Orientable vertex transitive embeddings of ${{\sf K}}_p$

1.   Introduction

If a graph can be embedded in a surface, then naturally there will be a problem: how many non-isomorphic ways can it be done. One of the main aims of topological graph theory is to enumerate all the symmetrical embeddings of a given class of graphs in closed surfaces, see ^[10,14,15]. We will restrict our attention here to the orientable vertex-transitive embeddings of ${{\sf K}}_p$, where $p\geq5$ is a prime.

An orientable map is a 2-cell embedding of a finite graph in an orientable surface. That is, 'drawing' a graph $\Gamma = (V, E)$ into an orientable surface $\mathcal{S}$ such that both any two edges do not intersect except for the end point and $E$ divides the surface $\mathcal{S}$ into discs. So an embedding divides the surface into open discs, called faces, the set of faces is denoted by $F$, and the map is denoted by $\mathcal{M} = (V, E, F)$. The graph of a map is called the underlying graph, and the orientable surface is called the supporting surface of the map. For convenience, a map $\mathcal{M}$ is called a complete map if its underlying graph is a complete graph ${{\sf K}}_n$.

An incident triple $(v, e, f)$ is called a flag. An automorphism of a map $\mathcal{M}$ is a permutation of the flags which preserves the incident relation. So it is exactly an automorphism of the underlying graph which preserves the supporting surface. All automorphisms of $\mathcal{M}$ form the automorphism group ${{\sf Aut}}(\mathcal{M})$.

A map $\mathcal{M}$ is said to be $G$-vertex-transitive (or a vertex-transitive embedding of its underlying graph) if $G\le{{\sf Aut}}(\mathcal{M})$ is transitive on the vertex set $V$; if in addition $G$ also preserves the orientation of the supporting surface, then $\mathcal{M}$ is called orientable vertex-transitive. Similarly, orientable arc-transitive maps are defined.

Recent development of the theory of maps was closely related to the theory of map colorings, with the topic of highly 'symmetrical' maps always at the center of interest and recent investigation began with Biggs ^[1,2]. In the past fifty years, plenty of results about 'symmetrical' maps have been obtained, see ^{[14,19,20,21]} and references therein. Particularly, see ^[1,6,14,15] for arc transitive maps, see ^{[17,20,23,24]} for vertex transitive maps, see ^[13,22] for edge transitive maps. Very recently, some special families of edge-transitive maps underlying complete bipartite graphs are classified in ^[7,8,9,25], and for more information about the embeddings of complete graphs, see ^[12,16,18].

Two maps $\mathcal{M}_1$ and $\mathcal{M}_2$ are isomorphic, denoted by $\mathcal{M}_1\cong\mathcal{M}_2$, if there is a one-to-one correspondence from the vertices of $\mathcal{M}_1$ to the vertices of $\mathcal{M}_2$ that maps flags to flags. It follows that ${{\sf Aut}}\mathcal{M}_1\cong{{\sf Aut}}\mathcal{M}_2$ if $\mathcal{M}_1\cong\mathcal{M}_2$. Recall that $\phi(n)$ is the Euler phi-function, i.e. the number of positive integers which is less than and coprime to $n$, where $n$ is a positive integer.

The purpose of this paper is to enumerate the number of orientable vertex-transitive maps with underlying graphs being complete graphs ${{\sf K}}_p$, where $p\geq5$ is a prime. The following theorem is the main result.

Theorem 1.1. Let $\mathcal{M}$ be an orientable vertex transitive map with underlying graph ${{\sf K}}_p$, where $p\geq5$ is a prime. Let $G = {{\sf Aut}}(\mathcal{M})$. Then $\mathcal{M}$ is a Cayley map of ${\mathbb Z}_p$, $G\cong{\mathbb Z}_p:G_\alpha$ is a Frobenius group, where $G_\alpha$ is a cyclic group for each $\alpha\in V$.

Further, if $G_\alpha\cong{\mathbb Z}_k$ acting on the neighborhood of $\alpha$ has $r$ orbits with $(k, p) = 1$, $rk = p-1$ and $r\geq2$ a prime, then the number of non-isomorphic orientable vertex transitive maps of ${{\sf K}}_p$ equals

where $|\mathcal{A}_r| = (r-1)!k^{r-1}\phi(k)$ and $|\mathcal{A}_1| = \phi(p-1).$

With regard to the Theorem 1.1, we can deduce the following similar conclusions when $r$ is a composite integer.

Corollary 1.2. If $r = p_1p_2$ with $p_i$ different prime and $i = 1, 2$, then the number of non-isomorphic orientable vertex transitive maps of ${{\sf K}}_p$ equals $\frac{|\mathcal{A}_{p_1p_2}|-|\mathcal{A}_{p_1}|-|\mathcal{A}_{p_2}|+|\mathcal{A}_1|}{p_1p_2}$.

Corollary 1.3. If $r = p_1^2p_2$ with $p_i$ different prime and $i = 1, 2$, then the number of non-isomorphic orientable vertex transitive maps of ${{\sf K}}_p$ equals $\frac{|\mathcal{A}_{p_1^2p_2}|-|\mathcal{A}_{p_1p_2}|-|\mathcal{A}_{p_1^2}|+|\mathcal{A}_{p_1}|}{p_1^2p_2}$.

This paper is organized as follows. After this introductory section, some preliminary results are given in Section 2, then the enumeration of the different and non-isomorphic orientable vertex-transitive complete maps is given in Section 3 and Section 4, respectively. We give the complete proof of Theorem 1.1 in Section 5. Finally, we present conclusions for the paper in Section 6.

2.   Preliminaries

In this section, we need some notations for convenience which will be used in the following discussion.

Let $F = {\mathbb F}_p$ be the field of order $p$ with $p\geq5$ a prime. Let $F^+ = {\mathbb F}^+_p$ and $F^\times = {\mathbb F}^\times_p$ be the additive group and the multiplicative group of $F$, respectively. It follows that

Let ${\textbf 0}$ be the identity of $F^+$. Let $F^{\#}$ be the set of all non-identity elements of $F^+$, namely, $F^{\#} = F^+\setminus\{{\textbf 0}\}$. Then the complete graph ${{\sf K}}_p$ may be represented as a Cayley graph

A Cayley map $\mathcal{M}$ is an embedding of a Cayley graph $\Sigma = {{\sf Cay}}(H, S)$ into a surface such that ${{\sf Aut}}(\mathcal{M})$ contains a subgroup $N$ acting regularly on the vertices, and $\mathcal{M}$ is called a Cayley map of $N$ (or a Cayley embedding of $\Sigma$ with respect to $N$).

For a vertex $v$, a cyclic permutation of the neighbor set ${{\it \Gamma}}(v)$ of $v$ is called a rotation at $v$ and denoted by $R_v$. A rotation system $R({{\it \Gamma}})$ of a graph ${{\it \Gamma}}$ is the set of rotations at all vertices, that is $R({{\it \Gamma}}) = \{R_v\}_{v\in V}$. Hence each rotation system $R({{\it \Gamma}})$ defines an orientable embedding of ${{\it \Gamma}}$, refer to [3, pp.104-108]. Noting that the vertex rotations $R_v$ can be regarded as permutations not only of the neighborhood ${{\it \Gamma}}(v)$ but also the generating set $S$, so Cayley maps have another equivalent definitions ^[11]. A map with underlying graph being Cayley graph $\Sigma = {{\sf Cay}}(H, S)$ is a Cayley map if the induced local cyclic permutations of $S$ are all the same. Moreover, each circular permutation $\rho$ of $F^{\#}$ gives rise to a unique orientable Cayley embedding of ${{\sf K}}_p$ with the underlying graph $\Gamma = {{\sf Cay}}(F^+, F^{\#})$.

3.   Enumeration of different embeddings

In this section, we determine enumeration of different vertex transitive embeddings of ${{\sf K}}_p$. Now, we begin by citing the well-known conclusion about vertex transitive maps.

Lemma 3.1. ([17, Lemma 2.2]) Let $\mathcal{M}$ be an orientable vertex transitive map. Let $G = {{\sf Aut}}(\mathcal{M})$. Then the stabilizer $G_\alpha\cong {\mathbb Z}_k$ or ${{\sf D}}_{2k}$ for a vertex $\alpha$, and each orbit of $G_\alpha$ acting on the neighborhood of $\alpha$ has length $k$.

Next, by [17, Theorem 1.1] and [3, Lemma 5.4.1], we can obtain the following lemma.

Lemma 3.2. Let $\mathcal{M}$ be an orientable vertex transitive embedding with underlying graph ${{\sf K}}_p$, where $p\geq5$ is a prime. Let $G = {{\sf Aut}}(\mathcal{M})$. Then $\mathcal{M}$ is a Cayley map of ${\mathbb Z}_p$, $G\cong{\mathbb Z}_p{:}{\mathbb Z}_k$ is a Frobenius group, and $G_\alpha\cong{\mathbb Z}_k$ such that $(k, p) = 1$.

Assume that $G_\alpha\cong {\mathbb Z}_k: = \langle a\rangle$ with $o(a)\geq2$. Then by Lemma 3.2, $G\cong F^+{:}\langle a\rangle$ is a Frobenius group. It follows that $\langle a\rangle$ is half-transitive on $F^+$, and $|\langle a\rangle| = k$ is a divisor of $|F^+|-1 = p-1$. Let $r = (p-1)/k$. Thus $G_\alpha$ acting on $\Gamma(\alpha)$ has $r$ orbits with $r\geq2$, and we get the lemma as follows.

Lemma 3.3. If $G\cong F^+{:}{\mathbb Z}_k$, then there are exactly $(r-1)!k^{r-1}\phi(k)$ different orientable vertex-transitive embeddings of ${{\sf K}}_p$.

Proof. Taking $\alpha = {\textbf 0}$ for convenience with ${\textbf 0}$ the identity element of $F^+$, then $G_{\textbf 0}$ partitions $\Gamma({\textbf 0})$ into $r$ orbits, and by Lemma 3.1, the length of each orbit is $k$. Since $F^{\#}$ be the set of all non-identity elements of $F^+$, we have $|F^{\#}| = p-1$ and

where $\Delta_i$ is an orbit of $G_{\textbf 0}$ acting on $\Gamma({\textbf 0})$, $|\Delta_i| = k$ with $1\leq i\leq r$ and $r\geq2$.

Note that the vertex ${\textbf 0}$ and the neighbors can be lied on a disc such that ${\textbf 0}$ is in the centre and the neighbors of ${\textbf 0}$ are around ${\textbf 0}$. Without loss of generality, we may assume that the $p-1$ neighbors of ${\textbf 0}$ (i.e. all the elements of $F^{\#}$) are in clockwise order around ${\textbf 0}$, say $\beta_1, \beta_2, \cdots, \beta_{p-1}.$ Viewing

as a circular permutation of $F^{\#}$. Since the number of the circular permutations of $F^{\#}$ equals the number of arrangements of $\beta_i$, we can obtain that to determine the number of the orientable vertex transitive complete maps is only need to determine the different choices of $\beta_i$ with $1\leq i\leq p-1$.

Setting $\beta_1 = {\textbf 1}$ and $\beta_1\in\Delta_1$ for convenience, where ${\textbf 1}$ is the identity element of $F^\times$. If $\beta_2\in\Delta_1$, then ${\textbf 1}^{a^j} = \beta_2$ for some $a^j$, where $0 < j < k$. Correspondingly, $a^j: \beta_2\mapsto\beta_3\mapsto\beta_4\mapsto\cdots\mapsto{\textbf 1}$. Then $G_{\textbf 0}$ acting on $\Gamma({\textbf 0})$ has only one orbit which is a contradiction. Thus $\beta_2\not\in\Delta_1$. Hence $\beta_2\in\Delta_i$ with $2\leq i\leq r$. It follows that $\beta_2$ has $(r-1)k$ different choices.

Setting $\beta_2\in\Delta_2$ for convenience. If $\beta_3\in\Delta_2$, then $\beta_2^{a^s} = \beta_3$ for some $a^s$, where $0 < s < k$. Correspondingly, there has $a^s: \beta_2\mapsto\beta_3\mapsto\beta_4\mapsto\cdots\mapsto{\textbf 1}$. Then $\beta_1\in\Delta_2$, and $G_{\textbf 0}$ acting on $\Gamma({\textbf 0})$ has only one orbit which is a contradiction. Thus $\beta_3\not\in\Delta_2$.

If $\beta_3\in\Delta_1$, then ${\textbf 1}^{a^t} = \beta_3$ for some $a^t$, where $0 < t < k$. Correspondingly, there are $a^t: {\textbf 1}\mapsto\beta_3\mapsto\beta_5\mapsto\cdots\mapsto{\textbf 1}$, and $a^t: \beta_2\mapsto\beta_4\mapsto\beta_6\mapsto\cdots\mapsto\beta_2$. Thus $G_{\textbf 0}$ acting on $\Gamma({\textbf 0})$ has two orbits, namely, $r = 2$. Since the number of generators of $G_{\textbf 0}$ is $\phi(k)$, we obtain that $a^t$ has $\phi(k)$ different choices. Notice that $G_{\textbf 0}$ is cyclic, then except ${\textbf 1}$, $\beta_2$, the remaining vertices of $\Delta_1$, $\Delta_2$ can be obtained by ${\textbf 1}$, $\beta_2$ through the conjugate action of $a$, $a^2$, $\cdots$, $a^{k-1}$, respectively. So

is a circular permutation of $F^{\#}$. It follows that the number of $\rho|_{r = 2}$ is determined by the choices of ${\textbf 1}$, $\beta_2$, $a$. Thus the number of $\rho|_{r = 2}$ equals $(r-1)k\cdot\phi(k) = k\phi(k).$ Let the corresponding maps generated by $\rho|_{r = 2}$ be

Hence the number of vertex transitive maps of ${{\sf K}}_p$ equals $k\phi(k)$ if $r = 2$.

Now, suppose that $r\geq3$. Then $\beta_3\not\in\Delta_1$, $\beta_3\not\in\Delta_2$ and $\beta_3\in\Delta_i$ with $3\leq i\leq r$. Taking $\beta_3\in\Delta_3$ for convenience, and $\beta_3$ has $(r-2)k$ different choices. If $\beta_4\in\Delta_3$, then $\beta_3^{a^l} = \beta_4$ for some $a^l$, where $0 < l < k$. It follows that there has $a^l: \beta_3\mapsto\beta_4\mapsto\beta_5\mapsto\cdots\mapsto{\textbf 1}$. So $G_{{\textbf 0}}$ acting on $\Gamma({\textbf 0})$ has one orbit which is a contradiction as $r\geq3$. Thus $\beta_4\not\in\Delta_3$.

If $\beta_4\in\Delta_2$, then $\beta_2^{a^{l'}} = \beta_4$ for some $a^{l'}$, where $0 < l' < k$. Correspondingly, there have $a^{l'}: \beta_2\mapsto\beta_4\mapsto\beta_6\mapsto\cdots\mapsto\beta_2$ and $a^{l'}: {\textbf 1}\mapsto\beta_3\mapsto\beta_5\mapsto\cdots\mapsto{\textbf 1}$. We obtain that $G_{{\textbf 0}}$ acting on $\Gamma({\textbf 0})$ has two orbits which is a contradiction as $r\geq3$. Thus $\beta_4\not\in\Delta_2$.

If $\beta_4\in\Delta_1$, then ${\textbf 1}^{a^{l''}} = \beta_4$ for some $a^{l''}$, where $0 < l'' < k$. Correspondingly, there have $a^{l''}: {\textbf 1}\mapsto\beta_4\mapsto\beta_7\mapsto\cdots\mapsto{\textbf 1}$, $a^{l''}: \beta_2\mapsto\beta_5\mapsto\beta_8\mapsto\cdots\mapsto\beta_2$ and $a^{l''}: \beta_3\mapsto\beta_6\mapsto\beta_9\mapsto\cdots\mapsto\beta_3$. Thus $G_{{\textbf 0}}$ acting on $\Gamma({\textbf 0})$ has three orbits, that is, $r = 3$. Note that $G_{{\textbf 0}}$ is cyclic, then except ${\textbf 1}$, $\beta_2$ and $\beta_3$, the remaining vertices of $\Delta_1$, $\Delta_2$ and $\Delta_3$ can be obtained by ${\textbf 1}$, $\beta_2$ and $\beta_3$ through the conjugate action of $a$, $a^2$, $\cdots$, $a^{k-1}$, respectively. So

is a circular permutation of $F^{\#}$. It follows that the number of $\rho|_{r = 3}$ is determined by the choices of ${\textbf 1}$, $\beta_2$, $\beta_3$ and $a$. Further, the number of $\rho|_{r = 3}$ equals

Let the corresponding maps generated by $\rho|_{r = 3}$ be

Thus the number of orientable vertex transitive maps of ${{\sf K}}_p$ equals $2!k^2\phi(k)$ if $r = 3$.

Next, suppose that $r\geq4$. Then $\beta_4\not\in\Delta_1$, $\beta_4\not\in\Delta_2$, $\beta_4\not\in\Delta_3$ and $\beta_4\in\Delta_i$ with $4\leq i\leq r$. Taking $\beta_4\in\Delta_4$ for convenience, and $\beta_4$ has $(r-3)k$ different choices. If $\beta_5\in\Delta_4$, then $\beta_4^{a^m} = \beta_5$ for some $a^m$, where $0 < m < k$. It follows that there has $a^m: \beta_4\mapsto\beta_5\mapsto\beta_6\mapsto\cdots\mapsto{\textbf 1}$. So $G_{{\textbf 0}}$ acting on $\Gamma({\textbf 0})$ has one orbit which is a contradiction as $r\geq5$. Thus $\beta_5\not\in\Delta_4$.

If $\beta_5\in\Delta_3$, then $\beta_3^{a^{m'}} = \beta_5$ for some $a^{m'}$, where $0 < m' < k$. Correspondingly, there have $a^{m'}: \beta_3\mapsto\beta_5\mapsto\beta_7\mapsto\cdots\mapsto{\textbf 1}$ and $a^{m'}: \beta_2\mapsto\beta_4\mapsto\beta_6\mapsto\cdots\mapsto\beta_2$. It is easy to see that $G_{{\textbf 0}}$ acting on $\Gamma({\textbf 0})$ has two orbits which is a contradiction as $r\geq5$. Thus $\beta_5\not\in\Delta_3$.

If $\beta_5\in\Delta_2$, then $\beta_2^{a^{m''}} = \beta_5$ for some $a^{m''}$, where $0 < m'' < k$. Correspondingly, there have $a^{m''}: \beta_2\mapsto\beta_5\mapsto\beta_8\mapsto\cdots\mapsto\beta_2$, $a^{m''}: \beta_3\mapsto\beta_6\mapsto\beta_9\mapsto\cdots\mapsto\beta_3$ and $a^{m''}: \beta_4\mapsto\beta_7\mapsto\beta_{10}\mapsto\cdots\mapsto{\textbf 1}$. Then we have that $G_{{\textbf 0}}$ acting on $\Gamma({\textbf 0})$ has three orbits which is a contradiction as $r\geq5$. Thus $\beta_5\not\in\Delta_2$.

If $\beta_5\in\Delta_1$, then ${\textbf 1}^{a^s} = \beta_5$ for some $a^s$, where $0 < s < k$. Correspondingly, there have $a^s: {\textbf 1}\mapsto\beta_5\mapsto\beta_9\mapsto\cdots\mapsto{\textbf 1}$, $a^s: \beta_2\mapsto\beta_6\mapsto\beta_{10}\mapsto\cdots\mapsto\beta_2$, $a^s: \beta_3\mapsto\beta_7\mapsto\beta_{11}\mapsto\cdots\mapsto\beta_3$ and $a^s: \beta_4\mapsto\beta_8\mapsto\beta_{12}\mapsto\cdots\mapsto\beta_4$. Thus $G_{{\textbf 0}}$ acting on $\Gamma({\textbf 0})$ has four orbits, equivalently, $r = 4$. So

is a circular permutation of $F^{\#}$, and the number of $\rho|_{r = 4}$ is determined by the choices of ${\textbf 1}$, $\beta_2$, $\beta_3$, $\beta_4$ and $a$. It follows that the number of $\rho|_{r = 4}$ equals

Let the corresponding maps generated by $\rho|_{r = 4}$ be

Thus the number of orientable vertex transitive complete maps of ${{\sf K}}_p$ equals $3!k^3\phi(k)$ if $r = 4$.

Here $r$ can be generalized to any integer. Similarly, if $G_\alpha = \langle a\rangle$ acting on $\Gamma(\alpha)$ has $r$ orbits, then $\beta_i\in\Delta_i$ such that $\beta_1 = {\textbf 1}$ and $1\leq i\leq r$, and

is a circular permutation of $F^{\#}$. Since $\beta_i$ has $(r-i+1)k$ different choices with $2\leq i\leq r$, and $a$ has $\phi(k)$ different choices, it follows that the number of $\rho|_r$ equals

Let the corresponding maps generated by $\rho|_r$ be

Hence if $G_\alpha\cong {\mathbb Z}_k$, then the number of different orientable vertex transitive maps of ${{\sf K}}_p$ equals $(r-1)!k^{r-1}\phi(k).$ □

Recall that a Cayley map ${\sf CayM}(G, S)$ is called balanced if $s$ and $-s$ are placed on the antipodal points for all elements $s\in S$, see ^[21]. The map $\mathcal{M}_r({\textbf 1}, \beta_2, \beta_3, \cdots, \beta_r, a)$ is a Cayley map of the group $F^+$. Let $\eta$ be the unique involution of ${{\sf GL}}(1, p)\cong{\mathbb Z}_{p-1}$. Then

is an automorphism of $\mathcal{M}_r({\textbf 1}, \beta_2, \beta_3, \cdots, \beta_r, a)$.

Lemma 3.4. A map $\mathcal{M}_r({\textbf 1}, \beta_2, \beta_3, \cdots, \beta_r, a)$ is balanced if and only if $\beta_i^{-1} = \beta_{i+\frac{p-1}{2}}$ with $1\leq i\leq r$.

Proof. Assume that $\mathcal{M}_r({\textbf 1}, \beta_2, \beta_3, \cdots, \beta_r, a)$ is balanced. Then the vertex $\beta_i^{-1}$ is placed at the antipodal position of the vertex $\beta_i$ with $p$ an odd prime and $1\leq i\leq{p-1\over2}$. Thus $\beta_i^{-1} = \beta_{i+\frac{p-1}{2}}$ with $1\leq i\leq r$.

Conversely, assume that $\beta_i^{-1} = \beta_{i+\frac{p-1}{2}}$ with $p$ odd prime and $1\leq i\leq r$. Then for any $1\leq l\leq k-1$, we have

reading the subscripts modulo $(p-1)$. So $\beta_j^{-1} = \beta_{\frac{p-1}{2}+j}$ is at the antipodal position of $\beta_j$ for all $j$ with $1\leq j\leq{p-1\over2}$, and therefore $\mathcal{M}_r({\textbf 1}, \beta_2, \beta_3, \cdots, \beta_r, a)$ is balanced. □

4.   Enumeration of non-isomorphic embeddings

We notice that many different orientable vertex transitive maps of ${{\sf K}}_p$ may be isomorphic. The complete Cayley maps (that is, complete map and its automorphism group is regular on the vertices) of non-isomorphic groups were not isomorphic. However, to determine the number of non-isomorphic maps, we need the following lemma.

Lemma 4.1. Let $\mathcal{M}$ be an orientable vertex transitive map with underlying graph ${{\sf K}}_p$ and $p\geq5$ a prime. Let $G = {{\sf Aut}}(\mathcal{M})$. If $G\cong{\mathbb Z}_p{:}G_\alpha$, where $G_\alpha$ is a cyclic group for any $\alpha\in V$, then $\mathcal{M}^\sigma\cong\mathcal{M}$ for each $\sigma\in {{\sf Aut}}(G)$. On the contrary, $\sigma\in {{\sf Aut}}(G)$ if $\mathcal{M}^\sigma\cong\mathcal{M}$.

Proof. Since $G\cong{\mathbb Z}_p{:}G_\alpha$ such that $G_\alpha$ is a cyclic group for any $\alpha\in V$, it follows that by [4, Lemma 4.5],

Suppose that $\sigma$ fixes $\alpha$ for each $\sigma\in{{\sf Aut}}(G)$. Since ${\mathbb Z}_p$ is a regular and normal subgroup of ${{\sf Aut}}(G)$, we have that for any $1\neq x\in {\mathbb Z}_p$,

So $\alpha^x\neq\alpha$, namely, $x$ does not fix $\alpha$. It follows that $\sigma\in {\mathbb Z}_{p-1}$, and then $G_\alpha^\sigma = G_\alpha$ for each $\sigma\in {\mathbb Z}_{p-1}$. Further, for each $\tau\in{{\sf Aut}}(G)$, since

such that $\tau = x\sigma$, where $x\in {\mathbb Z}_p$ and $\sigma\in {\mathbb Z}_{p-1}$. Hence $\mathcal{M}^\sigma\cong\mathcal{M}$ for each $\sigma\in {{\sf Aut}}(G)$ by arbitrariness of $x$.

On the contrary, if $\mathcal{M}^\sigma\cong\mathcal{M}$, then ${{\sf Aut}}(\mathcal{M}^\sigma)\cong{{\sf Aut}}(\mathcal{M}) = G$. It follows that $({{\sf Aut}}(\mathcal{M}))^\sigma = G^\sigma = G\cong{{\sf Aut}}(\mathcal{M}^\sigma)$ for each $\sigma\in{{\sf Aut}}(\mathcal{M})$. Note that ${{\sf Z}}(G) = 1$, then $G = G/{{\sf Z}}(G)\cong {{\sf Inn}}(G)\lhd {{\sf Aut}}(G)$. Hence $\sigma\in {{\sf Aut}}(G)$. □

Now, we determine the number of non-isomorphic vertex transitive embeddings of ${{\sf K}}_p$ if $G_\alpha\cong{\mathbb Z}_k$. Let

Then $\mathcal{A}_r$ is a finite non-empty set, and $|\mathcal{A}_r| = (r-1)!k^{r-1}\phi(k)$. Let $X = {{\sf Aut}}(G)$. Then

Let $\langle z\rangle = {\mathbb Z}_{p-1}$. Since ${\mathbb Z}_p\lhd X$, and ${\mathbb Z}_p$ acting on $V$ is regular, we by the 'Frattini Argument' (or [5, Exercise 1.4.1]) have that $X_\alpha\cong{\mathbb Z}_{p-1} = \langle z\rangle$, and further $G\lhd X$.

Lemma 4.2. If $r$ is a prime, then the number of non-isomorphic orientable vertex transitive maps of ${{\sf K}}_p$ equals

Proof. Since $o(a) = k = \frac{p-1}{r}$, it follows that $z^r = a$. Let

be a circular permutation of $F^{\#}$ such that $\beta_i = {\textbf 1}^{z^i}$ and $2\leq i\leq r$. Then there have $z^i: {\textbf 1}\mapsto\beta_i\mapsto\beta_{2i}\cdots\mapsto{\textbf 1}$, $z^i: \beta_2\mapsto\beta_{i+2}\mapsto\beta_{2i+2}\cdots\mapsto\beta_2$, $\cdots$, $z^i: \beta_{i-1}\mapsto\beta_{2i-1}\mapsto\beta_{3i-1}\cdots\mapsto\beta_{i-1}$. Furthermore, $z^i$ can be identified with the permutation

Thus $G_\alpha = \langle z^i\rangle$ and $a = z^r\in\langle z^i\rangle$, which is a contradiction as $i\not|r$.

Let $({\textbf 1}, \beta_2, \beta_3, \cdots, \beta_r, {\textbf 1}^a, \beta_2^a, \beta_3^a, \cdots, \beta_r^a, \cdots, {\textbf 1}^{a^{k-1}}, {\beta}_2^{a^{k-1}}, {\beta}_3^{a^{k-1}}, \cdots, {\beta}_r^{a^{k-1}})$ be a circular permutation of $F^{\#}$ such that $\beta_2 = {\textbf 1}^z$, and gives rise to a unique Cayley embedding $\mathcal{M}_1$ of ${{\sf K}}_p$. Then there have

and $z^i$ can be identified with the permutation

It follows that ${{\sf Aut}}(\mathcal{M}_1) = F^+{:}\langle z\rangle = G.r > G$, and $\mathcal{M}_1$ is arc transitive. Thus $\mathcal{M}_1\in\mathcal{A}_1$, $\mathcal{A}_1\subset\mathcal{A}_r$ and

So $X\setminus G$ contains no element which is an automorphism of $\mathcal{M}'_r$ for $\mathcal{M}'_r\in\mathcal{A}_r\setminus\mathcal{A}_1$. Since $G\lhd X$ and

we have that $X/G\cong {\mathbb Z}_r$ acting on $\mathcal{A}_r\setminus\mathcal{A}_1$ is semiregular. Let $X$ act on $\mathcal{A}_r\setminus\mathcal{A}_1$. Then $(\mathcal{M}'_r)^X$ is an orbit of this action, and the length of this orbit equals

It follows that by Lemma 4.1, there are

non-isomorphic orientable vertex transitive maps of ${{\sf K}}_p$. □

Furthermore, we can obtain the following results by the proof of Lemma 4.2.

Lemma 4.3. If $r = p_1p_2$ with $p_i$ different primes and $i = 1, 2$, then the number of non-isomorphic vertex-transitive maps of ${{\sf K}}_p$ equals

For example, when $r = 15 = 3\cdot5$, then the number of non-isomorphic vertex-transitive complete maps equals $\frac{|\mathcal{A}_{15}\setminus(\mathcal{A}_3\cup\mathcal{A}_5)|}{15} = \frac{|\mathcal{A}_{15}|-|\mathcal{A}_3|-|\mathcal{A}_5|+|\mathcal{A}_1|}{15}$.

Lemma 4.4. If $r = p_1^2p_2$ with $p_i$ different prime and $i = 1, 2$, then the number of non-isomorphic vertex-transitive maps of ${{\sf K}}_p$ equals

For example, when $r = 28 = 2^2\cdot7$, then the number of non-isomorphic vertex-transitive complete maps equals $\frac{|\mathcal{A}_{28}\setminus(\mathcal{A}_{14}\cup\mathcal{A}_4)|}{28} = \frac{|\mathcal{A}_{28}|-|\mathcal{A}_{14}|-|\mathcal{A}_4|+|\mathcal{A}_2|}{28}$.

Furthermore, if $r = p_1^{l_1}p_2^{l_2}\cdots p_t^{l_t}$, where $p_i$ are different from each other primes, $1\leq l_i$ and $1\leq i\leq t$, then here can obtain generalization of the above results.

5.   Proof of Theorem

In this section, we complete the proof of Theorem 1.1 in view of the above series of results.

Proof of Theorem 1.1.

Let $\mathcal{M} = (V, E, F)$ be an orientable vertex transitive map with underlying graph ${{\sf K}}_p = (V, E)$, where $p\geq5$ is a prime. Let $G = {{\sf Aut}}(\mathcal{M})$. We by Lemma 3.2 have that $\mathcal{M}$ is a Cayley map of ${\mathbb Z}_p$, and $G = {\mathbb Z}_p{:}G_\alpha$ is a Frobenius group, where $G_\alpha$ is a cyclic group for each $\alpha\in V$. Further, if $G_\alpha\cong{\mathbb Z}_k$ acting on the neighborhood of $\alpha$ has $r$ orbits with $(k, p) = 1$, $rk = p-1$ and $r\geq2$ a prime, then by Lemma 3.3 and Lemma 4.2 there are exactly

non-isomorphic orientable vertex transitive maps of ${{\sf K}}_p$. □

6.   Conclusions

Determining and enumerating all the 2-cell embeddings of a given class of graphs is one of the main research topics in topological graph theory. Complete maps have always been a focus of attention for many scholars. Li characterized the classification of vertex-transitive embeddings of complete graphs in ^[17]. The manuscript obtained accurate counting results of non-isomorphic orientable vertex-transitive complete maps with $p$ vertices, where $p\geq5$ is a prime.

Acknowledgments

The authors would like to thank the anonymous referees for their valuable comments. The first author was supported by the NSFC (11861076), the NSF of Henan Province (232300420357) and key R & D and promotion projects of Henan Province (212102210543). The second author was supported by Youth Talent Promotion Project of Henan Province (2019HYTP035).

Conflict of interest

All authors declare no conflicts of interest in this paper.