Classification of edge-transitive Nest graphs

Abstract

A finite simple graph $\Gamma$ 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).

1 Introduction

All groups in this paper will be finite and all graphs will be finite and simple. Let n be an integer with $n \ge 4$ and let ${\mathbb {Z}}_n$ denote the ring of residue classes of integers modulo n. Fix $a, b, c, k \in {\mathbb {Z}}_n$ such that each of them is distinct from 0 (the zero element of ${\mathbb {Z}}_n$), the elements a, b and c are pairwise distinct, and in the case when n is even, $k \ne n/2$. Then the Nest graph ${\mathcal {N}}(n;a,b,c;k)$ is defined to have vertex set $\{ u_i: i \in {\mathbb {Z}}_n\} \cup \{ v_i: i \in {\mathbb {Z}}_n\}$, and six types of edges such as

• $\{u_i,u_{i+1}\}$ for $i \in {\mathbb {Z}}_n$ (rim edges),

• $\{v_i,v_{i+k}\}$ for $i \in {\mathbb {Z}}_n$ (hub edges),

• $\{u_i,v_i\}$ for $i \in {\mathbb {Z}}_n$ ( 0-spoke edges),

• $\{u_i,v_{i+a}\}$ for $i \in {\mathbb {Z}}_n$ (a-spoke edges),

• $\{u_i,v_{i+b}\}$ for $i \in {\mathbb {Z}}_n$ (b-spoke edges),

• $\{u_i,v_{i+c}\}$ for $i \in {\mathbb {Z}}_n$ (c-spoke edges),

where the sums in the subscripts are computed in ${\mathbb {Z}}_n$. It is clear that the permutation $\rho$ of the set $\{u_i: i \in {\mathbb {Z}}_n\} \cup \{v_i: i \in {\mathbb {Z}}_n\}$ defined as

$$\begin{aligned} \rho :=(u_0,u_1,\ldots ,u_{n-1})(v_0,v_1,\ldots ,v_{n-1}) \end{aligned}$$

is an automorphism of ${\mathcal {N}}(n;a,b,c;k)$, the set $\{u_i: i \in {\mathbb {Z}}_n\}$ is an $\langle \rho \rangle$-orbit, and the subgraph induced by this orbit is a cycle. In fact, this property characterises the Nest graphs. More precisely, if $\Gamma$ 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 $\Gamma$ is isomorphic to a Nest graph ${\mathcal {N}}(n;a,b,c;k)$. In what follows, the term Nest graph will also be used for the graph $\Gamma$.

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.

Table 1 All edge-transitive Nest graphs ${\mathcal {N}}(n;a,b,c;k)$

Theorem 1

Up to graph isomorphism, the edge-transitive Nest graphs are exactly the Nest graphs listed in Table 1.

Remark 1

Some words about the sporadic examples in rows no. 1–6. The first three are well-known strongly regular graphs: the complement of the Petersen graph, the Hamming graph H(2, 4), and the Shrikhande graph. The last three are not strongly-regular, 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).

Remark 2

Due to [5, Theorem 8], the edge-transitive Nest graphs of girth 3 are the graphs: ${\mathcal {N}}(4;1,2,3;1)$, ${\mathcal {N}}(12;1,3,10;5)$, the graphs in rows no. 1–3 and 5 in Table 1; and the graphs described in the following families.

• (a) ${\mathcal {N}}(n;1,2l+1,2l+2;1)$, where $l \ge 1$ and n is an even divisor of $2(l^2+l+1)$ with $n \ge 4l+2$.

• (b) ${\mathcal {N}}(2m;1,b,b+m+1;m-1)$, where $b=4b_0-1$ for some $b_0 > 1$ and m is a divisor $b^2+3$ with $m \equiv 2\!\! \pmod 4$ and $b < 2m$.

All these graphs are covered by Theorem 1. The graph ${\mathcal {N}}(4;1,2,3;1)$ is the graph in row no. 8 with $m=2$ and $s=1$, and the graph ${\mathcal {N}}(12;1,3,10;5)$ 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 $2s=(b-1)^2$.

2 Preliminaries

We collect all facts needed in the proof of Theorem 1. Given a graph $\Gamma$, let $V(\Gamma )$, $E(\Gamma )$, $A(\Gamma )$ and $\textrm{Aut}(\Gamma )$ denote its vertex set, edge set, arc set and automorphism group, respectively. The number $|V(\Gamma )|$ is called the order of $\Gamma$. The set of vertices adjacent with a given vertex v is denoted by $\Gamma (v)$. If $G \le \textrm{Aut}(\Gamma )$ and $v \in V(\Gamma )$, 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 $\textrm{Orb}(G,V(\Gamma ))$. If G is transitive on $V(\Gamma )$, then $\Gamma$ is said to be G-vertex-transitive and $\Gamma$ is called vertex-transitive when it is $\textrm{Aut}(\Gamma )$-vertex-transitive; (G-)edge- and (G-)arc-transitive graphs are defined correspondingly.

Let $\pi =\{C_1,\ldots ,C_k\}$ be an arbitrary partition of $V(\Gamma )$ with parts $C_i$. The quotient graph of $\Gamma$ with respect to $\pi$, denoted by $\Gamma /\pi$, is defined to have vertex set $\pi$, and two distinct vertices $C_i$ and $C_j$ are adjacent if and only if there is an edge $\{u,v\} \in E(\Gamma )$ such that $u \in C_i$ and $v \in C_j$. Now, if there exists a constant r such that

$$\begin{aligned} \forall \{u,v\} \in E(\Gamma ):~\textrm{if}~v \in C_i~\textrm{then}~u \not \in C_i~\textrm{and}~ |\Gamma (u) \cap C_i|=r, \end{aligned}$$

then $\Gamma$ is called an r-cover of $\Gamma /\pi$. The term cover will also be used instead of 1-cover. In the special case when $\pi =\textrm{Orb}(N,V(\Gamma ))$ for an intransitive normal subgroup $N \lhd \textrm{Aut}(\Gamma )$, $\Gamma /N$ will also be written for $\Gamma /\pi$ and when $\Gamma$ is also an r-cover (cover, respectively) of $\Gamma /N$, then the term normal r-cover (normal cover, respectively) will also be used. It is well-known that, if $\Gamma$ is a G-edge-transitive graph, $\Gamma$ is regular with valency $\kappa$, $N \lhd G$ and N is intransitive, then $\Gamma$ is a normal r-cover of $\Gamma /N$ for some r such that r divides $\kappa$.

The next result establishes some obvious isomorphisms.

Lemma 1

[5, Lemma 5] The Nest graph ${\mathcal {N}}(n;a,b,c;k)$ is isomorphic to ${\mathcal {N}}(n;a',b',c';k)$, where $\{a,b,c\}=\{a',b',c'\}$, as well as to any of the graphs:

$$\begin{aligned} {\mathcal {N}}(n;a,b,c;-k),~{\mathcal {N}}(n;-a,-b,-c;k)~\textrm{and}~{\mathcal {N}}(n;-a,b-a,c-a;k). \end{aligned}$$

Let B be a finite group. If $A \le B$, then the core of A in B, denoted by $\textrm{core}_B(A)$, 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 graph $\Gamma ={\mathcal {N}}(n;a,b,c;k)$ is core-free if $\langle \rho \rangle$ is core-free in $\textrm{Aut}(\Gamma )$, where $\rho$ is the permutation of $V(\Gamma )$ defined as $\rho =(u_0,u_1,\ldots ,u_{n-1})(v_0,v_1,\ldots ,v_{n-1})$.

Theorem 2

[7, Theorem 1.2] A Nest graph is edge-transitive and core-free if and only if it is isomorphic to one of the following graphs:

$$\begin{aligned} {\mathcal {N}}(5;1,2,3;2),~{\mathcal {N}}(8;1,3,4;3), ~{\mathcal {N}}(8;1,2,5;3)~\textrm{and}~{\mathcal {N}}(12;2,4,8;5). \end{aligned}$$

Dealing with non-core-free Nest graphs the following lemma will be useful.

Lemma 2

(cf. [9, Lemma 12]) Let $\Gamma ={\mathcal {N}}(n;a,b,c;k)$ be a G-edge-transitive Nest graph and let $X < C$ be a subgroup such that $X \lhd G$, where

$$\begin{aligned} C < G,~C=\langle \rho \rangle ~\textrm{and}~ \rho =(u_0,u_1,\ldots ,u_{n-1})(v_0,v_1,\ldots ,v_{n-1}). \end{aligned}$$

If $|X| < n/2$, then the following hold.

• (1) The kernel of the action of G on $\textrm{Orb}(X,V(\Gamma ))$ is equal to X.

• (2) $\Gamma$ is a normal cover of $\Gamma /X$.

• (3) $\Gamma /X$ is a ${\bar{G}}$-edge-transitive Nest graph, where ${\bar{G}}$ is the image of G under its action on $\textrm{Orb}(X,V(\Gamma ))$.

Let n and m be positive integers such that m divides n. There is a unique homomorphism from the group $({\mathbb {Z}}_n,+)$ onto the group $({\mathbb {Z}}_m,+)$ which maps $1 \in {\mathbb {Z}}_n$ to $1 \in {\mathbb {Z}}_m$, denote this homomorphism by $\theta _{n,m}$. Then define the mapping ${\hat{\theta }}_{n,m}: \{u_i: i \in {\mathbb {Z}}_n\} \cup \{v_i: i \in {\mathbb {Z}}_n\} \rightarrow \{u_i: i \in {\mathbb {Z}}_m\} \cup \{v_i: i \in {\mathbb {Z}}_m\}$ as

$$\begin{aligned} u_i^{{\hat{\theta }}_{n,m}}=u_{i^{\theta _{n,m}}}~\textrm{and}~ v_i^{{\hat{\theta }}_{n,m}}=v_{i^{\theta _{n,m}}}~\mathrm{for~each}~i \in {\mathbb {Z}}_n. \end{aligned}$$

(1)

Observe that, with the notations in Lemma 2, the X-orbits are exactly the preimages of the mapping ${\hat{\theta }}_{n,m}$, where $m=n/|X|$. Therefore, we have the following corollary.

Corollary 1

With the notations in Lemma 2, let $m=n/|X|$. Then

$$\begin{aligned} \Gamma /N \cong \Delta := {\mathcal {N}}\big (m;a^{\theta _{n,m}}, b^{\theta _{n,m}},c^{\theta _{n,m}};k^{\theta _{n,m}}\big ). \end{aligned}$$

Furthermore, the mapping ${\hat{\theta }}_{n,m}$ defined in (1) is a homomorphism from $\Gamma$ onto $\Delta$.

The following theorem is a special case of the main result of [9].

Theorem 3

(cf. [9, Theorem 1]) Let $\Gamma ={\mathcal {N}}(n;a,b,c;k)$ be a Nest graph with n is odd. Then $\Gamma$ is edge-transitive if and only if $\Gamma \cong {\mathcal {N}}(5;1,2,3;2)$.

We conclude the section with some necessary conditions for a Nest graph to be edge-transitive.

Lemma 3

[9, Lemma 11] Let $\Gamma$ 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.

• (1) $HgH = Hg^{-1}H$ and $|H|=6|H \cap H^g|= \frac{1}{2}|H \langle g \rangle \cap HgH|$.

• (2) $HgH \ne Hg^{-1}H$ and $|H|=3|H \cap H^g|=|H\langle g \rangle \cap HgH|$.

3 Proof of Theorem 1

Throughout this section we keep the following notation.

Hypothesis 1

• $\Gamma ={\mathcal {N}}(n;a,b,c;k)$ is a Nest graph of order 2n, $n \ge 4$,

• $C=\langle \rho \rangle$, where $\rho =(u_0,u_1,\ldots ,u_{n-1})(v_0,v_1,\ldots ,v_{n-1})$,

• $G=\textrm{Aut}(\Gamma )$ and $N=\textrm{core}_G(C)$.

Recall that $\Gamma$ is core-free if C has trivial core in G, i.e., $N=1$. In this case, due to Theorem 2, $\Gamma$ 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 $N \ne 1$.

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 $\Gamma$ is edge-transitive, then it is isomorphic to one of the graphs in rows no. 7–8 in Table 1.

Proof

If $n=4$, then $\Gamma \cong {\mathcal {N}}(4;1,2,3;1)$ 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 $\Gamma /N \cong K_4$, hence $u_0$ has two neighbours in each N-orbit distinct from $u_0^N$. 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 $\langle 2 \rangle$ as the even elements, and to the elements in the coset $\langle 2 \rangle +1$ as the odd elements, where $\langle 2 \rangle$ is the additive subgroup of ${\mathbb {Z}}_n$ generated by 2.

There are two vertex-orbits under $\langle \rho \rangle$. As $\Gamma$ is edge-transitive, these orbits are merged to one vertex-orbit under G, so $\Gamma$ is vertex-transitive. Also, since $\Gamma$ is edge-transitive, it follows that $G_{u_0}$ is either transitive on $\Gamma (u_0)$ or it has two orbits of length 3. Consequently, there is an element $g \in \Gamma (u_0)$ whose restriction to $\Gamma (u_0)$ has order 3. Denote this restriction by $g|_{\Gamma (u_0)}$. The intersection of the N-orbits with $\Gamma (u_0)$ is a $\Gamma (u_0)$-invariant partition, which consists of the sets:

$$\begin{aligned} \{u_1,u_{-1}\},~\{v_0,v_a\},~\{v_b,v_c\}. \end{aligned}$$

Therefore, $g|_{\Gamma (u_0)}$ 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

$$\begin{aligned} g|_{\Gamma (u_0)}=(u_{-1},v_0,v_b)(u_1,v_a,v_c)~\textrm{or}~ (u_1,v_0,v_b)(u_{-1},v_a,v_c). \end{aligned}$$

(2)

Assume for the moment that $g|_{\Gamma (u_0)}=(u_1,v_0,v_b)(u_{-1},v_a,v_c)$. It is straightforward to check that the permutation f of $V(\Gamma )$ defined as $u_i^f=u_i$ and $v_i^f=v_{i-a}$ is an isomorphism from $\Gamma$ to the Nest graph $\Gamma ':={\mathcal {N}}(n;-a,b-a,c-a;k)$ (compare this with Lemma 1). Moreover, $f^{-1}gf$ is an automorphism of $\Gamma '$ and $(f^{-1}gf)|_{\Gamma (u_0)}=(u_1,v_{-a},v_{b-a})(u_{-1},v_0,v_{c-a})$. 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 $1 \le s \le m-1$ such that $\gcd (s,m)=1$ and $(\rho ^i)^g=\rho ^{si}$ for every $i \in {\mathbb {Z}}_n$ such that i is even. Note that g is uniquely determined. Indeed, if $i \in {\mathbb {Z}}_n$ such that i is even, then $u_i^g=u_0^{\rho ^i g}=u_0^{g \rho ^{si}}=u_{si}$ and $v_i^g=v_0^{\rho ^i g}=v_0^{g \rho ^{si}}=v_{si+b}$; whereas if i is odd, then $u_i^g=u_{-1}^{\rho ^{i+1} g}=u_{-1}^{g \rho ^{si+s}}=v_{si+s}$, and $v_i^g=v_b^{\rho ^{i-b} g}=v_b^{g \rho ^{si-sb}}=u_{si-sb-1}$. To sum up,

$$\begin{aligned} u_i^g={\left\{ \begin{array}{ll} u_{si} &{} \textrm{if}~i~\mathrm{is~even} \\ v_{si+s} &{} \textrm{if}~i~\mathrm{is~odd}, \end{array}\right. }\quad v_i^g={\left\{ \begin{array}{ll} v_{si+b} \quad \text{ if } i \text{ is } \text{ even }, \\ u_{si-sb-1} \quad \text{ if } i \text{ is } \text{ odd }. \end{array}\right. } \end{aligned}$$

(3)

Then $u_1=u_1^{g^3}=v_{2s}^{g^2}=v_{2s^2+b}^g=u_{2s^3-1}$, implying that $s^3 \equiv 1 \!\! \pmod m$.

There are 12 N-orbits on the set of edges of $\Gamma$, a representative for each orbit is shown in the third column of Table 2. Clearly, g maps each representative to an edge of $\Gamma$. In some cases, this condition can be translated to a condition in terms of the elements $a, b, c, k \in {\mathbb {Z}}_n$, which is shown in the last column. We explain below these latter conditions case by case.

Table 2 Representatives of the N-orbits on the edges

• Row no. 1 $v_{2s}=u_1^g=v_a$, and so $a=2s$.

• Row no. 2 $\{u_0,u_{-1}\}^g=\{u_0,v_0\}$, no condition.

• Row no. 3 $\{u_0,v_0\}^g=\{u_0,v_b\}$, no condition.

• Row no. 4 $\{u_{-1},v_{-1}\}^g=\{v_0,u_{-s-sb-1}\}$. Since $-s-sb-1$ is odd, we obtain that $s+sb+1=b$ or c. We show later (see row no. 6) that it must be equal to c.

• Row no. 5 $v_c=v_a^g=v_{sa+b}$, and so $sa+b=c$. Note that, then

  $$\begin{aligned} sc-sb=s^2a=2s^3=2. \end{aligned}$$

  (4)

• Row no. 6 $\{u_{-1},v_{a-1}\}^g=\{v_0,u_{sa-s-sb-1}\}$. Since $sa-s-sb-1$ is odd, we obtain that $-sa+s+sb+1=b$ or c. Moreover,

  $$\begin{aligned} \{ s+sb+1, -sa+s+sb+1\}=\{ b, c\}. \end{aligned}$$

  Furthermore, $(s+sb+1)-(-sa+s+sb+1)=sa=c-b$ and $c-b \ne b-c$ (use (4) and that $n > 4$), we conclude that $s+sb+1=c$ and $-sa+s+sb+1=b$.

• Row no. 7 $\{u_0,v_b\}^g=\{u_0,u_{-1}\}$, no condition.

• Row no. 8 $\{u_{-1},v_{b-1}\}^g=\{v_0,v_{sb-s+b}\}$. We may assume w.l.o.g. that $sb-s+b=-k$.

• Row no. 9 $\{u_0,v_c\}^g=\{u_0,u_{sc-sb-1}\}$, no condition due to (4).

• Row no. 10 $\{u_{-1},v_{c-1}\}^g=\{v_0,v_{sc-s+b}\}$, and in view of the condition obtained by row no.8, we have $sc-s+b=k$.

• Row no. 11 $\{v_0,v_k\}^g=\{v_b,u_{sk-sb-1}\}$. Since $sk-sb-1$ is odd, it follows that $sk-sb-1=b/b-a$.

• Row no. 12 $\{v_b,v_{b+k}\}^g=\{u_{-1},v_{sb+sk+b}\}$. Again, as $sb+sk+b$ is odd, $sb+sk+b=a-1/-1$. Moreover,

  $$\begin{aligned} (sk-sb-1,sb+sk+b)=(b,a-1)~\textrm{or}~(b-a,-1). \end{aligned}$$

  (5)

  Observe next that (4) combined with the conditions in rows no.8 and 10 yield $2k=sc-sb=2$. Thus $k=1$ or $m+1$.

Case 1. $k=1$.

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., $\Gamma$ is the graph in row no. 7 in Table 1.

Case 2. $k=m+1$.

In this case we get $2b=2s-2s^2+m$. Using that b is odd, we find in turn that $m \equiv 2 \!\! \pmod 4$, and either $b=s-s^2+m/2$ and $c=s+s^2+m/2$, i.e., $\Gamma$ 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,

$$\begin{aligned}{} & {} \Gamma \cong {\mathcal {N}}(2m;c-b,-b,a-b;m+1)\\{} & {} \qquad ={\mathcal {N}}(2m;2s^2,s^2-s+m/2,s^2+s+m/2;m+1). \end{aligned}$$

The latter graph is isomorphic to the graph in row no. 8 in Table 1 with letting s to be $s^2$ because $s^4= \pm s$ holds in ${\mathbb {Z}}_n$. $\square$

We continue by showing that the graphs in rows no. 7–8 in Table 1 are edge-transitive.

Lemma 5

Let $\Delta$ be the Nest graph given in row no. 7 in Table 1, i.e.,

$$\begin{aligned} \Delta ={\mathcal {N}}(2m;2s,s-s^2+m,s+s^2+m;1), \end{aligned}$$

where m is odd and $s^3 \equiv 1 \!\! \pmod m$. Then $\Delta$ is edge-transitive.

Proof

Let $\rho =(u_0,u_1,\ldots ,u_{n-1})(v_0,v_1,\ldots ,v_{n-1})$, let $N=\langle \rho ^2 \rangle$, and let g be the permutation of $V(\Delta )$ defined in (3). Let us consider again the 12 representatives of the N-orbits on the edges of $\Delta$ that are given in Table 2.

We claim that g maps any of these edges to an edge of $\Delta$. 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

$$\begin{aligned} s+s(s-s^2+m)+1\equiv & {} s+s^2+m \pmod 2, \\ s+s(s-s^2+m)+1\equiv & {} s+s^2+m \pmod m. \end{aligned}$$

The first congruence holds because m is odd, while the second follows from the fact that $s^3 \equiv 1 \pmod m$. 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

$$\begin{aligned} (s+1)(s-s^2+m)-s\equiv & {} -1 \pmod 2, \\ (s+1)(s-s^2+m)-s\equiv & {} -1 \pmod m. \end{aligned}$$

Again, the first congruence holds because m is odd, while the second follows from the fact that $s^3 \equiv 1 \pmod m$. 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 $\Delta$. Then $e=(e_*)^{\rho ^i}$, where $e_*$ is one of the representatives in Table 2 and $i \in {\mathbb {Z}}_n$ such that i is even. A straightforward computation shows that $\rho ^i g=g \rho ^{is}$, and hence $e^g=(e_*)^{\rho ^i g}=(e_*)^{g\rho ^{is}}$, which shows that $e^g \in E(\Delta )$. We have shown that g is an automorphism of $\Delta$. Clearly, the group $\langle \rho , g \rangle$ acts transitively on $V(\Delta )$.

Consider the arc-orbit of $(u_0,u_1)$ under $\langle \rho , g \rangle$. This contains the arcs

$$\begin{aligned} (u_0,v_a),~(u_0,v_c),~(u_{-1},u_0),~(v_0,u_0),~(v_b,u_0). \end{aligned}$$

This shows that all edges incident with $u_0$ belong to the same orbit under $\langle \rho , g \rangle$, hence $\Delta$ is indeed edge-transitive. $\square$

The proof of the next lemma goes exactly in the same way as the proof of the previous lemma, hence it is omitted.

Lemma 6

Let $\Delta$ be the Nest graph given in row no. 8 in Table 1, i.e.,

$$\begin{aligned} \Delta ={\mathcal {N}}(2m;2s,s-s^2+m/2,s+s^2+m/2;1), \end{aligned}$$

where $m \equiv 2 \pmod 4$ and $s^3 \equiv 1 \pmod m$. Then $\Delta$ is edge-transitive.

We turn to the case when $1< |N| < n/2$.

Lemma 7

Assuming Hypothesis 1, suppose that $\Gamma$ is edge-transitive and $|N| < n/2$ and let $X \le N$. Then $X \lhd G$ and $\Gamma /X$ 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 $N \lhd G$, we obtain that $X \lhd G$. Now, the lemma follows from Lemma 2(3). $\square$

As the next step, we describe the quotient graph $\Gamma /N$.

Lemma 8

Assuming Hypothesis 1, suppose that $\Gamma$ is edge-transitive and $|N| < n/2$. Then the quotient graph $\Gamma /N$ is isomorphic to one of the following graphs:

$$\begin{aligned} {\mathcal {N}}(5;1,2,3;2),~{\mathcal {N}}(8;1,3,4;3),~{\mathcal {N}}(8;1,2,5;3) ~\textrm{and}~{\mathcal {N}}(12;2,4,8;5). \end{aligned}$$

Proof

By Lemma 2(3), $\Gamma /N$ is a ${\bar{G}}$-edge-transitive Nest graph, where ${\bar{G}}$ is the image of G under its action on $\textrm{Orb}(N,V(\Gamma ))$. 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 $N \le M \le C$ and ${\bar{M}}=\textrm{core}_{{\bar{G}}}({\bar{C}})$. As M is normal in G, we find that $M=N$, so ${\bar{C}}$ has trivial kernel in ${\bar{G}}$. This yields that $\Gamma /N$ is a core-free graph, and the lemma follows from Theorem 2. $\square$

We proceed with the case when $1< |N| < n/2$ and $\Gamma /N \cong {\mathcal {N}}(5;1,2,3;2)$.

Lemma 9

Assuming Hypothesis 1, suppose that $\Gamma$ is edge-transitive, $1< |N| < n/2$, and $\Gamma /N \cong {\mathcal {N}}(5;1,2,3;2)$. Then $\Gamma$ is isomorphic to one of the graphs in rows no. 5–6 in Table 1.

Proof

We are going to show that $\Gamma \cong {\mathcal {N}}(10;1,3,4;3)$ or $\Gamma \cong {\mathcal {N}}(10;2,4,6;3)$.

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, $\Gamma /M$ is an edge-transitive Nest graph of order 10p. This is impossible due to Theorem 3, hence we find that $|N|=2^\ell$ for some $\ell \ge 1$ and $n=5 \cdot 2^\ell$.

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 $\ell =1$ or 2.

If $\ell =2$, then $\Gamma \cong {\mathcal {N}}(20;2,5,7;9)$. Using Magma [2] we check that the core of C in $\textrm{Aut}(\Gamma )$ has order 10, which contradicts that $|N|=4$.

Thus $\ell =1$, and $\Gamma$ is isomorphic to one of the following graphs:

$$\begin{aligned} {\mathcal {N}}(10;1,3,4;3),~{\mathcal {N}}(10;2,4,6;3)~\textrm{and}~{\mathcal {N}}(10;2,5,7;1). \end{aligned}$$

Using Magma [2], one can quickly check that the corresponding core N has order 2, 2 and 5, respectively. Finally, if $\Gamma$ is one of the first two graphs, then $\Gamma /N \cong {\mathcal {N}}(5;1,2,3;1)$, which can be deduced directly if one uses the fact that $|N|=2$ and combines Corollary 1 with Lemma 1. $\square$

In view of Theorem 2 and the Lemmas 4–6, 8 and 9, Theorem 1 follows if we show the lemma below.

Lemma 10

Assuming Hypothesis 1, suppose that $\Gamma$ is edge-transitive and $1< |N| < n/2$. Then $\Gamma /N \cong {\mathcal {N}}(5;1,2,3;2)$.

Proof

Assume on the contrary that $\Gamma /N \not \cong {\mathcal {N}}(5;1,2,3;2)$. Then by Lemma 8, $\Gamma /N$ is isomorphic to one of the following graphs:

$$\begin{aligned} {\mathcal {N}}(8;1,3,4;3),~{\mathcal {N}}(8;1,2,5;3)~\textrm{and}~{\mathcal {N}}(12;2,4,8;5). \end{aligned}$$

We discuss the case when $\Gamma /N \cong {\mathcal {N}}(8;1,3,4;3)$ in details and give only a sketch for the other two cases.

Case 1. $\Gamma /N \cong {\mathcal {N}}(8;1,3,4;3)$.

In this case $n=8|N|$. If |N| is even, then Lemma 7 yields the existence of an edge-transitive Nest graph of order 32. This is impossible due to [5, Table 1], thus |N| is odd.

Let p be a prime divisor of |N| and let Q be the subgroups of N such that $|Q|=|N|/p$. By Lemma 7, $\Gamma /Q$ has order 16p. Using this and [5, Table 1], we find that $p > 3$. Let

$$\begin{aligned} \Delta ={\mathcal {N}}\big (8p;a^{\theta _{n,8p}},b^{\theta _{n,8p}}, c^{\theta _{n,8p}};k^{\theta _{n,8p}}\big ), \end{aligned}$$

By Corollary 1, $\Delta =\Gamma ^{{\hat{\theta }}_{n,8p}} \cong \Gamma /Q$. It follows from Lemma 2 that there is a subgroup ${\bar{G}} \le \textrm{Aut}(\Delta )$ such that ${\bar{G}}$ acts transitively on $E(\Delta )$, ${\bar{G}} \cong G/Q$, and ${\bar{G}}$ contains the group ${\bar{C}}$, where ${\bar{C}}=\langle (u_0,u_1,\ldots ,u_{8p-1})(v_0,v_1,\ldots ,v_{8p-1}) \rangle$. Let P be the Sylow p-subgroup of ${\bar{C}}$. Since $N \lhd G$, it follows that $P \lhd {\bar{G}}$. By the Schur–Zassenhaus theorem (see, e.g., [6, Chapter 1, Theorem 18.1]), there exists a subgroup $R < {\bar{G}}$ such that ${\bar{G}}=P: R$ (the semidirect product of P with R). Moreover, conjugating R with a suitable element, we may also assume that $|{\bar{C}} \cap R|=8$.

Using again Corollary 1 and the fact that the composition of $\theta _{n,8p}$ with $\theta _{8p,8}$ is equal to $\theta _{n,8}$, we have that,

$$\begin{aligned} \Delta /P \cong {\mathcal {N}}\big (8;a^{\theta _{n,8}},b^{\theta _{n,8}}, c^{\theta _{n,8}};k^{\theta _{n,8}}\big ) \cong \Gamma /N \cong {\mathcal {N}}(8;1,3,4;3). \end{aligned}$$

For the sake simplicity, write ${\mathcal {E}}={\mathcal {N}}\big (8;a^{\theta _{n,8}},b^{\theta _{n,8}}, c^{\theta _{n,8}};k^{\theta _{n,8}}\big )$ and $A=\textrm{Aut}({\mathcal {E}})$.

Claim. There is an injective homomorphism $\phi : R \rightarrow A$ such that $R^\phi$ acts transitively on $E({\mathcal {E}})$ and $R^\phi$ contains the permutation $\sigma$, where $\sigma =(u_0,\ldots ,u_7)(v_0,\ldots ,v_7)$.

For $x \in R$, denote by $x^*$ the image of x under its action on $\textrm{Orb}(P,V(\Delta ))$. Note that, for any $w \in \{u_0,\ldots ,u_7\} \cup \{v_0,\ldots ,v_7\}$, the preimage of w under ${\hat{\theta }}_{8p,8}$ is equal to a P-orbit. This allows us to define the permutation $x'$ of $V({\mathcal {E}})$ as follows:

$$\begin{aligned} (u_i)^{x'}=\Big (\big (u_i^{({\hat{\theta }}_{8p,8})^{-1}}\big )^{x^*} \Big )^{{\hat{\theta }}_{8p,8}}~\textrm{and}~ (v_i)^{x'}=\Big (\big (v_i^{({\hat{\theta }}_{8p,8})^{-1}}\big )^{x^*} \Big )^{{\hat{\theta }}_{8p,8}}~\mathrm{for~each}~i \in {\mathbb {Z}}_8. \end{aligned}$$

Now, define $\varphi$ to be the mapping $\varphi : {\bar{G}} \rightarrow \textrm{Sym}(V({\mathcal {E}}))$, the group of all permutations of $V({\mathcal {E}})$, by letting $x^\varphi =x'$. It is straightforward to check that $\varphi$ is a homomorphism from ${\bar{G}}$ to A. Let $\phi$ be the restriction of $\varphi$ to R.

Suppose that $x, y \in R$ such that $x^\phi =y^\phi$. Then $x^*=y^*$, so $xy^{-1}$ is in the kernel of the action of R on $\textrm{Orb}(P,V(\Delta ))$. By Lemma 2(1), the kernel of the action of ${\bar{G}}$ on $\textrm{Orb}(P,V(\Delta ))$ is equal to P. Now, since ${\bar{G}}=P: R$, it follows that $x=y$, and so $\phi$ is injective. It can be easily checked that this also implies that $R^\phi$ acts transitively on $E({\mathcal {E}})$. Finally, the fact that $\sigma \in R^\phi$ follows from the assumption that $|{\bar{C}} \cap R|=8$. Indeed, if follows from the definition of $\varphi$ that $((u_0,u_1,\ldots ,u_{8p-1})(v_0,v_1,\ldots ,v_{8p-1}))^\varphi =\sigma$, and as $(u_0,u_1,\ldots ,u_{8p-1})^p(v_0,v_1,\ldots ,v_{8p-1})^p \in R$, $\sigma ^p \in R^\phi$, so $\sigma \in R^\phi$. This completes the proof of the claim.

Let $B=\textrm{ncl}_A(\langle \sigma \rangle )$, the normal closure of $\langle \sigma \rangle$ in A (i.e., the smallest normal subgroup of A containing in $\langle \sigma \rangle$). A computation with Magma [2] shows that B is the only proper subgroup of A acting transitively on $E({\mathcal {E}})$ and containing $\sigma$ and that $|A|=2|B|$. Consequently, $B \le R^\phi$ and therefore, there is a subgroup $L \le R$ such that $L^\phi =B$.

Let us consider the centraliser $C_L(P)$. As $P \lhd {\bar{G}}$, $C_L(P) \lhd L$, hence $(C_L(P))^\phi \lhd B$. On the other hand, ${\bar{C}} \le C_{{\bar{G}}}(P)$, hence $\sigma \in (C_L(P))^\phi$, implying that $\textrm{ncl}_B(\langle \sigma \rangle ) \le (C_L(P))^\phi$. A computation with Magma [2] shows that $B=\textrm{ncl}_B(\langle \sigma \rangle )$, hence $(C_L(P))^\phi =B=L^\phi$, so $C_L(P)=L$, or equivalently, $P: L=P \times L$.

Now, let $H=(P \times L)_{u_0}$, the stabiliser of the vertex $u_0$ in $P \times L$ (here $0 \in {\mathbb {Z}}_{8p}$). Notice that $H < L$ and that $H^\phi =B_{u_0}$, the stabiliser of the vertex $u_0$ in B (here $0 \in {\mathbb {Z}}_{8}$). To sum up, $\Delta$ is a $(P \times L)$-edge-transitive Nest graph of order 8p and H is a vertex stabiliser in $P \times L$ such that $H^\phi =B_{u_0}$.

The desired contradiction will arise after applying Lemma 3 to $P \times L$ and H. Due to this lemma, there is an element $g \in P \times L$ of order 8p satisfying all conditions in either part (1) or (2) of Lemma 3. Then $g=z \lambda$, where z is a generator of P and $\lambda$ is an element of order 8 in L.

Using that z commutes with every element of H, it is easy to see that $g^{-1} \notin H g H$. Thus $HgH \ne Hg^{-1}H$, and so we have $|H|=3 |H \cap H^g|=|H\langle g \rangle \cap HgH|$. Since $H^z=H$, it follows that $H^g=H^\lambda$. Also,

$$\begin{aligned} H \langle g \rangle =\bigcup _{i=0}^{p-1}z^i H \langle \lambda \rangle ~\textrm{and}~HgH=z H \lambda H. \end{aligned}$$

All these yield that the equalities $|H|=3 |H \cap H^g|=|H\langle g \rangle \cap HgH|$ reduce to the following equalities in L:

$$\begin{aligned} |H|=3 |H \cap H^\lambda | = |H\langle \lambda \rangle \cap H \lambda H|. \end{aligned}$$

Now, applying $\phi$ and letting $\mu =\lambda ^\phi$, we deduce from these the following equalities in B:

$$\begin{aligned} |B_{u_0}|=3 |B_{u_0} \cap (B_{u_0})^\mu | = |B_{u_0} \langle \mu \rangle \cap B_{u_0} \mu B_{u_0}|. \end{aligned}$$

Recall that ${\mathcal {E}}$ is the Hamming graph H(2, 4), $B \le \textrm{Aut}({\mathcal {E}})$ such that B acts transitively on the arcs of ${\mathcal {E}}$ and $2|B|=|A|$. As $A \cong (S_4 \times S_4): {\mathbb {Z}}_2$, 3 must divide $|B_{u_0} \cap (B_{u_0})^\mu |$. If $u_0$ and $u_{0}^\mu$ are non-adjacent, then one can easily find that $B_{u_0} \cap (B_{u_0})^\mu$ is a 2-group. This contradicts that $|B_{u_0} \cap (B_{u_0})^\mu |$ is divisible by 3, hence $u_0$ and $u_0^\mu$ are adjacent. Using also that $B_{u_0}$ is transitive on ${\mathcal {E}}(u_0)$, we have that $|B_{u_0}|=6|B_{u_0} \cap B_{u_0}^\mu |$, a contradiction.

Case 2. $\Gamma /N \cong {\mathcal {N}}(8;1,2,5;3)$.

The argument used in Case 1 can be copied literally. We show that |N| must be odd with a prime divisor $p > 3$. Then we let $\Delta =\Gamma ^{{\hat{\theta }}_{n,8p}}$ and define the subgroups ${\bar{G}}, {\bar{C}}, P$ and R of $\textrm{Aut}(\Delta )$ in the same way as in Case 1, in particular, ${\bar{G}}=P: R$. In this case,

$$\begin{aligned} \Delta /P \cong {\mathcal {N}}\big (8;a^{\theta _{n,8}},b^{\theta _{n,8}}, c^{\theta _{n,8}};k^{\theta _{n,8}}\big ) \cong \Gamma /N \cong {\mathcal {N}}(8;1,2,5;3). \end{aligned}$$

Next, we let ${\mathcal {E}}={\mathcal {N}}\big (8;a^{\theta _{n,8}},b^{\theta _{n,8}}, c^{\theta _{n,8}};k^{\theta _{n,8}}\big )$, $A=\textrm{Aut}({\mathcal {E}})$ and $B=\textrm{ncl}_A(\langle \sigma \rangle )$, where $\sigma =(u_0,u_1,\ldots ,u_7)(v_0,v_1,\ldots ,v_7)$. A computation with Magma [2] shows that $\textrm{ncl}_B(\langle \sigma \rangle )=B$. An injective homomorphism $\phi : R \rightarrow A$ is constructed as in Case 1 such that $R^\phi =A$ or $R^\phi =B$. Then we define L as the subgroup of R for which $L^\phi =B$ and show that $P: L=P \times L$. Finally, a contradiction arises after applying Lemma 3 to $P \times L$ and H, where $H=(P \times L)_{u_0}$.

Case 3. $\Gamma /N \cong {\mathcal {N}}(12;2,4,8;5)$.

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 $\Delta =\Gamma ^{{\hat{\theta }}_{n,8p}}$ and define the subgroups ${\bar{G}}, {\bar{C}}, P$ and R of $\textrm{Aut}(\Delta )$ in the same way as in Case 1, in particular, ${\bar{G}}=P: R$. In this case,

$$\begin{aligned} \Delta /P \cong {\mathcal {N}}\big (12;a^{\theta _{n,12}},b^{\theta _{n,12}}, c^{\theta _{n,12}};k^{\theta _{n,12}}\big ) \cong \Gamma /N \cong {\mathcal {N}}(12;2,4,8;5). \end{aligned}$$

Next, we let ${\mathcal {E}}={\mathcal {N}}\big (12;a^{\theta _{n,12}},b^{\theta _{n,12}}, c^{\theta _{n,12}};k^{\theta _{n,12}}\big )$, $A=\textrm{Aut}({\mathcal {E}})$, $B=\textrm{ncl}_A(\langle \sigma \rangle )$ and $D=\textrm{ncl}_B(\langle \sigma \rangle )$, where $\sigma =(u_0,u_1,\ldots ,u_{12})(v_0,v_1,\ldots ,v_{12})$. Then $D< B < A$ and $|A|=2|B|=4|D|$. We compute by Magma [2] that $\textrm{ncl}_D(\langle \sigma \rangle )=D$. An injective homomorphism $\phi : R \rightarrow A$ is constructed as in Case 1 such that $R^\phi =A$ or $R^\phi = B$ or $R^\phi$ 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 $L^\phi =D$. We show that $P: L=P \times L$. Finally, a contradiction arises after applying Lemma 3 to $P \times L$ and H, where $H=(P \times L)_{u_0}$. $\square$