Classification of edge-transitive Nest graphs

A finite simple graph Γ\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\Gamma $$\end{document} is called a Nest graph if it is regular of valency 6 and admits an automorphism with two orbits of the same length such that at least one of the subgraphs induced by these orbits is a cycle. In this paper, we complete classification of the edge-transitive Nest graphs and by this solve the problem posed by Jajcay et al. (Electron J Comb 26:#P2.6, 2019).


Introduction
All groups in this paper will be finite and all graphs will be finite and simple. Let n be an integer with n ≥ 4 and let Z n denote the ring of residue classes of integers modulo n. Fix a, b, c, k ∈ Z n such that each of them is distinct from 0 (the zero element of Z n ), the elements a, b and c are pairwise distinct, and in the case when n is even, k = n/2. Then the Nest graph N (n; a, b, c; k) is defined to have vertex set {u i : i ∈ Z n } ∪ {v i : i ∈ Z n }, and six types of edges such as   is an automorphism of N (n; a, b, c; k), the set {u i : i ∈ Z n } is an ρ -orbit, and the subgraph induced by this orbit is a cycle. In fact, this property characterises the Nest graphs. More precisely, if is any regular graph of order 2n and valency 6, which admits an automorphism with two orbits of the same length such that at least one of the subgraphs induced by these orbits is a cycle, then is isomorphic to a Nest graph N (n; a, b, c; k). In what follows, the term Nest graph will also be used for the graph . Nest graphs were introduced by Jajcay et al. [5] (see also [11]) and these graphs can be regarded as the hexavalent analogues of the generalised Petersen graphs [12]. The tetravalent analogues are the Rose Window graphs [13] and the pentavalent analogues are Tabačjn graphs [1]. Symmetry properties of the generalised Petersen, Rose Window and Tabačjn graphs have attracted considerable attention [1,3,4,8,10], in particular, the question which of them are edge-transitive has been answered in the articles [1,4,8].
Jajcay et al. [5] initiated the study of the edge-transitive Nest graphs. They classified those of girth 3 and the task to classify all was posed as an open problem (see [5,Problem 2]). Recently, the author and Ruff [9] showed that the complement of the Petersen graph is the only edge-transitive Nest graph of twice odd order. Furthermore, the so called core-free edge-transitive Nest graphs were determined in [7] (for a definition of a core-free Nest graph, we refer to the paragraph preceding Theorem 2). In this paper, we aim to complete the classification of the edge-transitive Nest graphs. Our main result is the following theorem.
Hamming graph H (2,4), and the Shrikhande graph. The last three are not stronglyregular, each can be described as a normal r -cover of a strongly regular graph: a normal 2-cover of the complete bipartite graph K 3,3 and two normal covers of the complement of the Petersen graph (for the definition of a normal r -cover, see the 2nd paragraph in Sect. 2).
All these graphs are covered by Theorem 1. The graph N (4; 1, 2, 3; 1) is the graph in row no. 8 with m = 2 and s = 1, and the graph N (12; 1, 3, 10; 5) is isomorphic to the graph in row no. 8 with m = 6 and s = 1. To see this, one can use the basic isomorphisms described in Lemma 1. Using the same basic isomorphisms, one can show that the graph in family (a) is isomorphic to the graph in row no. 7 with m = n/2 and s = l; and the graph in family (b) is isomorphic to the graph in row no. 8 with either 2s = b − 1 or 2s = (b − 1) 2 .

Preliminaries
We collect all facts needed in the proof of Theorem 1. Given a graph , let V ( ), E( ), A( ) and Aut( ) denote its vertex set, edge set, arc set and automorphism group, respectively. The number |V ( )| is called the order of . The set of vertices adjacent with a given vertex v is denoted by (v). If G ≤ Aut( ) and v ∈ V ( ), then the stabiliser of v in G is denoted by G v , the orbit of v under G by v G , and the set of all G-orbits by Orb(G, V ( )). If G is transitive on V ( ), then is said to be Gvertex-transitive and is called vertex-transitive when it is Aut( )-vertex-transitive; (G-)edge-and (G-)arc-transitive graphs are defined correspondingly. Let π = {C 1 , . . . , C k } be an arbitrary partition of V ( ) with parts C i . The quotient graph of with respect to π , denoted by /π, is defined to have vertex set π , and two distinct vertices C i and C j are adjacent if and only if there is an edge {u, v} ∈ E( ) such that u ∈ C i and v ∈ C j . Now, if there exists a constant r such that then is called an r -cover of /π. The term cover will also be used instead of 1cover. In the special case when π = Orb(N , V ( )) for an intransitive normal subgroup N Aut( ), /N will also be written for /π and when is also an r -cover (cover, respectively) of /N , then the term normal r -cover (normal cover, respectively) will also be used. It is well-known that, if is a G-edge-transitive graph, is regular with valency κ, N G and N is intransitive, then is a normal r -cover of /N for some r such that r divides κ.
The next result establishes some obvious isomorphisms. Let B be a finite group. If A ≤ B, then the core of A in B, denoted by core B (A), is the largest normal subgroup of B contained in A. In the case when A has trivial core in B, it is also called core-free. Following [7, Definition 1.1], we say that a Nest Dealing with non-core-free Nest graphs the following lemma will be useful.
If |X | < n/2, then the following hold.
(1) The kernel of the action of G on Orb(X , V ( )) is equal to X .
(2) is a normal cover of /X.

(3) /X is aḠ-edge-transitive Nest graph, whereḠ is the image of G under its action
on Orb(X , V ( )).
Let n and m be positive integers such that m divides n. There is a unique homomorphism from the group (Z n , +) onto the group (Z m , +) which maps 1 ∈ Z n to 1 ∈ Z m , denote this homomorphism by θ n,m . Then define the mappingθ n,m : Observe that, with the notations in Lemma 2, the X -orbits are exactly the preimages of the mappingθ n,m , where m = n/|X |. Therefore, we have the following corollary. Furthermore, the mappingθ n,m defined in (1) is a homomorphism from onto .
The following theorem is a special case of the main result of [9]. We conclude the section with some necessary conditions for a Nest graph to be edge-transitive.

Lemma 3 [9, Lemma 11]
Let be a G-edge-transitive Nest graph of order 2n, and let H be a vertex stabiliser of G. Then G contains an element g of order n satisfying one of the following sets of conditions.

Proof of Theorem 1
Throughout this section we keep the following notation.
Recall that is core-free if C has trivial core in G, i.e., N = 1. In this case, due to Theorem 2, is edge-transitive if and only if it is isomorphic to one of the graphs in rows no. 1-4 in Table 1, therefore, we need to consider the case when N = 1.
In our first lemma we deal with the case when n is even and |N | = n/2.

Lemma 4
Assuming Hypothesis 1, suppose that n = 2m and |N | = m. If is edgetransitive, then it is isomorphic to one of the graphs in rows no. 7-8 in Table 1.
Proof If n = 4, then ∼ = N (4; 1, 2, 3; 1) due to [5, Table 1]. This graph is described in row no. 8 in Table 1 with s = 1 (see also Lemma 1). For the rest of the proof we assume that n > 4. The quotient graph /N ∼ = K 4 , hence u 0 has two neighbours in each N -orbit distinct from u N 0 . Therefore, we may assume w.l.o.g. that a is even and each of b, c and k is odd. In what follows, we refer to the elements in 2 as the even elements, and to the elements in the coset 2 + 1 as the odd elements, where 2 is the additive subgroup of Z n generated by 2.
There are two vertex-orbits under ρ . As is edge-transitive, these orbits are merged to one vertex-orbit under G, so is vertex-transitive. Also, since is edgetransitive, it follows that G u 0 is either transitive on (u 0 ) or it has two orbits of length 3. Consequently, there is an element g ∈ (u 0 ) whose restriction to (u 0 ) has order 3. Denote this restriction by g| (u 0 ) . The intersection of the N -orbits with (u 0 ) is a (u 0 )-invariant partition, which consists of the sets: Therefore, g| (u 0 ) permutes these sets in a 3-cycle, and we may assume w.l.o.g. that g maps v 0 to v b . Then v b is mapped to either u 1 or u −1 , and correspondingly we have that Assume for the moment that g| . Thus in the rest of the proof we assume that the first possibility holds in (2). Now, as g normalises N , there is an integer 1 ≤ s ≤ m − 1 such that gcd(s, m) = 1 and (ρ i ) g = ρ si for every i ∈ Z n such that i is even. Note that g is uniquely determined. Indeed, if i ∈ Z n such that i is even, then u g i = u Then = v g 2s 2 +b = u 2s 3 −1 , implying that s 3 ≡ 1 (mod m). There are 12 N -orbits on the set of edges of , a representative for each orbit is shown in the third column of Table 2. Clearly, g maps each representative to an edge of . In some cases, this condition can be translated to a condition in terms of the elements a, b, c, k ∈ Z n , which is shown in the last column. We explain below these latter conditions case by case.
Row no. 1 v 2s = u g 1 = v a , and so a = 2s. Row no.
We show later (see row no. 6) that it must be equal to c.   (4) and that n > 4), we conclude that s + sb + 1 = c and −sa + s + sb Observe next that (4) combined with the conditions in rows no.8 and 10 yield 2k = sc − sb = 2. Thus k = 1 or m + 1.
The equations in rows no. 1, 4 and 5 yield s + sb + 1 = 2s 2 + b. On the other hand, we have sb − s + b = −1 in row no. 8, and these yield 2b = 2s − 2s 2 . Using that b is odd, we find in turn that m is odd, b = s − s 2 + m and c = s + s 2 + m, i.e., is the graph in row no. 7 in Table 1.
In this case we get 2b = 2s − 2s 2 + m. Using that b is odd, we find in turn that m ≡ 2 (mod 4), and either b = s − s 2 + m/2 and c = s + s 2 + m/2, i.e., is the graph in row no. 8 in Table 1, or b = s − s 2 + m/2 + m and c = s + s 2 + m/2 + m.
In the latter case, by Lemma 1, The latter graph is isomorphic to the graph in row no. 8 in Table 1 with letting s to be s 2 because s 4 = ±s holds in Z n .
We continue by showing that the graphs in rows no. 7-8 in Table 1 are edgetransitive. Table 1, i.e.,

where m is odd and s 3 ≡ 1 (mod m). Then is edge-transitive.
Proof Let ρ = (u 0 , u 1 , . . . , u n−1 )(v 0 , v 1 , . . . , v n−1 ), let N = ρ 2 , and let g be the permutation of V ( ) defined in (3). Let us consider again the 12 representatives of the N -orbits on the edges of that are given in Table 2.
We claim that g maps any of these edges to an edge of . This is equivalent to show that all conditions in the 5th column of Table 2 holds. The condition in row no. 1 clearly holds. Next, the condition in row no. 4 reads as The first congruence holds because m is odd, while the second follows from the fact that s 3 ≡ 1 (mod m). The condition in row no. 5 clearly holds. Then using this and the condition in row no. 4, we obtain that the condition in row no. 6 also holds. The condition in row no. 8 reads as Again, the first congruence holds because m is odd, while the second follows from the fact that s 3 ≡ 1 (mod m). This then imply that also the condition in row no. 10 holds. Finally, we have to check the last two rows. Taking b at the right side of the equation in row no. 11, we obtain the same condition as in row no. 8. Finally, due to row no. 11 and (5), the equation in row no. 12 reduces to 2sk − 1 = a − 1, which clearly holds. Now, let e be an arbitrary edge of . Then e = (e * ) ρ i , where e * is one of the representatives in Table 2 and i ∈ Z n such that i is even. A straightforward computation shows that ρ i g = gρ is , and hence e g = (e * ) ρ i g = (e * ) gρ is , which shows that e g ∈ E( ). We have shown that g is an automorphism of . Clearly, the group ρ, g acts transitively on V ( ).
Consider the arc-orbit of (u 0 , u 1 ) under ρ, g . This contains the arcs This shows that all edges incident with u 0 belong to the same orbit under ρ, g , hence is indeed edge-transitive.
The proof of the next lemma goes exactly in the same way as the proof of the previous lemma, hence it is omitted.
We turn to the case when 1 < |N | < n/2.

Lemma 7 Assuming Hypothesis 1, suppose that is edge-transitive and |N | < n/2 and let X ≤ N . Then X G and /X is an edge-transitive Nest graph of order 2n/|X |.
Proof Since N is a cyclic group, its subgroup X is characteristic. Using this and the fact that N G, we obtain that X G. Now, the lemma follows from Lemma 2(3).
As the next step, we describe the quotient graph /N . Proof By Lemma 2(3), /N is aḠ-edge-transitive Nest graph, whereḠ is the image of G under its action on Orb(N , V ( )). Since the kernel of the latter action is equal to N (see Lemma 2(1)), it follows that there is unique subgroup M of G such that N ≤ M ≤ C andM = coreḠ(C). As M is normal in G, we find that M = N , sō C has trivial kernel inḠ. This yields that /N is a core-free graph, and the lemma follows from Theorem 2.
In this case n = 5|N |. Assume for the moment that |N | has an odd prime divisor, say p and denote by M the subgroup of N of order |N |/ p. By Lemma 7, /M is an edge-transitive Nest graph of order 10 p. This is impossible due to Theorem 3, hence we find that |N | = 2 for some ≥ 1 and n = 5 · 2 . Now, [5, Table 1] shows that there are three graphs with n = 10, one with n = 20, and none with n = 40. This and Lemma 7 yield that = 1 or 2.
Using Magma [2], one can quickly check that the corresponding core N has order 2, 2 and 5, respectively. Finally, if is one of the first two graphs, then /N ∼ = N (5; 1, 2, 3; 1), which can be deduced directly if one uses the fact that |N | = 2 and combines Corollary 1 with Lemma 1.
In view of Theorem 2 and the Lemmas 4-6, 8 and 9, Theorem 1 follows if we show the lemma below.
For x ∈ R, denote by x * the image of x under its action on Orb(P, V ( )). Note that, for any w ∈ {u 0 , . . . , u 7 } ∪ {v 0 , . . . , v 7 }, the preimage of w underθ 8 p,8 is equal to a P-orbit. This allows us to define the permutation x of V (E) as follows: Now, define ϕ to be the mapping ϕ :Ḡ → Sym(V (E)), the group of all permutations of V (E), by letting x ϕ = x . It is straightforward to check that ϕ is a homomorphism fromḠ to A. Let φ be the restriction of ϕ to R.
Let B = ncl A ( σ ), the normal closure of σ in A (i.e., the smallest normal subgroup of A containing in σ ). A computation with Magma [2] shows that B is the only proper subgroup of A acting transitively on E(E) and containing σ and that |A| = 2|B|. Consequently, B ≤ R φ and therefore, there is a subgroup L ≤ R such that L φ = B.
Let us consider the centraliser C L (P). As P Ḡ , C L (P) L, hence (C L (P)) φ B. On the other hand,C ≤ CḠ(P), hence σ ∈ (C L (P)) φ , implying that ncl B ( σ ) ≤ (C L (P)) φ . A computation with Magma [2] shows that B = ncl B ( σ ), hence (C L (P)) φ = B = L φ , so C L (P) = L, or equivalently, P : L = P × L. Now, let H = (P × L) u 0 , the stabiliser of the vertex u 0 in P × L (here 0 ∈ Z 8 p ). Notice that H < L and that H φ = B u 0 , the stabiliser of the vertex u 0 in B (here 0 ∈ Z 8 ). To sum up, is a (P × L)-edge-transitive Nest graph of order 8 p and H is a vertex stabiliser in P × L such that H φ = B u 0 .
The desired contradiction will arise after applying Lemma 3 to P × L and H . Due to this lemma, there is an element g ∈ P × L of order 8 p satisfying all conditions in either part (1) or (2) of Lemma 3. Then g = zλ, where z is a generator of P and λ is an element of order 8 in L.
Using that z commutes with every element of H , it is easy to see that g − All these yield that the equalities |H | = 3|H ∩ H g | = |H g ∩ Hg H| reduce to the following equalities in L: Now, applying φ and letting μ = λ φ , we deduce from these the following equalities in B: Recall that E is the Hamming graph H (2, 4), B ≤ Aut(E) such that B acts transitively on the arcs of E and 2|B| = |A|. As A ∼ = (S 4 × S 4 ) : Z 2 , 3 must divide |B u 0 ∩ (B u 0 ) μ |. If u 0 and u μ 0 are non-adjacent, then one can easily find that B u 0 ∩ (B u 0 ) μ is a 2-group. This contradicts that |B u 0 ∩ (B u 0 ) μ | is divisible by 3, hence u 0 and u μ 0 are adjacent. Using also that B u 0 is transitive on E(u 0 ), we have that |B u 0 | = 6|B u 0 ∩ B μ u 0 |, a contradiction. We follow again the argument used in Case 1. We show that |N | must be odd with a prime divisor p > 3. Then we let = θ n,8 p and define the subgroupsḠ,C, P and R of Aut( ) in the same way as in Case 1, in particular,Ḡ = P : R. In this case, /P ∼ = N 12; a θ n,12 , b θ n,12 , c θ n, 12 ; k θ n,12 ∼ = /N ∼ = N (12; 2, 4, 8; 5).
Next, we let E = N 12; a θ n,12 , b θ n,12 , c θ n, 12 ; k θ n,12 , A = Aut(E), B = ncl A ( σ ) and D = ncl B ( σ ), where σ = (u 0 , u 1 , . . . , u 12 )(v 0 , v 1 , . . . , v 12 ). Then D < B < A and |A| = 2|B| = 4|D|. We compute by Magma [2] that ncl D ( σ ) = D. An injective homomorphism φ : R → A is constructed as in Case 1 such that R φ = A or R φ = B or R φ is conjugate to D in A. Then we replace R with a suitable conjugate in A such that R contains a subgroup L for which L φ = D. We show that P : L = P × L. Finally, a contradiction arises after applying Lemma 3 to P×L and H , where H = (P×L) u 0 .