a graph and let X˜ = Cov( X , ζ ) be a regular cover of X with respect to ¯ extends to an automorphism of the voltage assignment ζ . Then, an automorphism α of X lifts if and only if α the group of covering transformations. Proposition 2.4. (See [31].) Let X˜ be a regular K -cover of X . A group G  Aut X , acting semiregularly on V ( X ), lifts as a direct product G × K if and only if there exists a voltage assignment ζ : A ( X ) → K such that for each α ∈ G and each walk W of X we have that ζ W = ζ W α . Let ℘1 : X˜ 1 → X and ℘2 : X˜ 2 → X be covering projections. The two covering projections ℘1 : X˜ 1 → X ˜ : X˜ 1 → X˜ 2 and and ℘2 : X˜ 2 → X are said to be isomorphic if there exists a graph isomorphism α β ∈ Aut X such that α˜ ℘1 = ℘2 β . In this case, two covering graphs X˜ 1 and X˜ 2 are also isomorphic. The following proposition is well known. Proposition 2.5. (See [24].) Two connected regular Zr -coverings Cov( X , ζ ) and Cov( X , ζ  ) where ζ and ζ  are T -reduced, are isomorphic if and only if there exists an automorphism σ ∈ Aut(Zr ) such that ζ(σu , v ) = ζ(u , v ) for any cotree arc (u , v ) of X . We end this subsection with Lorimer’s result [28, Theorem 9] about regular coverings of cubic symmetric graphs. Proposition 2.6. (See [28].) Let X be a connected cubic symmetric graph and let G be an s-regular subgroup of Aut X for some s  1. Let N be a normal subgroup of G and let W be the set of its orbits. If |W | > 2 then N is the kernel of G acting on W and G / N is an s-regular subgroup of Aut X W . Furthermore, X is a regular covering of X W with the covering transformation group N. 3. The exceptional graphs By [47, Theorem 3.1], an arc-transitive graph of girth g can be at most ( g + 2)/2-arc-transitive. Hence a cubic symmetric graph of girth 6 is at most 4-arc-transitive. By Proposition 2.1 the only cubic symmetric graphs of girth 6 that are 3-regular or 4-regular are the Heawood graph F014A, the Pappus graph F018A and the Desargues graph F020B. These three graphs together with the Moebius–Kantor graph F016A, the unique cubic 2-regular graph of order 16, will be refereed to as the exceptional graphs of girth 6. These graphs are exceptional in the sense that they are the only cubic symmetric graphs of girth 6 in which a 2-arc lies on more than one 6-cycle, as is seen by the following result proven by Feng and Nedela [16, Lemma 4.2].

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Fig. 2. The Heawood graph F014A.

Fig. 3. The Moebius–Kantor graph F016A.

Proposition 3.1. (See [16].) Let X be a connected cubic symmetric graph of girth 6 and let c be the number of 6-cycles passing through an edge in X . Then, c = 2, 4, 6, or 8. If c > 2 then X is isomorphic to one of the following four graphs: the Heawood graph F014A, the Moebius–Kantor graph F016A, the Pappus graph F018A, and the Desargues graph F020B. Certain properties of the four exceptional graphs that will be needed in the subsequent sections, are considered in Examples 3.2–3.5 below. Example 3.2. The Heawood graph F014A, shown in Fig. 2, is the unique cubic symmetric graph of order 14. It is 4-regular and of girth 6. Note that F014A is the Cayley graph Cay( D 14 , {x, xy , xy 3 }) of the dihedral group D 14 = x, y | x2 = y 7 = 1, y x = y −1 . Since the permutations, whose cycle decompositions are, respectively,

(u 0 u 8 u 7 u 11 u 12 u 13 u 10 u 5 )(u 1 u 9 u 6 u 4 )(u 2 u 3 ) and

(u 0 u 1 u 2 u 3 u 4 u 5 )(u 8 u 12 u 6 u 9 u 11 u 10 )(u 7 u 13 ), are automorphisms of F014A, it follows that the lengths of consistent cycles are 6 and 8. For more information on the Heawood graph see [2,4,23]. Example 3.3. The Moebius–Kantor graph F016A, also known as the generalized Petersen graph GP(8, 3), shown in Fig. 3, is the unique cubic symmetric graph of order 16. It is 2-regular, has girth 6 and it is the Cayley graph Cay( D 16 , {x, xy , xy 3 }) of D 16 = x, y | x2 = y 8 = 1, y x = y −1 . (It is the only generalized Petersen graph except for the prisms GP(n, 1), n  3, with that property.) It is also a Cayley graph of the group (Z4 × Z2 )  Z2 . It has a hexagonal embedding in a torus and an octagonal embedding in a double torus (see [10]). Since the permutations, whose cycle decompositions are, respectively,

(u 0 u 8 u 7 u 6 u 11 u 14 u 4 u 5 )(u 1 u 9 u 13 u 2 u 15 u 12 u 3 u 10 ) and

(u 0 u 1 u 12 u 13 u 7 u 6 u 11 u 15 u 9 u 3 u 4 u 5 )(u 8 u 2 u 14 u 10 ), are automorphisms of F016A, it follows that the lengths of consistent cycles are 8 and 12. For more information on this graph see [2,4,33]. Example 3.4. The Pappus graph F018A, shown in Fig. 4, is the unique cubic symmetric graph of order 18. It is 3-regular, has girth 6 and it is the Cayley graph Cay((Z3 × Z3 )  Z2 , {xz, yz, z}) of the

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Fig. 4. The Pappus graph F018A.

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Fig. 5. The Desargues graph F020B.

group H = (Z3 × Z3 )  Z2 = x, y , z | x3 = y 3 = z2 = 1, xz = x−1 , y z = y −1 , xy = yx. Its automorphism group, isomorphic to Z3  (( S 3 × S 3 )  Z2 ), has the unique 1-regular subgroup which is isomorphic to H  Z3 . Since the permutations, whose cycle decompositions are, respectively,

(u 0 u 1 u 13 u 16 u 10 u 5 )(u 8 u 2 u 17 u 11 u 14 u 4 )(u 7 u 3 u 12 u 6 u 9 u 15 ) and

(u 11 u 15 u 12 u 17 u 9 u 14 u 10 u 5 u 0 u 1 u 2 u 6 )(u 16 u 4 u 8 u 13 u 3 u 7 ), are automorphisms of F018A, it follows that the lengths of consistent cycles are 6 and 12. Observe that the above shunt automorphism of a consistent oriented 6-cycle of F018A is (3, 6)-semiregular. Considering the corresponding quotient graph it may be seen that F018A is the I 36 (3)-path (see Fig. 9). For more information on this graph we refer the reader to [2,4,14,15,26]. Example 3.5. The Desargues graph F020B, also known as the generalized Petersen graph GP(10, 3), shown in Fig. 5, is one of the two cubic symmetric graphs of order 20. It is 3-regular, has girth 6 and it is, by [39, Theorem 1], a non-Cayley graph. Since the permutations, whose cycle decompositions are, respectively,

(u 0 u 1 u 2 u 6 u 7 u 8 )(u 5 u 13 u 3 u 11 u 14 u 12 )(u 4 u 19 )(u 10 u 17 u 9 u 18 u 16 u 15 ) and

(u 0 u 1 u 2 u 6 u 11 u 18 u 16 u 15 u 10 u 5 )(u 8 u 13 u 3 u 7 u 19 u 9 u 14 u 12 u 17 u 4 ), are automorphisms of F020B, it follows that the lengths of consistent cycles are 6 and 10. For more information on this graph see [4,26]. 4. Proving Theorem 1.1 The following result may be extracted from [16, Lemma 6.1]. Proposition 4.1. (See [16].) Let X be a cubic symmetric graph of girth 6 and let G be a 1-regular subgroup of Aut X . If G = α , β | α 3 = β 2 = (α β α −1 β −1 )3 = 1, . . . then G is a proper subgroup of Aut X . Propositions 3.1 and 4.1 combined together enable us to prove the following result about consistent oriented cycles in cubic symmetric graphs of girth 6, needed throughout the rest of this paper.

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Proposition 4.2. Let X be a cubic symmetric graph of girth 6. Then the following statements hold: (i) if X is the Heawood graph F014A, the Moebius–Kantor graph F016A, the Pappus graph F018A, or the Desargues graph F020B, then consistent oriented cycles are, respectively, of length 6 and 8, of length 8 and 12, of length 6 and 12, and of length 6 and 10; (ii) if X is 2-regular different form the Moebius–Kantor graph F016A then consistent oriented cycles are of length 6 and n > 6; (iii) if X is 1-regular then all consistent oriented cycles are of length 6. Proof. Part (i) follows by Examples 3.2–3.5. We may therefore assume that X is neither of these graphs. Let C = u 0 u 1 u 2 u 3 u 4 u 5 u 0 be an oriented 6-cycle in X , and let v i ∈ N (u i ) \ V (C ), i ∈ Z6 . Suppose first that X is 2-regular. Then Proposition 3.1 implies that each 2-arc in X is contained in exactly one 6-cycle. Since X is 2-regular, there exists an automorphism α ∈ Aut X that maps the 2-arc (u 0 , u 1 , u 2 ) to the 2-arc (u 1 , u 2 , u 3 ). Since the underlying cycle of C is the only 6-cycle passing = u i +1 for every i ∈ Z6 . It follows that through the 2-arc (u 1 , u 2 , u 3 ), we have C α = C , and so u α i oriented 6-cycles in X are consistent. Moreover, since X is 2-regular there exists an automorphism of X which maps C to its opposite oriented 6-cycle u 0 u 5 u 4 u 3 u 2 u 1 u 0 , and therefore oriented 6-cycles in X form a single orbit of consistent oriented cycles. Part (ii) now follows by Proposition 2.2. Finally, to prove part (iii), assume that X is 1-regular. By Proposition 3.1 there exists a 6-cycle C  , different from C such that C and C  are the only 6-cycles containing the arc (u 5 , u 0 ). Let α ∈ Aut X u 0 act on the neighbors of u 0 according to the rule ( v 0 u 5 u 1 ) and let β act on the vertices incident with the edge u 5 u 0 according to the rule (u 4 u 1 )( v 5 v 0 ) or the rule (u 4 v 0 )(u 1 v 5 ). In other words, β fixes C and C  , or interchanges C and C  . If β fixes C and C  then one can easily see that α β α −1 β −1 is of order 3, contradicting Proposition 4.1. Hence β interchanges C and C  , and so β α fixes C and takes the arc (u 4 , u 5 ) to the arc (u 5 , u 0 ). It follows that C is an oriented consistent 6-cycle. Moreover, using the fact that X is 1-regular one can easily see that the two oriented cycles of any 6-cycle in X belong to different orbits of consistent oriented cycles. Again the result follows from Proposition 2.2. 2 Using Proposition 4.2 we can now prove the following three lemmas. Lemma 4.3. Let X be a cubic 2-regular graph of girth 6 different from the Moebius–Kantor graph F016A. Then

n > 6, is (k, n)-semiregular where k  2 and X is the shunt automorphism of a consistent oriented n-cycle D, an I kn (t )-path. Moreover, if k = 2 then X is the generalized Petersen graph GP(12, 5) ∼ = GP(12, 7).

in such a way that D

= u 0 u 1 u 2 · · · un−1 u 0 and let Proof. Denote the vertices of D 0 0 0 0 0

ρ be a shunt

in other words, (u s )ρ = u s+1 for every s ∈ Zn . Since X is different automorphism associated with D; 0 0 from F016A, Proposition 3.1 implies that each 2-arc in X lies on exactly one 6-cycle. Note that ρ is 0 1 n of order n, for otherwise ρ = 1 fixes the (n − 1)-arc (u 0 , u 0 , u 20 , . . . , un0−1 ), contradicting 2-regularity

) ∈ W. of X . Let W be the set of orbits of ρ in its action on V ( X ) and let |W | = k. Clearly, V ( D Moreover, since the only cubic arc-transitive circulants are the Cayley graphs Cay(Zn , {±1, n/2}) which are of girth 4, we have that |W | = k  2. The lemma will be proven by induction on k. In view of the structure of the graphs in question the basis of induction encompasses all values of k from {2, 3, 4, 5}.

) and W 1 is the orbit adjacent to W 0 . Let the orbits in W be labeled in such a way that W 0 = V ( D Since X is cubic and ρ is of order n, a standard counting argument shows that the orbits in W can be of lengths n, n/2 and n/3. Clearly, | W 0 | = n. Let the labeling of vertices in W 1 = {u 1s | s ∈ Zn } follow the action of ρ , that is, (u 1s )ρ = u 1s+1 and let u 0s ∼ u 1s for every s ∈ Zn . Suppose first that W 1 is of length n/3. Then |W | = 2, and for s+n/3

every s ∈ Zn we have that u 1s = u 1 since n/3 > 2, the 2-arc 1+n/3 2+n/3 2 2 1 u 10 u 11 u 0 u0 u1 u0 u0 , s ∈ Zn we have that u 1s =

s+2n/3

= u1

1+n/3 (u 10 , u 11 , u 0 )

s+n/3

s+2n/3 , u0 }. But then n / 3 1+n/3 1 1 0 u 00 u 01 u 0 u 0 u 1 u 0 u 0 and

, and thus N (u 1s ) = {u 0s , u 0

lies on two different 6-cycles

a contradiction. Now suppose that W 1 is of length n/2. Then, for every s+n/2

u1

s+n/2

, and thus u 0s , u 0

1+n/2

∈ N (u 1s ). But then the 2-arc (u 10 , u 11 , u 0

) lies

K. Kutnar, D. Marušiˇc / Journal of Combinatorial Theory, Series B 99 (2009) 162–184 n/2 1+n/2

171

1+n/2 2+n/2

on two different 6-cycles u 00 u 01 u 0 u 0 u 11 u 10 u 00 and u 10 u 11 u 0 u0 u 21 u 20 u 10 , a contradiction. Hence W 1 is of length n, d( W 0 , W 1 ) = 1 and [ W 0 , W 1 ] = nK 2 . If |W | = k = 2 then ρ is (2, n)-semiregular, and so X is the generalized Petersen graph. It follows that X is isomorphic to GP(12, 5) ∼ = GP(12, 7) which is clearly the I 212 (7)-path (see Section 2.2). Thus we may assume that |W |  3. Note that in this case the valency of the subgraph  W 1  is either 0 or 1, and that there exists an orbit in W which is adjacent to W 1 and different from W 0 . Now consider the 2-arc (u 00 , u 10 , u 20 ). It lies on the unique 6-cycle, call it C . The uniqueness of this 6-cycle implies that u 00 , u 10 and u 20 are the only vertices from W 0 contained in C , and thus u 01 and u 21 are also contained in C . Also, since the valency of the subgraph  W 1  is either 0 or 1, the sixth vertex u of C , adjacent to both u 01 and u 21 , must belong to an orbit adjacent to W 1 different from W 0 , call it W 2 . It also follows that d( W 1 ) = 0, d( W 1 , W 2 ) = 2 and | W 2 | = n. Moreover, without loss of generality we may label the vertices in W 2 = {u 2s | s ∈ Zn } in such a way that (u 2s )ρ = u 2s+1 and

u 1s ∼ u 2s for every s ∈ Zn , and that u = u 22 . These imply that N (u 1s ) = {u 0s , u 2s , u 2s+2 } for every s ∈ Zn . Hence, if k = 3 then ρ is (3, n)-semiregular. In addition, since X is cubic and d( W 1 , W 2 ) = 2, we must s+n/2

} for every s ∈ Zn , which implies that X is an have that n is even and that N (u 2s ) = {u 1s , u 1s−2 , u 2 I n3 (n/2)-path. Now we may assume that |W | > 3, and so vertices in W 2 are also adjacent to vertices in some orbit different from W 1 , call it W 3 . As with the orbits W i , i  2 we may assume that the labeling of the vertices in W 3 = {u 3s | s ∈ Zn } is done in such a way that (u 3s )ρ = u 3s+1 and u 2s ∼ u 3s for s+n/3

every s ∈ Zn . We claim that | W 3 | = n. Suppose first that | W 3 | = n/3. Then u 3s = u 3 s+n/3 s+2n/3 {u 2s , u 2 , u2 }.

s+2n/3

= u3

for every s ∈ Zn , and thus = Consider the 2-arc It lies on the unique 6-cycle, say C , different from the two cycles passing, respectively, through the 2-arcs 2 0 0 2 2 2 (u 01 , u 02 , u − 1 ) and (u 1 , u 2 , u 1 ). Therefore we must have that both u 3 and u 3 are contained in C . But 0 2 since N (u 3 ) ∩ N (u 3 ) = ∅, this is impossible, and so | W 3 | = n/3. The assumption that | W 3 | = n/2 leads to a contradiction in a similar way, and so | W 3 | = n as claimed. Furthermore, the 6-cycle containing the 2-arc (u 02 , u 01 , u 22 ) clearly contains u 03 and u 23 and since d( W 2 , W 3 ) = 1, either there exists t ∈ Zn \ {0} such that 2t ≡ 2 (mod n) and u 3s ∼ u 3s+t for every s ∈ Zn , or there exists u ∈ W 4 , where W 4 ∈ W is different from W i , i ∈ {0, 1, 2, 3}, adjacent to both u 03 and u 23 . In the first case k = 4, and so X is an I n4 (t )-path. In the second case the vertices in W 4 = {u 4s | s ∈ Zn } may be laN (u 3s )

(u 02 , u 01 , u 22 ).

beled in such a way that (u 4s )ρ = u 4s+1 and u 3s ∼ u 4s for every s ∈ Zn and that u = u 24 . It follows that

N (u 3s ) = {u 2s , u 4s , u 4s+2 } for every s ∈ Zn . This completes the basis of our induction. For the inductive step, let us now assume that for every j < i, where i ∈ Zk \ {0, 1, 2, 3, 4}, the orbit W j ∈ W is of length n and that N (u sj −1 )∩ W j = {u sj }, s ∈ Zn , if j is odd and N (u sj −1 )∩ W j = {u sj , u sj+2 }, s ∈ Zn , if j is even. Moreover, if j = 0 then W j is an independent set of vertices. (In other words, the

i −1

induced subgraph  j =0 W j  of X is almost identical with the I ni (t )-path, with the only difference in the last orbit.) Without loss of generality we may label the vertices in the orbit W i = {u is | s ∈ Zn } in

such a way that (u is )ρ = u is+1 and u is−1 ∼ u is for every s ∈ Zn . Suppose that W i is of length different from n. First suppose that W i is such that d( W i −2 , W i −1 ) = 1 (that is, i is odd). Then, by hypothesis, d( W i −3 , W i −2 ) = 2. The 2-arc (u 0i −2 , u 0i −3 , u 2i −2 ) lies on the unique 6-cycle, say C , which is different

2 ) and (u 0i −3 , u 2i −2 , u 2i −3 ), respectively. from the two cycles passing through the 2-arcs (u 0i −3 , u 0i −2 , u − i −3 0 2 Therefore, both u i −1 and u i −1 are contained in C , and so N (u 0i −1 ) ∩ N (u 2i −1 ) = ∅. Moreover, since

by hypothesis i  k − 1, we have that N (u 0i −1 ) ∩ N (u 2i −1 ) ⊆ V ( W i ). Therefore when | W i | = n/3 or | W i | = n/2 we have n/3 = 2 or n/2 = 2, respectively, both contradicting the fact that n > 6. Hence −2 s s n | W i | = n and u is− 1 , u i −1 ∈ N (u i ) for every s ∈ Zn . If i = k − 1 then X is an I k (n/2)-path. Otherwise there exists an orbit W i +1 adjacent to W i that is different from W i −1 . Clearly, the vertices in this orbit may be labeled in such a way that W i +1 = {u is+1 | s ∈ Zn } and u is ∼ u is+1 , where (u is+1 )ρ =

+1 u is+ . 1 Now suppose that W i is such that d( W i −2 , W i −1 ) = 2 (that is, i is even). Then d( W i −3 , W i −2 ) = 1. Using the fact that X is cubic we get that W i −2 and W i are the only orbits in W adjacent to W i −1 ,

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K. Kutnar, D. Marušiˇc / Journal of Combinatorial Theory, Series B 99 (2009) 162–184 s+n/3

and that  W i −1  is an independent set of vertices. Suppose that | W i | = n/3. Then u is = u i s+2n/3 ui

and

N (u is )

=

s+n/3 s+2n/3 {u is−1 , u i −1 , u i −1 }

it must lie on a 6-cycle and

=

n /3 for every s ∈ Zn . Consider the 2-arc (u 0i −1 , u 0i , u i −1 ). Since n /3 0 d( W i −3 , W i −2 ) = 1, we get that ∅ = N (u i −2 ) ∩ N (u i −2 ) ⊆ W i −1 or ∅ = n/3−2 n/3−2 2 ∅ = N (u i −2 )∩ N (u 0i −2 ) ⊆ W i −1 or ∅ = N (u i −2 )∩ N (u − ) ⊆ W i −1 . But i −2

2 N (u i −2 )∩ N (u − ) ⊆ W i −1 or i −2 since n > 6, one can easily see that none of these can occur. Hence, we may assume that | W i | = n/2, n /3

s+n/2

s+n/2

n /2

and u is−1 , u i −1 ∈ N (u is ) for every s ∈ Zn . Then (u 0i −1 , u 0i , u i −1 ) is a 2-arc, and and so u is = u i the facts that it lies on a 6-cycle and that d( W i −3 , W i −2 ) = 1 combined together imply that ∅ = n/2−2

2 N (u i −2 ) ∩ N (u 0i −2 ) ⊆ W i −1 or ∅ = N (u i −2 ) ∩ N (u − i −2 ) ⊆ W i −1 or ∅ = N (u i −2 n /2

n /2

n/2−2 2 ∅ = N (u i −2 ) ∩ N (u − ) ⊆ W i −1 . i −2 [ W i −1 , W i ] = nK 2 . By induction

as required.

2

) ∩ N (u 0i −2 ) ⊆ W i −1 or

Again, since n > 6 none of these can occur, and so | W i | = n and

ρ is (k, n)-semiregular and it easily follows that X is an I kn (t )-path,

Lemma 4.4. Let X be a cubic 2-regular graph of girth 6 different from the Moebius–Kantor graph F016A with

n > 6. Then X is an I n (t )-path and n is even. Moreover the following hold: a consistent oriented n-cycle D, k (i) if k is odd then n ≡ 2 (mod 4); (ii) if k is even and t = n/2 + 1 then n ≡ 0 (mod 4);

where k is the number of orbits of a shunt automorphism of D. Proof. Since X is different from F016A, Proposition 3.1 implies that each 2-arc lies on a unique

is (k, n)-semiregular and 6-cycle. Moreover, Lemma 4.3 implies that the shunt automorphism ρ of D that X is an I kn (t )-path. Therefore X ρ is a path of length k  2. For k = 2 we get that X is the generalized Petersen graph GP(12, 7) ∼ = GP(12, 5), and so (ii) holds. We may therefore assume that k  3. Let W i , i ∈ Zk , be the labeling of the orbits of ρ such that W i is adjacent to W i +1 in X ρ for every i ∈ Zk \ {k − 1}. Since X is an I kn (t )-path, the vertices of X may be labeled in such a way that u is ∈ W i ,

= u 0 u 1 · · · un−1 u 0 , and N (u s ) = {u s , u s , u s+2 } for every where (u is )ρ = u is+1 , for i ∈ Zk and s ∈ Zn , D 0 2 0 0 0 1 0 2 s ∈ Zn . By Proposition 2.1, n is even. Now, to prove part (i), assume that k is odd and that n ≡ 0 (mod 4). Then n/2 is even and since X is an I kn (t )-path, the subgraph  W k−2 , W k−1  is a disjoint union of two graphs Y and Y  of order n each having vertices of valency 2 (vertices from W k−2 ) and valency 3 (vertices from W k−1 ). Since X is 2-regular, there exists an automorphism γ mapping the 2-arc (u 00 , u 10 , u 20 ) to the 2-arc (uk0−2 , uk0−1 , uk−−22 ). Using the fact that each 2-arc in X lies on the unique 6-cycle one can see that

γ = Y or D

γ = Y  . But since  W 0  = D is of valency 2 and none of Y and Y  is regular this is D impossible. Hence n ≡ 2 (mod 4) when k is odd. Finally, to prove part (ii), assume that k is even, n ≡ 2 (mod 4) and that t = n/2 + 1. Then t is even and  W k−1  is a disjoint union of two cycles uk0−1 ukt −1 uk2t−1 · · · uk−−2t1 uk−−t 1 uk0−1 +t 1+2t −2t 1−t 1 and uk1−1 uk1− u · · · uk1− u u , both of length n/2. But since each 2-arc in X lies on the 1 k−1 1 k−1 k−1 unique 6-cycle, one can see that the automorphism that maps the 2-arc (u 00 , u 10 , u 20 ) to the 2-arc

to u 0 ut u 2t · · · u −2t , u −t u 0 , a contradiction which completes the (uk0−1 , ukt −1 , uk2t−1 ) maps D k−1 k−1 k−1 k−1 k−1 k−1 proof of Lemma 4.4. 2 Observe that Lemma 4.4 implies that if k is even the subgraph induced by the vertices in the last orbit of the 2-regular I kn (t )-path is a n-cycle. Given an abelian group A, the generalized dihedral group, we denote it here by Dih( A ), is the semidirect product A  Z2 , where Z2 is the cyclic group of order 2, and the generator of Z2 maps elements of A to their inverses. In particular, Dih(Zr ) is the dihedral group D 2r of order 2r. We will however use the standard notation for dihedral groups. Then, next lemma gives a more detail information on the structure of cubic 2-regular graphs of girth 6.

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173

Lemma 4.5. Let X be a cubic 2-regular graph of girth 6 different from the Moebius–Kantor graph F016A. Then X is the I kn (t )-path, where n > 6 is even, k = n/2 or k = n/6, and t = n/2 when k is odd and t = n/2 + 1 when k is even. Moreover, X is a normal Cayley graph of Dih(Zn/2 × Zk ) = x, y , z | x2 = y 2 = z2 = (xyz)2 = (xy )n/2 = ( yz)n/2 = (zx)n/2 = 1, (xy )k = ( yz)k = (zx)k .

be a consistent oriented n-cycle, n > 6, of X . Lemmas 4.3 and 4.4 combined together Proof. Let D

is (k, n)-semiregular and X is the I n (t )-path for imply that n is even, a shunt automorphism ρ of D k

is the unique oriented n-cycle associated with the an appropriate t. Observe also that for odd k, D shunt automorphism ρ . Let W i , i ∈ Zk , be the labeling of the orbits of ρ such that W i is adjacent to W i +1 in X ρ for every i ∈ Zk \ {k − 1}. Further the vertices of X may be labeled in such a way that u is ∈ W i , where (u is )ρ = u is+1 , for i ∈ Zk and s ∈ Zn . Let Ω be the set of nonoriented consistent n-cycles of X . Consider the natural action of Aut X on Ω . Since X is 2-regular and | V ( X )| = nk, the automorphism group Aut X is of order 6kn. Moreover, one can easily see that Aut X D = D 2n , and so the standard counting argument shows that   OrbAut X ( D ) = | Aut X : Aut X D | = 3k.

Applying Proposition 2.2 we get that there are precisely 3k consistent n-cycles in X (and so 6k consistent oriented n-cycles). By Lemma 4.4, n is even and hence [ W i , W i +1 ] is a union of two n-cycles for every odd i ∈ Zk \ {k − 1}. For k even let C be the induced subgraph  W k−1 . We identify a particular set of k cycles of length n of X as follows:



C=

if k odd, {[ W i , W i +1 ] | i ∈ Zk \ {k − 1} odd} ∪ { D } {[ W i , W i +1 ] | i ∈ Zk \ {k − 1} odd} ∪ { D } ∪ {C } if k even.

We claim that C is a block of imprimitivity for the action of Aut X on Ω . To prove this we first need to show that all n-cycles in C are consistent. For this purpose let us label the n-cycles in C in such a way that C 0 = D, and the two n-cycles induced by the vertices of even labels and of odd labels in [ W i , W i +1 ], i ∈ Zk \ {k − 1} odd, are, respectively, C i and C i +1 . Moreover, when k is even we let C k−1 = C . Since X is 2-regular, for each 2-arc ai = (u 0i +1 , u 0i , u 2i +1 ), i ∈ Zk \ {k − 1} odd, there exists

βi ∈ Aut X mapping the 2-arc (u 00 , u 10 , u 20 ) to ai . Furthermore if k is even, the 2-regularity of X implies that there also exists an automorphism β ∈ Aut X which maps (u 00 , u 10 , u 20 ) to (uk0−1 , ukt −1 , uk2t−1 ). Using β

the fact that each 2-arc in X lies on a unique 6-cycle one can easily see that C 0 i = C i , and that when k β

ρ

is even C 0 = C k−1 . Moreover, since C i = C i +1 , for every odd i ∈ Zk \ {k − 1}, it follows that C i ∈ Ω for every i ∈ Zk . Therefore C ⊆ Ω , and it remains to show that C is a block of imprimitivity for the action of Aut X on Ω . First observe that C is a 2-factor of X , and so also C γ must be a 2-factor of X for every γ ∈ Aut X . Now assume that there exists γ ∈ Aut X such that C γ ∩ C = ∅. Then for some γ i , j ∈ Zk we have that C i = C j . Assume first that j ∈ / {0, k − 1} is odd. Then C j = u 0j u 0j+1 u −j 2 · · · u −j+21 u 0j γ and since C is a 2-factor, one can see that for even s, none of the edges u sj −l u sj −l−1 ,

(2)

where l ∈ {0, 1, . . . , j − 1} and none of the edges u sj +l u sj +l+1 ,

(3)

where l ∈ {1, 2, . . . , k − j − 2} if k is even and l ∈ {1, 2, . . . , k − j − 3} if k is odd, is contained in an n-cycle in C γ . This implies that C 0 , C 2r +1 ∈ C γ , where 2r + 1 ∈ {1, 3, 5, . . . , k − 3} ∪ {k − 1} if k is even and 2r + 1 ∈ {1, 3, 5, . . . , k − 2} if k is odd. Now the fact that C 0 ∈ C γ implies that also for odd s, none of the edges in (2) and (3) is contained in an n-cycle in C γ , and thus C γ = C . Using the same line of argument it may be seen that also when j ∈ {0, k − 1} or when j is even we get that C γ = C , and hence C is a block as claimed. Furthermore, since |Ω| = 3k, it follows that there exist two more blocks, say D and E , each containing k consistent n-cycles, such that Ω = C ∪ D ∪ E .

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Clearly each 2-arc in X lies on a consistent n-cycle. Moreover, since |Ω| = 3k and | V ( X )| = kn, each vertex lies on exactly three consistent n-cycles, one from each of the three sets C , D and E . As C , D and E are blocks of imprimitivity of Aut X it follows that Aut X acts on the set {C , D , E } as a group of degree 3. Therefore there exists a normal subgroup G of Aut X such that either Aut X /G ∼ = S 3 or Aut X /G ∼ = Z3 . But the latter cannot occur. Namely, since each vertex of X lies on three consistent n-cycles and the stabilizer of a vertex is isomorphic to S 3 , there exists an automorphism that fixes C and interchanges D and E . Hence Aut X /G ∼ = S 3 and since | Aut X | = 6kn, the group G is of order nk. Since G is a kernel of the action of Aut X on {C , D , E }, it fixes C , D and E setwise. But then the vertex stabilizer G u = 1 is trivial for each vertex u ∈ V ( X ). The standard counting argument shows that |OrbG (u )| = |G : G u | = |G | = nk. Therefore G is regular on X and since it is normal in Aut X , the graph X is a normal Cayley graph of G. Since X is symmetric, the observations on symmetric normal Cayley graphs given in Section 2.5 imply that X = Cay(G , S ) where S is a set of three involutions, say S = {x, y , z}, and there exists α ∈ Aut G such that xα = y, y α = z and zα = x. Let X C be the quotient graph of X relative to C , that is, the graph with vertex set C and edge set induced by the edge set E ( X ) in a natural way. Clearly, G acts on X C . Further, applying Lemma 4.4 one can see that X C is a cycle of length k. It follows that G acts on X C as a dihedral group of order 2k, and so there exists a normal subgroup K of G such that G / K ∼ = D 2k . Since ρ 2 fixes each n-cycle in C , 2 2 we have that ρ ∈ K . Also, since |G | = kn, | D 2k | = 2k and ρ is of order n/2, it follows that K = ρ 2 . Let u ∈ V ( X ). Then without loss of generality we may assume that u ρ = L xy u, and so ρ −2 L xy ∈ G u (where L G denote the left regular representation of G). But since G is regular on V ( X ), it follows that ρ 2 = L xy . Using the fact that ρ 2 is of order n/2 we get that (xy )n/2 = 1. (Observe that the group relation (xy )n/2 gives rise to n-cycles in C .) By symmetry (using the automorphism α ) we get 2

(xy )n/2 = 1,

( yz)n/2 = 1 and (zx)n/2 = 1.

The 6-cycles in X arise either from xyzyxz or from (xyz)2 . In the first case xzx = yzy and applying the automorphism α we get that yxy = zxz. But if the vertices of X are relabel in such a way that u 10 = 1 ∈ G then u 40 = yxy = zxz ∈ W 0 which implies that k = 2 and n/2 + 1 = 3, a contradiction. (Namely, if k > 2 then zxz ∈ W 2 ∪ W 3 .) Therefore 6-cycles in X must arise from the relation (xyz)2 , and so







G = x, y , z  x2 = y 2 = z2 = (xy )n/2 = ( yz)n/2 = ( zx)n/2 = (xyz)2 = 1, . . . .

(4)

Depending on whether k is odd or even, Lemma 4.4 implies that the structure of X is as shown in Figs. 6–8. Let a = xy, b = yz and c = zx. Using (4) it may be seen that a, b and c commute pairwise, and that furthermore the following relations hold: aα = b, bα = c, c α = a and ab = c −1 . Let H be the abelian subgroup of G generated by a and b. Since ab = c −1 , we have c ∈ H . It may be check that ax = a y = a z = a−1 and b x = b y = b z = b−1 , and so H is normal G. Furthermore, since G is generated by x, y and z, it follows that G / H = xH , y H , zH . Since a = xy ∈ H and b = yz ∈ H , we also get that xH = y H = zH , and so G / H = xH  ∼ = Z2 . (Note that xH = H for otherwise x ∈ H , and hence also y = x · xy ∈ H which implies that x and y commute. But since x and y are involutions and xy is of order n/2, this is clearly impossible.) Using the fact that |G | = kn we get that | H | = kn/2. Moreover, since H is an abelian group generated by two elements of order n/2, it follows that H ∼ = = ab ∼ (a × b)/(a ∩ b). Note that this implies that k divides n/2. It remains to find the intersection a ∩ b. Depending on whether k is odd or even and whether t = 1 or t = n/2 + 1 three different cases need to be considered. Case 1. Let k be odd. Then n /2

n /2

n /2

n/2 n/2 n/2−1 n/2−2 u0 · · · u 20 u 10 u 00

u 00 u 01 u 02 · · · uk0−1 uk−1 uk−2 uk−3 · · · u 1 u 0 u 0

is a cycle in X (see Fig. 6 where a subpath of this cycle is shown). As n/2 is odd (recall that, n ≡ 2 (mod 4) by Lemma 4.4) the corresponding relation in the group G is

(zx)(k−1)/2 z( yz)(k−1)/2 (xy )(n−2)/4 x, and therefore

K. Kutnar, D. Marušiˇc / Journal of Combinatorial Theory, Series B 99 (2009) 162–184

Fig. 6. The structure of X if k is odd.

Fig. 7. The structure of X if k is even and t = 1.

175

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Fig. 8. The structure of X if k is even and t = n/2 + 1.

xz(xz)(k−1)/2 ( yz)(k−1)/2 = ( yx)(n−2)/4 .

(5)

Since yx is of order n/2, and a, b and c commute pairwise, Eq. (5) implies that (xz)k ( yz)k = 1, and so bk = ck . Moreover, since bα = c and c α = a, we get that ak = bk = ck . Now the fact that ab = c −1 implies that a3k = b3k = c 3k = 1. Since a is of order n/2, it follows that n/2 divides 3k. Using the fact that k divides n/2 we get that either k = n/2 or k = n/6. Therefore either H ∼ = Zn2/2 or H ∼ = Zn/2 ×Zn/6 . Case 2. Let k be even and let t = 1. Then u 00 u 01 u 02 · · · uk0−1 uk1−1 uk1−2 uk1−3 · · · u 11 u 10 u 10 is a 2k-cycle in X (see Fig. 7 where a subpath of this cycle is shown). The corresponding relation in G is ( zx)k/2−1 ( zy )k/2 zx. Since a, b and c commute pairwise and bα = c, we get that ak/2 = bk/2 = ck/2 . Moreover, the fact that ab = c −1 implies that a3k/2 = b3k/2 = c 3k/2 = 1. Since a is of order n/2, it follows that n/2 divides 3k/2. But since k divides n/2, one can see that this is impossible, and so this case cannot occur. Case 3. Let k be even and let t = n/2 + 1. Then n/2+1 n/2+1 n/2+1

u 00 u 01 u 02 · · · uk0−1 uk−1 uk−2 uk−3

n/2+1 n/2+1 n/2 n/2−1 u0 u0 u0 · · · u 20 u 10 u 00

· · · u1

is a cycle in X (see Fig. 8 where a subpath of this cycle is shown). Since, by Lemma 4.4, n ≡ 0 (mod 4), the corresponding relation in G is ( zx)k/2−1 zy ( zy )k/2−1 z(xy )n/4 x which implies that x( zx)k/2−1 ( zy )k/2 z = ( yx)n/4 , and therefore (xz)k/2 ( yz)k/2 = ( yx)n/4 . Since yx is of order n/2 and a, b and c commute pairwise, it follows that (xz)k ( yz)k = 1. Applying the automorphism α we get that ak = bk = ck , and since ab = c −1 , it follows that a3k = b3k = c 3k = 1. Now using the same arguments as in Case 1 it turns out that either k = n/2 or k = n/6, and so either G ∼ = Dih(Zn2/2 ) or ∼ G = Dih(Zn/2 × Zn/6 ). This completes the proof of Lemma 4.5. 2 We are now ready to prove Theorem 1.1.

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177

Proof of Theorem 1.1. If X is a cubic 2-regular graph of girth 6 different from F016A then Lemma 4.5 implies that X is I kn (t )-path, a normal Cayley graph of the group G = Dih(Zn/2 × Zk ) with the presentation





x, y , z  x2 = y 2 = z2 = (xyz)2 = (xy )n/2 = ( yz)n/2 = ( zx)n/2 = 1, (xy )k = ( yz)k = ( zx)k , . . .



with respect to the set S = {x, y , z} where n > 6, either k = n/2 or k = n/6, and t = n/2 if k odd and t = n/2 + 1 if k even. Setting m = n/2 we see that the first part of Theorem 1.1 holds. For the converse assume that X = Cay(G , S ) where G = Dih(Zm × Zk ) = x, y , z | x2 = y 2 = z2 = (xyz)2 = (xy )m = ( yz)m = (zx)m = 1, (xy )k = ( yz)k = (zx)k , . . ., and S = {x, y , z}, m > 3, and either k = m or k = m/3. Since x, y and z are involutions and (xyz)2 = 1, it is clear that X is cubic of girth 6. Observe also that X is an I k2m (t )-path with respect to the orbits of the subgroup x, y . Hence, it remains to show that X is 2-regular. Note that each element of G can be written in the form ai b j z where a = xy, b = yz, i , j ∈ Zm , and  ∈ {0, 1}. Let c = zx, then H = a, b, c  = a, b ∼ = Zm × Zk is an abelian group. Let α : G → G be defined by the rule



ai b j z

α

 − j b − j +i = a− j + 1 − j +i +1 a

We claim that

b

   α ai b j z ai b j z =



 = 0,  = 1.

(6)







(ai −i b j − j z1+ )α if  = 1,    if  = 0 (ai +i b j + j z )α ⎧ − j + j  − j + j  +i −i  a b ⎪ ⎨ − j + j  +1 − j+ j  +i −i  +1 if  = 1 = a− j − j  +1 b− j− j  +i +i  +1 z if  = 1 ⎪ b z if  = 0 ⎩a    a − j − j b − j − j +i +i if  = 0

and that ai b j z

if if

α ∈ Aut G. Namely, since a z = a−1 and b z = b−1 , one can easily see that





z

α 

   α ai b j z =











and and and and 

a− j +1 b− j +i +1 za− j + b− j +i + z       a− j b− j +i a− j + b− j +i + z

 − j + j  +1−  − j + j  +i −i  +1−  1+  = a− j − j  +  −bj− j  +i +i  +    z a

=

b

⎧ − j + j  − j + j  +i −i  b ⎪ ⎨ a− j + j  + 1 − j + j  +i −i  +1

z

a b z − j − j  +1 b − j − j  +i +i  +1 z ⎪ ⎩a    a − j − j b − j − j +i +i

if if if if

 =1  =1  =0  =0



  = 1,   = 0,   = 1,  = 0 if if

 = 1, =0

 = 1, =0 and   = 1, and   = 0, and   = 1, and   = 0, if if

where i , i  , j , j  ∈ Zm and  ,   ∈ {0, 1}. Hence α is an endomorphism of G. Now assume that (ai b j z )α = 1 for some i , j ∈ Zm and  ∈ {0, 1}. If  = 1 then (6) implies that a− j +1 b− j +i +1 = z ∈ H , and so also x = zc ∈ H and y = xa ∈ H , contradicting the fact that G is not abelian. If however  = 0 then (6) gives a j = b− j +i . When k = m we clearly have j = − j + i = 0, and thus i = j = 0. When k = m/3 we get j = i − j ≡ δk (mod m) where δ ∈ {0, 1, 2}. Therefore, i = j = 0, or i = 2k and j = k, or i = k and j = 2k. Since ak = bk , in all three cases we get that ai b j = 1, which implies that the endomorphism α is injective, and thus α ∈ Aut G as claimed. Since xα = (abz)α = a1−1 b−1+1+1 z = y, y α = (bz)α = a−1+1 b−1+1 z = z and zα = abz = x, we have that α ∈ Aut X 1 permutes the neighbors of 1, and so X is symmetric. Also, because of the relation (xyz)2 = 1 the girth of X is 6. Moreover, since X is an I k2m (t )-path, there exist consistent oriented 2m-cycles in X , and so part (iii) of Proposition 4.2 implies that X is 2-regular. 2 Corollary 4.6. Let m be an integer. Then the following hold. (i) If m = 1 then there is no cubic 2-regular graph of girth 6 and of order 2m2 or 6m2 . (ii) If m = 2 then there exists no cubic 2-regular graph of girth 6 and of order 2m2 and there exists a unique cubic 2-regular graph of girth 6 and of order 6m2 , it is isomorphic to the I 212 (7)-path.

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(iii) If m = 3 then there exists no cubic 2-regular graph of girth 6 and of order 2m2 and there exists a unique cubic 2-regular graph of girth 6 and of order 6m2 , it is isomorphic to the I 318 (9)-path. (iv) If m  4 then there is a unique cubic 2-regular graph of girth 6 and of order 2m2 , it is isomorphic to the 2m (t )-path where t = m if m is odd and t = m + 1 if m is even; and there is a unique cubic 2-regular Im 6m graph of girth 6 and of order 6m2 , it is isomorphic to the I m (t )-path where t = 3m if m is odd and t = 3m + 1 if m is even. 5. Proving Theorem 1.2 The automorphism group of a cubic 1-regular graph X with consistent oriented 6-cycles has a presentation of the form Aut X = a, b | a2 = b6 = (ab)3 = 1, . . ., see for example [11,20]. This fact and Proposition 4.2 combined together imply the following result. Proposition 5.1. Let X be a cubic 1-regular graph of girth 6. Then Aut X = a, b | a2 = b6 = (ab)3 = 1, . . .. Given a group G the commutator subgroup [G , G ] of G is define as G  = [G , G ] = [a, b] | a, b ∈ G  where [a, b] = a−1 b−1 ab. (Clearly, [G , G ]  G and G /[G , G ] is abelian, see for example [44, Theorem 2.23].) To prove that cubic 1-regular graphs of girth 6 are normal Cayley graphs of generalized dihedral groups we require the following old group-theoretic result due to Miller [37]. Proposition 5.2. (See [37].) If a group is generated by two elements a and b its commutator subgroup is generated by the commutators a−i b− j ai b j of the powers of these two elements. Proposition 5.3. Let X be a cubic 1-regular graph of girth 6. Then Aut X ∼ = (Zrm × Zm )  Z6 , for some m, r ∈ Z and X is a normal Cayley graph of the generalized dihedral group Dih(Zrm × Zm ). Moreover, r = 1, 3. Proof. By Proposition 5.1, G = Aut X = a, b | a2 = b6 = (ab)3 = 1, . . .. Let c = ab. Then G is also generated by a and c. Now Proposition 5.2 implies that the commutator subgroup G  of G is generated by the commutators of the powers of a and c. Since a is an involution and c is of order 3, G  = ¯ w, and ¯ = acac −1 . A short computation shows that w w ¯ =w ac −1 ac , acac −1 . Let w = ac −1 ac and w  so G is an abelian group. (Note that this implies that G  is a proper subgroup of G.) Moreover, since −1 ¯ −1 , we have that w and w ¯ / w  ∩  w ¯∼ ¯ are of the same order, and so G  ∼ w ab = w = = w × w ¯ rm = 1. (Note that when m = 1 the commutator subgroup G  ∼ Zrm × Zm where w rm = w is cyclic.) Z = r Since G = a, c | a2 = c 3 = (ac )6 = 1, . . ., the quotient group G /G  = aG  , cG  | (aG  )2 = (cG  )3 = 1, . . ., is abelian and isomorphic to a subgroup of Z6 . We claim that aG  = G  . Namely, if aG  = G  , and so a ∈ G  , then since G  is abelian, we have w a = w. But then equality



awa = aac −1 aca = c −1 aca = ac −1 ac

−1

= w −1

implies that w −1 = w, and so w is an involution. It follows that G  is isomorphic to a subgroup of Z2 × Z2 , which implies that |G |  24, contradicting the fact that the smallest cubic 1-regular graph is of order 26 (see [2,4]). Hence aG  = G  , as claimed. ¯ −1 which implies that Now suppose that c ∈ G  . Then c w = wc, and by computation w = w G  =  w  is cyclic and G /G  = aG   ∼ = Z2 . It may be checked that w a = w −1 , and so G is a dihedral group. Therefore we have ca = c −1 which implies that b2 = 1, a contradiction. Hence c ∈ / G  , and consequently G /G  ∼ = Z6 and G ∼ = (Zrm × Zm )  Z6 . ¯ −1 , and G  is normal and abelian in H , H is a ¯a=w / G  , w a = w −1 , w Let H = G  , a. Since a ∈  ∼ generalized dihedral group Dih(G ) = (Zrm × Zm )  Z2 . Moreover, since w b = ab3 = [a, b][a, b−1 ] ∈ G  , ¯ b = w ∈ G  and ab = b−1 ab = aab−1 ab ∈ H we have that H a = H and H b = H which implies that w H  G. The subgroup H acts on X by left translation. Let us assume that there exist h ∈ H , h = 1, such that h ∈ H v for some vertex v ∈ V ( X ). The graph X may also be seen as an orbital graph of the left

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179

action of G on the set C of left cosets of the subgroup C = c  = ab = Z3 , arising from the suborbit {aH , abaH , ababaH } of length 3. Hence v = gC for some g ∈ G. Thus if hgC = gC then g −1 hgC = C , and so g −1 hg ∈ C . But, since H is normal in G, we also have that g −1 hg ∈ H , a contradiction. Therefore, the vertex stabilizer H v is trivial for every vertex v ∈ V ( X ). Moreover, since | H | = | V ( X )|, by orbit-stabilizer property, H act regularly on X . Since H is normal in G, it follows that X is a normal Cayley graph of H ∼ = Dih(Zrm × Zm ). Finally, since X is 1-regular, Theorem 1.1 implies that r = 1, 3. 2 The next four propositions will be needed in the proof of Theorem 1.2. Proposition 5.4. (See [30].) Let r > 3 be an integer. There is a subgroup k of order 3 in Zr∗ such that k2 + e e k + 1 ≡ 0 (mod r ) if and only if r has the prime decomposition r = 3s p 11 . . . pt t , where s ∈ {0, 1}, t  1, and p i ≡ 1 (mod 3). Since for a dihedral group D 2r = τ , ρ | τ 2 = ρ r = 1, (τρ )2 = 1 cubic Cayley graphs relative to the set of generators {τρ i , ρ j }, i , j ∈ Zr , are of girth 4, the following proposition completely determines cubic 1-regular Cayley graphs of dihedral groups of girth 6. Proposition 5.5. (See [34].) Let r be a positive integer and S = {i 1 , i 2 , i 3 } be a subset of Zr . Then the graph X = Cay( D 2r , {τρ i 1 , τρ i 2 , τρ i 3 }) is connected and 1-regular where D 2r = τ , ρ | τ 2 = ρ r = 1, (τρ )2 = 1 if and only if r  11 is odd and there exists a nonidentity element k ∈ Zr∗ such that k2 + k + 1 ≡ 0 (mod r ), S ∼ {0, 1, k + 1} and X ∼ = Cay( D 2r , {τ , τρ , τρ k+1 }). Moreover, X is of girth 6. To state the next result we need to introduce certain covers of the cube Q 3 , the so-called C Q (k, r ) graphs introduced in [17]. Let V ( Q 3 ) = {u 1 , u 2 , u 3 , u 4 , u 5 , u 6 , u 7 , u 8 }. For any two nonnegative integers k and r with 1  k  r − 1 and (k, r ) = 1, the graph C Q (k, r ) is defined to have vertex set V (C Q (k, r )) = V ( Q 3 ) × Zr and edge set





E C Q (k, r ) = (u 1 , i )(u 6 , i ), (u 1 , i )(u 7 , i ), (u 1 , i )(u 8 , i ), (u 2 , i )(u 5 , i ),

(u 2 , i )(u 8 , i ), (u 3 , i )(u 8 , i ), (u 4 , i )(u 7 , i ),

  (u 2 , i )(u 7 , i + 1), (u 3 , i )(u 5 , i + k), (u 3 , i ) u 6 , i − k−1 ,     (u 4 , i ) u 5 , i − k−1 − 1 , (u 4 , i )(u 6 , i + k)  i ∈ Zr .

Proposition 5.6. (See [17, Theorem 1.1].) Let X˜ be a connected Zr -cover of the cube Q 3 . Then X˜ is 1-regular of girth 6 if and only if X˜ is isomorphic to C Q (k, r ) where r  7 divides (k2 + k + 1). j

Let X = F018A be the Pappus graph and let V ( X ) = {u i | i ∈ Z3 , j ∈ Z6 } as shown in Fig. 9. Then for any two nonnegative integers k and r with 1  k  r − 1 and (k, r ) = 1, the graph C P (k, r ) is defined to have vertex set V (C P (k, r )) = V ( X ) × Zr and edge set





E C P (k, r ) =

  , i  j ∈ Z6 , i ∈ Zr

          ∪ u 01 , i u 22 , i + 1 , u 21 , i u 42 , i + k , u 41 , i u 02 , i + k2  i ∈ Zr

          ∪ u 11 , i u 32 , i − k2 , u 31 , i u 52 , i − 1 , u 51 , i u 12 , i − k  i ∈ Zr

          ∪ u 02 , i u 32 , i − k2 , u 12 , i u 42 , i + k , u 22 , i u 52 , i − 1  i ∈ Zr .



j



j

 

j



j

 

j



j +1

u0 , i u1 , i , u1 , i u2 , i , u0 , i u0

Proposition 5.7. Let X˜ be a connected Zr -cover of the Pappus graph X = F018A. Then (i) X˜ is 1-regular of girth 6 if and only if X˜ is isomorphic to C P (k, r ) where r  7 divides (k2 + k + 1); (ii) X˜ is 2-regular of girth 6 if and only if X˜ is isomorphic to F054A, the I 318 (9)-path.

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Fig. 9. The Pappus graph with voltage assignment ζ . The spanning tree consists of bold edges all carrying voltage 0.

Proof. The Pappus graph is a Cayley graph for the group H = Dih(Z3 ×Z3 )=x, y , z | x3 = y 3 = z2 = 1, xz = x−1 , y z = y −1 , xy = yx with respect to the generating set {xz, yz, z}. Moreover, its automorphism group has the unique 1-regular subgroup isomorphic to G = H  Z3 (see Example 3.4). j Let V ( X ) = {u i | i ∈ Z3 , j ∈ Z6 } as shown in Fig. 9. Then 6-cycles of the form u 00 u 10 u 20 u 30 u 40 u 50 u 00 , u 01 u 22 u 21 u 42 u 41 u 02 u 01 and u 11 u 32 u 31 u 52 u 51 u 12 u 11 correspond to the group relation (xz · yz)3 in H , and 6-cycles j

j +1

j +2

j +2

j +2

j

j

of the form u 0 u 0 u 0 u 1 u 2 u 1 u 0 , j ∈ Z6 , correspond to the group relation (xz · yz · z)2 . Let the vertex u 00 correspond to the identity element u 00 = 1 ∈ H , let α ∈ G 1 (the stabilizer of u 00 = 1) be such that (1, xz)α = (1, yz), (1, yz)α = (1, z) and (1, z)α = (1, xz), that is, α acts on the neighbors of u 00 = 1 according to the rule (u 10 u 50 u 01 )). Further let σ = xz · yz = xy −1 ∈ G. (Note that σ is a two-step rotation of the cycle u 00 u 10 u 20 u 30 u 40 u 50 u 00 .) Let ζ : A ( X ) → Zr be a voltage assignment such that arcs of a spanning tree T consisting of the j j j j edges {u 0 u 1 , u 1 u 2 , u 00 u 10 , u 10 u 20 , u 20 u 30 , u 30 u 40 , u 40 u 50 | j ∈ Z6 }, as shown in Fig. 9, have voltage 0 and its corresponding covering graph X˜ = Cov( X , ζ ) is symmetric graph of girth 6. There are ten fundamental cycles u 00 u 01 u 22 u 21 u 20 u 10 u 00 ,

u 10 u 11 u 32 u 31 u 30 u 20 u 10 ,

u 20 u 21 u 42 u 41 u 40 u 30 u 20 ,

u 40 u 41 u 02 u 01 u 00 u 10 u 20 u 30 u 40 ,

u 50 u 51 u 12 u 11 u 10 u 20 u 30 u 40 u 50 ,

u 10 u 11 u 12 u 42 u 41 u 40 u 30 u 20 u 10 ,

u 20 u 21 u 22 u 52 u 51 u 50 u 40 u 30 u 20

u 30 u 31 u 52 u 51 u 50 u 40 u 30 ,

u 00 u 01 u 02 u 32 u 31 u 30 u 20 u 10 u 00 , and

u 00 u 10 u 20 u 30 u 40 u 50 u 00

in X , which are generated, respectively, by ten cotree arcs (u 01 , u 22 ), (u 11 , u 32 ), (u 21 , u 42 ), (u 31 , u 52 ), (u 41 , u 02 ), (u 51 , u 12 ), (u 02 , u 32 ), (u 12 , u 42 ), (u 22 , u 52 ) and (u 50 , u 00 ). Under the action of α and σ each of these cycles C is mapped to a cycle of the same length (see Table 1 where all these cycles and their voltages are listed). ¯ and σ¯ from the set { zi | i ∈ Z10 } of voltages of the ten fundaNow let us consider the mappings α mental cycles of X to the cyclic group Zr which are defined by ζCα¯ = ζC α and ζCσ¯ = ζC σ , respectively. ¯ and σ¯ Since G is the unique 1-regular subgroup of Aut X , Proposition 2.3 implies that the mappings α are extended to automorphisms of Zr . Let these extended automorphisms be denoted by α ∗ and σ ∗ , respectively. From Table 1 we can see that ∗

z1σ = z3 ,

∗

z3σ = z5 − z9 ,

z0σ = z2 , z2σ = z4 − z9 ,

∗ ∗

σ∗

z7σ = − z6 − z9 ,

∗

z4α = − z9 − z8 .

z6 = z8 , z0α = − z5 − z9 ,

∗

∗

(7)

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181

Table 1 Fundamental cycles and their images with corresponding voltages

ζC

Cα

ζC α

Cσ

ζC σ

u 00 u 01 u 22 u 21 u 20 u 10 u 00

z0

u 00 u 10 u 11 u 12 u 51 u 50 u 00

− z5 + z9

u 20 u 21 u 42 u 41 u 40 u 30 u 20

z2

u 10 u 11 u 32 u 31 u 30 u 20 u 10

z1

u 50 u 40 u 30 u 31 u 52 u 51 u 50

z3

u 30 u 31 u 52 u 51 u 50 u 40 u 30

z3

u 20 u 21 u 42 u 41 u 40 u 30 u 20

z2

u 51 u 12 u 42 u 21 u 22 u 52 u 51

− z5 + z7 − z2 + z8

u 40 u 41 u 02 u 01 u 00 u 50 u 40

z4 − z9

u 30 u 31 u 52 u 51 u 50 u 40 u 30

z3

u 52 u 31 u 32 u 02 u 01 u 22 u 52

− z3 − z6 + z0 + z8

u 50 u 51 u 12 u 11 u 10 u 00 u 50

z5 − z9

u 40 u 41 u 02 u 01 u 00 u 10 u 20 u 30 u 40

z4

u 22 u 21 u 20 u 10 u 00 u 50 u 51 u 52 u 22

− z9 − z8

u 00 u 01 u 22 u 21 u 20 u 30 u 40 u 50 u 00

z0 + z9

u 50 u 51 u 12 u 11 u 10 u 20 u 30 u 40 u 50

z5

u 01 u 02 u 41 u 40 u 50 u 51 u 52 u 22 u 01

− z4 − z8 − z0

u 10 u 11 u 32 u 31 u 30 u 40 u 50 u 00 u 10

z1 + z9

u 00 u 01 u 02 u 32 u 31 u 30 u 20 u 10 u 00

z6

u 00 u 10 u 20 u 30 u 31 u 52 u 51 u 50 u 00

z3 − z9

u 20 u 21 u 22 u 52 u 51 u 50 u 40 u 30 u 20

z8

u 10 u 11 u 12 u 42 u 41 u 40 u 30 u 20 u 10

z7

u 50 u 40 u 41 u 42 u 21 u 22 u 52 u 51 u 50

− z2 + z7

u 30 u 31 u 32 u 02 u 01 u 00 u 50 u 40 u 30

− z6 − z9

u 20 u 21 u 22 u 52 u 51 u 50 u 40 u 30 u 20

z8

u 51 u 12 u 11 u 32 u 02 u 01 u 22 u 52 u 51

z5 + z1 − z6 + z0 + z8

u 40 u 41 u 42 u 12 u 11 u 10 u 00 u 50 u 40

− z7 − z9

z9

u 00 u 50 u 51 u 52 u 22 u 01 u 00

− z9 − z8 − z0

u 20 u 30 u 40 u 50 u 00 u 10 u 20

z9

C

u 00 u 10 u 20 u 30 u 40 u 50 u 00

Since X˜ should be of girth 6, at least one of the two 6-cycles D = u 00 u 10 u 20 u 21 u 22 u 01 u 00 and D  =

u 00 u 10 u 20 u 30 u 40 u 50 u 00 passing through the 2-arc (u 00 , u 10 , u 20 ) lifts to a 6-cycle in X˜ . If D lifts to a 6-

cycle in X˜ then z0 = 0 and equations in Table 1 imply that zi = 0 for every i ∈ Z10 , a contradiction. Thus z0 = 0 and D  = u 00 u 10 u 20 u 30 u 40 u 50 u 00 lifts to a 6-cycle, and so we have that z9 = 0. Now equations in (7) imply that z0 , z1 , z2 , z3 , z4 , z5 , z6 , z7 and z8 have the same order in Zr . Moreover, since Zr =  zi | i ∈ Z10  each of zi , i ∈ Z10 \ {9}, generates the group Zr . By Proposition 2.5 we may ∗ now assume that z0 = 1 and z2 = k for some k ∈ Zr∗ . Since z0σ = z2 , σ ∗ is an automorphism of Zr ∗

∗

induced by the mapping 1 → k. Therefore, equations z2σ = z4 − z9 = z4 and z4σ = z0 + z9 = z0 = 1 imply that z4 = k2 and k3 = 1, respectively. Since the 6-cycle u 01 u 22 u 21 u 42 u 41 u 02 u 01 must lift to a 6-cycle,

∗ ∗ we get that 1 + k + k2 = 0 (mod r ). Since z1α = z3 = z1σ (see Table 1), also

α ∗ is an automorphism

∗ of Zr induced by 1 → k. Therefore, z9α = − z9 − z8 − z0 = − z8 − z0 implies that z8 = − z0 = −1. Now ∗ ∗ σ z8 = − z7 − z9 = − z7 implies that z7 = k, and thus z7σ = − z6 − z9 = − z6 gives us that z6 = −k2 . The ∗ ∗ ∗ α σ σ relations z = z − z , z = z − z = z and z = z + z = z now gives us that z = −1, z = −k

6

3

9

3

5

9

5

5

1

9

1

3

5

and z1 = −k2 . It follows that X˜ is isomorphic to the graph C P (k, r ). Conversely, for any positive integers k, r satisfying that r divides k2 + k + 1, the graph C P (k, r ) is symmetric graph of girth 6 because two automorphisms α and ρ of X lifts by Proposition 2.3. Now we need to see when the constructed Zr -cover of X is also 2-arc-transitive. For this purpose let us assume that the automorphism β ∈ Aut X u 0 , that fixes the arc (u 00 , u 01 ) and interchanges (u 00 , u 10 ) 0

and (u 00 , u 50 ), lifts. Considering the action of β on the cycle u 00 u 01 u 22 u 21 u 20 u 10 u 00 one can easily see that β∗

β∗

z0 = − z4 = −k2 and z4 = z0 = 1. Hence, k4 = 1 which combined together with k3 = 1 imply that

k = 1. But then r = 3, and so X˜ is isomorphic to F054A, the unique cubic symmetric graph of order 54 which is 2-regular. To finish the proof, we need to show that if X˜ = C P (k, r ) is 1-regular and r divides k2 + k + 1 then r  7. Assume that X˜ is 1-regular and r divides k2 + k + 1. Since k2 + k + 1 is odd, r should be odd. Also, one can easily check that 5 cannot divide k2 + k + 1 for any integer k. If r = 3 then k = 1, and so X˜ is isomorphic to F054A. Therefore, r  7. 2 Let m > 3, t = m if m odd and t = m + 1 if m even. And further let k and r be nonnegative integers 2m such that 1  k  r − 1 and (k, r ) = 1. We introduce certain covers C I (m, k, r ) of the I m (t )-path in 2m the following way. Let X be the I m (t )-path, a Cayley graph Cay(G , {x, y , z}) of a generalized dihedral group G = Dih(Zm × Zm ) = x, y , z | x2 = y 2 = z2 = (xyz)2 = (xy )m = ( yz)m = ( zx)m = 1. Then the graph C I (m, k, r ) is defined to have vertex set V (C I (m, k, r )) = V ( X ) × Zr (where V ( X ) is identified with G) and edge set

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E C I (m, k, r ) = ( g , i )( gx, i − k), ( g , i ) g y , i − k2 , ( g , i )( gz, i − 1)  g ∈ G , i ∈ Zr . We are now ready to prove Theorem 1.2. Proof Theorem 1.2. Let X be a cubic 1-regular graph of girth 6. Then, by Proposition 5.3, Aut X ∼ = (Zrm × Zm )  Z6 and X is a normal Cayley graph of a generalized dihedral group H = Dih(Zrm × Zm ) = ¯ | a2 = w rm = w ¯ rm = 1, w a = w −1 , w ¯a = w ¯ −1 , w w ¯ w , . . . where r , m ∈ Z and r = 1, 3. ¯ =w a, w , w ¯∼ H Z is normal in . Therefore, in view of Proposition 2.6, X is a Zr -cover of Observe that  w  ∩  w = r a cubic symmetric Cayley graph of order 2m2 . If m = 1 then X is a Cayley graph of D 2r , and Propositions 5.4 and 5.5 combined together imply that part (i) holds. If m = 2 then X is a Zr -cover of a graph isomorphic to a Cayley graph of the group Dih(Z2 × Z2 ) which is clearly the cube Q 3 . Then Proposition 5.6 implies that X is isomorphic to C Q (k, r ) as claimed in part (ii). If m = 3 then X is a Zr -cover of a cubic symmetric graph of order 18, and hence of the Pappus graph F018A. Then Proposition 5.7 implies that part (iii) holds. We may therefore assume that m  4. Then X is a Zr -cover of a cubic symmetric graph Y of order 2m2 and of girth at most 6. But since the order of none of the five cubic symmetric graphs of girth less than 6 is of the form 2m2 , the girth of Y is precisely 6. Now Theorem 1.1 implies that Y is a normal Cayley graph of a group G = Dih(Zm × Zm ) = x, y , z | x2 = y 2 = z2 = (xyz)2 = (xy )m = ( yz)m = (zx)m = 1 relative to the set of generators S = {x, y , z}. Moreover, by observations on cubic symmetric normal Cayley graphs given in Section 2.5, the automorphism α ∈ Aut G defined by xα = y, y α = z and zα = x gives rise to an automorphism of Y . Let ax , a y and a z denote, respectively, three arbitrary arcs in Y of the form ( g , gx), ( g , g y ) and ( g , gz), where g ∈ G. Then there are six different types of 2-arcs in Y : (ax a y ), (a y ax ), (a y a z ), (a z a y ), (ax a z ) and (a z ax ). Let Axy , A yx , A yz , Azy , Azx and Axz , respectively, be the corresponding sets of all such 2-arcs. Clearly, these sets are orbits of the subgroup M = xy , yz, zx = xy , yz of G isomorphic α α to Zm × Zm . Observe also that Aα xy = A yz , A yz = A zx and A zx = Axy , and that any 6-cycle in Y is of the form (ax a y a z )2 . Now we need to determine all possible 1-regular Zr -covers X = Cov(Y ) of Y of girth 6. By Proposition 4.2, X has consistent 6-cycles which are lifts of consistent 6-cycles in Y . Of course, by our results in Section 4, Y also has consistent 2m-cycles. Therefore the 1-regular subgroup of Aut Y which lifts must have consistent 6-cycles (and no consistent 2m-cycles). Such a 1-regular subgroup is the group G  α   Aut Y . Namely, each element of this group either fixes the sets of x-edges, y-edges and z-edges, or cyclically permutes these three sets. Consequently, this group does not contain a shunt automorphism corresponding to consistent 2m-cycles as every such cycle arises from one of the relations (xy )m , ( yz)m and ( zx)m . Therefore this is the desired 1-regular subgroup of Aut Y that lifts to Aut X . Moreover, since M centralizes the group of covering transformations isomorphic to Zr , Proposition 2.4 implies that there exists a voltage assignment ζ : A (Y ) → Zr such that for each σ ∈ M and each walk W of Y we have that ζ W = ζ W σ . As Axy , A yz and Azx are orbits of M, it follows that all 2-arcs of the form (ax a y ) have the same voltage, say ζ(ax a y ) (the same holds for the 2-arcs of the forms (a y a z ) and (a z ax ) whose voltages we denote by ζ(a y a z ) and ζ(a z ax ) , respectively). The fact that 6-cycles in Y lifts to 6-cycles in X implies that ζ(ax a y a z )2 = ζ(ax a y )(a z ax )(a y a z ) = 0, and thus

ζ(az ax ) = −ζ(ax a y ) − ζ(a y az ) .

(8)

Let C xy = (ax a y ), C yz = (a y a z ) and C zx = (a z ax ) be the 2-arcs originating at the vertex of Y corresponding to the identity element in G. Recall that the automorphism α ∈ Aut Y cyclically permutes these 2-arcs. Moreover, since α induces an automorphism α ∗ of the group of covering transformations (identified here with Zr ), we have that ∗

ζCαxy = ζC yz ,

∗

ζCαyz = ζC zx

and

∗

ζCαzx = ζC xy .

(9)

Since X is connected, we must have ζ(ax a y ) , ζ(a y a z )  = Zr , and so each of ζ(ax a y ) and ζ(a y a z ) generates the group Zr . Hence we may assume that ζ(ax a y ) = 1 and ζ(a y a z ) = k for some k ∈ Zr∗ . Therefore (9) implies that

∗2

α ∗ maps 1 to k, and thus ζ(az ax ) = ζ(αax a y ) = k2 . Combining together Eq. (8) and Proposi-

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183

tion 5.4 we get that k2 + k + 1 = 0 and that r is of desired form. Moreover, letting ζax = −k, ζa y = −k2 and ζa z = −1 one can easily see that X is isomorphic to the graph C I (m, k, r ). To finish the proof we need to show that the constructed Zr -cover of Y is indeed 1-regular. For this purpose let as assume on the contrary that the automorphism β ∈ Aut Y v , that fixes the arc a z and interchanges ax and a y lifts. Then since (ax a y )β = (a y ax ), the induced automorphism β ∗ of Zr∗ maps 1 to −1. Thus, since (a y a z )β = (ax a z ), we have that −k = −k2 . But then this fact and k2 + k + 1 = 0 combined together imply that r = 3, a contradiction. This completes the proof of Theorem 1.2. 2 e

e

Corollary 5.8. Let m be an integer and let r > 3 be an integer with prime decomposition r = 3s p 11 . . . pt t > 3, where s ∈ {0, 1}, t  1, and p i ≡ 1 (mod 3). Then the following hold. (i) If m = 1 and r = 7 then there exists no cubic 1-regular graph of girth 6 and of order 2rm2 . (ii) If m = 1 and r  11 then there exists a cubic 1-regular graph of girth 6 and of order 2rm2 . (iii) If m > 1 and r > 3 then there exists a cubic 1-regular graph of girth 6 and of order 2rm2 . 6m 2m (t )-path is a Z3 -cover of the 2-regular I m (t )Corollary 5.9. Let m > 3 be an integer. Then the 2-regular I m path.

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