Classification of 14-Valent 1-Regular Core-Free Cayley Graphs

Abstract

A Cayley graph $\Sigma =Cay(G,S)$ is called 1-regular core-free if G is core-free in some $Y⩽Aut\Sigma$ and $Aut\Sigma$ acts regularly on the set of 1-arcs of $\Sigma$. In this paper, we classify the 14-valent 1-regular core-free Cayley graphs. In particular, we discover a non-normal Cayley graph on a non-abelian simple group. That is, 14-valent 1-regular Cayley graph on the alternating group $A_6$, with full automorphism group isomorphic to $S_7$. To our knowledge, this is the first example of a non-normal Cayley graph on a non-abelian simple group with even valency greater than 10.

1. Introduction

The concept of Cayley graphs can be naturally extended to infinite groups, directed graphs, and weighted graphs, serving as an important model for constructing transitive graphs, symmetric graphs, and distance-regular graphs. The study of their normality and core-freeness is of great significance for characterizing the structure of automorphism groups, analyzing regular covers of graphs, and the primitivity of group actions. Meanwhile, Cayley graphs also serve as an important model for interconnection networks. For further applications, see [1,2,3,4,5].

All graphs considered in this paper are assumed to be finite, connected, simple and undirected, unless otherwise stated. Let $\Sigma$ be a graph, its vertex set, edge set, arc set and full automorphism group are denoted by $V\Sigma$, $E\Sigma$, $Arc\Sigma$ and $Aut\Sigma$, respectively. Let Y be a subgroup of $Aut\Sigma$, denoted $Y⩽Aut\Sigma$. The valency of $\Sigma$ is denoted by $val\Sigma$, and let t be a positive integer. If Y acts transitively on the set of t-arcs of $\Sigma$, then $\Sigma$ is called a $(Y,t)$-arc-transitive graph, where a t-arc is a $(t+1)$-tuple $(v_0,v_1,\dots ,v_t)$ of vertices such that $(v_{i-1},v_i)\in E\Sigma$ for each i and $v_{i-1}\ne v_{i+1}$. If $\Sigma$ is $(Y,t)$-arc-transitive but not $(Y,t+1)$-arc-transitive, then it is called a $(Y,t)$-transitive graph. In particular, when $Y=Aut\Sigma$, $(Aut\Sigma ,t)$-arc-transitive graphs and ($Aut\Sigma ,t$)-transitive graphs are simply called t-arc-transitive graphs and t-transitive graphs, respectively. A 0-arc-transitive graph is called a vertex-transitive graph, and a 1-arc-transitive graph is called an arc-transitive graph or a symmetric graph. For an arc-transitive graph $\Sigma$, if it is $(Y,t)$-arc-transitive and satisfies $\left|Y\right|=|V\Sigma |·val\Sigma {(val\Sigma -1)}^{t-1}$, then $\Sigma$ is called a $(Y,t)$-regular graph. When $Y=Aut\Sigma$, a $(Y,t)$-regular graph is simply called a t-regular graph. If for each vertex $v\in V\Sigma$, the stabiliser subgroup $Y_v$ acts primitively on the neighbourhood $\Sigma \left(v\right)=\{u\mid (v,u)\in E\Sigma \}$, then the graph is called Y-locally primitive.

Let G be a finite group with identity denoted by 1. If there exists a subset S of G such that $1\notin S$ and $S=S^{-1}$, where $S^{-1}=\{s^{-1}\mid s\in S\}$, and the graph $\Sigma$ has vertex set $V\Sigma =G$ and edge set $E\Sigma =\left\{\right(g,sg)\mid g\in G,s\in S\}$, then $\Sigma$ is called a Cayley graph of the group G, denoted $\Sigma =Cay(G,S)$, and S is called its connection set. By definition, $Cay(G,S)$ is $\left|S\right|$-regular, and G acts regularly on the vertex set by right multiplication and can be regarded as a regular subgroup of $Aut\Sigma$; thus, Cayley graphs are vertex-transitive. Conversely, the Sabidussi theorem states that a graph $\Sigma$ is a Cayley graph of some group G if and only if $Aut\Sigma$ contains a subgroup that is regular on $V\Sigma$ and isomorphic to G. Moreover, $Cay(G,S)$ is connected if and only if S generates G. If G is a normal subgroup of $Aut(Cay(G,S\left)\right)$, then the Cayley graph is called normal, and in this case $Aut(Cay(G,S\left)\right)$ can be expressed as a semidirect product of G with some vertex stabiliser. If there exists a subgroup $Y⩽Aut(Cay(G,S\left)\right)$ such that ${Core}_Y\left(G\right):={\bigcap }_{x\in Y}{G}^x=1$, where $G^x=x^{-1}Gx$, then the Cayley graph is called core-free Cayley graph. The constructions and concrete examples of core-free Cayley graphs can be found in [6].

Normal and non-normal Cayley graphs are complementary in the analysis of the structure of the automorphism group, and core-freeness provides another approach to the study of normality of Cayley graphs. In recent years, significant progress has been made in the study of normality of Cayley graphs. For cubic Cayley graphs on non-abelian simple groups, related results on normality can be found in [7,8,9,10]. Ref. [7] proved that, except for a few candidates of groups G, such $\Sigma$ are normal. Building on this result, Ref. [8] verified the normality of connected cubic Cayley graphs over several finite simple groups. Ref. [9] further proved that, except when $G=A_{47}$, all such Cayley graphs $\Sigma$ are normal. In addition, Ref. [10] showed that when $G=A_{47}$, $\Sigma$ is 5-arc-transitive. Results on the normality of tetravalent Cayley graphs appear in [11,12,13]. It is shown in [11] that such a graph $\Sigma$ is normal except for several exceptional groups G. Reference [12] proved that such 2-transitive Cayley graphs $\Sigma$ are normal except for 7 possible candidates for G. Subsequently, Ref. [13] refined the result for the case $s=2$ based on the results in [11]. Results on the normality of pentavalent Cayley graphs appear in [14,15,16]. It is demonstrated in [16] that if $\Sigma$ is 1-transitive, then $\Sigma$ is normal. Furthermore, Ref. [15] (Theorem 1.4) provided a general characterization of the normality of $\Sigma$. Recently, Ref. [14] further refined this result. Research on the normality of heptavalent Cayley graphs is developed in [17,18].

However, directly investigating the normality of Cayley graphs on non-abelian simple groups presents certain challenges. Consequently, many studies approach the normality of Cayley graphs on non-abelian simple groups by classifying core-free Cayley graphs. Specifically, research on cubic core-free Cayley graphs can be found in [6,19,20]. All cubic 5-transitive core-free Cayley graphs are systematically classified in [19]. References [6,20] provided another proof for the result in [9]. Studies of tetravalent core-free Cayley graphs can be found in [21]. By classifying connected tetravalent core-free s-regular Cayley graphs, [21] proved that if $\Sigma$ is s-regular, then $\Sigma$ is normal except for $G=A_{35}$. Studies of pentavalent core-free Cayley graphs are contained in [22,23]. It is established in [22] that every pentavalent 2-regular Cayley graph $\Sigma$ over non-abelian simple group G is normal. Reference [23] proves that a pentavalent 2-transitive Cayley graph $\Sigma$ over non-abelian simple group G with a solvable vertex stabilizer is normal with the only exception of the alternating group $A_5$. In this paper, we introduce a new approach combining theoretical and computational methods to classifying core-free Cayley graphs, explore the normality of 14-valent 1-regular Cayley graphs, and the main result is the following theorem.

Theorem 1.

Let $\Sigma =Cay(G,S)$ be a 14-valent 1-regular core-free Cayley graph. Then either

(1) 

the pairs $(G,Aut\Sigma )$ are listed in the following

Table 1. Pairs $(G,Aut\Sigma )$.

G	$Aut\Sigma$	$\Sigma$
$\mathrm{PSL}(3,2)\times S_4$	$\mathrm{PSL}{(3,2)}^2⋊C_2$	Lemmas 1 and 2
$A_6$	$S_7$	Lemma 7
$A_6⋊C_2$	$C_2\times S_7$	Lemma 7

.

(2) 

or there exists a subgroup Y of $Aut\Sigma$ such that $G\le Y$ and G is core-free in Y. Moreover, the pairs $(G,Y)$ are listed in the following

Table 2. Pairs $(G,Y)$.

G	Y	$\Sigma$
$A_7\times A_6$	$A_7^2⋊C_2$	Lemma 3
$A_7\times S_6$	$A_7^2⋊C_2^2$	Lemma 4
$S_7\times S_6$	$A_7^2⋊D_8$	Lemma 5
$S_{13}$	$S_{14}$	Lemma 6
$(C_2^5⋊A_6)⋊C_2$	$C_2^6⋊S_7$	Lemma 8
$C_2^6⋊S_6$	$C_2^7⋊S_7$	Lemma 9
$A_7\times A_6$	$A_7^2⋊C_2$	Lemma 10
$A_7⋊S_6$	$A_7^2⋊C_2^2$	Lemma 11
$S_7\times S_6$	$A_7^2⋊D_8$	Lemma 12
$S_{13}$	$S_{14}$	Lemma 13

.

By Theorem 1, we find that all 14-valent 1-regular Cayley graph on non-abelian simple groups are normal. In addition, a review of recent literature confirms that all 10-valent 1-regular Cayley graphs on non-abelian simple groups are normal [24]. Previously, no non-normal examples with even valency greater than 10 were known. Hence, the following Remark 1 holds.

Remark 1.

Let G be a non-abelian simple group and Σ a 14-valent 1-regular Cayley graph on G. Then not all such Σ are normal. In our results, we find a non-normal 14-valent 1-regular Cayley graph, namely a Cayley graph on $A_6$ whose automorphism group is $S_7$.

2. Materials and Methods

In algebraic graph theory, a coset graph is a highly symmetric graph whose construction is based on the action of a group on cosets and the decomposition into double cosets, providing a unified framework for the study of arc-transitive graphs. Specifically, let Y be a finite group and H be a core-free subgroup of Y. First, this condition guarantees that the right multiplication action of Y on the set of right cosets $[Y:H]=\{Hx\mid x\in Y\}$ is faithful, thereby allowing Y to be regarded as a permutation group on the vertex set. Second, take an arbitrary element $g\in Y\setminus H$, define the graph $\Sigma =Cos(Y,H,g)$ with vertex set $[Y:H]$, where two vertices $Hx,Hy$ are adjacent if and only if $y{x}^{-1}\in HgH$, with $HgH$ a double coset. This definition is independent of the choice of representatives, if $Hx=H{x}^{\prime }$, then $x^{\prime }=hx$ for some $h\in H$, and hence $y{\left(x^{\prime }\right)}^{-1}=y{x}^{-1}{h}^{-1}\in HgH·H=HgH$. In this setting, Y acts on the vertex set by right multiplication via $\left(Hx\right)·t=H\left(xt\right)$, and this action preserves adjacency, because if $y{x}^{-1}\in HgH$, then $\left(yt\right){\left(xt\right)}^{-1}=y{x}^{-1}\in HgH$, thus Y is a subgroup of the automorphism group $Aut\Sigma$. Such graphs play a central role in the study of symmetric graphs. Third, the graph $\Sigma$ is connected if and only if $\langle H,g\rangle =Y$, since the vertices reachable from H via adjacency relations are precisely the right cosets of $H\langle H,g\rangle$, if this subgroup is proper, the graph is disconnected. Fourth, the valency of the graph equals $|H:H\cap H^g|$, where $H^g=g^{-1}Hg$. In particular, this result follows from counting the distinct cosets adjacent to H, the adjacent vertices are of the form $HhgH$ and $hg$, which varies over a set of double coset representatives, and their number equals the index of H in $H^g$. The construction of coset graphs and their typical examples are available in [25].

If Y contains a regular subgroup G (that is, G acts regularly on $[Y:H]$, in which case $Y=GH$ and $G\cap H=\left\{1\right\}$), then the graph $\Sigma$ is isomorphic to the Cayley graph $Cay(G,G\cap (HgH\left)\right)$. Specifically, by fixing the coset H to correspond to the identity, the vertices are identified with elements of G, and the adjacency relation translates to the Cayley graph generated by the set $S=G\cap \left(HgH\right)$. First, since the original graph is undirected, S satisfies the symmetry condition $S=S^{-1}$. Second, this construction unifies coset graphs and Cayley graphs: on the one hand, Cayley graphs are the special case where $H=\left\{1\right\}$, on the other hand, coset graphs can be viewed as quotients or covers of Cayley graphs under the action of a larger group. Third, coset graphs have wide applications in algebraic graph theory, many classical arc-transitive graphs and distance-transitive graphs, such as the Petersen graph, the Heawood graph, and the Tutte–Coxeter graph, can be realized by suitably choosing an almost simple group Y, a subgroup H, and an element g. Their properties, including girth, diameter, and automorphism group structure, can be derived from group-theoretic considerations, and they play an important role in the classification and construction of symmetric graphs. For a more detailed theory of coset graphs, one may refer to Sabidussi’s classical work on vertex-transitive graphs and Cayley graphs, Li’s systematic classification of finite edge-transitive graphs, and the algebraic graph theory texts by Biggs and others [26,27,28].

Let $\Sigma =Cay(G,S)$ be a core-free Y-arc-transitive Cayley graph, where $G⩽Y⩽Aut\Sigma$. Fix a vertex $v\in V\Sigma$, denote $H=Y_v$ the stabiliser of v in Y, and set $\left|H\right|=n$. First, consider the right multiplication action of Y on the set of right cosets $[Y:G]$, that is, for any $x\in Y$ and coset $Gg$, define ${\left(Gg\right)}^x=Ggx$. Since Y acts transitively on the vertex set of $\Sigma$, by the orbit-stabiliser theorem we have $|Y:Y_v|=|V\Sigma |=|G|$. Second, $\left|Y\right|=|Y_v|·|Y:Y_v|=|H|·|G|$, and $|Y:G|=\left|Y\right|/\left|G\right|$, hence $\left|H\right|=|Y:G|$, which means that the order of H equals the size of the coset space $[Y:G]$. Third, to show that H acts regularly on $[Y:G]$, we first prove that the action is transitive. Take arbitrary $g\in Y$ and $x,y\in H$. If ${\left(Gg\right)}^x={\left(Gg\right)}^y$, then $Ggx=Ggy$, that is, $gx{y}^{-1}{g}^{-1}\in G$, or equivalently $x{y}^{-1}\in g^{-1}Gg$. Fourth, to deduce $x=y$, we identify the vertex set with the coset space. Since $\Sigma$ is a Cayley graph, we may choose g such that the coset $Gg$ corresponds to the vertex v, and then H is precisely the subgroup stabilising this coset. In this case, if $x,y\in H$ satisfy ${\left(Gg\right)}^x={\left(Gg\right)}^y$, then $x{y}^{-1}$ belongs to both H and $g^{-1}Gg$. Because G is core-free in Y (that is, ${\bigcap }_{x\in Y}{x}^{-1}Gx=\left\{1\right\}$) and by the stabilising property of H, we obtain $x{y}^{-1}=1$, so $x=y$. Finally, we have H acts transitively on $[Y:G]$, and since $\left|H\right|=\left|\right[Y:G\left]\right|$, this action is necessarily regular. Consequently, H is a regular subgroup of $S_n$, and Y acts transitively on $[Y:G]$. In this setting, G is the stabiliser in Y of a vertex $i\in \{1,2,\dots ,n\}$. Without loss of generality, we may assume that G fixes 1.

Assume further that the generating set S contains an involution $\tau$. By [6] (Proposition 3.2), under the core-free and Y-arc-transitive conditions, $\tau$ satisfies $\tau \in N_{S_n}(H\cap H^{\tau })\setminus \left({\bigcup }_{1\ne K⊴H}{N}_{S_n}\left(K\right)\right)$, that is, $\tau$ normalises $H\cap H^{\tau }$ but does not normalise any nontrivial normal subgroup of H. This condition ensures that the construction of the coset graph is appropriate and avoids degenerate cases. In this situation, the graph $\Sigma$ is isomorphic to the coset graph $Cos(Y,H,\tau )$, where $Y=\langle H,\tau \rangle$ is the subgroup of $S_n$ generated by H and $\tau$. The vertex set of this coset graph is $[Y:H]$, and the edge set is induced by the double coset $H\tau H$, that is, two vertices $Ha$ and $Hb$ are adjacent if and only if $a^{-1}b\in H\tau H$. Moreover, $G=\{\sigma \in Y\mid 1^{\sigma }=1\}$ and $S=\{\sigma \in H\tau H\mid 1^{\sigma }=1\}$. In other words, the group G of the original Cayley graph is precisely the subgroup of Y fixing the vertex 1, and the generating set S consists of those elements in the double coset $H\tau H$ that map 1 to 1. This representation lifts the automorphism group of the graph to Y and completely translates the symmetry of the graph into the combinatorial structure of the group Y and its subgroup H, providing a unified algebraic framework for subsequent constructions and classifications.

Based on the above representation, all connected core-free arc-transitive Cayley graphs whose vertex stabiliser is isomorphic to a regular subgroup H of $S_n$ can be constructed. Up to isomorphism, $Cos(Y,H,\tau )$ is independent of the specific choice of H and depends only on the conjugacy class of H and an equivalence class of $\tau$. First, let ${\tau }_1\in N_{S_n}\left(P_1\right)$, ${\tau }_2\in N_{S_n}\left(P_2\right)$, where $P_1=H\cap H^{{\tau }_1}$ and $P_2=H\cap H^{{\tau }_2}$. If there exists $h\in H$ such that $P_1^h=P_2$, then $Cos(Y_1,H,{\tau }_1^h)\cong Cos(Y_2,H,{\tau }_2)$, where $Y_1=\langle H,{\tau }_1\rangle$ and $Y_2=\langle H,{\tau }_2\rangle$. From $P_1^h=P_2$ we obtain $H\cap H^{{\tau }_1^h}=P_2$, and $Y_1^h=\langle H^h,{\tau }_1^h\rangle =\langle H,{\tau }_1^h\rangle$. Since both ${\tau }_2$ and ${\tau }_1^h$ satisfy $H\cap H^{{\tau }_2}=H\cap H^{{\tau }_1^h}=P_2$ and both lie in $N_{S_n}\left(P_2\right)$, by adjusting elements in H they generate conjugate groups, thereby inducing an isomorphism between the corresponding coset graphs. Second, if the subgroup $H\cap H^{\tau }$ is conjugate in H, then $Cos(Y,H,\tau )$ is independent of the choice of this subgroup. Finally, for any $\sigma \in N_{S_n}\left(H\right)$, we have $Cos(Y,H,\tau )\cong Cos(Y^{\sigma },H,{\tau }^{\sigma })$. This isomorphism is directly given by the conjugation action induced by $\sigma$. Specifically, $\sigma$ fixes H, maps Y to $Y^{\sigma }$ and $\tau$ to ${\tau }^{\sigma }$, while preserving the coset graph structure. The above results systematically characterise the algebraic invariants of such graphs, showing that the free parameters in the construction essentially reduce to the conjugacy class of H in $S_n$ and the conjugacy class of $H\cap H^{\tau }$ in H, thereby laying a theoretical foundation for the complete classification of connected core-free arc-transitive Cayley graphs.

Up to isomorphism, a connected 1-regular core-free Cayley graph $Cay(G,S)$ with vertex stabiliser H, arc stabiliser P, and where S contains an involution can be constructed by determining the involutions satisfying the required conditions. The specific construction procedure is as follows (see [6], p. 6017):

• Step 1: Compute the set $I:=\{\tau \in N_{S_n}\left(P\right)\setminus {\cup }_{1\ne K⊴H}{N}_{S_n}\left(K\right)\mid {\tau }^2=1,1^{\tau }=1\}$.
• Step 2: Determine the set $I(n,H)$ of representatives of the orbits of involutions in I under the action of $N_{S_n}\left(H\right)$.
• Step 3: For each $\tau \in I(n,H)$, construct the group $Y=\langle \tau ,H\rangle$, and then obtain $G=\{\sigma \in Y\mid 1^{\sigma }=1\}$ and $S=\{\sigma \in H\tau H\mid 1^{\sigma }=1\}$.

3. The Proof of Theorem 1

Let $\Sigma =Cay(G,S)$ be a 14-valent $(Y,1)$-regular core-free Cayley graph satisfying $G⩽Y⩽Aut\Sigma$. Take an arbitrary vertex $v\in V\Sigma$ and denote $H=Y_v$ the vertex stabiliser subgroup of v in Y. By the definition of a $(Y,1)$-regular graph, Y acts regularly on the arc set of $\Sigma$. Hence, for a vertex v, the stabiliser H acts regularly on the neighbourhood $\Sigma \left(v\right)$. A regular action is both transitive and semi-regular, so $\left|H\right|=|\Sigma (v\left)\right|$. Since $\Sigma$ is known to be 14-valent, we have $|\Sigma (v\left)\right|=14$, and thus $\left|H\right|=14$.

Since $14=2\times 7$, by Sylow theorem, let $n_7$ be the number of Sylow 7-subgroups of H, then $n_7\equiv 1 (mod 7)$ and $n_7\mid 2$, so $n_7=1$. Hence the Sylow 7-subgroup is unique, denote it by N, and we have $N⊴H$ with $N\cong C_7$. Let P be any Sylow 2-subgroup of H, then $\left|P\right|=2$, thus, $P\cong C_2$. By the normality of N, we have $H=N⋊P$. In this case, the structure of the semidirect product is uniquely determined by the homomorphism $\phi :P\to Aut\left(N\right)$.

Let $P=\langle \tau \rangle$, where ${\tau }^2=1$. Since $Aut\left(N\right)\cong Aut\left(C_7\right)\cong C_6$, which is a cyclic group of order 6 whose elements have orders 1, 2, 3, and 6, it follows from $\phi {\left(\tau \right)}^2=\phi \left({\tau }^2\right)=\phi \left(1\right)=id$ that $\phi \left(\tau \right)$ can only have order 1 or 2 in $Aut\left(C_7\right)$. In $C_6$, the unique automorphism of order 2 is the inversion map $x\mapsto x^{-1}$. If $\phi \left(\tau \right)=id$, then the semidirect product becomes a direct product, so $H\cong C_7\times C_2\cong C_{14}$. If $\phi \left(\tau \right)$ is the automorphism of order 2, then $H\cong C_7⋊C_2$, which is the dihedral group $D_{14}$. Hence the vertex stabiliser H is isomorphic to either the cyclic group $C_{14}$ or the dihedral group $D_{14}$.

Next, we discuss these two cases in detail in Section 3.1 and Section 3.2.

3.1. H Is Isomorphic to the Cyclic Group $C_{14}$

We next consider the case when $H\cong C_{14}$.

Let $x=(1,2)(3,4)(5,6)(7,8)(9,10)(11,12)(13,14)$ and $y=(1,3,5,7,9,11,13)(2,4,6,8,10,12,14)$. It is easy to verify that $H=\langle x,y\rangle$. By the assumption of the theorem, we have $P=1$. With the help of the discrete algebra system GAP [29], using the code in the Appendix A: CoreFree(SymmetricGroup(14), List(AllSmallGroups(14), x->Image(RegularActionHomomorphism(x)))[1], TrivialSubgroup(List(AllSmallGroups(14), x->Image(RegularActionHomomorphism(x)))[1])), we compute that there are 27,408 possible 14-valent 1-regular core-free Cayley graphs. For convenience, we denote these graphs by ${\Sigma }_i$ where $i=1,2,\dots$, 27,408. Furthermore, the corresponding $\tau$, G, S and Y for ${\Sigma }_i$ are denoted by ${\tau }_i$, $G_i$, $S_i$ and $Y_i$, respectively. Hence, ${\Sigma }_i=Cay(G_i,S_i)$. In this case, according to the order of the graph ${\Sigma }_i$, we can divide them into seven classes. In increasing order, their orders are 28, 156, 4032, 907,200, 1,814,400, 3,628,800 and 6,227,020,800, respectively. Below we first deal with the first three classes of graphs.

With the help of the discrete algebra system GAP [29], we obtain that none of the 7 graphs in the first and second classes is a 14-valent 1-regular core-free Cayley graph, although they are all 14-valent core-free Cayley graphs. Meanwhile, among the 11 graphs in the third class, the 2nd, 5th, 6th, 8th, 10th and 11th are all 14-valent 1-regular core-free Cayley graphs. By means of the discrete algebra system GAP [29], we obtain that the corresponding indices of these six 14-valent 1-regular core-free Cayley graphs are 1507, 2840, 3523, 3540, 3637 and 3661, let $i\in \{1507,2840,3523,3540,3637,3661\}$. We now present their corresponding ${\tau }_i$ as follows:

$$\begin{array}{ccccc}\hfill {\tau }_{1507}& =(5,7)(6,10)(8,14)(9,13),\hfill & & \hfill {\tau }_{2840}& =(4,6)(5,11)(8,14)(9,13),\hfill \\ \hfill {\tau }_{3523}& =(4,8)(5,7)(6,12)(9,13),\hfill & & \hfill {\tau }_{3540}& =(4,8)(5,9)(6,14)(11,13),\hfill \\ \hfill {\tau }_{3637}& =(4,8)(5,13)(9,11)(10,12),\hfill & & \hfill {\tau }_{3661}& =(4,8)(5,13)(6,12)(7,9).\hfill \end{array}$$

Based on the above results, we obtain the following lemma.

Lemma 1.

If $i\in \{1507,2840,3523,3540,3637,3661\}$, then ${\Sigma }_i=Cay(G_i,S_i)$ is a 14-valent 1-regular core-free Cayley graph, where $G_i\cong \mathrm{PSL}(3,2)\times S_4$ and

$$\begin{array}{cc}\hfill S_{1507}=& \{z,yzy,xz{x}^{-1},yxz{x}^{-1}y,x^{-2}yzy{x}^2,x^{-2}z{x}^2,x^{-1}yz{x}^{-3}y,x^{2}z{x}^{-3},y{x}^{3}zyx,\hfill \\ & x^{-3}zx,z{x}^{-1}{\left(zx\right)}^{2}x,x^{3}z{x}^{-2},x^{-1}{\left(x^{-1}z\right)}^{2}xz,x^{-1}z{x}^3\mid z={\tau }_{1507}\},\hfill \\ \hfill S_{2840}=& \{z,yzy,xz{x}^{-1},x^{-1}yz{x}^{-3}y,x^{-3}yzy{x}^3,x^{-2}z{x}^{-2},x^{3}z{x}^{-3},x^{2}z{x}^2,y{x}^{3}zyx,\hfill \\ & y{x}^{2}z{x}^{-1}y,x^{-2}yzy{x}^2,x^{-3}zx,x^{-1}z{x}^3,yxz{x}^{-2}y\mid z={\tau }_{2840}\},\hfill \\ \hfill S_{3523}=& \{z,yzy,y{x}^{3}z{x}^{-1}y,xz{x}^{-1},x^{-1}z{x}^3,x^{-3}yzy{x}^3,x^{2}z{x}^{-3},x^{-2}yz{x}^{-2}y,\hfill \\ & x^{-2}z{x}^2,x^{-1}yzyx,yxz{x}^{-3}y,y{x}^{2}zy{x}^2,x^{-3}zx,x^{3}z{x}^{-2}\mid z={\tau }_{3523}\},\hfill \\ \hfill S_{3540}=& \{z,yzy,x^{-2}yzy{x}^2,y{x}^{3}z{x}^{-1}y,xz{x}^{-1},x^{-3}z{x}^{-2},x^{2}z{x}^3,x^{-1}yz{x}^{-2}y,\hfill \\ & x^{-1}z{x}^2,x^{-3}yzy{x}^3,y{x}^{2}zyx,yxz{x}^{-3}y,x^{3}z{x}^{-3},x^{-2}zx\mid z={\tau }_{3540}\},\hfill \\ \hfill S_{3637}=& \{z,x^{-2}yzy{x}^3,yzy,x^{-2}z{x}^3,x^{-3}yzy{x}^2,x^{-3}z{x}^2,y{x}^{3}z{x}^{-1}y,xz{x}^{-1},\hfill \\ & x^{-1}yzyx,x^{2}zx,x^{-1}z{x}^{-2},y{x}^{2}z{x}^{-2}y,x^{3}z{x}^{-3},yxz{x}^{-3}y\mid z={\tau }_{3637}\},\hfill \\ \hfill S_{3661}=& \{z,yzy,x^{-3}z{x}^{-3},y{x}^{2}zy{x}^2,xz{x}^{-1},x^{-1}yzyx,y{x}^{3}z{x}^{-1}y,x^{-2}yz{x}^{-2}y,\hfill \\ & x^{-2}z{x}^2,x^{-1}z{x}^{-2},x^{3}z{x}^3,yxz{x}^{-3}y,x^{-3}yzy{x}^3,x^{2}zx\mid z={\tau }_{3661}\}.\hfill \end{array}$$

Proof. 

By the assumption of the theorem, ${\Sigma }_i$ is a $(Y_i,1)$-regular 14-valent core-free Cayley graph. Through computation, we obtain $|Aut{\Sigma }_i|=|Y_i|$. It follows that $Aut{\Sigma }_i=Y_i$. Therefore, ${\Sigma }_i=Cay(G_i,S_i)$ is a 14-valent 1-regular core-free Cayley graph. We now determine the structure of $G_i$.

Let $G_i=\langle a_1,a_2,b_1,b_2,b_3,b_4\rangle$. Computation with GAP [29] shows that $|G_i|=4032=168\times 24$. Let $N_1=\langle a_1,a_2\rangle$, where $a_1=(2,10)(4,6)$ and $a_2=(4,8,14)(6,10,12)$, then $N_1\cong \mathrm{PSL}(3,2)$ and $|N_1|=168$. Let $M=\langle b_1,b_2,b_3,b_4\rangle$, where $b_1=(5,7)(9,13)$, $b_2=(3,11)(5,7)$, $b_3=(5,9)(6,10)(7,13)(8,14)$ and $b_4=(3,5,9)(4,6,10)(7,13,11)(8,14,12)$. Consider the subgroup $N_2=\langle b_1,b_2\rangle$. Direct verification shows that $b_1^2=b_2^2=1$ and $b_1{b}_2=b_2{b}_1$, hence $\langle b_1\rangle \cong \langle b_2\rangle \cong C_2$ and $\langle b_1\rangle$ centralizes $\langle b_2\rangle$. Note that $\langle b_1\rangle \cap \langle b_2\rangle =1$, we have $N_2\cong C_2\times C_2\cong V_4$. One further verifies that $b_3,b_4$ normalize $N_2$, thus, $N_2⊲M$. Let $K=\langle b_3,b_4\rangle$, since $b_3^2=1$, $b_4^3=1$ and $b_3{b}_4{b}_3=b_4^{-1}$, we have $K\cong S_3$. Moreover, $K\cap N_2=\left\{1\right\}$ and $M=N_{2}K$; thus by the definition of the semidirect product [30], we obtain $M=N_2⋊K\cong V_4⋊S_3\cong S_4$.

We now show that $N_1$ and M are both normal subgroups of $G_i$. From the structure of the generators, $a_1,a_2$ move only the set $X=\{2,4,6,8,10,12,14\}$, while $b_1,b_2,b_3,b_4$ move only the set $Y=\{3,5,6,7,8,9,10,11,12,13,14\}$. Direct verification shows that $N_1$ and M normalize each other in $G_i$, and by construction $G_i=N_{1}M$. By the product formula $|N_{1}M|=|N_1\left|\right|M|/|N_1\cap M|$, we obtain $|N_1\cap M|=1$. Hence $N_1\cap M=\left\{1\right\}$. Thus by the definition of the direct product [30], we obtain $G_i\cong N_1\times M\cong \mathrm{PSL}(3,2)\times S_4$.    □

According to Lemma 1, with the help of the discrete computational algebra system GAP [29], we can determine the number of 14-valent 1-regular core-free Cayley graphs in the third class up to isomorphism, and we can also determine their full automorphism groups.

Lemma 2.

If $i\in \{1507,2840,3523,3540,3637,3661\}$, then up to isomorphism there are three graphs among the ${\Sigma }_i$, and $Aut{\Sigma }_i\cong \mathrm{PSL}{(3,2)}^2⋊C_2$. Furthermore, we have ${\Sigma }_{1507}\cong {\Sigma }_{3637}$, ${\Sigma }_{2840}\cong {\Sigma }_{3661}$ and ${\Sigma }_{3523}\cong {\Sigma }_{3540}$.

Proof. 

Let $Aut{\Sigma }_i=\langle {\alpha }_1,{\alpha }_2,{\alpha }_3,{\alpha }_4,\beta \rangle$. Computation with GAP [29] shows that $|Aut{\Sigma }_i|=$ 56,448 = 2 × 28,224 = $2\times {168}^2$. Let $N=\langle {\alpha }_1,{\alpha }_2,{\alpha }_3,{\alpha }_4\rangle$, where ${\alpha }_1=(1,11)(2,10,6)(5,9,13,7)(8,14,12)$, ${\alpha }_2=(5,13)(6,14)(7,9)(8,10)$, ${\alpha }_3=(1,5)(2,14)(6,12)(11,13)$, ${\alpha }_4=(1,13,5,11)(2,8,6)(3,9)(4,12,14)$. By analyzing the structure of the N, we obtain two normal subgroups $N_1$ and $N_2$ of N such that $N_1\cap N_2=\left\{1\right\}$ and $N_1{N}_2=N$. Computation shows that $N_1\cong N_2\cong \mathrm{PSL}(3,2)$, therefore $N\cong \mathrm{PSL}(3,2)\times \mathrm{PSL}(3,2)$. Furthermore, we have $\left|\right[Aut{\Sigma }_i:N\left]\right|=2$ and $N⊲Aut{\Sigma }_i$. Let $\beta$ be a generator of a complement of N in $Aut\Sigma$, and let $K=\langle \beta \rangle$, where $\beta =(1,14)(2,11)(3,10)(4,9)(5,12)(6,13)(7,8)$. Since $\beta$ is a product of disjoint transpositions, ${\beta }^2=1$ and $K\cong C_2$. Computation gives $K\cap N=\left\{1\right\}$ and $NK=Aut{\Sigma }_i$, hence $Aut{\Sigma }_i=N⋊K$. That is, $Aut{\Sigma }_i\cong \mathrm{PSL}{(3,2)}^2⋊C_2$.    □

For convenience, we only present the first graph in each class below. The other graphs can be obtained by computation using the code in the Appendix A. We next deal with the fourth class. By the discrete computational algebra system GAP [29], there are 137 graphs in the fourth class that are 14-valent core-free Cayley graphs, but we do not know which of them are 14-valent 1-regular core-free Cayley graphs. This is because we are unable to determine the full automorphism groups of these graphs in this case. Through computation, we have ${\tau }_{22}=(8,10)(12,14)$. Based on this result, we obtain the following lemma.

Lemma 3.

When $i=22$, $G_{22}\cong A_7\times A_6$, $Y_{22}\cong A_7^2⋊C_2$ and $S_{22}=\{z,yzy,xz{x}^{-1},yxz{x}^{-1}y,x^{2}z{x}^{-2},y{x}^{2}z{x}^{-2}y,x^{-2}z{x}^2,x^{3}z{x}^{-3},x^{-1}yzy{x}^2,x^{-3}yz{x}^{-3}y,x^{-1}zx,x^{-3}z{x}^3,x^{-2}yzyx,y{x}^{3}zy{x}^3\mid z={\tau }_{22}\}$.

Proof. 

Let $G_{22}=\langle a_1,a_2,b_1,b_2\rangle$. Computation with GAP [29] shows that $|G_{22}|$ = 907,200 = $2520\times 360$. Let $N_1=\langle a_1,a_2\rangle$, where $a_1=(2,10,14,12)(3,13,11,5,9)(4,8)$ and $a_2=(2,12,10,4,14,6,8)(3,7,5,13,9)$, then $N_1\cong A_7$ and $|N_1|=2520$. Let $N_2=\langle b_1,b_2\rangle$, where $b_1=(3,5,11,13)(7,9)$ and $b_2=(3,7,9,11)(5,13)$, then $N_2\cong A_6$ and $|N_2|=360$. Computation with GAP [29] shows that $N_1⊲G_{22}$, $N_2⊲G_{22}$, and $G_{22}=N_1{N}_2$. Note that $|G_{22}|=|N_1\left|\right|N_2|/|N_1\cap N_2|$ and $|G_{22}|=|N_1\left|\right|N_2|=2520\times 360$, hence $|N_1\cap N_2|=1$. Thus $N_1\cap N_2=1$. It follows that $G_{22}\cong N_1\times N_2\cong A_7\times A_6$. We now determine the structure of $Y_{22}$.

Let $Y_{22}=\langle {\alpha }_1,{\alpha }_2,{\beta }_1,{\beta }_2,\gamma \rangle$. Calculations show that $|Y_{22}|$ = 12,700,800 = 2 × 6,350,400 = 2 × ${2520}^2$. Let $H=\langle {\alpha }_1,{\alpha }_2,{\beta }_1,{\beta }_2\rangle$, where ${\alpha }_1=(1,11)(3,7)(5,9,13)$, ${\alpha }_2=(1,13,11,5,7,9,3)$, ${\beta }_1=(1,11)(3,7)(5,9,13)$ and ${\beta }_2=(1,13,11,5,7,9,3)$. Using GAP [29], we obtain $\langle {\alpha }_1,{\alpha }_2\rangle \cong A_7$, $\langle {\beta }_1,{\beta }_2\rangle \cong A_7$. Furthermore, $\langle {\alpha }_1,{\alpha }_2\rangle$ centralizes $\langle {\beta }_1,{\beta }_2\rangle$, that is $\langle {\alpha }_1,{\alpha }_2,{\beta }_1,{\beta }_2\rangle =\langle {\alpha }_1,{\alpha }_2\rangle \times \langle {\beta }_1,{\beta }_2\rangle$. Consequently $H\cong A_7^2$ and $\left|H\right|=$ 6,350,400. Direct verification show that $\left|\right[Y_{22}:H\left]\right|=2$ and thus $H⊲Y_{22}$, that is, H is a normal subgroup of $Y_{22}$ of index 2. Let $\gamma =(1,12)(2,7)(3,14)(4,9)(5,8)(6,13)(10,11)$ and set $K=\langle \gamma \rangle$. We compute ${\gamma }^2=1$, so $K\cong C_2$, and $K\cap H=\left\{1\right\}$, $HK=Y_{22}$, thus, $Y_{22}=H⋊K$. It follows that $Y_{22}\cong A_7^2⋊C_2$.    □

With the aid of the discrete computational algebra system GAP [29], we obtain that there are 205 $(Y_i,1)$-regular core-free Cayley graphs in class 5. Here $i=7$ is the first graph in class 5. Accordingly, we have ${\tau }_7=(10,12)(11,13)$. We then obtain the following lemma.

Lemma 4.

If $i=7$, then $G_7\cong A_7\times S_6$, $Y_7\cong A_7^2⋊C_2^2$ and $S_7=\{z,yzy,xz{x}^{-1},yxz{x}^{-1}y,x^{2}z{x}^{-2},y{x}^{2}z{x}^{-2}y,x^{3}z{x}^{-3},y{x}^{3}z{x}^{-3}y,x^{-3}z{x}^3,x^{-1}z{x}^2,x^{-2}yzy{x}^3,x^{-2}zx,x^{-1}yzyx,x^{-3}yzy{x}^2\mid z={\tau }_7\}$.

Proof. 

We first determine the structure of $G_7$. Let $G_7=\langle a_1,a_2,b_1,b_2,c\rangle$. Calculations show that $|G_7|=$ 1,814,400. Let $S=Soc\left(G_7\right)$. Then we obtain $\left|S\right|=$ 907,200. To determine the structure of S, set $A=\langle a_1,a_2\rangle$, where $a_1=(2,14,4,8,10,6,12)$, $a_2=(2,12,10,6,14,8,4)$. Then $\langle a_1,a_2\rangle \cong A_7$. Let $B=\langle b_1,b_2\rangle$, where $b_1=(3,13,5)$, $b_2=(3,5,9)(7,11,13)$. Then $\langle b_1,b_2\rangle \cong A_6$. From the permutation representation, A and B act on disjoint vertex sets, hence they commute with each other, so $A\times B\cong A_7\times A_6$. We compute $|A\times B|$ = 2520 × 360 = 907,200 = $\left|S\right|$, and $A\times B⊲G$, thus $S=A\times B\cong A_7\times A_6$. Further calculations show that $\left|\right[G_7:S\left]\right|=2$, that is, S is a normal subgroup of $G_7$ of index 2. Let $c=(5,11)(6,10)$ and set $K=\langle c\rangle$. Then $c^2=1$, so $K\cong C_2$, and $K\cap S=\left\{1\right\}$, $KS=G_7$, thus, $G_7=S⋊K$. We now analyse this conjugation action. Computing the images of the generators under conjugation by c yields $c{a}_1{c}^{-1}=a_1$, $c{a}_2{c}^{-1}=a_2$, $c{b}_1{c}^{-1}=b_1^{-1}$, $c{b}_2{c}^{-1}=b_2^{-1}$. Hence c centralises A and induces an anti-automorphism on B. Since $B\cong A_6$ and its outer automorphism group is $C_2\times C_2$, c realises a nontrivial outer automorphism, specifically an involutory automorphism of $A_6$. Thus $K\cong C_2$ acts trivially on A and acts on B via an involutory outer automorphism, so it acts on $S=A\times B$ accordingly. Consequently $G_7\cong (A_7\times A_6)⋊C_2$. Note that $A_6⋊C_2\cong S_6$, and K acts trivially on A, therefore $G_7\cong A_7\times (A_6⋊C_2)\cong A_7\times S_6$.

Since $|Y_7|$ = 25,401,600 = 4 × 6,350,400. Let $S=Soc\left(Y_7\right)$. Then computations show that $S\cong A_7\times A_7$ and $\left|S\right|=$ 6,350,400. By definition, S is the product of all minimal normal subgroups of $Y_7$, and $S⊲Y_7$. Further calculations show that $\left|\right[Y_7:S\left]\right|=4$, that is, S is a normal subgroup of $Y_7$ of index 4. Let $C=\langle {\gamma }_1,{\gamma }_2\rangle$, where ${\gamma }_1=(2,4)(3,13)(5,9)(7,11)(8,10)(12,14)$, ${\gamma }_2=(1,6)(2,13)(3,4)(5,8)(7,12)(9,10)(11,14)$. Using GAP [29] we obtain $C\cong C_2\times C_2$ and $C\cap S=\left\{1\right\}$. Moreover $CS=Y_7$, so $Y_7=S⋊C$, where the semidirect product is given by the conjugation action of C on S. We now analyse this conjugation action. Let $A=\langle {\sigma }_1,{\sigma }_2\rangle$, where ${\sigma }_1=(1,11,5,3,13,7,9)$, ${\sigma }_2=(1,9,3,13,11,5,7)$. Then $A\cong A_7$. From the structure $S\cong A_7\times A_7$, S has two normal subgroups isomorphic to $A_7$, denoted $S_1$ and $S_2$, respectively, and $S=S_1\times S_2$. By computing the images of the generators under the action of C, we find that ${\gamma }_1$ and ${\gamma }_2$ together generate a $C_2\times C_2$ action on S that swaps $S_1$ and $S_2$ and induces nontrivial automorphisms within each factor. Specifically, ${\gamma }_1$ induces an involutory automorphism on $S_1$ while simultaneously swapping the two factors, and ${\gamma }_2$ likewise induces a swapping action, together they generate an elementary abelian subgroup of order four. Consequently $C\cong C_2\times C_2$ acts on S by swapping the two direct factors and applying an involution on each factor. In summary, we have $S\cong A_7\times A_7$, $C\cong C_2\times C_2$, and the action of C decomposes as a direct product of two $C_2$ factors, each of which swaps the two $A_7$ factors. Hence $Y_7\cong A_7^2⋊C_2^2$.    □

By computation, there are 335 graphs in class 6 and 26,713 graphs in class 7, all of which are $(Y_i,1)$-regular core-free Cayley graphs. Here $i=2$ and 1 are the first graphs in classes 6 and 7, respectively. Accordingly, we have ${\tau }_2=(12,14)$ and ${\tau }_1=(2,3)$. We then obtain the following two lemmas.

Lemma 5.

If $i=2$, then $G_2\cong S_7\times S_6$, $Y_2\cong A_7^2⋊D_8$, and $S_2=\{z,xzx,xz{x}^{-1},xxz{x}^{-1}x,x^{2}z{x}^{-2},x{x}^{2}z{x}^{-2}x,x^{3}z{x}^{-3},x{x}^{3}z{x}^{-3}x,x^{-3}z{x}^3,x^{-3}xzx{x}^3,x^{-2}z{x}^2,x^{-1}xzx{x}^2,x^{-1}zx,x^{-2}xzxx\mid z={\tau }_2\}$.

Proof. 

Let $G_2=\langle a,b\rangle$, where $a=(2,4,10,14)(3,5,13,9,7)$ and $b=(2,10,6)(4,12,14,8)(7,11,9,13)$. Using GAP [29] we obtain $|G_2|$ = 3,628,800 = 5040 × 720. Let $N_1=\langle (2,4,10,14),(2,1,6)(4,12,14,8)\rangle$ and $N_2=\langle (3,5,13,9,7),(7,11,9,13)\rangle$. Direct computation in GAP [29] shows that $N_1\cong S_7$ with $|N_1|=5040$, and $N_2\cong S_6$ with $|N_2|=720$. Further calculations show that $N_1,N_2⊲G_2$, $N_1\cap N_2=\left\{1\right\}$, and $N_1{N}_2=G_2$. Moreover, since the permutations in $N_1$ act only on the vertex set $\{2,4,6,8,10,12,14\}$ while those in $N_2$ act only on $\{3,5,7,9,11,13\}$, and these two vertex sets are disjoint, it follows that for any $x\in N_1$ and $y\in N_2$, we have $xy=yx$. By the direct product criterion, if $H,K⊲G_2$ satisfy $H\cap K=\left\{1\right\}$, $HK=G_2$, and the elements of H commute with those of K, then $G_2\cong H\times K$. Hence, taking $H=N_1$ and $K=N_2$, we obtain $G_2\cong N_1\times N_2\cong S_7\times S_6$.

We next prove the structure of $Y_2$. Let $Y_2=\langle {\alpha }_1,{\alpha }_2,{\beta }_1,{\beta }_2,\gamma ,\delta \rangle$, where ${\alpha }_1=(1,13,11)(3,5,9)$, ${\alpha }_2=(5,11,13,7,9)$, ${\beta }_1=(4,12,10)$, ${\beta }_2=(2,12)(4,6)(8,10,14)$, $\gamma =(2,12)(4,10)(5,9)(6,14)(7,13)$, $\delta =(1,14)(2,13)(3,8)(4,9)(5,12)(6,11)(7,10)$. Set $H=\langle {\alpha }_1,{\alpha }_2\rangle$ and $K=\langle {\beta }_1,{\beta }_2\rangle$. Direct computation yields ${\alpha }_1^3={\alpha }_2^5={\left({\alpha }_1{\alpha }_2\right)}^2=1$, and ${\alpha }_1,{\alpha }_2$ act transitively on the set ${\mathrm{\Omega }}_1=\{1,3,5,7,9,11,13\}$, hence $H\cong A_7$. Similarly, ${\beta }_1^3={\beta }_2^6={\left({\beta }_1{\beta }_2\right)}^2=1$, and ${\beta }_1,{\beta }_2$ act transitively on ${\mathrm{\Omega }}_2=\{2,4,6,8,10,12,14\}$, therefore $K\cong A_7$. Since ${\mathrm{\Omega }}_1\cap {\mathrm{\Omega }}_2=⌀$, every permutation in H fixes all vertices in ${\mathrm{\Omega }}_2$ and every permutation in K fixes all vertices in ${\mathrm{\Omega }}_1$, consequently H and K commute elementwise and $H\cap K=\left\{1\right\}$. Thus $HK\cong H\times K\cong A_7\times A_7$. Computing conjugations we obtain $\gamma {\alpha }_1{\gamma }^{-1}={\beta }_1$, $\gamma {\alpha }_2{\gamma }^{-1}={\beta }_2$, $\gamma {\beta }_1{\gamma }^{-1}={\alpha }_1$, $\gamma {\beta }_2{\gamma }^{-1}={\alpha }_2$, and $\delta {\alpha }_1{\delta }^{-1}={\beta }_2$, $\delta {\alpha }_2{\delta }^{-1}={\beta }_1$, $\delta {\beta }_1{\delta }^{-1}={\alpha }_2$, $\delta {\beta }_2{\delta }^{-1}={\alpha }_1$. Moreover ${\gamma }^2={\delta }^2={\left(\gamma \delta \right)}^4=1$ and $\gamma \delta =\delta {\gamma }^{-1}$, so $\langle \gamma ,\delta \rangle \cong D_8$. Since $\gamma$ and $\delta$ both normalise $HK$, and $\left(HK\right)\cap \langle \gamma ,\delta \rangle =\left\{1\right\}$, all generators belong to $HK\langle \gamma ,\delta \rangle$, hence $Y_2=HK\langle \gamma ,\delta \rangle$. Therefore, $Y_2\cong A_7^2⋊D_8$.    □

Lemma 6.

Assume $i=1$. Then we have $G_1\cong S_{13}$, $Y_1\cong S_{14}$, and $S_1=\{(12,13),(11,14),(10,11),(9,12),(8,9),(7,10),(6,7),(5,8),(4,5),(3,6),(2,3),(2,3,6,7,10,11,14,4,5,8,9,12,13),(2,13,12,9,8,5,4,14,11,10,7,6,3),(2,13)\}$.

Proof. 

By the discrete computational algebra system GAP [29], we have $G_1=\langle (2,3),(2,3,6,7,10,11,14,4,5,8,9,12,13)\rangle$. This group is generated by two permutations, a transposition $(2,3)$ and a 13-cycle $(2,3,6,7,10,11,14,4,5,8,9,12,13)$. Note that the vertex 1 is fixed, so the group $G_1$ acts on the set $\{2,3,\dots ,14\}$ consisting of 13 vertices. By relabelling, we may transform the 13-cycle into the standard cycle $(1,2,\dots ,13)$ and the transposition into $(1,2)$. The group generated by $(1,2,\dots ,13)$ and $(1,2)$ contains all adjacent transpositions, hence generates the full symmetric group $S_{13}$. Therefore $G_1\cong S_{13}$. We now determine the structure of $Y_1$.

Let $\alpha =(1,8,4,6,11,13,12,5,2,10,9,14,7),\beta =(1,12,6,10,14)(2,5,4,8,11,9,3,13)$. By computation, the group $Y_1=\langle \alpha ,\beta \rangle$. Note that $\alpha$ is a 13-cycle fixing the vertex 3, and $\beta$ fixes the vertex 7. We relabel the vertex set so that $\alpha$ becomes a standard cycle. Define a mapping $\phi :\{1,2,\dots ,14\}\to \{1,2,\dots ,14\}$ according to the rule

$$\left\{\begin{array}{c}1\mapsto 1, 8\mapsto 2, 4\mapsto 3, 6\mapsto 4, 11\mapsto 5, 13\mapsto 6, 12\mapsto 7,\hfill \\ 5\mapsto 8, 2\mapsto 9, 10\mapsto 10, 9\mapsto 11, 14\mapsto 12, 7\mapsto 13, 3\mapsto 14.\hfill \end{array}\right.$$

Under this mapping, ${\alpha }^{\prime }=\phi \alpha {\phi }^{-1}=(1,2,3,4,5,6,7,8,9,10,11,12,13)$ and ${\beta }^{\prime }=\phi \beta {\phi }^{-1}=(1,7,4,10,12)(2,3,5,6,8,9,11,14)$, with ${\alpha }^{\prime }$ fixing 14 and ${\beta }^{\prime }$ fixing 13. Hence $Y_1\cong \langle {\alpha }^{\prime },{\beta }^{\prime }\rangle$. Note that ${\beta }^{\prime 5}={(2,3,5,6,8,9,11,14)}^5=(2,9,5,14,8,3,11,6)=:d$ and ${\beta }^{\prime 8}={(1,7,4,10,12)}^3=(1,10,7,12,4)=e$. Furthermore, conjugating e by ${\alpha }^{\prime }$ yields $f:={\alpha }^{\prime }e{\alpha }^{\prime -1}=(2,11,8,13,5)$. Consider the product $g=fd$ (first f then d). A direct calculation gives $g=(2,6)(3,11)(5,9)(8,13,14)$. Consequently $g^3=(2,6)(3,11)(5,9)$, and therefore ${\alpha }^{\prime }{g}^3{\alpha }^{\prime -1}=(3,7)(4,12)(6,10)$. Observe that $h=g^3\left({\alpha }^{\prime }{g}^3{\alpha }^{\prime -1}\right)=(2,10,6)(3,11,7)(4,12)(5,9)$. This shows that h contains the transposition $(4,12)$. Since $({\alpha }^{\prime }{g}^3{\alpha }^{\prime -1})(4,12)=(3,7)(4,12)(6,10)(4,12)=(3,7)(6,10)$, we obtain $(3,7)=(({\alpha }^{\prime }{g}^3{\alpha }^{\prime -1})(4,12)){(6,10)}^{-1}$, where $(6,10)$ is a factor of ${\alpha }^{\prime }{g}^3{\alpha }^{\prime -1}$, hence $(3,7)\in Y_1$.

Conjugating $(3,7)$ by ${\alpha }^{\prime }$ yields all transpositions of the form $(3+k,7+k)$ (mod 13). Since 3 and 7 differ by 4 and $gcd(4,13)=1$, these transpositions generate the full symmetric group $S_{13}$ acting on $\{1,\dots ,13\}$. Moreover, ${\beta }^{\prime }$ maps the vertex 14 to 2, so the group $Y_1$ acts transitively on $\{1,\dots ,14\}$ and contains $S_{13}$, therefore the whole group is $S_{14}$. Hence $Y_1\cong S_{14}$.    □

3.2. H Is Isomorphic to the Dihedral Group $D_{14}$

Let $x=(1,2)(3,14)(4,13)(5,12)(6,11)(7,10)(8,9)$ and $y=(1,7,13,5,11,3,9)(2,8,14,6,12,4,10)$, it is easy to verify that $H=\langle x,y\rangle$. By the hypothesis of the theorem, we have $P=1$. With the aid of the discrete algebra system GAP [29], using the code in the Appendix A: CoreFree(SymmetricGroup(14), List(AllSmallGroups(14), x->Image(RegularActionHomom orphism(x)))[2], TrivialSubgroup(List(AllSmallGroups(14), x->Image(RegularActionHomo morphism(x)))[2])), we compute that there are 4028 possible 14-valent 1-regular core-free Cayley graphs. For convenience, we denote these graphs by ${\Sigma }_i$ (where $i=1,2,\dots ,4028$). According to the orders of the graphs ${\Sigma }_i$, they can be divided into 13 classes, in increasing order, their orders are 28, 64, 128, 156, 360, 720, 4032, 23,040, 46,080, 907,200, 1,814,400, 3,628,800 and 6,227,020,800. We first deal with the first seven classes of graphs.

With the aid of the discrete algebra system GAP [29], we obtain that the eight graphs in classes 1 to 4 and class 7 are not 14-valent 1-regular core-free Cayley graphs. Meanwhile, we obtain that the first graph in class 5 and the first graph in class 6 are 14-valent 1-regular core-free Cayley graphs. Using the discrete algebra system GAP [29], we find that the corresponding indices of these two graphs are 527 and 2752, that is, $i\in \{527,2752\}$. We now present their corresponding ${\tau }_i$ as follows:

$$\begin{array}{cc}\hfill {\tau }_{527}& =(5,7)(6,8)(9,13)(10,14),\hspace{0.33em}\hspace{0.33em}{\tau }_{2752}=(3,5)(4,6)(7,11)(8,12)(9,13)(10,14).\hfill \end{array}$$

Based on the above results, we obtain the following lemma.

Lemma 7.

If $i\in \{527,2752\}$, then ${\Sigma }_i=Cay(G_i,S_i)$ is a 14-valent 1-regular core-free Cayley graph, where $G_{527}\cong A_6$, $Aut{\Sigma }_{527}\cong S_7$, $G_{2752}\cong A_6⋊C_2$, $Aut{\Sigma }_{2752}\cong C_2\times S_7$ and

$$\begin{array}{cc}\hfill S_{527}=& \{z,y^{2}xz{y}^{2}x,y^{-2}z{y}^2,y^{-3}xz{y}^{-1}x,xzx,y^{2}zy,yz{y}^{-3},yxz{y}^{-2}x,y^{-1}xz{y}^{-3}x,\hfill \\ & yz{y}^{-3}z,y^{3}z{y}^{-1},y^{-1}xz{y}^{-3}zx,y^{-1}z{y}^{-2},y^{-2}xzyx\mid z={\tau }_{527}\},\hfill \\ \hfill S_{2752}=& \{y^{-3}z{y}^{-1},yxz{y}^{-2}x,yz{y}^3,y^{-2}xzyx,z,y^{2}xz{y}^{-3}x,xzx,y^{2}zy,y^{-1}xz{y}^{3}x,\hfill \\ & y^{-1}z{y}^{-2},y^{3}xz{y}^{-1}x,y^{3}z{y}^2,y^{-3}xz{y}^{2}x,y^{-2}z{y}^{-3}\mid z={\tau }_{2752}\}.\hfill \end{array}$$

Proof. 

By the hypothesis of the theorem, we have that ${\Sigma }_i$ is a $(Y_i,1)$-regular 14-valent core-free Cayley graph. Through computation, we obtain $|Aut{\Sigma }_i|=|Y_i|$, which implies $Aut{\Sigma }_i=Y_i$. Therefore, ${\Sigma }_i=Cay(G_i,S_i)$ is a 14-valent 1-regular core-free Cayley graph. We now determine the structures of $G_i$ and $Aut{\Sigma }_i$ respectively.

For $i=527$. Let $G_{527}=\langle a,b\rangle$, where $a=(3,5,9,7)(4,6,10,8)(11,13)(12,14)$, $b=(5,7,11,9,13)(6,8,12,10,14)$. Define $\mathrm{\Omega }=\left\{\{3,4\},\{5,6\},\{7,8\},\{9,10\},\{11,12\},\{13,14\}\right\}$ and establish a bijection $\psi :\mathrm{\Omega }\to \{1,2,3,4,5,6\}$ given by $\{3,4\}\mapsto 1, \{5,6\}\mapsto 2, \{7,8\}\mapsto 3, \{9,10\}\mapsto 4, \{11,12\}\mapsto 5, \{13,14\}\mapsto 6$. Direct verification shows the action of a and b on $\mathrm{\Omega }$: a such that vertex $\{3,4\}\to \{5,6\}\to \{9,10\}\to \{7,8\}\to \{3,4\}$ and $\{11,12\}$ with $\{13,14\}$, thus the induced permutation on $\{1,\dots ,6\}$ is $a^{\prime }=(1,2,4,3)(5,6)$. The element b fixes $\{3,4\}$ and such that $\{5,6\}\to \{7,8\}\to \{11,12\}\to \{9,10\}\to \{13,14\}\to \{5,6\}$, so the induced permutation is $b^{\prime }=(2,3,5,4,6)$. Since both $a^{\prime }$ and $b^{\prime }$ are even permutations, we have $G_{527}\cong \langle a^{\prime },b^{\prime }\rangle \subseteq A_6$. Compute the commutator $c=a^{\prime }{b}^{\prime }{a}^{\prime -1}{b}^{\prime -1}$, since $a^{\prime }=(1,2,4,3)(5,6)$ and $b^{\prime }=(2,3,5,4,6)$, we obtain $a^{\prime }{b}^{\prime }{a}^{\prime -1}=(1,3,6,5)(4,2)$, and $c=(1,3,6,5)(4,2)·b^{\prime -1}=(1,3,6,5)(4,2)·(2,6,4,5,3)=(1,3,6)(2,4,5)$, that is, c is a product of two disjoint 3-cycles, in particular containing the 3-cycle $(1,3,6)$. Moreover, $a^{\prime }$ and $b^{\prime }$ act transitively on $\{1,\dots ,6\}$ (since $b^{\prime }$ is a 5-cycle and $a^{\prime }$ sends 1 to 2); hence $\langle a^{\prime },b^{\prime }\rangle$ is a transitive subgroup of $A_6$ containing a 3-cycle, and therefore must be the whole $A_6$. Consequently $G_{527}\cong A_6$. We now determine the structure of $Y_{527}$.

Let $Y_{527}=\langle \alpha ,\beta ,\gamma \rangle$, where $\alpha =(1,3,5,7,9,11,13)(2,4,6,8,10,12,14)$, $\beta =(3,7,5)(4,8,6)(9,13,11)(10,14,12)$, $\gamma =(1,2)(3,4)(5,6)(7,8)(9,14)(10,13)(11,12)$. Define $\mathrm{\Omega }=\{1,3,5,7,9,11,13\}$, $\Delta =\{2,4,6,8,10,12,14\}$, then $\alpha ,\beta ,\gamma$ all preserve both $\mathrm{\Omega }$ and $\Delta$. Let $N=\langle \alpha ,\beta \rangle$ and consider the restriction homomorphism $\rho :N\to Sym\left(\mathrm{\Omega }\right)$. Direct computation gives $\rho \left(\alpha \right)=(1,3,5,7,9,11,13)$ and $\rho \left(\beta \right)=(3,7,5)(9,13,11)$. Since a 7-cycle together with a 3-cycle generates $A_7$, and $\rho$ is injective (if $h\in N$ satisfies $\rho \left(h\right)=1$, then h acts trivially on $\mathrm{\Omega }$, because h can be expressed as a product of $\alpha$ and $\beta$, and the action of $\alpha ,\beta$ on $\Delta$ is conjugate to that on $\mathrm{\Omega }$, it follows that h also acts trivially on $\Delta$, so $h=1$), we obtain $N\cong A_7$. We compute ${\gamma }^2=1$, and $\gamma \alpha {\gamma }^{-1}={\alpha }^{-1}$, $\gamma \beta {\gamma }^{-1}={\beta }^{-1}$; hence $\gamma$ normalises N. Since $\gamma$ is an odd permutation while every element of N is even, we have $\gamma \notin N$. Consequently $Y_{527}=N\langle \gamma \rangle$ with $N\cap \langle \gamma \rangle =\left\{1\right\}$, and thus $Y_{527}\cong N⋊\langle \gamma \rangle \cong A_7⋊C_2\cong S_7$.

For $i=2752$. We first determine the structure of $G_{2752}$. Let $G_{2752}=\langle a,b,c\rangle$, where $a=(3,5,13,9,7)(4,6,14,10,8)$, $b=(3,7,13,9)(4,8,14,10)(5,11)(6,12)$, $c=(5,11)(6,12)$. Set $M=\langle a,b\rangle$, and take $\mathrm{\Omega }=\{3,5,7,9,11,13\}$, $\Delta =\{4,6,8,10,12,14\}$, then a and b preserve both $\mathrm{\Omega }$ and $\Delta$. Define a bijection $\phi :\mathrm{\Omega }\to \{1,\dots ,6\}$ by $3\mapsto 1,5\mapsto 2,7\mapsto 3,9\mapsto 4,11\mapsto 5,13\mapsto 6$, and map the restrictions of a and b to $\mathrm{\Omega }$ to $a^{\prime }=(1,2,6,4,3)$ and $b^{\prime }=(1,3,6,4)(2,5)$. Direct computation yields $a^{\prime 5}=b^{\prime 4}={\left(a^{\prime }{b}^{\prime }\right)}^2=1$, and the commutator $a^{\prime }{b}^{\prime }{a}^{\prime -1}{b}^{\prime -1}=(1,2,3)(4,5,6)$ contains a 3-cycle; hence $\langle a^{\prime },b^{\prime }\rangle$ is a transitive subgroup of $A_6$ containing a 3-cycle, so $\langle a^{\prime },b^{\prime }\rangle =A_6$. The restriction homomorphism $\rho :M\to Sym\left(\mathrm{\Omega }\right)$ is injective (if $h\in M$ acts trivially on $\mathrm{\Omega }$, then by conjugacy it also acts trivially on $\Delta$, so $h=1$); therefore $M\cong A_6$. Computing conjugations gives $ca{c}^{-1}=a^{-1}$ and $cb{c}^{-1}=b^{-1}$, thus c normalises M. Moreover, ${c|}_{\mathrm{\Omega }}=(5,11)$ is an odd permutation, while every element of ${M|}_{\mathrm{\Omega }}\cong A_6$ is even, so $c\notin M$. Consequently $G_{2752}=M\langle c\rangle$ with $M\cap \langle c\rangle =\left\{1\right\}$, and hence $G_{2752}\cong M⋊\langle c\rangle \cong A_6⋊C_2$. We now determine the structure of $Y_{2752}$.

Let $Y_{2752}=\langle \alpha ,\beta ,\gamma \rangle$, where $\alpha =(1,2)(3,4)(5,6)(7,8)(9,10)(11,12)(13,14)$, $\beta =(1,2)(3,4)(5,6)(7,8)(9,10)(11,14)(12,13)$, $\gamma =(1,3,5,7,9,11,13)(2,4,6,8,10,12,14)$. Direct verification shows ${\alpha }^2=1$, and $\alpha$ commutes with both $\beta$ and $\gamma$, hence $\langle \alpha \rangle \cong C_2$ lies in the centre of $Y_{2752}$. Let $H=\langle \beta ,\gamma \rangle$. Consider the sets $\mathrm{\Omega }=\{1,3,5,7,9,11,13\}$ and $\Delta =\{2,4,6,8,10,12,14\}$, then $\gamma$ preserves both $\mathrm{\Omega }$ and $\Delta$, acting as the 7-cycle $(1,3,5,7,9,11,13)$ on $\mathrm{\Omega }$ and as the 7-cycle $(2,4,6,8,10,12,14)$ on $\Delta$. The element $\beta$ acts as the identity on $\mathrm{\Omega }$ and as $(2,4,6,8,10,14,12)$ on $\Delta$. Via the bijection $\phi :\mathrm{\Omega }\to \{1,2,3,4,5,6,7\}$ defined by $\phi \left(1\right)=1,\phi \left(3\right)=2,\phi \left(5\right)=3,\phi \left(7\right)=4,\phi \left(9\right)=5,\phi \left(11\right)=6,\phi \left(13\right)=7$, the restriction ${\gamma |}_{\mathrm{\Omega }}$ maps to the 7-cycle $(1,2,3,4,5,6,7)$, while ${\beta |}_{\Delta }$ gives a transposition under the conjugation by $\phi$. Therefore the action of $\langle \beta ,\gamma \rangle$ on $\{1,\dots ,7\}$ generates $S_7$, and this action is faithful, consequently $H\cong S_7$. Moreover $\langle \alpha \rangle \cap H=\left\{1\right\}$ and all generators lie in $\langle \alpha \rangle H$, so $Y_{2752}=\langle \alpha \rangle H\cong C_2\times S_7$.    □

For convenience, we present only the first graph in each class below. The other graphs can be obtained using the code provided in the Appendix A. We now deal with class 8. By the discrete computational algebra system GAP, there are 43 graphs in class 8 that are 14-valent core-free Cayley graphs, but we do not know which of them are 14-valent 1-regular core-free Cayley graphs. This is because we are unable to determine the full automorphism groups of these graphs. Here $i=12$ is the first graph in class 8. Accordingly, we have ${\tau }_{12}=(9,10)(11,13)(12,14)$. We then obtain the following lemma.

Lemma 8.

If $i=12$, then $G_{12}\cong (C_2^5⋊A_6)⋊C_2$, $Y_{12}\cong C_2^6⋊S_7$, and $S_{12}=\{z,y^{-1}xz{y}^{-1}x,y^{-2}z{y}^2,y^{-3}xz{y}^{-3}x,y^{3}z{y}^{-3},y^{2}xz{y}^{2}x,yz{y}^{-1},xzx,y^{2}z{y}^3,y^{3}xz{y}^{-2}x,y^{-1}zyx,y^{-3}z{y}^{-2},y^{-2}xz{y}^{3}x,yxzy\mid z={\tau }_{12}\}$.

Proof. 

We now determine the structure of $G_{12}$. Let $a_1=(5,6)(13,14)$, $a_2=(11,12)(13,14)$, $a_3=(7,8)(11,12)$, $a_4=(3,4)(11,12)$, $a_5=(9,10)(11,12)$, $b=(3,13,12,10,6)(4,14,11,9,5)$, $c=(3,13,12,8,6)(4,14,11,7,5)$, $d=(7,14)(8,13)(9,10)$, and set $G_{12}=\langle a_1,\dots ,a_5,b,c,d\rangle$. Direct verification shows that ${\left\{a_i\right\}}_{i=1}^5$ pairwise commute and are involutions, hence $V=\langle a_1,\dots ,a_5\rangle \cong C_2^5$. We compute $b^5=c^5={\left(bc\right)}^2=1$, and the action of b and c on the set $\{3,4,5,6,7,8,9,10,11,12,13,14\}$ partitions the vertices into two 6-orbits ${\mathrm{\Omega }}_1=\{3,5,6,10,12,13\}$ and ${\mathrm{\Omega }}_2=\{4,7,8,9,11,14\}$. Via the bijection $\psi :{\mathrm{\Omega }}_1\to \{1,\dots ,6\}$ defined by $\psi \left(3\right)=1,\psi \left(5\right)=2,\psi \left(6\right)=3,\psi \left(10\right)=4,\psi \left(12\right)=5,\psi \left(13\right)=6$, we obtain ${b|}_{{\mathrm{\Omega }}_1}=(1,6,5,4,2)$ and ${c|}_{{\mathrm{\Omega }}_1}=(1,6,5,3,2)$. These generate a transitive subgroup of $S_6$ whose commutator contains a 3-cycle, so $\langle b,c\rangle \cong A_6$. Computing the conjugation action yields $b{a}_1{b}^{-1}=a_2$, $b{a}_2{b}^{-1}=a_3$, $b{a}_3{b}^{-1}=a_4$, $b{a}_4{b}^{-1}=a_5$, $b{a}_5{b}^{-1}=a_1$, $c{a}_1{c}^{-1}=a_3$, $c{a}_2{c}^{-1}=a_4$, $c{a}_3{c}^{-1}=a_5$, $c{a}_4{c}^{-1}=a_1$, $c{a}_5{c}^{-1}=a_2$. Therefore b and c normalise V, and consequently $N=V\langle b,c\rangle$ is a semidirect product $N\cong C_2^5⋊A_6$. Finally, $d^2=1$, $db{d}^{-1}=c$, $dc{d}^{-1}=bc$, and d permutes the $\left\{a_i\right\}$, hence d normalises N and $d\notin N$. Thus $G_{12}=N\langle d\rangle \cong N⋊C_2\cong (C_2^5⋊A_6)⋊C_2$.

Let $Y_{12}=\langle {\alpha }_1,{\alpha }_2,{\alpha }_3,{\alpha }_4,{\alpha }_5,{\alpha }_6,\beta ,\gamma \rangle$, where ${\alpha }_1=(1,2)(13,14)$, ${\alpha }_2=(3,4)(13,14)$, ${\alpha }_3=(5,6)(13,14)$, ${\alpha }_4=(7,8)(13,14)$, ${\alpha }_5=(9,10)(13,14)$, ${\alpha }_6=(11,12)(13,14)$, $\beta =(1,7,14,6,4,11)(2,8,13,5,3,12)(9,10)$, $\gamma =(1,12,4,5)(2,11,3,6)(7,10)(8,9)$. Direct verification shows that the ${\alpha }_i$ pairwise commute and are involutions, and ${\left\{{\alpha }_i\right\}}_{i=1}^6$ are linearly independent, hence $V=\langle {\alpha }_1,\dots ,{\alpha }_6\rangle$ is an elementary abelian 2-group of order $2^6$, thus, $V\cong C_2^6$. We compute ${\beta }^6=1$, ${\gamma }^4=1$, and $\langle \beta ,\gamma \rangle \cong S_7$, denote $K=\langle \beta ,\gamma \rangle$. Computing the conjugation action, we obtain $\beta {\alpha }_1{\beta }^{-1}={\alpha }_2$, $\beta {\alpha }_2{\beta }^{-1}={\alpha }_3$, $\beta {\alpha }_3{\beta }^{-1}={\alpha }_4$, $\beta {\alpha }_4{\beta }^{-1}={\alpha }_5$, $\beta {\alpha }_5{\beta }^{-1}={\alpha }_6$, $\beta {\alpha }_6{\beta }^{-1}={\alpha }_1$, $\gamma {\alpha }_1{\gamma }^{-1}={\alpha }_1$, $\gamma {\alpha }_2{\gamma }^{-1}={\alpha }_3$, $\gamma {\alpha }_3{\gamma }^{-1}={\alpha }_5$, $\gamma {\alpha }_4{\gamma }^{-1}={\alpha }_4$, $\gamma {\alpha }_5{\gamma }^{-1}={\alpha }_6$, $\gamma {\alpha }_6{\gamma }^{-1}={\alpha }_2$, thus $\beta$ and $\gamma$ normalise V, so $V⊴Y_{12}$. Moreover $V\cap K=\left\{1\right\}$ and all generators lie in $VK$, consequently $Y_{12}=VK$, and therefore $Y_{12}\cong V⋊K\cong C_2^6⋊S_7$.    □

By computation, there are 52 graphs in class 9 that are 14-valent core-free Cayley graphs, and $i=108$ is the first graph in class 9. Accordingly, we have ${\tau }_{108}=(7,8)(9,10)(11,13)(12,14)$. We then obtain the following lemma.

Lemma 9.

If $i=108$, then $G_{108}\cong C_2^6⋊S_6$, $Y_{108}\cong C_2^7⋊S_7$, and $S_{108}=\{z,y^{-3}xz{y}^{-3}x,y^{-2}z{y}^2,y^{2}xz{y}^{2}x,y^{3}z{y}^{-3},xzx,y^{2}z{y}^3,y^{3}xz{y}^{-2}x,y^{-1}zyx,y^{-3}z{y}^{-2},y^{-2}xz{y}^{3}x,yz{y}^{-1}x,yxzy,y^{-1}xz{y}^{-1}\mid z={\tau }_{108}\}$.

Proof. 

We first determine the structure of $G_{108}$. Let $G_{108}=\langle a_1,a_2,a_3,a_4,a_5,a_6,b,c\rangle$, where $a_1=(13,14)$, $a_2=(7,8)$, $a_3=(9,10)$, $a_4=(11,12)$, $a_5=(3,4)$, $a_6=(5,6)$, $b=(5,9,13,7,11)(6,10,14,8,12)$, $c=(3,5,7,11,13,9)(4,6,8,12,14,10)$. Direct verification shows that the $a_i$ pairwise commute and are involutions, and ${\prod }_{i=1}^6{a}_i^{{\epsilon }_i}=1$ if and only if all ${\epsilon }_i=0$, hence $V=\langle a_1,\dots ,a_6\rangle \cong C_2^6$. We compute $b^5=c^6={\left(bc\right)}^2=1$, and via a bijection $\phi :\{1,\dots ,6\}\to \{a_1,\dots ,a_6\}$ the actions of b and c can be transformed into standard generators of $S_6$, thus $K=\langle b,c\rangle \cong S_6$. Conjugation calculations show that b and c normalise V, so $V⊴G_{108}$. Consequently $V\cap K$ is a normal subgroup of K and is a 2-group. Since the only normal 2-subgroup of $S_6$ is trivial, we have $V\cap K=\left\{1\right\}$. Moreover all generators lie in $VK$, so $G_{108}=VK$, and therefore $G_{108}\cong V⋊K\cong C_2^6⋊S_6$.

Let $Y_{108}=\langle {\alpha }_1,{\alpha }_2,{\alpha }_3,{\alpha }_4,{\alpha }_5,{\alpha }_6,{\alpha }_7,\beta ,\gamma \rangle$, where ${\alpha }_1=(13,14)$, ${\alpha }_2=(11,12)(13,14)$, ${\alpha }_3=(3,4)(11,12)$, ${\alpha }_4=(5,6)(11,12)$, ${\alpha }_5=(9,10)(11,12)$, ${\alpha }_6=(7,8)(11,12)$, ${\alpha }_7=(1,2)(3,4)(5,6)(7,8)(9,10)(13,14)$, $\beta =(1,6,8,13,11)(2,5,7,14,12)(3,9)(4,10)$, $\gamma =(1,6,10,11)(2,5,9,12)(7,14)(8,13)$. Direct verification shows that the ${\alpha }_i$ pairwise commute and are involutions, and their product generates an elementary abelian 2-group $V=\langle {\alpha }_1,\dots ,{\alpha }_7\rangle$ of order $2^7$, that is, $V\cong C_2^7$. We compute ${\beta }^5={\gamma }^4=1$, and via a bijection $\psi :\{1,\dots ,7\}\to \{{\alpha }_1,\dots ,{\alpha }_7\}$ (for instance $\psi \left(1\right)={\alpha }_1,\dots ,\psi \left(7\right)={\alpha }_7$) the actions of $\beta$ and $\gamma$ can be transformed into standard generators of $S_7$, hence $K=\langle \beta ,\gamma \rangle \cong S_7$. Conjugation calculations show that $\beta$ and $\gamma$ permute the set $\{{\alpha }_1,\dots ,{\alpha }_7\}$, thus normalise V, therefore $V⊴Y_{108}$. Since $S_7$ is simple and has no nontrivial normal 2-subgroup, and $V\cap K$ is a normal subgroup of K which is a 2-group, we have $V\cap K=\left\{1\right\}$. Moreover all generators lie in $VK$, so $Y_{108}=VK$. Consequently $Y_{108}\cong V⋊K\cong C_2^7⋊S_7$.    □

By computation, there are 25 graphs in class 10 that are 14-valent core-free Cayley graphs, and $i=8$ is the first graph in class 10. Accordingly, we have ${\tau }_8=(8,10)(12,14)$. We then obtain the following lemma.

Lemma 10.

If $i=8$, then $G_8\cong A_7\times A_6$, $Y_8\cong A_7^2⋊C_2$, and $S_8=\{z,y^{-1}xz{y}^{-3}x,y^{-2}z{y}^2,y^{2}xz{y}^{2}x,y^{3}z{y}^{-3},xzx,y^{-3}z{y}^3,yz{y}^{-1},y^{-1}xzyx,y^{3}xz{y}^{-2}x,y^{2}z{y}^{-2},y^{-1}zy,yxz{y}^{-1}x,y^{-2}xz{y}^{3}x\mid z={\tau }_8\}$.

Proof. 

We first determine the structure of $G_8$. Let $G_8=\langle a,b,c,d\rangle$, where $a=(2,6,4,14,10,12,8)$, $b=(2,14,12,8,6,4,10)$, $c=(3,13,9,5,11)$, $d=(3,9)(7,11)$. It is easy to see that $a,b$ act on the set ${\mathrm{\Omega }}_1=\{2,4,6,8,10,12,14\}$, while $c,d$ act on ${\mathrm{\Omega }}_2=\{3,5,7,9,11,13\}$, and ${\mathrm{\Omega }}_1\cap {\mathrm{\Omega }}_2=⌀$, hence each of $a,b$ commutes with each of $c,d$. Direct computation shows $a^7=b^7=1$, and a and b generate $A_7$. Similarly, $c^5=d^2=1$, and c and d generate $A_6$. Let $H=\langle a,b\rangle \cong A_7$ and $K=\langle c,d\rangle \cong A_6$. Then H and K commute elementwise and $H\cap K=\left\{1\right\}$, therefore $HK\cong H\times K\cong A_7\times A_6$. Since all generators lie in $HK$, we have $G_8\subseteq HK$, and clearly $HK\subseteq G_8$, consequently $G_8=HK\cong A_7\times A_6$.

Let $Y_8=\langle {\alpha }_1,{\alpha }_2,{\beta }_1,{\beta }_2,\gamma \rangle$, where ${\alpha }_1=(1,13)(5,9,7,11)$, ${\alpha }_2=(1,9,3,13)(5,11)$, ${\beta }_1=(4,12,14,6)(8,10)$, ${\beta }_2=(2,14,8,6,4)$, $\gamma =(1,14)(2,9)(3,12)(4,5)(6,11)(7,10)(8,13)$. Set $H=\langle {\alpha }_1,{\alpha }_2\rangle$ and $K=\langle {\beta }_1,{\beta }_2\rangle$. Direct computation gives ${\alpha }_1^2={\alpha }_2^3={\left({\alpha }_1{\alpha }_2\right)}^3=1$, and ${\alpha }_1,{\alpha }_2$ act transitively on ${\mathrm{\Omega }}_1=\{1,3,5,7,9,11,13\}$, hence $H\cong A_7$. Similarly, ${\beta }_1^2={\beta }_2^5={\left({\beta }_1{\beta }_2\right)}^3=1$, and ${\beta }_1,{\beta }_2$ act transitively on ${\mathrm{\Omega }}_2=\{2,4,6,8,10,12,14\}$, therefore $K\cong A_7$. Since ${\mathrm{\Omega }}_1\cap {\mathrm{\Omega }}_2=⌀$, the elements of H commute with those of K, and $H\cap K=\left\{1\right\}$, thus $HK\cong H\times K\cong A_7\times A_7$. Computing conjugations, we have $\gamma {\alpha }_1{\gamma }^{-1}={\beta }_1$, $\gamma {\alpha }_2{\gamma }^{-1}={\beta }_2$, $\gamma {\beta }_1{\gamma }^{-1}={\alpha }_1$, $\gamma {\beta }_2{\gamma }^{-1}={\alpha }_2$, and ${\gamma }^2=1$. Hence $\gamma$ normalises $HK$ and swaps H and K. Moreover $\gamma \notin HK$ (since $\gamma$ maps ${\mathrm{\Omega }}_1$ to ${\mathrm{\Omega }}_2$), so $Y_8=\left(HK\right)\langle \gamma \rangle$ is a semidirect product, where $\langle \gamma \rangle$ acts by swapping the two factors. Consequently $Y_8\cong A_7^2⋊C_2$.    □

By computation, there are 33 graphs in class 11 that are 14-valent core-free Cayley graphs, and $i=6$ is the first graph in class 11. Accordingly, we have ${\tau }_6=(10,14)(11,13)$. We then obtain the following lemma.

Lemma 11.

If $i=6$, then $G_6\cong A_7⋊S_6$, $Y_6\cong A_7^2⋊C_2^2$, and $S_6=\{z,y^{-1}xz{y}^{-1}x,y^{-2}z{y}^2,y^{-3}xz{y}^{-3}x,y^{3}z{y}^{-3},y^{2}xz{y}^{2}x,yz{y}^{-1},xzx,y^{-1}zy,yxz{y}^{-2}x,y^{2}z{y}^3,y^{3}xz{y}^{3}x,y^{-3}z{y}^{-2},y^{-2}xzyx\mid z={\tau }_6\}$.

Proof. 

Let $G_6=\langle a,b,c,d\rangle$, where $a=(1,3)(5,9,11)(7,13)$, $b=(1,11,9)(3,13,7)$, $c=(2,8,14,10,12,6,4)$, $d=(2,6,14,8,10,4,12)$. Set $N=\langle a,b\rangle$, direct computation shows $N\cong A_7$. Set $M=\langle c,d\rangle$, then $M\cong A_6$. Consider $H=\langle N,M\rangle$, it is easy to prove that $H\cong A_7\times A_6$ and $H⊲G_6$. Furthermore, $G_6/H\cong C_2$, so $G_6$ is an extension of H of index 2. Note that the action of $G_6$ swaps N and M, that is, there exists $g\in G_6\setminus H$ such that $gN{g}^{-1}=M$ and $gM{g}^{-1}=N$. In particular, $G_6$ contains an involutory automorphism $\tau$ satisfying $\tau \left(N\right)=M$ and $\tau \left(M\right)=N$, and the inner automorphisms induced by $\tau$ on N and M correspond to odd permutations in $S_6$ and $S_7$, respectively. Hence $G_6$ is isomorphic to $(A_7\times A_6)⋊C_2$, where $C_2$ swaps the two direct factors. Since $Aut\left(A_7\right)\cong S_7$ and $Aut\left(A_6\right)\cong S_6\times C_2$, the action of this $C_2$ corresponds to the action of $S_6$ on $A_7$ via the embedding $S_6\cong A_6⋊C_2$ into $Aut\left(A_7\right)$, together with the corresponding action of $S_7$ on $A_6$. Consequently the structure of $G_6$ is an extension of $A_7$ by $S_6$, denoted $A_7⋊S_6$, and this extension is nonsplit.

Let $Y_6=\langle \alpha ,\beta ,\gamma ,\delta ,\epsilon \rangle$, where $\alpha =(1,3,5,7,9,11,13)(2,4,6,8,10,12,14)$, $\beta =(3,7)(4,6)(10,14)(11,13)$, $\gamma =(2,6)(3,5)(10,14)(11,13)$, $\delta =(1,13)(2,8)(3,9)(5,11)(6,12)(10,14)$, $\epsilon =(1,8)(2,13)(3,12)(4,7)(5,14)(6,9)(10,11)$. Let $S=\langle \alpha ,\beta ,\gamma \rangle$. Direct computation shows that S is the socle of $Y_6$, with $S\cong A_7\times A_7$ and $S⊲Y_6$, and the quotient $Y_6/S$ has order 4. Take a complement $H=\langle \delta ,\epsilon \rangle$ of S in $Y_6$, then $H\cong C_2\times C_2$ and $Y_6=S⋊H$. Decompose S as $S=N_1\times N_2$, where $N_1,N_2\cong A_7$. Computing the conjugation action of the generators $\delta$ and $\epsilon$ on S, we find that both $\delta$ and $\epsilon$ map $N_1$ to $N_2$ and $N_2$ to $N_1$, that is, they swap the two direct factors. Moreover, the inner automorphisms induced on $N_1$ by each generator are distinct, so $\delta \epsilon$ preserves $N_1$ and $N_2$ but induces a nontrivial inner automorphism on each factor. Hence H embeds as $C_2\times C_2$ into $Aut(A_7\times A_7)$, and its image contains an involution swapping the two factors as well as an involution preserving each factor; thus the semidirect product is nontrivial. Consequently $Y_6\cong A_7^2⋊C_2^2$.    □

By computation, there are 65 graphs in class 12 that are 14-valent core-free Cayley graphs, and $i=1$ is the first graph in class 12. Accordingly, we have ${\tau }_1=(12,14)$. We then obtain the following lemma.

Lemma 12.

If $i=1$, then $G_1\cong S_7\times S_6$, $Y_1\cong A_7^2⋊D_8$, and $S_1=\{z,yxzyx,y^{-2}z{y}^2,y^{-1}xz{y}^{-1}x,y^{3}z{y}^{-3},y^{-3}xz{y}^{-3}x,yz{y}^{-1},y^{2}xz{y}^{2}x,y^{-1}zy,xzx,y^{-3}z{y}^3,y^{3}xz{y}^{-2}x,y^{2}z{y}^{-2},y^{-2}xz{y}^{3}x\mid z={\tau }_1\}$.

Let $G_1=\langle a,b,c,d\rangle$, where $a=(2,6,8,4,14,12,10)(5,9,13,7,11)$, $b=(3,11,5,9,13)(4,8,14,6)(10,12)$, $c=(7,11)$. Here $V_4$ denotes the Klein four-group, which acts by swapping the two factors and inducing nontrivial inner automorphisms on each factor. Set $N=\langle a,b\rangle$. Direct computation shows that a and b generate $A_7\times A_6$, that is, $N\cong A_7\times A_6$, and it is easy to verify that $N⊲G_1$. Since $|G_1/N|=4$, consider a complement of N in $G_1$. Computation shows that N has 36 complements in $G_1$, each isomorphic to the Klein four-group $V_4$. Choose one such complement $H=\langle c,d\rangle$, where $d=(4,10)(7,11)$, then $H\cong V_4$ and $G_1=N⋊H$. Write $N=N_1\times N_2$, where $N_1\cong A_7$ and $N_2\cong A_6$. Examining the action of the generators of H on N, we find that $c=(7,11)$ preserves each factor but induces a nontrivial inner automorphism on each factor, while $d=(4,10)(7,11)$ swaps $N_1$ and $N_2$ and simultaneously induces corresponding inner automorphisms on them. Hence H embeds faithfully into $Aut(A_7\times A_6)$ with image isomorphic to $V_4$, and the semidirect product is nontrivial. Therefore $G_1\cong (A_7\times A_6)⋊V_4\cong S_7\times S_6$.

Let $Y_1=\langle a,b,c\rangle$, where $a=(1,2)(3,14)(4,13)(5,12)(6,11)(7,10)(8,9)$, $b=(1,7,13,5,11,3,9)(2,8,14,6,12,4,10),c=(12,14)$. We first compute the socle $S=Soc\left(Y_1\right)$ of $Y_1$. Direct computation yields $S=\langle (2,14,12)(3,7,5),(2,14,12)(4,8,6),(2,14,4)(5,9,7),(2,14)(4,10,8,6)(7,11,9),(2,14,4,12,10,6,8)(7,9)(11,13),(1,13)(2,14,12,8,10,4,6)(7,9)\rangle$. It is easy to verify that $S\cong A_7\times A_7$ and $S⊲Y_1$. Since $|Y_1/S|=8$, consider the complements of S in $Y_1$, there are six such complements, each isomorphic to the dihedral group $D_8$. Choose one such complement $H=\langle (1,3)(2,12)(4,10)(8,14)(9,11),(1,4)(2,9)(3,12)(5,14)(6,13)(7,8)(10,11)\rangle$, then $H\cong D_8$ and $Y_1=S⋊H$. Write $S=N_1\times N_2$, where $N_1,N_2\cong A_7$. Examining the action of the generators of H on S, we find that the elements of the cyclic subgroup of order 4 in H preserve each $N_i$, while the involutions in H either preserve both factors or swap $N_1$ and $N_2$. Specifically, two of the involutions swap $N_1$ and $N_2$, and the other two, while swapping, also induce a nontrivial inner automorphism on one of the factors. Hence H embeds faithfully into $Aut(A_7\times A_7)$ with image isomorphic to $D_8$, and the semidirect product is nontrivial. Consequently $Y_1\cong A_7^2⋊D_8$.

By computation, there are 3799 graphs in class 13 that are 14-valent core-free Cayley graphs, and $i=2$ is the first graph in class 13. Accordingly, we have ${\tau }_2=(10,11)(13,14)$. We then obtain the following lemma.

Lemma 13.

If $i=2$, then $G_2\cong S_{13}$, $Y_2\cong S_{14}$, and $S_2=\{(10,11)(13,14),(9,10)(12,13),(8,9)(11,12),(7,8)(10,11),(6,7)(9,10),(5,6)(8,9),(4,5)(7,8),(3,4)(6,7),(3,14)(4,12,5,13)(6,11)$$(7,10)(8,9),(3,14)(4,13,5,12)(6,11)(7,10)(8,9),(2,3)(13,14),(2,3)(5,6),(2,13)(3,11,4,12)(5,10)(6,9)(7,8),(2,13)(3,12,4,11)(5,10)(6,9)(7,8)\}$.

Proof. 

Let $G_2=\langle a,b,c\rangle$, where $a=(3,4,9,6,11,5,10,7,12)(8,14,13),b=(2,9,10)(4,5,14,8,7,13,12,6)$. We first compute the socle $S=Soc\left(G_2\right)$ of $G_2$ and obtain that the set of even permutations in $\langle a,b\rangle$ forms $A_{13}$, that is, $S\cong A_{13}$, and $S⊲G_2$. Since $|G_2/S|=2$, consider a complement of S in $G_2$. Computation shows that S has 4 complements in $G_2$, each isomorphic to the cyclic group $C_2$ of order 2. Choose one such complement $H=\langle c\rangle$, where c is an odd permutation among a and b, specifically $c=(3,4,9,6,11,5,10,7,12)(8,14,13)$ itself is odd. Then $H\cong C_2$ and $G_2=S⋊H$. Examining the action of the generator c on S, we see that its conjugation action on $A_{13}$ corresponds to an odd permutation, hence H embeds faithfully into $Aut\left(A_{13}\right)$ with image isomorphic to $C_2$, and the semidirect product is nontrivial. Consequently $G_2\cong A_{13}⋊C_2\cong S_{13}$.

Let $Y_2=\langle \alpha ,\beta ,\gamma \rangle$, where $\alpha =(1,2,3,4,5,6,7,8,9,10,11,12,13)$, $\beta =(12,13,14)$, $\gamma =(13,14)$. Set $S=\langle \alpha ,\beta \rangle$. Direct computation shows that $\alpha$ and $\beta$ generate the alternating group $A_{14}$, that is, $S\cong A_{14}$, and it is easy to verify that $S⊲Y_2$. Since $|Y_2/S|=2$, consider a complement of S in $Y_2$. Computation shows that S has 4 complements in $Y_2$, each isomorphic to the cyclic group $C_2$ of order 2. Choose one such complement $H=\langle \gamma \rangle$, where $\gamma =(13,14)$, then $H\cong C_2$ and $Y_2=S⋊H$. Examining the action of the generator $\gamma$ on S, we find that $\gamma =(13,14)$ is a transposition, and its conjugation action on $A_{14}$ corresponds to an odd permutation; hence H embeds faithfully into $Aut\left(A_{14}\right)$ with image isomorphic to $C_2$, and the semidirect product is nontrivial. Consequently $Y_2\cong A_{14}⋊C_2\cong S_{14}$.    □

Proof of Theorem 1.

Let $\Sigma =Cay(G,S)$ be a 14-valent 1-regular core-free Cayley graph. Then from the discussion of Lemmas 1–13, we directly obtain that Theorem 1 holds.

4. Conclusions

In this paper, we carry out a systematic classification of 14-valent 1-regular core-free Cayley graphs. By combining the theory of core-free graph classification with the discrete computational algebra system GAP, we systematically characterise the structural properties of such graphs.

Our results show that the vertex stabiliser of a 14-valent 1-regular core-free Cayley graph is isomorphic to either the cyclic group $C_{14}$ or the dihedral group $D_{14}$. For each case, we determine the corresponding graph structures and their full automorphism groups.

In particular, we discover a 14-valent 1-regular Cayley graph on the non-abelian simple group $A_6$, whose full automorphism group is $S_7$. To our knowledge, this is the first example of a non-normal Cayley graph on a non-abelian simple group with even valency greater than 10.