Classification of cubic vertex-transitive tricirculants

TL;DR

This paper classifies all finite connected cubic vertex-transitive tricirculant graphs, revealing their structure and categorizing them into specific families with a few small exceptions.

Contribution

It provides a complete classification of finite connected cubic vertex-transitive tricirculants, identifying their structure and infinite families, extending previous understanding of symmetric graphs.

Findings

Most such graphs are prisms or Möbius ladders.

Two infinite families of graphs are characterized for orders 6k with k odd.

Small exceptions of order less than 54 are explicitly listed.

Abstract

A finite graph is called a tricirculant if admits a cyclic group of automorphism which has precisely three orbits on the vertex-set of the graph, all of equal size. We classify all finite connected cubic vertex-transitive tricirculants. We show that except for some small exceptions of order less than 54, each of these graphs is either a prism of order 6k with k odd, a M\"obius ladder, or it falls into one of two infinite families, each family containing one graph for every order of the form 6k with k odd.

Tables

Table 1. Table 1. Graphs of small order

Label	Type	Label	Type	Label	Type	Label	Type	Label	Type
[6,1]	$1,3$	[18,1]	$3$	[24,1]	$3$	[30,7]	$3$	[42,4]	$1$
[6,2]	$3$	[18,3]	$3$	[30,1]	$2$	[30,8]	$4$	[42,9]	$3$
[12,1]	$3$	[18,4]	$2$	[30,2]	$1$	[36,1]	$3$	[42,10]	$3$
[12,2]	$1$	[18,5]	$1$	[30,5]	$3$	[42,3]	$2$	[48,1]	$3$

Table 2. Table 2. Net voltages of closed walks of length 8 8 8 in Δ 1 subscript Δ 1 \Delta_{1} .

Label	Net voltage	Starting at $u$	Starting at $v$
I	$0$	$12$	$10$
II	$\pm \left(2r\right)$	$8$	$8$
III	$\pm \left(2s\right)$	$8$	$8$
IV	$\pm \left(r+s\right)$	$8$	$4$
V	$\pm \left(r-s\right)$	$8$	$4$
VI	$\pm \left(k+2r-s\right)$	$12$	$10$
VII	$\pm \left(k+2s-r\right)$	$12$	$10$
VIII	$\pm \left(k+3r-2s\right)$	$4$	$6$
IX	$\pm \left(k+3s-2r\right)$	$4$	$6$
X	$\pm \left(3r-s\right)$	$4$	$6$
XI	$\pm \left(3s-r\right)$	$4$	$6$
XII	$\pm \left(2r-2s\right)$	$4$	$6$
XIII	$\pm \left(4r-4s\right)$	$0$	$2$

Table 3. Table 3. Number of 8 8 8 -cycles through each edge-type of T 1 ​ ( k , r , s ) subscript 𝑇 1 𝑘 𝑟 𝑠 T_{1}(k,r,s)

Congruence	$0$-edge	$R$-edge	$S$-edge	$K$-edge	$8$-signature
$3s-2r+k$	$5$	$5$	$6$	$6$	$(5,5,6)$
$3r-rs+k$	$5$	$6$	$5$	$6$	$(5,5,6)$
$3r-s$	$6$	$6$	$4$	$4$	$(4,6,6)$
$3s-r$	$6$	$4$	$6$	$4$	$(4,6,6)$
$4r-4s$	$4$	$4$	$4$	$4$	$(4,4,4)$

Table 4. Table 4. Net voltages of closed walks of length 7 7 7 in Δ 1 subscript Δ 1 \Delta_{1} .

Label	Net voltage	Starting at $u$	Starting at $v$
I	$\pm \left(k+2r-2s\right)$	$8$	$10$
II	$\pm \left(k+r+s\right)$	$12$	$8$
III	$\pm \left(k+2r\right)$	$6$	$4$
IV	$\pm \left(k+2s\right)$	$6$	$4$
V	$\pm \left(3r-2s\right)$	$2$	$6$
VI	$\pm \left(3s-2r\right)$	$2$	$6$

Table 5. Table 5. Net voltages of closed walks of length 6 6 6 in Δ 2 subscript Δ 2 \Delta_{2} .

Label	Net voltage	Starting at $u$	Starting at $v$	Starting at $w$
I	$0$	$2$	$0$	$4$
II	$\pm \left(k-s\right)$	$8$	$8$	$8$
III	$\pm \left(k-r-s\right)$	$4$	$4$	$4$
IV	$\pm \left(k+r-s\right)$	$4$	$4$	$4$
V	$\pm \left(3r\right)$	$2$	$0$	$2$
VI	$\pm \left(2r\right)$	$2$	$0$	$4$
VII	$\pm \left(6s\right)$	$0$	$2$	$0$
VII	$\pm \left(r-2s\right)$	$4$	$6$	$2$
IX	$\pm \left(r+2s\right)$	$4$	$6$	$2$

(A1)	$k+2r+s=0$	(B1)	$k+2s+r=0$
(A2)	$k-2r+s=0$	(B2)	$k-2s+r=0$
(A3)	$r+3s=0$	(B3)	$s+3r=0$
(A4)	$r-3s=0$	(B4)	$s-3r=0$

Equations

$$E_K$$

$$E_0$$

$$E_R$$

$$E_S$$

$$r^{*}=\left\{\begin{array}[]{ll}\frac{k+3}{2},&\hbox{ if }k\equiv 1\>(\hbox{\rm mod}\>4)\\ \frac{k+3}{2}+k,&\hbox{ if }k\equiv 3\>(\hbox{\rm mod}\>4)\end{array}\right.$$

$$T_1(k,r,s)\cong T_1(k,s,r),$$

$$T_1(k,ar,as)\cong T_1(k,r,s).$$

$$3s-2r+k\equiv 0$$

$$3r-2s+k\equiv 0$$

$$3r-s\equiv 0$$

$$3s-r\equiv 0$$

$$4r-4s\equiv 0$$

$$3s$$

$$2r=d^{\prime }(a{k}^{\prime }+3s^{\prime })$$

$${\varphi }_0$$

$${\varphi }_1$$

$$2r-3s$$

$$3r-3s+r(k-1)$$

$$r-s+r(k-1)/3$$

$$E_K$$

$$E_0$$

$$E_R$$

$$E_S$$

$$r\equiv 2s$$

$$r\equiv -2s$$

$$x_{R}r+x_{S}s+x_{K}k\equiv 0.$$

$$x_{R}r+x_{S}s\equiv 0$$

$$x_{R}2s+x_{S}s\equiv 0,$$

$$s(2x_R+x_S)\equiv 0.$$

$$2x_R+x_S\ge k.$$

$$x_{R}r+x_{S}s+x_{K}k\equiv 0.$$

$$x_{R}r+x_{S}s+k\equiv 0,$$

$$x_{R}r+x_{S}s\equiv k,$$

$$s(2x_R+x_S)\equiv k.$$

$$E_K$$

$$E_0$$

$$E_R$$

$$E_S$$

$$T_4(k,r,s)\cong T_4(k,-r,s)\cong T_4(k,r,-s)\cong T_4(k,s,r)$$

Full text

1. Introduction

All graphs in this paper are finite. A connected graph $\Gamma$ admitting a cyclic group of automorphisms $H$ having $k$ orbits of vertices of equal size larger than $1$ is called a $k$-circulant and a generator of $H$ is then called a $k$-circulant automorphism of $\Gamma$. In particular, $1$-, $2$- and $3$-circulants are generally called circulants, bicirculants and tricirculants, respectively. A graph is called cubic if it is connected and regular of valence $3$.

A famous and longstanding polycirculant conjecture claims that every vertex-transitive graph and digraph is a $k$-circulant for some $k$ (see [2, 11]). It is particularly interesting to study conditions under which vertex-transitive graphs admit $k$-circulant automorphisms of large order (and thus relatively small $k$); see, for example, [3, 16]. On the other hand, existence of a $k$-circulant automorphism with small $k$ often restricts the structure of a vertex-transitive graph to the extent that allows a complete classification. There is a series of classification results about cubic arc-transitive $k$-circulant for $k\leq 5$ (see [5, 8]). In particular, cubic arc-transitive tricirculants where completely classified in [8], where it was shown that only $4$ such graphs exist: $K_{3,3}$, the Pappus Graph, Tutte’s $8$-cage and a graph on $54$ vertices. This work culminated in a beautiful paper [7], where it was shown that for all square-free values of $k$ coprime to $6$ there exist only finitely many cubic arc-transitive $k$-circulants.

In this paper we widen the focus to a much wider and structurally richer class of cubic vertex-transitive graphs that are not necessarily arc-transitive. It is know that every cubic vertex-transitive graph has a $k$-circulant automorphisms [12] and that no fixed $k$ exists such that every cubic vertex-transitive graph is a $k$-circulant [16]. As for the classification results, it is clear that a cubiv graph is a vertex-transitive $1$-circulant if and only if it is a cubic Cayley graph on a cyclic group. Furthermore, vertex-transitive cubic bicirculants were classified in [13]. The goal of this paper is to provide a complete classification of cubic, vertex-transitive tricirculants.

The following theorem and remarks are a brief summary of the contents of this paper.

Theorem 1**.**

Let $\Gamma$ be a cubic vertex-transitive tricirculant. Then the order of $\Gamma$ is $6k$ for some positive integer $k$ and one of the following holds:

1. (1)

$k$* is odd and $\Gamma$ is isomorphic to one of the graphs $\textrm{X}(k)$ or $\textrm{Y}(k)$, described in Section 4 (Definition 7) and Section 5 (Definition 26), respectively.* 2. (2)

$\Gamma$* is a prism and $k$ is odd, or $\Gamma$ is a Möbius ladder.* 3. (3)

the order of $\Gamma$ is less than $54$ and $\Gamma$ is one of the twenty graph described in Section 3.

Remark 2**.**

Prisms and Möbius Ladders are circulant graphs. Meanwhile $\textrm{X}(k)$ and $\textrm{Y}(k)$ are bicirculants if $k$ is odd and $3\nmid k$. In this case, $\textrm{X}(k)$ is isomorphic to the Generalized Petersen graph $\textrm{GP}(3k,k+(-1)^{\alpha})$ where $\alpha\in\{1,2\}$ and $\alpha\equiv k$ (mod $3$).

Remark 3**.**

$\textrm{Y}(k)$ can be embedded on the torus yielding the toroidal map $\{6,3\}_{\alpha,3}$ where $\alpha=\frac{1}{2}(k-1)$ (see [4] for the definitions pertaining to the maps on the torus).

2. Diagrams

Even though this paper is about simple graphs, it will be convenient in the course of the analysis to have a slightly more general definition, which allows the graphs to have loops, parallel edges and semi-edges. To distinguish between a simple graph (which we will refer to simply as a graph) and these more general objects, that we shall call pregraphs. In what follows, we briefly introduce the concept of a pregraph and refer the reader to [9, 10] for more detailed explanation.

A pregraph is an ordered $4$-tuple $(V,D;\mathop{{\rm beg}},\mathop{{\rm inv}})$ where $D$ and $V\neq\emptyset$ are disjoint finite sets of darts and vertices, respectively, $\mathop{{\rm beg}}:D\to V$ is a mapping which assigns to each dart $x$ its initial vertex $\mathop{{\rm beg}}\,x$, and $\mathop{{\rm inv}}:D\to D$ is an involution which interchanges every dart $x$ with its inverse dart, also denoted by $x^{-1}$. The neighbourhood of a vertex $v$ is defined as the set of darts that have $v$ for its initial vertex and the valence of $v$ is the cardinality of the neighbourhood. Note that a simple graph $\Gamma$ can be represented as a pregraph by letting $V={\rm V}(\Gamma)$ be the vertex-set of $\Gamma$, letting $D=\{(u,v):u,v\in{\rm V}(\Gamma),u\sim_{\Gamma}v\}$ be the arc-set of $\Gamma$, and by letting $\mathop{{\rm beg}}(u,v)=u$ and $\mathop{{\rm inv}}(u,v)=(v,u)$. Notions such as morphism, isomorphism and automorphism of pregraphs are obvious generalisations of those for (simple) graphs and precise definitions can be found in [9, 10].

The orbits of $\mathop{{\rm inv}}$ are called edges. The edge containing a dart $x$ is called a semi-edge if $\mathop{{\rm inv}}\,x=x$, a loop if $\mathop{{\rm inv}}\,x\neq x$ while $\mathop{{\rm beg}}\,(x^{-1})=\mathop{{\rm beg}}\,x$, and is called a link otherwise. The endvertices of an edge are the initial vertices of the darts contained in the edge. Two links are parallel if they have the same endvertices. When we present a pregraph as a drawing, the links are drawn in the usual way as a line between the points representing its envertices, a loops is drawn as a closed curve at its unique endvertex and a semi-edge is drawn as a segment attached to its unique envertex. The following lemma, which will serve as the starting point of our analysis of cubic tricirculants, is an easy exercise illustrating the definition of a pregraph.

Lemma 4**.**

Up to isomorphism there are precisely four non-isomorphic cubic pregraphs on three vertices such that no vertex is the initial vertex of more than one semi-edge, namely the pregraphs $\Delta_{1},\Delta_{2},\Delta_{3}$ and $\Delta_{4}$ depicted in Figure 2.1.

Given a pregraph $\Gamma$ and a group $H\leq\hbox{\rm Aut}(\Gamma)$ we define the quotient $\Gamma/H$ to be the pregraph $(V^{\prime},D^{\prime};\mathop{{\rm beg}}^{\prime},\mathop{{\rm inv}}^{\prime})$ where $V^{\prime}$ and $D^{\prime}$ are the sets ${\rm V}(\Gamma)/H$ and ${\rm D}(\Gamma)/H$ of orbits of $H$ on $V$ and $D$, respectively, and $\mathop{{\rm beg}}^{\prime}$ and $\mathop{{\rm inv}}^{\prime}$ are defined by $\mathop{{\rm beg}}^{\prime}(x^{H})=(\mathop{{\rm beg}}x)^{H}$ and $\mathop{{\rm inv}}^{\prime}(x^{H})=(\mathop{{\rm inv}}x)^{H}$ for every $x\in{\rm D}(\Gamma)$. The mapping $\wp_{H}\colon{\rm V}(\Gamma)\cup{\rm D}(\Gamma)\to V^{\prime}\cup D^{\prime}$ that maps a vertex or a dart $x$ to its $H$-orbit $x^{H}$ is then an epimorphism of pregraphs, called the quotient projection with respect to $H$. If $H$ acts semiregularly on ${\rm V}(\Gamma)$ (that is, if the vertex-stabiliser $H_{v}$ is trivial for every vertex $v\in{\rm V}(\Gamma)$), then the quotient projection $\wp_{H}$ is a regular covering projection and in particular, it preserves the valence of the vertices. Moreover, in this case the graph $\Gamma$ is isomorphic to the derived covering graph of $\Gamma/H$, a notion which we now define.

Let $\Gamma^{\prime}=(V^{\prime},D^{\prime},\mathop{{\rm beg}}^{\prime},\mathop{{\rm inv}}^{\prime})$ be an arbitrary connected pregraph, let $N$ be a group and let $\zeta\colon D^{\prime}\to N$ be a mapping (called a voltage assignment) satisfying the condition $\zeta(x)=\zeta(\mathop{{\rm inv}}^{\prime}x)^{-1}$ for every $x\in D^{\prime}$. Then $\hbox{\rm Cov}(\Gamma^{\prime},\zeta)$ is the graph with $D^{\prime}\times N$ and $V^{\prime}\times N$ as the sets of darts and vertices, respectively, and the functions $\mathop{{\rm beg}}$ and $\mathop{{\rm inv}}$ defined by $\mathop{{\rm beg}}(x,a)=(\mathop{{\rm beg}}^{\prime}x,a)$ and $\mathop{{\rm inv}}(x,a)=(\mathop{{\rm inv}}^{\prime}x,a\zeta(x))$. If, for a voltage assignment $\zeta\colon{\rm D}(\Gamma^{\prime})\to N$, there exists a a spanning tree $T$ in $\Gamma^{\prime}$ such that $\zeta(x)$ is the trivial element of $N$ for every dart $x$ in $T$, then we say that $\zeta$ is normalised; note that in this case $\hbox{\rm Cov}(\Gamma^{\prime},\zeta)$ is connected if and only if the images of $\zeta$ generate $N$. The following lemma is a well-known fact in the theory of covers:

Lemma 5**.**

Let $H$ be a group of automorphisms acting semiregularly on the vertex-set of a connected graph $\Gamma$, let $\Gamma^{\prime}=\Gamma/H$ and let $T$ be a spanning tree in $\Gamma^{\prime}$. Then there exists a voltage assignment $\zeta\colon{\rm D}(\Gamma^{\prime})\to N$ which is normalised with respect to $T$ and such that $\Gamma\cong\hbox{\rm Cov}(\Gamma^{\prime},\zeta)$.

Let us now move our attention back to cubic tricirculants. In the lemma below, the voltage assignments $\zeta_{i}\colon\Delta_{i}\to\mathbb{Z}_{n}$ are given in Figure 2.2 by writing the value $\zeta_{i}(x)$ next to the drawing of each dart $x\in{\rm D}(\Delta_{i})$.

Lemma 6**.**

Let $\Gamma$ be a cubic tricirculant, let $\rho$ be a corresponding tricirculant automorphism and let $n$ be the order of $\rho$. Then $n=2k$ for some positive integer $k$ and there exist elements $r,s\in\mathbb{Z}_{n}$ and $i\in\{1,2,3,4\}$ such that $\Gamma\cong\hbox{\rm Cov}(\Delta_{i},\zeta_{i})$ where $\zeta_{i}\colon{\rm D}(\Delta_{i})\to\mathbb{Z}_{n}$ is the voltage assignment defined by Figure 2.2.

Proof.

Note first that the quotient $\Gamma/\langle\rho\rangle$ is a pregraph with three vertices of valency $3$, one for each orbit under the action of $\langle\rho\rangle$. Since $\rho$ is a semiregular automorphism of $\Gamma$ of order $n$, the group $\langle\rho\rangle$ is isomorphic to $\mathbb{Z}_{n}$ and acts semiregularly on ${\rm V}(\Gamma)$. By Lemma 5, $\Gamma\cong\hbox{\rm Cov}(\Gamma/\langle\rho\rangle,\zeta)$ for some voltage assignment $\zeta\colon{\rm D}(\Gamma/\langle\rho\rangle,\mathbb{Z}_{n})$. Note that by definition of voltage assignments, it follows that $\zeta(x)=-\zeta(x^{-1})$ for every $x\in{\rm D}(\Gamma/\langle\rho\rangle)$, implying that if $x$ is a semi-edge, then $\zeta(x)$ is an element of order at most $2$ in $\mathbb{Z}_{n}$, and since $\Gamma$ has no semi-edges, $\zeta(x)$ must in fact have order $2$ in $\mathbb{Z}_{n}$. Since every pregraph with three vertices in which every vertex has valence $3$ contains at least one semi-edge, it follows that $n=2k$ for some positive integer $k$, and moreover, $\zeta(x)=k$ for every semi-edge $x$ of $\Gamma/\langle\rho\rangle$. Since $\Gamma$ has no parallel edges, this also implies that every vertex of $\Gamma/\langle\rho\rangle$ is the initial vertex of at most one semi-edge. By Lemma 4, $\Gamma/\langle\rho\rangle\cong\Delta_{i}$ for some $i\in\{1,2,3,4\}$ and we may thus assume that $\Gamma\cong\hbox{\rm Cov}(\Delta_{i},\zeta_{i})$ for some $i\in\{1,2,3,4\}$ and some voltage assignment $\zeta_{i}\colon\Delta_{i}\to\mathbb{Z}_{n}$. Finally, in view of Lemma 5, we may assume that $\zeta_{i}(x)=0$ for every edge $x$ belonging to a chosen spanning tree of $\Delta_{i}$. In particular, $\zeta_{i}$ can be chosen as shown in Figure 2.2. ∎

A cubic tricirculant isomorphic to $\hbox{\rm Cov}(\Delta_{i},\zeta_{i})$ is said to be of Type $i$. Note that in principle, a cubic vertex-transitive tricirculant could be of more than one type; however, it can be shown that this is not the case except for the complete bipartite graph $K_{3,3}$ which is both of Type 1 as well as of Type 3.

The following sections are devoted to the analysis of the tricirculants graphs arising from the voltage assignments $\zeta_{i}$, and in particular, to determining sufficient and necessary conditions for vertex-transitivity. We will consider the graphs of order at most $48$ separately in Section 3 and as for the graphs of larger order, we will show that no graph of Type 4 is vertex-transitive, that Type 3 yields prisms and Möbius ladders, and that Types 1 and 2 each yield one infinite family of vertex-transitive graphs, namely the graphs $\textrm{X}(k)$ defined in Definition 7 and the graphs $\textrm{Y}(k)$ defined in Definition 26.

Finally, to facilitate discussion in the forthcoming sections, we specify the notion of a walk. A walk $\omega$ in a (pre)graph is a sequence of darts $(x_{1},x_{2},...,x_{n})$ such that $\mathop{{\rm beg}}(x_{i+1})=\mathop{{\rm beg}}(x_{i}^{-1})$ for all $i\in\{1..,n-1\}$; that is, the initial vertex of $x_{i+1}$ is the end vertex of $x_{i}$. We say that $\omega$ is closed if the initial vertex of $x_{1}$ is the end vertex of $x_{n}$. If additionally $\mathop{{\rm beg}}(x_{i})\neq\mathop{{\rm beg}}(x_{j})$ for all $i\neq j$, then we say that $\omega$ is a cycle. We define the inverse of $\omega$ as the walk $\omega^{-1}=(x_{n}^{-1},x_{n-1}^{-1},\ldots,x_{1}^{-1})$. We say that a walk $\omega$ is a reduced walk if no two consecutive arcs are inverse to one another (if $\omega$ is closed, we will consider the first and last arcs of $\omega$ to be consecutive). It is pertinent to point out that, from this definition, all walks and cycles are directed and have an initial vertex. Therefore, no reduced walk is equal to its inverse.

3. Graphs of small order

To make this classification as general and as neat as possible, we will treat cubic tricirculant graphs of small order separately, as if we allow graphs that are “too small”, special cases and exceptions will inevitably occur. Since a cubic tricirculant must have order $6k$, for some positive $k$, we present, in Table 1, the complete list of all cubic vertex-transitive tricirculants of order at most $48$, along with their Types, obtained from the census [14] of cubic vertex-transitive graphs. Each graph in the census is labelled with a pair of integers in square brackets, the first being the order of the graph and the second being just an identifier.

The graphs [6,1], [18,4] and [30,8] are arc-transitive and correspond to $K_{3,3}$, $F018A$ (the Pappus graph) and $F030A$ (Tutte’s $8$-cage) respectively (see [1, 8] for further details). The fourth arc-transitive cubic tricirculant, denoted $F054A$ in [1], does not appear in Table 1, as it has $54$ vertices; it corresponds to the graph $\textrm{Y}(9)$ defined in Section 5. Since a cubic vertex-transitive graph that is also edge-transitive is automatically arc-transitive [17], we have that no vertex-transitive cubic tricirculant is edge-transitive, with the exception of the four aforementioned graphs.

For the remainder of this paper, all cubic tricirculants will have order at least $54$.

4. Type 1

Let $k\geq 9$ be an integer, and let $r$ and $s$ be two distinct elements of $\mathbb{Z}_{2k}$. Let $T_{1}(k,r,s)$ be the covering graph arising from $\zeta_{1}$ given in Figure 2.2; for convenience, we repeat the drawing here (see Figure 4.1).

By the definition of a derived cover, the vertices of $T_{1}(k,r,s)$ are then the pairs $(x,i)$ where $x\in\{u,v,w\}$ is a vertex of $\Delta_{1}$ and $i\in\mathbb{Z}_{2k}$. Note that $T_{1}(k,r,s)$ is connected if and only if $k$, $r$ and $s$ generate $\mathbb{Z}_{2k}$, that is, if and only if $\gcd(k,r,s)=1$.

To simplify notation, we write $u_{i}$ instead of $(u,i)$ and similarly for $v_{i}$ and $w_{i}$. Then $U=\{u_{0},u_{1},...,u_{2k-1}\}$, $V=\{v_{0},v_{1},...,v_{2k-1}\}$ and $W=\{w_{0},w_{1},...,w_{2k-1}\}$ are the respective fibres of vertices $u$, $v$ and $w$, and the edge-set of $T_{1}(k,r,s)$ is then the union of the sets

[TABLE]

Definition 7**.**

For an odd positive integer $k$, let

[TABLE]

and let $\textup{X}(k)=T_{1}(k,r^{*},1)$.

We can now state the main theorem of this section.

Theorem 8**.**

Let $\Gamma$ be a connected cubic tricirculant of Type 1 with at least $54$ vertices. Then $\Gamma$ is vertex-transitive if and only if it is isomorphic to $\textup{X}(k)$ for some odd integer $k$, $k\geq 9$. In this case, $\Gamma$ is a bicirculant if $3\nmid k$ and is isomorphic to the Generalized Petersen graph $\textrm{GP}(3k,k+(-1)^{\alpha})$ where $\alpha\in\{1,2\}$ and $\alpha\equiv k$ (mod $3$), and thus a bicirculant.

The rest of the section is devoted to the proof of Theorem 8. Clearly, as a tricirculant of Type 1, $\Gamma$ is isomorphic to $T_{1}(k,r,s)$ for some integer $k$ and elements $r,s\in\mathbb{Z}_{2k}$ (see Lemma 6). We shall henceforth assume that $\Gamma=T_{1}(k,r,s)$ for some $k\geq 9$ and that $\Gamma$ is connected. For a symbol $X$ from the set of symbols $\{0,R,S,K\}$, edges in $E_{X}$ will be called edges of type $X$, or simply $X$-edges.

Define $\rho$ as the permutation given by $\rho(u_{i})=u_{i+1}$, $\rho(v_{i})=v_{i+1}$ and $\rho(w_{i})=w_{i+1}$, and note that $\rho$ is a $3$-circulant automorphism of $T_{1}(k,r,s)$. Further, observe that

[TABLE]

and if $a\in\mathbb{Z}$ is such that $\gcd(2k,a)=1$, then

[TABLE]

Lemma 9**.**

Let $k$ be an odd integer, $k\geq 9$, and let $r^{*}$ and $\textup{X}(k)$ be as in Definition 7. Then the graph $X(k)$ is vertex-transitive.

Proof.

Recall that $\textup{X}(k)\cong T_{1}(k,r^{*},1)$. Define $\phi$ as the mapping given by:

[TABLE]

That $\phi$ is indeed an automorphism follows from the fact that $k$ is odd, $r^{*}$ is even and the congruence $k+3-2r^{*}\equiv 0$ holds. Since $\phi$ transitively permutes the three $\langle\rho\rangle$-orbits $U$, $V$ and $W$, the group $\langle\rho,\phi\rangle$ acts transitively on the vertices of $T_{1}(k,r^{*},1)$. ∎

Now, recall that $\Gamma$ is connected and equals $T_{1}(k,r,s)$, for some $k\geq 9$, $r,s\in\mathbb{Z}_{2k}$. We may also assume that $\Gamma$ is vertex-transitive, however, for some of the results in this section vertex-transitivity is not needed. When possible, we will not assume the graph to be vertex-transitive, but rather only to have some weaker form of symmetry, that we will define in the following paragraphs.

A simple graph $\Gamma$ is said to be $c$-vertex-regular, for some $c\geq 3$, if there are the same number of $c$-cycles through every vertex of $\Gamma$.

For an edge $e$ of $\Gamma$ and a positive integer $c$ denote by $\epsilon_{c}(e)$ the number of $c$-cycles that pass through $e$. For a vertex $v$ of $\Gamma$, let $\{e_{1},e_{2},e_{3}\}$ be the set of edges incident to $v$ ordered in such a way that $\epsilon_{c}(e_{1})\leq\epsilon_{c}(e_{2})\leq\epsilon_{c}(e_{3})$. The triplet $(\epsilon_{c}(e_{1}),\epsilon_{c}(e_{2}),\epsilon_{c}(e_{3}))$ is then called the $c$-signature of $v$. If for a $c\geq 3$ all vertices in $\Gamma$ have the same $c$-signature, then we say $\Gamma$ is $c$-cycle-regular and the signature of $\Gamma$ is the signature of any of its vertices. Note for every $c\geq 3$, a vertex-transitive graph is necessarily $c$-cycle-regular, and every $c$-cycle-regular graph is $c$-vertex-regular.

If $c$ equals the girth of the graph, then following [15], a $c$-cycle-regular graph will be called girth-regular.

Lemma 10**.**

If $\Gamma$ is $4$-vertex-regular, then neither $r$ nor $s$ equals $k$.

Proof.

Suppose $r=k$. Since $\Gamma$ has no parallel edges, we see that $s\neq k$. Observe that $(u_{0},v_{0},w_{r},u_{r})$ and $(u_{0},w_{0},v_{-r},u_{-r})$ are the only $4$ cycles of $\Gamma$ through $u_{0}$.

Meanwhile, there exists a unique $4$-cycle through $v_{0}$, namely $(v_{0},w_{r},u_{r},u_{0})$, which contradicts $\Gamma$ being $4$-vertex-regular. Hence $r\not=k$, and view of the isomorphism $T_{1}(k,s,r)\cong T_{1}(k,r,s)$ (see (4.6)), this also shows that $s\not=k$. ∎

For the sake of simplicity denote a dart in $\Delta_{1}$ starting at vertex $a$, pointing to vertex $b$ and having voltage $x$ by $(ab)_{x}$. Recall that a walk in $\Delta_{1}$ is a sequence of darts $(x_{1},x_{2},...,x_{n})$, for instance $((vw)_{r},(wu)_{0},(uu)_{k})$ is a walk of length $3$. However, since $\Gamma$ is a simple graph a walk in $\Gamma$ will be denoted as a sequence of vertices, as it is normally done.

Lemma 11**.**

If $\Gamma$ is $8$-cycle-regular, then neither $r$ nor $s$ equals [math].

Proof.

Suppose $r=0$. Since $\Gamma$ has no parallel edges, we see that $s\neq 0$. Suppose there is an $8$-cycle $C$ in $\Gamma$. Such a cycle, when projected to $\Delta_{1}$, yields a reduced closed walk $\omega$ in $\Delta_{1}$ whose $\zeta_{1}$-voltage is [math] . Note that $\omega$ cannot trace three darts with voltage [math] consecutively, as this would lift into a $3$-cycle contained in $C$. This implies $\omega$ necessarily visits the dart $(vw)_{s}$ or its inverse at least once. Furthermore, since $\gcd(k,s)=1$ and $8<9\leq k$, $\omega$ must trace $(wv)_{-s}$ as many times as it does $(vw)_{s}$. By observing Figure 4.1, the reader can see that if $\omega$ traces the dart $(vw)_{s}$, then it must also trace the semi-edge $(uu)_{k}$ before tracing $(wv)_{-s}$, as $\omega$ cannot trace $3$ consecutive darts with voltage [math]. Since $\omega$ has net voltage [math], it necessarily traces $(uu)_{k}$ an even amount of times. Moreover, $\omega$ must trace a dart in $\{(uv)_{0},(vu)_{0},(uw)_{0},(wu)_{0}\}$ immediately before and immediately after tracing $(uu)_{k}$. Hence, $\omega$ must visit the set $\{(vw)_{s},(wv)_{-s}\}$ at least twice; the semi-edge $(uu)_{k}$ at least twice; and the set $\{(uv)_{0},(vu)_{0},(uw)_{0},(wu)_{0}\}$ at least $4$ times. This already amounts to $8$ darts, none of which is $(vw)_{r}$ or its inverse. Therefore, no $8$-cycle in $\Gamma$ visits an $R$-edge. However, for $X\neq R$ there is at least one $8$-cycle through every $X$-edge, as $(u_{0},v_{0},w_{s},u_{s},u_{s+k},w_{s+k},v_{k},u_{k})$ and $(u_{0},w_{0},v_{-s},u_{-s},u_{k-s},v_{k-s},w_{k},u_{k})$ are $8$-cycles in $\Gamma$. This contradicts our hypothesis of $\Gamma$ being $8$-cycle-regular. The proof when $s=0$ is analogous. ∎

Now, observe that $(u_{0},u_{k},v_{k},w_{k+r},u_{k+r},u_{r},w_{r},v_{0},u_{0})$ is a cycle of length $8$ in $\Gamma$ starting at $u_{0}$. In what follows we will study the $8$-cycle structure of $\Gamma$ to determine conditions for vertex-transitivity. Recall that each $8$-cycle in $\Gamma$ quotients down into a closed walk of length $8$ having net voltage [math] in $\Delta_{1}$. We can thus, provided we are careful, determine how many $8$-cycles pass through any given vertex of $\Gamma$ by counting closed walks of length $8$ and net voltage [math] in the quotient $\Delta_{1}$. Note that only closed walks that are reduced will lift into cycles of $\Gamma$. Therefore, we may safely ignore non-reduced walks and focus our attention exclusively on reduced ones. Define $\mathcal{W}_{8}$ as the set of all reduced closed walks of length $8$ in $\Delta_{1}$.

Every element of $\mathcal{W}_{8}$ having net voltage [math] lifts into a closed walk of length $8$ in $\Gamma$. In principle, the latter walk might not be a cycle of $\Gamma$. However, we will show that this never happens in our particular case. For this, it suffices to show that $\Gamma$ does not admit cycles of length $4$ or smaller, as any closed walk of length $8$ that is not a cycle is a union of smaller cycles, one of which will have length smaller or equal to $4$.

Lemma 12**.**

$\Gamma$* has girth at least 5.*

Proof.

Suppose $\Gamma$ has a $3$-cycle. Since $\Gamma$ is vertex-transitive, this would mean that there is a reduced closed walk with net voltage [math] of length $3$ in $\Delta_{1}$ that visits $u$. From Figure 4.1 we get that such a closed walk must be either $((uv)_{0},(vw)_{r},(wu)_{0})$, $((uv)_{0},(vw)_{s},(wu)_{0})$ or one of their inverses. This would imply that either $r=0$ or $s=0$, a contradiction. Similarly, the existence of $4$-cycles in $\Gamma$ would imply there is a closed walk with net voltage [math] of length $4$ visiting $u$. The only such walks are $((uv)_{0},(vw)_{r},(wu)_{0},(uu)_{k}))$, $((uu)_{k},(uv)_{0},(vw)_{r},(wu)_{0})$, $((uv)_{0},(vw)_{s},(wu)_{0},(uu)_{k}))$, $((uv)_{0},(vw)_{s},(wu)_{0},(uu)_{k}))$ and their inverses. In all eight cases, we have that $r=k$ or $s=k$, again a contradiction. ∎

Now, let $N$ be the set of the net voltages of walks in ${\mathcal{W}}_{8}$, expressed as linear combinations of $k$, $r$ and $s$, where we view $r$ and $s$ as indeterminants over $\mathbb{Z}_{2k}$. For instance, the closed walk $((uv)_{0},(vw)_{r},(wv)_{-s},(vw)_{r},(wu)_{0},(uv)_{0},(vw)_{r},(wu)_{0})$ has net voltage $3r-s$. For an element $\nu\in N$, let ${\mathcal{W}}(\nu)$ be the set of walks in ${\mathcal{W}}$ with net voltage $\nu$ and for a vertex $x$ of $\Delta_{1}$, let ${\mathcal{W}}_{8}^{x}(\nu)$ be the set of walks in ${\mathcal{W}}(\nu)$ starting at $x$.

For each $\nu\in N$, we have computed the number of elements in ${\mathcal{W}}_{8}^{u}(\nu)$ and in ${\mathcal{W}}_{8}^{v}(\nu)$. The result is displayed in Table 2. Notice that, if $\omega\in{\mathcal{W}}_{8}(\nu)$, then $\omega^{-1}\in{\mathcal{W}}_{8}^{x}(-\nu)$. This means $\omega$ has voltage [math] if and only if $\omega^{-1}$ does too. For this reason, and since we are interested in walks with net voltage [math], we have grouped walks with net voltage $\nu$ along with their inverses having net voltage $-\nu$. This computation is straightforward but somewhat lengthy. For this reason and to avoid human error, we have done it with the help of a computer programme.

From the first row of Table 2 we know that there are always $12$ walks starting at $u$ in ${\mathcal{W}}_{8}$ that have net voltage [math], regardless of the values of $r$ and $s$. Similarly, there are always $10$ such walks starting at $v$. Since $\Gamma$ is vertex-transitive, the amount of walks in ${\mathcal{W}}_{8}$ having net voltage [math] (after evaluating $r$ and $s$ in $\mathbb{Z}_{2k}$) starting at $u$ and those starting at $v$ must be the same. It follows that, for some $\nu\in N$ with $|{\mathcal{W}}_{8}^{v}(\nu)|>|{\mathcal{W}}_{8}^{u}(\nu)|$, the equation $\nu\equiv 0$ (mod $2k$) holds when we evaluate $r$ and $s$ in $\mathbb{Z}_{2k}$. We thus see that at least one expression in VIII–XIII is congruent to [math] modulo $2k$. In fact, if $\Gamma$ is vertex-transitive, then at most one of these expressions can be congruent to [math].

Lemma 13**.**

If $\Gamma$ is $4$-vertex-regular and $8$-cycle-regular, then exactly one of the following equations modulo $2k$ holds:

[TABLE]

Proof.

We will slightly abuse notation and, for a label $x\in\{\textrm{I},\textrm{II},...,\textrm{XIII}\}$, we will refer by $x$ to the congruence equation modulo $2k$ obtained by making the expression labelled $x$ in Table 2 congruent to [math]. For instance, the equation $2r\equiv 0$ will be referred to as II.

First note that if II or III holds then $r=k$ or $r=k$, contradicting Lemma 10. If V holds then $(v_{0},w_{r},v_{0})$ is a $2$-cycle in $\Gamma$, which is not possible. Similarly, XII implies the existence of a $4$-cycle in $\Gamma$, contradicting Lemma 12. Now, if XIII holds, then either $2r-2s\equiv 0$, which is not possible, or $2r-2s+k\equiv 0$.

If VI holds, then neither X nor XI can hold as this would imply $2r-2s\equiv 0$ (subtracting IV from X or from XI, respectively). Therefore, IV excludes X and XI. If IV and I are the only equations to hold, we would have $6$ more elements of ${\mathcal{W}}_{8}$ through $u$ than through $v$. This means that if IV holds then necessarily VII, IX and XII also hold. However, XII implies $2r-2s+k\equiv 0$ and subtracting this from VIII we get $r=0$, a contradiction. Hence, VII, IX and XII cannot all hold at the same time and so IV can never hold.

Suppose VI holds, then one of the equations in $\{\textrm{VIII},\textrm{IX},\textrm{X},\textrm{XI},\textrm{XIII}\}$ must also hold. Note that VI and VIII imply III; VI and IX imply V; VI and X imply $r=k$; VI and XIII imply $s=k$. Therefore only XI can hold. But VI and XI imply VII, and so we would have $40$ elements of ${\mathcal{W}}_{8}$ starting at $u$ but only $36$ starting at $v$. This would contradict vertex-transitivity. Thus VI cannot hold. An analogous reasoning, where the roles of $r$ and $s$ are interchanged, shows that VII cannot hold. We have shown that the only equations that can hold are in $\{\textrm{VIII},\textrm{IX},\textrm{X},\textrm{XI},\textrm{XIII}\}$. However, if two or more of these equations hold, then we would have more elements of ${\mathcal{W}}_{8}$ through $v$ than we would have through $u$. It follows that exactly one equation in $\{\textrm{VIII},\textrm{IX},\textrm{X},\textrm{XI},\textrm{XIII}\}$ holds. ∎

Lemma 13 tells us, for each of the $5$ possible equations, exactly which walks in ${\mathcal{W}}_{8}$ will lift to $8$-cycles and so we can count exactly how many $8$-cycles pass through a given vertex or edge of $\Gamma$. For instance, if $3s-2r+k\equiv 0$, then there are $16$ walks in ${\mathcal{W}}_{8}$ through $u$ that will lift to an $8$-cycle (see Table 2). Since a closed walk in ${\mathcal{W}}_{8}$ and its inverse lift to the same cycle in $\Gamma$, we see that there are $8$ cycles of length $8$ through every vertex in $\Gamma$. Moreover, there are exactly $5$ cycles of length $8$ through every edge of type $R$ or [math] while there are $6$ such cycles through each edge of type $K$ or $S$. It follows that the $8$-signature of a vertex $x$ of $\Gamma$ is $(5,5,6)$ whenever $3s-2r+k\equiv 0$ and thus, in this case, $\Gamma$ is $8$-cycle-regular with $8$-signature $(5,5,6)$. Table 3 shows the number of $8$-cycles through each vertex and through each edge, depending on its type, for each of the $5$ cases described in Lemma 13.

We will now show that equations 4.10, 4.11 and 4.12 cannot hold when $\Gamma$ is vertex-transitive. This will be proved in Lemmas 15, 16 and 17. But first we need to show a result about the subgraph induced by the [math]- and $R$-edges.

Recall that that $\Gamma=T_{1}(k,r,s)$ with $k\geq 9$ and that it is connected. Henceforth we also assume that $\Gamma$ is vertex-transitive. Denote by $\Gamma_{0,R}$ the subgraph that results from deleting the edges of type $S$ and $K$ from $\Gamma$. Equivalently, $\Gamma_{0,R}$ is the subgraph induced by [math]- and $R$-edges. We see that $\Gamma_{0,R}$ is $2$-valent and that it has $\gcd(2k,r)$ connected components, each of which is a cycle of length $6k/\gcd(2k,r)$.

Since $\Gamma_{0,R}$ only has edges of type [math] and $R$, any two vertices in the same connected components must have indices that differ in a multiple of $r$. It is not hard to see that, indeed, each connected component consist precisely of all those vertices whose indices are congruent modulo $\gcd(2k,r)$. We now prove an auxiliary result that will used both in the case where (4.10) as well as in the case where (4.8) or (4.9) holds.

Suppose that no automorphism of $\Gamma$ maps an $S$-edge to a [math]-edge. Since $\Gamma$ is vertex-transitive, there is an automorphism that maps a vertex from $V$ to a vertex in $U$. Such an automorphism then maps an $S$-edge to a $K$-edge. Since all $S$-edges as well as all the $K$-edges are in the same orbit, this then implies that $E_{S}\cup E_{K}$ forms a single edge-orbit of $\hbox{\rm Aut}(\Gamma)$.

Lemma 14**.**

Suppose that no automorphism of $\Gamma$ maps an $S$-edge to a [math]-edge, then $\Gamma_{0,R}$ is disconnected, and the set of connected components induce a block system for the vertex-set of $\Gamma$.

Proof.

By the discussion above the lemma, $E_{S}\cup E_{K}$ forms a single edge-orbit in $\Gamma$. Now note that $\hbox{\rm Aut}(\Gamma)$ also acts as a group of automorphism on $\Gamma_{0,R}$ and that this action is transitive, since the edges removed consist of an edge-orbit of $\Gamma$. It follows that connected components of $\Gamma_{0,R}$ form a block system for the vertex set of $\Gamma$. Now, suppose $\Gamma_{0,R}$ consists of a single cycle of length $6k$, $(w_{0},u_{0},v_{0},w_{r},u_{r},v_{r}\dots v_{(2k-1)r})$. Notice that the vertex antipodal to $u_{0}$ in this cycle is $u_{kr}$ which is the same as $u_{k}$, as $r$ is odd. We see that edges of type $K$ in $\Gamma$ join vertices that are antipodal in $\Gamma_{0,R}$, while $S$-edges do not. Since this $6k$-cycle is a block of imprimitivity, $K$-edges can only be mapped into $K$-edges and thus $U$ is an orbit of $\hbox{\rm Aut}(\Gamma)$, contradicting the assumption that $\Gamma$ is vertex-transitive. Hence, $\Gamma_{0,R}$ is disconnected. ∎

Lemma 15**.**

If $3r-s\equiv 0$ $(\hbox{\rm{mod }}2k)$, $k\geq 9$, then $T_{1}(k,r,s)$ is not vertex-transitive.

Proof.

Let $\Gamma=T_{1}(k,r,s)$ where $3r-s\equiv 0$ $(\hbox{\rm{mod }}2k)$. A computer assisted counting shows that $S$-edges and $K$-edge have exactly $4$ distinct $8$-cycles passing through them, while [math]- and $R$-edges have $6$ such cycles (see Table 3). It follows that any automorphism sending a vertex $u\in U$ to a vertex $v\in V$ must necessarily map the $K$-edge incident to $u$ into the $S$-edge incident to $v$. Thus $E_{S}\cup E_{K}$ is an orbit of edges under the action of $\hbox{\rm Aut}(\Gamma)$.

Since we are assuming that $3r-s\equiv 0$ we get that any number dividing $k$ and $r$ must also divide $s$, but since $\gcd(k,r,s)=1$, we have that $\gcd(k,r)=1$ and then $\gcd(2k,r)\in\{1,2\}$. In light of Lemma 14, $\Gamma_{0,R}$ is disconnected, implying $\gcd(2k,r)\neq 1$ and therefore $\gcd(2k,r)=2$. From the congruence $3r-s\equiv 0$ we also get that $r$ and $s$ must have the same parity thus making $s$ even and $k$ odd. Recall that each connected component of $\Gamma_{0,R}$ consists of all the vertices whose indices are congruent modulo $\gcd(2k,r)$, so $\Gamma_{0,R}$ has two connected components: one containing all the vertices with even index, and the other containing those with odd index. Since $s$ is even and $k$ is odd, $S$-edges in $\Gamma$ join vertices with same parity, while $K$-edges join vertices with distinct parity. We know that each of the two connected components of $\Gamma_{0,R}$ is a block of imprimitivity of $\Gamma$, an so any automorphism of $\Gamma$ must either preserve the parity of all indices, or of none at all. It follows that no automorphism can send $S$-edges into $K$-edges, and therefore no automorphism can send a vertex in $V$ to a vertex in $U$. We conclude $\Gamma$ cannot be vertex-transitive. ∎

Lemma 16**.**

If $3s-r\equiv 0$ $(\hbox{\rm{mod }}2k)$, $k\geq 9$, then $T_{1}(k,r,s)$ is not vertex-transitive.

Proof.

Set $\Gamma=T_{1}(k,r,s)$ and $\Gamma^{\prime}=T_{1}(k,r^{\prime},s^{\prime})$, where $r^{\prime}=s$ and $s^{\prime}=r$. Notice that the triplet $(k,r^{\prime},s^{\prime})$ satisfies equation 4.10 and so $\Gamma^{\prime}$ cannot be vertex-transitive. By observation 4.6, $\Gamma\cong\Gamma^{\prime}$. Therefore $\Gamma$ is not vertex-transitive. ∎

Lemma 17**.**

If $4r-4s\equiv 0$ $(\hbox{\rm{mod }}2k)$, $k\geq 9$, then $T_{1}(k,r,s)$ is not vertex-transitive.

Proof.

Let $\Gamma=T_{1}(k,r,s)$, with $4r-4s\equiv 0$ $(\hbox{\rm{mod }}2k)$ and $k\geq 9$. From the congruence $4r-4s\equiv 0$, we see that either $2r-2s\equiv 0$ or $2r-2s+k\equiv 0$, but the former contradicts Lemma 12. Hence $2r-2s+k\equiv 0$ and the closed walk $((uv)_{0},(vw)_{r},(wv)_{-s},(vw)_{r},(wv)_{-s},(vu)_{0},(uu)_{k})$ in $\Delta_{1}$ lifts to a $7$-cycle in $\Gamma$. We will show that $\Gamma$ cannot be $7$-vertex-regular. Following a similar procedure as the one used to obtain Table 2, we have counted all closed walks of length $7$ in $\Delta_{1}$ starting at $u$ and those starting at $v$, and we have computed their net voltage. The result is displayed in Table 4.

It is plain to see that if, in addition to I, any one of the voltages in rows III–VI is congruent to [math], then either $r=0$, $r=k$, $s=0$ or $s=k$, contradicting Lemmas 10 or 11. We see that only the voltages in row I or II can hold, and thus $\Gamma$ is not $7$-vertex-regular and hence cannot be vertex-transitive. ∎

In what follow we deal with the case where (4.8) holds, that is, when $3s-2r+k\equiv 0$ (mod $2k$) and $k\geq 9$. From Table 3 we see that $S$-edges and $K$-edge have exactly $6$ distinct $8$-cycles passing through them, while [math]- and $R$-edges have only $5$. In particular, no automorphism of $\Gamma$ maps an $S$-edge to a [math], implying that $E_{S}\cup E_{K}$ is an edge-orbit of $\Gamma$ (see the discussion above Lemma 14).

Lemma 18**.**

Suppose that $3s-2r+k\equiv 0$ (mod $2k$), $k\geq 9$. If $e$ is an edge in $E_{K}\cup E_{S}$, then the endpoints of $e$ belong to different connected components of $\Gamma_{0,R}$.

Proof.

Suppose a $K$-edge $e$ has both its endpoints in the same connected component, $C$, of $\Gamma_{0,R}$. Then, because $\Gamma$ is vertex-transitive and $C$ is a block of imprimitivity, both of the endpoints of any $K$-edge must belong to the same connected component of $\Gamma_{0,R}$. This means that the subgraph induced by [math]-edges, $R$-edges and $K$-edges is disconnected. Recall that $K$-edges and $S$-edges belong to a single edge-orbit in $\Gamma$ and thus, for every $S$-edge, $e^{\prime}$, there exists $\phi\in\hbox{\rm Aut}(\Gamma)$ such $\phi(e)=e^{\prime}$. Since $\phi$ also acts as an automorphism on $\Gamma_{0,R}$, it follows that the endpoints of $\phi(e)$ are contained in the component $\phi(C)$, thus making $\Gamma$ a disconnected graph, which is contradiction. ∎

Lemma 19**.**

Suppose that $3s-2r+k\equiv 0\>(\hbox{\rm{mod }}2k)$, $k\geq 9$. Then the subgraph $\Gamma_{0,R}$ has an even number of connected components.

Proof.

Let $e$ be a $K$-edge whose endpoints are in two different connected components $C_{1}$ and $C_{2}$. Let $\rho^{r}\in\hbox{\rm Aut}(\Gamma)$ be an "$r$-fold" rotation. That is, $\rho^{r}$ adds $r$ to the index of every vertex. It is plain to see that $\rho^{r}$ fixes the connected components of $\Gamma_{0,R}$ set-wise and that it send $K$-edges into $K$-edges. Moreover, since it is transitive on $U\cap C_{1}$, we have that all $K$-edges having an endpoint in $C_{1}$ will have the other endpoint in $C_{2}$. Hence $K$-edges ’pair up’ connected components of $\Gamma_{0,R}$. The result follows ∎

Proposition 20**.**

If $T_{1}(k,r,s)$ is connected and vertex-transitive, then the following statements (or the analogous three statements obtained by interchagning $r$ and $s$) hold:

1. (1)

$3s-2r+k\equiv 0$* $(\hbox{\rm{mod }}2k)$* 2. (2)

$k$* and $s$ are odd and $\gcd(k,s)=1$* 3. (3)

$r$* is even and $\gcd(k,r)=\{1,3\}$*

Proof.

First, note that item (1) follows from the combination of Lemma 13 and Lemmas 15, 16 and 17.

We will continue by proving (3). Set $d=\gcd(k,r)$, so that $r=dr^{\prime}$ and $k=dk^{\prime}$ for some integers $k^{\prime}$ and $r^{\prime}$. Since $3s-2r+k\equiv 0\>(\hbox{\rm{mod }}2k)$, we see that, for some odd integer $a$:

[TABLE]

By connectedness, $\gcd(k,r,s)=1$, and hence $\gcd(d,s)=1$ implying that $d$ divides $3$. Hence $d=\gcd(k,r)\in\{1,3\}$. Moreover, since $\Gamma_{0,R}$ has $\gcd(2k,r)$ connected components and this number must be even by Lemma 19, we see that $\gcd(2k,r)\in\{2,6\}$, and therefore $r$ is even.

Now, from the congruence $3s-2r+k\equiv 0\>(\hbox{\rm{mod }}2k)$ we see that $s$ and $k$ are either both even, or both odd. Since $\gcd(2k,r,s)=1$ and $r$ is even, $s$ must necessarily be odd and then $k$ is also odd. Set $d^{\prime}:=\gcd(k,s)$ so that $s=d^{\prime}s^{\prime}$ and $k=d^{\prime}k^{\prime}$ for some integers $s^{\prime}$ and $k^{\prime}$. Again, from equation (4.8) we obtain

[TABLE]

for some odd integer $a$. Thus $d^{\prime}\in\{1,2\}$, but since $s$ is odd, we see that $\gcd(k,s)=1$. This proves item (2) and completes the proof. ∎

Remark 21**.**

Conditions (1)–(3) (or their analogues) in Proposition 20 are also sufficient to prove vertex-transitivity of a connected $T_{1}(k,r,s)$. This will be shown after the proof on Lemma 23. Moreover, observe that if $k$ and $s$ satisfy condition (2) of Lemma 20, then $\gcd(2k,1)=1$ and $T_{1}(k,r,s)$ is isomorphic to $T_{1}(k,rs^{-1},1)$ where $s^{-1}$ is the multiplicative inverse of $s$ in $\mathbb{Z}_{2k}$. Hence, in what follows we may assume that $s=1$.

Lemma 22**.**

Let $k$ be an odd number and set $s=1$. There exists a unique element $r\in\mathbb{Z}_{2k}$ such that conditions (1) and (3) of Proposition 20 are satisfied. This unique $r$ equals $r^{*}$ from Definition 7.

Proof.

Suppose that such an $r$ exists. From (1), we get $r=(3-ak)/2$ for some odd integer $a$. Since $r$ must be a positive integer smaller than $2k$, it follows that $a\in\{-1,-3\}$. Hence $(3+k)/2$ and $(3+3k)/2=(3+k)/2+k$ are the only possible candidates for $r$. Since $k$ is odd, the integers $(3+k)/2$ and $(3+k)/2+k$ have different parity. Set $r$ to be whichever one of these two expressions is even. Note that any odd number that divides both $r$ and $k$ must divide $3$ and that this is true whether $r$ equals $(3+k)/2$ or $(3+k)/2+k$. Thus conditions (3) is satisfied and $r$ equals $r^{*}$ from Definition 7. ∎

We have just proved that once an odd $k\geq 9$ is prescribed, then there is at most one graph $T_{1}(k,r,1)$ that is connected and vertex-transitive. The following lemma generalizes this to an arbitrary value of $s$.

Lemma 23**.**

Let $k\geq 9$ be an odd integer and let $T_{1}(k,r,s)$ be vertex-transitive. Then $T_{1}(k,r,s)\cong\textup{X}(k)$.

Proof.

Recall that $\textup{X}(k)\cong T_{1}(k,r^{*},1)$, where $r^{*}$ is as in Definition 7. Since $T_{1}(k,r,s)$ is connected and vertex-transitive, conditions (1)–(3) of Proposition 20 hold. In particular, $s$ is relatively prime to $2k$. Then $T_{1}(k,r,s)\cong T_{1}(k,s^{-1}r,1)$, where $s^{-1}$ is the multiplicative inverse of $s$ in $\mathbb{Z}_{2k}$. By Lemma 22, $T_{1}(k,s^{-1}r,1)=T_{1}(k,r^{*},1)$. ∎

Notice that in the previous proof vertex-transitivity is only used to ensure that the graph $T_{1}(k,r,s)$ satisfies conditions (1)–(3) of Proposition 20, and from there $T_{1}(k,r,s)$ is shown to be isomorphic to $T_{1}(k,r^{*},1)$. Since we know from Lemma 9 that $T_{1}(k,r^{*},1)$ is vertex-transitive, we have that conditions (1)–(3) of Proposition 20 are not only necessary, but also sufficient for vertex-transitivity, and thus Proposition 20 may be regarded a characterization theorem.

Now that we have a chracterization for vertex-transitivity, we would like to know when a cubic vertex-transitive tricirculant of Type $1$ is also a bicirculant. Note that the only cubic vertex-transitive circulants are prism and Möbius ladders, which have girth $4$. It follows from Lemma 12 that no cubic vertex-transitive tricirculant of Type $1$ is a circulant.

Lemma 24**.**

Let $\Gamma=T_{1}(k,r,s)$ be connected and vertex-transitive. If $3\nmid k$, then $\Gamma$ is a bicirculant.

Proof.

Let $\varphi:V(\Gamma)\to V(\Gamma)$ be the mapping given by:

[TABLE]

It can be readily seen that $\varphi$ is indeed a graph automorphism. Moreover, since $r$ is even and both $k$ and $s$ are odd, $\varphi$ preserves the parity of the index of all vertices. In fact, it is plain to see that $\varphi$ is the product of the following two disjoint permutation cycles $\phi_{1}$ and $\phi_{2}$ of length $3k$:

[TABLE]

Hence, $\Gamma$ is a bicirculant. ∎

Corollary 25**.**

If $T_{1}(k,r,s)$ is vertex-transitive with $3\nmid k$, then $T_{1}(k,r,s)\cong\hbox{{\rm GP}}(3k,k+(-1)^{\alpha})$ where $\alpha\in\{1,2\}$ and $\alpha\equiv k$ (mod $3$).

Proof.

Let $T_{1}(k,r,s)$ be vertex-transitive and suppose $k\equiv 1$ (mod $3$). Let $\varphi=\phi_{0}\phi_{1}$ the automorphism described in the proof of Lemma 24. We will show that $u_{k}$ is adjacent to $\varphi^{(k+1)}(u_{k})$. Notice that $\varphi^{(k+1)}(u_{k})=v_{k+2s-r+\frac{k-1}{3}r}$, but $k+2s-r\equiv r-s$ so we may write $\varphi^{(k+1)}(u_{k})=v_{r-s+\frac{k-1}{3}r}$. Recall that $\gcd(2k,3)=1$ and $r$ is even. From the congruence $k+2r-3s\equiv 0$ we see that:

[TABLE]

So that $\varphi^{(k+1)}(u_{k})=v_{k}$, which is adjacent to $u_{k}$. The case when $k\equiv 2$ follows from a similar argument. ∎

We have now completed the proof of Theorem 8. The characterization of vertex-transitivity for cubic tricirculants of Type $1$ with at least $54$ vertices follows from Lemmas 9 and 23. That a vertex-transitive tricirculant of Type $1$ is a Generalized Petersen graph when $3\nmid k$ follows from Lemma 24 and Corollary 25.

5. Type 2

Let $k\geq 9$ be an integer, and let $r,s\in\mathbb{Z}_{2k}$. Define $T_{2}(k,r,s)$ as the derived graph of $\Delta_{2}$ with the normalized voltage assignment for $\mathbb{Z}_{2k}$ shown in Figure 5.1.

Then, with the same notation as in Section 4, $U=\{u_{0},u_{1},...,u_{2k-1}\}$, $V=\{v_{0},v_{1},...,v_{2k-1}\}$ and $W=\{w_{0},w_{1},...,w_{2k-1}\}$ are the respective fibres of vertices $u$, $v$ and $w$ in $\Delta_{2}$. The set of edges of $T_{2}(k,r,s)$ can be expressed as the union $E_{K}\cup E_{R}\cup E_{S}\cup E_{0}$ where:

[TABLE]

Similarly as with Type $1$ cubic tricirculants, we see that every cubic tricirculant of Type $2$ is isomorphic to $T_{2}(k,r,s)$ for an appropriate choice of $k$ and $r,s\in\mathbb{Z}_{2k}$.

Definition 26**.**

For an odd positive integer $k$, let $\textup{Y}(k)=T_{2}(k,2,1)$.

Theorem 27**.**

Let $\Gamma$ be a connected cubic tricirculant of Type 2 with at least $54$ vertices. Then $\Gamma$ is vertex-transitive if and only if it is isomorphic to $\textup{Y}(k)$ for some odd integer $k$, $k\geq 9$. In this case, $\Gamma$ is a bicirculant if $3\nmid k$. Furthermore, $\Gamma$ is a map on the torus of type $\{6,3\}_{\alpha,3}$ where $\alpha=\frac{1}{2}(k-3)$.

Remark 28**.**

Note that $T_{2}(k,r,s)$ and $T_{2}(k,r,-s)$ are in fact the exact same graph. Then, we can safely assume $s<k$.

Lemma 29**.**

For an odd integer $k$, $k\geq 9$, $\textup{Y}(k)$ is vertex-transitive.

Proof.

Since $\textup{Y}=T_{2}(k,r,s)$, it suffices to provide an automorphism of $T_{2}(k,2,1)$ that mixes the sets $U$, $V$ and $W$. Let $\varphi$ be the mapping given by:

[TABLE]

Now consider a vertex $u_{i}\in U$ with $i$ even. Observe that $\varphi$ maps $u_{i}$ to $v_{i+1}$ and that the neighbourhood $N(u_{i}):=\{v_{i},w_{i},w_{i+2}\}$ is mapped to $\{u_{i+1},v_{i},v_{i+2}\}$, which is precisely the neighbourhood of $v_{1+i}$. That is, $\varphi$ maps the neighbourhood of any vertex $u_{i}$ into the neighbourhood of its image, when $i$ even. The reader can verify the remaining cases and see that $\varphi$ is indeed a graph automorphism. ∎

For the rest of this section, let $k\geq 9$ be an integer, let $r,s\in\mathbb{Z}_{2k}$ and suppose $\Gamma=T_{2}(k,r,s)$ is vertex-transitive. In what follows we will show that $\Gamma$ is isomorphic to $\textup{Y}(k)$. As with cubic tricirculants of Type 1, the strategy will be to count reduced closed walks in the quotient $\Delta_{2}$. Observe that the walk $((wu)_{0},(uw)_{r},(ww)_{k},(wu)_{-r},(uw)_{0},(ww)_{k})$ starting at $w$ is a closed walk of length $6$ having net voltage [math], regardless of what $r$ and $s$ evaluate to in $\mathbb{Z}_{2k}$. As was done with graphs of Type 1, we computed all closed walks in $\Delta_{2}$ along with their net voltages. Again, it would be convenient if closed walks of length $6$ with net voltage [math] always lifted into $6$-cycles. For this it suffices to show that the girth of $\Gamma$ is at least $4$.

Lemma 30**.**

$\Gamma$* has no cycles of length $3$.*

Proof.

Suppose to the contrary, that $\Gamma$ contains a triangle $T$. By observing Figure 5.1 we know that all three vertices of $T$ belong to $V$ or they belong to $U\cup W$. Since $\Gamma$ is vertex-transitive, there is at least one triangle through every vertex in $W$. We can thus assume without loss of generality that all three vertices of $T$ are in $U\cup W$. This implies that $r=k$ and further, that there are $2$ distinct triangles through every vertex in $W$: the lifts of $((wu)_{0},(uw)_{r},(ww)_{k})$ and $((wu)_{-r},(uw)_{0},(ww)_{k})$. However, since the subgraph induced by $V$ is $2$-valent, there is at most one triangle through every vertex in $V$, contradicting the vertex-transitivity of $\Gamma$. ∎

Note that since $\Gamma$ is simple, $s\neq k$ and if $r=k$, $\Gamma$ would contain a triangle. Thus

Corollary 31**.**

Neither $r$ nor $s$ equals $k$.

Now, let ${\mathcal{W}}_{6}$ be set of all walks of length $6$ in $\Delta_{2}$ and let $N$ be the set of net voltages of elements of ${\mathcal{W}}_{6}$, expressed in terms of $k$, $r$ and $s$. It follows from Lemma 30 that walks in ${\mathcal{W}}_{6}$ with net voltage [math] will lift into $6$-cycles, and not just closed walks. Table 5 shows, for each element $n$ of $N$, how many closed walks with net voltage $n$ start at each vertex of $\Delta_{2}$.

From row I of Table 5 we see that there are at least $4$ walks $\omega\in{\mathcal{W}}_{6}$ starting at $w$. Therefore, at least one expression from rows II–IX must be congruent to [math]. However, a careful inspection of Table 5 shows that if neither of the expressions in rows VIII and IX are congruent to [math], then there are at least $2$ more walks in ${\mathcal{W}}_{6}$ with net voltage [math] starting at $w$ than there are those starting at $v$. This means that if $\Gamma$ is vertex-transitive, then one of the two following equations modulo $2k$ hold:

[TABLE]

Suppose 5.6 holds. By Remark 28 we can in fact write $r=2s$. Then the mapping $\phi:T_{2}(k,2s,s)\to T_{2}(k,-2s,-s)$ given by $x_{i}\mapsto x_{-i}$, for all $x\in U\cup V\cup W$, is a graph isomorphism. But, again by Remark 28, $T_{2}(k,-2s,-s)=T_{2}(k,-2s,s)$ so that any graph $T_{2}(k,r,s)$ with $r\equiv 2s$ is isomorphic to a graph $T_{2}(k,r^{\prime},s)$ satisfying $r^{\prime}\equiv-2s$. We can therefore limit our analysis to the case when 5.6 holds.

Let $\Gamma=T_{2}(k,r,s)$ be connected and vertex-transitive, and suppose $r\equiv 2s$. Note that any number dividing both $k$ and $s$ must also divide $r$, and since by connectedness of $\Gamma$ $\gcd(k,r,s)=1$ we see that $\gcd(k,s)=1$ and $\gcd(2k,s)\in\{1,2\}$. We will show that $\gcd(2k,s)$ cannot be $2$.

Lemma 32**.**

If $\Gamma$ is vertex-transitive, $\gcd(2k,s)=1$ and $\gcd(2k,r)=2$.

Proof.

In order to get a contradiction, suppose $\gcd(2k,s)=2$. Then $s$ is even and $k$ is odd, as $\gcd(k,s)=1$. Furthermore, the $2$-valent subgraph of $\Gamma$ induced by the set $V$ is the union of two $k$-cycles. We will show that these are in fact the only two $k$-cycles in $\Gamma$ which will imply that $\Gamma$ cannot be vertex-transitive. Suppose there is a $k$-cycle $C^{\prime}$ that visits a vertex not in $V$. Define $\Gamma[U\cup W]$ as the subgraph of $\Gamma$ induced by the set $U\cup W$. Since $r$ is even and $k$ is odd, it is not difficult to see that $\Gamma[U\cup W]$ is bipartite, with sets \{w_{i}:\text{i is even}\}\cup\{u_{i}:\text{i is odd}\} and \{w_{i}:\text{i is odd}\}\cup\{u_{i}:\text{i is even}\}. Therefore no $k$-cycle of $\Gamma$ is contained in $\Gamma[U\cup W]$. This means that $C^{\prime}$ visits at least one vertex in each of sets $U$, $V$ and $W$.

Now, $C^{\prime}$ projects onto a closed walk $C$ in $\Delta_{2}$ that has net voltage [math]. Define $x_{S}$ as the number of times $C$ traces the dart $(vv)_{s}$ minus the number of times it traces $(vv)_{-s}$, so that $x_{S}s$ is the total voltage contributed to $C$ by darts in $\{(vv)_{\pm s}\}$. Define $x_{R}$ similarly, and define $x_{K}$ as the number of times $C$ traces the semi-edge $(ww)_{k}$. We obtain the following congruence modulo $2k$:

[TABLE]

Recall that both $r$ and $s$ are even, making both $x_{R}r$ and $x_{S}s$ even. It follows that $x_{K}k$ is even, but since $k$ is odd, $x_{K}$ must be even, and hence $x_{K}k\equiv 0$ $(\hbox{\rm{mod }}2k)$. We thus have

[TABLE]

and since $r\equiv 2s$ $(\hbox{\rm{mod }}2k)$,

[TABLE]

From this, and because $\gcd(k,s)=1$, we see that $2x_{R}+x_{S}=0$ or $2x_{R}+x_{S}\geq k$. To see that $2x_{R}+x_{S}$ cannot equal [math], define $A$ as the set of darts in $\Delta_{2}$ with voltage [math] or $r$ and observe that $C$ must visit $A$ an even number of times. Since $C$ has odd length, $C$ visits $\{(vv)_{\pm s}\}\cup\{(ww)_{k}\}$ an odd amount of times. Then, since $x_{K}$ is even, $x_{S}$ must be odd. But $2x_{R}$ is even so $2x_{R}+x_{S}\neq 0$. We thus have

[TABLE]

Now, notice that if $C$ traces the dart $(uw)_{r}$, then it must immediately trace a dart in $\{(wu)_{0},(ww)_{k}\}$. This means that $C$ traces a dart in the subgraph of $\Delta_{2}$ induced by $U\cup W$ at least $2x_{R}$ times. Then $2x_{R}+x_{S}<k$, since $C$ has length $k$ and it must visit $(vu)_{0}$ at least once. We thus have $k>2x_{R}+x_{S}\geq k$, a contradiction. The result follows. ∎

Lemma 33**.**

If $T_{2}(k,r,s)$ is vertex-transitive, $k$ is odd.

Proof.

Suppose to the contrary that $k$ is even. Since $s$ is odd, we have $sk\equiv k$ $(\hbox{\rm{mod }}2k)$, and thus $s(2\cdot\frac{1}{2})k\equiv k$. But $k$ is even, so we can rewrite that expression as $2s\cdot\frac{k}{2}\equiv k$. Recall that $r\equiv 2s$ and then $r\frac{k}{2}\equiv k$.

Now, consider the walk of length $k+1$ in $\Delta_{2}$ that starts in $w$, traces the semi-edge $(ww)_{k}$ once and then traces the $2$-path $((wu)_{r},(uw)_{0})$ exactly $\frac{k}{2}$ times. Note that this walk is closed and has net voltage $\frac{k}{2}r+k\equiv 0$. Furthermore, it is easy to see that it lifts into a $(k+1)$-cycle in $\Gamma$ and that it doesn’t visit any vertex in $V$.

Since $\Gamma$ is vertex-transitive, there must be a $(k+1)$-cycle $C^{\prime}$ through each vertex in $V$. Note that the subgraph of $\Gamma$ induced by $V$ is a single $2k$-cycle, and thus $C^{\prime}$ must visit all three set $U$, $V$ and $W$. It follows that $C^{\prime}$ projects into a walk $C$ of length $k+1$ having net voltage [math] and that it visits all three vertices of $\Delta_{2}$. Furthermore, since $k+1$ is odd, the number of times $C$ visits $\{(vv)_{\pm s}\}$ must be of different parity than the number of times it traces $(ww)_{k}$. Define $x_{S}$, $x_{R}$ and $x_{K}$ as in the proof of Lemma 19. Hence

[TABLE]

Now, $r$ and $k$ are even, and so $x_{R}r$ and $x_{K}k$ are also even. It follows that $x_{S}s$ is even, but since $s$ is odd, $s_{S}$ is necessarily even. This means $C$ visits $\{(vv)_{\pm s}\}$ an even amount of times. It follows that $C$ traces $(ww)_{k}$ and odd amount of times, that is, $x_{K}$ is odd. We thus have

[TABLE]

and since $r\equiv 2s$,

[TABLE]

This implies $2x_{R}+x_{S}\geq k$, as $s$ and $k$ are relatively prime. However, $2x_{R}+x_{R}\leq k-1$ since $C$ has length $k+1$ and $C$ visits the set $\{(vu)_{0},(uv)_{0}\}$ at least twice. This contradiction arises from the assumption that $k$ is even. We conclude that $k$ is odd. ∎

Lemma 34**.**

For an odd $k>9$, if $T_{2}(k,r,s)$ is vertex-transitive then $T_{2}(k,r,s)\cong T_{2}(k,2,1)$.

Proof.

Let $T_{2}(k,r,s)$ be vertex-transitive. Then, $k$ is an odd integer, $\gcd(2k,s)=1$ and $r\equiv 2s$. Denote by $s^{-1}$ the multiplicative inverse of $s$ in $\mathbb{Z}_{2k}$. Now, define $\varphi$ as the mapping between $T_{2}(k,2,1)$ and $T_{2}(k,r,s)$ given by $x_{i}\mapsto x_{s^{-1}i}$, for $x\in U\cup V\cup W$. It is plain to see that $\varphi$ is the desired isomorphism. ∎

This proves the first claim of Theorem 27. We now proceed to show sufficient conditions under which $\Gamma$ is a bicirculant and we show that $\Gamma$ can be seen as a map on the torus.

Lemma 35**.**

If $k\geq 9$ is an odd integer such that $3\nmid k$, then $T_{2}(k,2,1)$ is a bicirculant.

Proof.

Consider the mapping $\varphi$ defined by:

[TABLE]

We have that for a vertex $u_{i}\in U$, $\varphi^{l}(u_{i})\in U$ if and only if $3|l$. This is, if we start at $U$, every third iteration of $\varphi$ lands us in $U$ again. Moreover, we have $\varphi^{3}(u_{i})=u_{i+3+k}$. In general $\varphi^{3l}(u_{i})=u_{i+3l+lk}$. Hence $\varphi^{3l}(u_{i})=u_{i}$ if and only if $3l+lk\equiv 0$. If $3\nmid k$, then $l=k$ is the smallest value for $l$ that satisfies $3l+lk\equiv 0$. It follows that the orbit of $u_{i}$ has size $3k$. It is plain to see that the vertices not in this orbit form an orbit of size $3k$ on their own, namely, the orbit of $u_{i+1}$. It is worthwhile to mention that if $3$ does divide $k$, then $\varphi$ has $6$ orbits of size $k$. ∎

Proposition 36**.**

Let $k\geq 3$ be an odd integer, then $T_{2}(k,2,1)$ admits an embedding on the torus with hexagonal faces, yielding a map of type $\{6,3\}_{\alpha,3}$, where $\alpha=\frac{1}{2}(k-3)$ (see [4] for details about the notation).

Proof.

First, notice that vertices in $U\cup W$ with even index form a cycle of length $2k$, $C_{1}=(w_{0},u_{0},w_{2},u_{2},...,w_{2k-2},u_{2k-2})$. Likewise, vertices in $U\cup W$ with odd index form a $2k$-cycle, $C_{2}$. The vertices in $V$ form a third cycle of length $2k$, $C_{3}$.

Now, $w_{i}w_{i+k}\in E$ for all $i\in\mathbb{Z}_{2k}$, and $i+k$ has different parity than $i$. This means each vertex of $C_{1}$ of the form $w_{i}$ is adjacent, through a $K$-edge, to a vertex of $C_{2}$. Further, each vertex of the form $u_{i}$ in $C_{1}$ is adjacent to a vertex in $C_{3}$, namely $v_{i}$. Note that the vertices of $C_{2}$ that are not adjacent to a vertex of $C_{1}$ are precisely those of the form $u_{i}$ (with $i$ odd), and that $u_{i}v_{i}\in E$ for all odd $i$. This is, every other vertex in $C_{2}$ has a neighbour in $C_{3}$.

Therefore, we can think of $T_{2}(k,2,1)$ as three stacked $2k$-cycles $C_{1}$, $C_{2}$ and $C_{3}$ where every other vertex of $C_{i}$ has a neighbour in $C_{i-1}$ while each of the remaining vertices has a neighbour in $C_{i+1}$, where $i\pm 1$ is computed modulo $3$ (see Figure 5.2 for a detailed example). From here it is clear that $T_{2}(k,2,1)$ can be embedded on a torus, tessellating it with $3k$ hexagons. It can be readily verified that this wields the map $\{6,3\}_{\alpha,3}$, where $\alpha=\frac{1}{2}(k-3)$, following the notation in [4]. ∎

Observe that $T_{2}(9,2,1)$ corresponds to the map $\{6,3\}_{3,3}$, which is a regular map (see Chapter 8.4 of [4]). It follows that $T_{2}(9,2,1)$ is arc-transitive, making it one of the four possible cubic arc-transitive tricirculant graphs (called $F054A$ in [8], following Foster’s notation).

Theorem 27 now follows from Lemmas 29, 34, 35 and Proposition 36.

6. Type 3

Let $k\geq 9$ be an integer, and let $r\in\mathbb{Z}_{2k}$. Define $T_{3}(k,r)$ as the derived graph of $\Delta_{3}$ with the normalized voltage assignment for $\mathbb{Z}_{2k}$ shown in Figure 6.1

Let $U=\{u_{0},u_{1},...,u_{2k-1}\}$, $V=\{v_{0},v_{1},...,v_{2k-1}\}$ and $W=\{w_{0},w_{1},...,w_{2k-1}\}$ be the respective fibers of vertices $u$, $v$ and $w$ in $\Delta_{3}$. It is clear that any cubic tricirculant of Type $3$ is isomorphic to $T_{3}(k,r)$ for some $k$ and $r$.

Theorem 37**.**

If $T_{3}(k,r)$ is connected, then it is isomorphic to either a prism or a Möbius Ladder.

Proof.

First, recall that $T_{3}(k,r)$ is connected if and only if $\gcd(k,r)=1$. Then $\gcd(2k,r)\leavevmode\nobreak\ \in\leavevmode\nobreak\ \{1,2\}$, depending on whether $r$ is even or odd.

If $r$ is odd, then the subgraph induced by [math]- and $R$-edges is a single cycle of length $6k$, $(w_{0},u_{0},v_{0},w_{r},u_{r},v_{r},...,w_{2k-r},u_{2k-r},v_{2k-r},w_{0})$. It is straighforward to see that $K$-edges join antipodal vertices in this cycle. Hence, in this case $T_{3}(k,r)$ is a Möbius Ladder.

If $r$ is even, then the graph induced by [math]- and $R$-edges is the union of two disjoint cycles of length $3k$: one consisting of all the vertices with even index, and the other consisting on those with odd index. Observe that $K$-edges connect these two cycles creating a prism. ∎

7. Type 4

Let $k$ be a positive integer, and let $r$ and $s$ be two distinct integers in $\mathbb{Z}_{2k}$. Define $T_{4}(k,r,s)$ as the derived graph of $\Delta_{4}$ with the normalized voltage assignment for $\mathbb{Z}_{2k}$ shown in Figure 7.1.

Let $U=\{u_{0},u_{1},...,u_{2k-1}\}$, $V=\{v_{0},v_{1},...,v_{2k-1}\}$ and $W=\{w_{0},w_{1},...,w_{2k-1}\}$ be the respective fibers of vertices $u$, $v$ and $w$ in $\Delta_{4}$. Then, the set of edges of $T_{4}(k,r,s)$ can be expressed as the union $E_{K}\cup E_{R}\cup E_{S}\cup E_{0}$ where:

[TABLE]

For $X\in\{0,R,S,K\}$, edges in $E_{X}$ will be called edges of type $X$, or simply $X$-edges. Similarly as with tricirculants of Types $1$, $2$ and $3$, every cubic tricirculant of type $4$ with $6k$ vertices is isomorphic to $T_{4}(k,r,s)$ for an appropriate choice of $r$ and $s$.

Remark 38**.**

Note that for any $k$, $r$ and $s$ the following isomorphism holds:

[TABLE]

Theorem 39**.**

There are no cubic vertex-transitive tricirculants of Type $4$ of order greater or equal to $54$.

The rest of this section is devoted to prove Theorem 39. We will assume henceforth that $k\geq 9$ and thus the order of $T_{4}(k,r,s)$ is at least $54$.

Lemma 40**.**

If $T_{4}(k,r,s)$ is vertex-transitive, then $r\neq s$ and $r\neq-s$.

Proof.

Suppose that $r=s$ and consider the graph $\Gamma:=T_{4}(k,r,r)$, with $k\geq 9$ and $1\leq r\leq k-1$. Observe that $(v_{0},v_{r},u_{r},w_{r},w_{0},u_{0},v_{0})$ is a $6$-cycle of $\Gamma$ that does not contain any $K$-edge but does contain edges of all other types. For $\Gamma$ to be vertex-transitive, there must be at least one $6$-cycle through a $K$-edge; otherwise $K$-edges conform a single edge-orbit and thus $U$ is a single vertex-orbit. However, such a cycle would quotient down to a closed walk of length $6$ in $\Delta_{4}$ having voltage [math] and tracing the semi-edge $(uu)_{k}$. By observing Figure 7.1 we see that any closed walk of length $6$ visiting $(uu)_{k}$ has net voltage $k\pm 3r$. This is, if $\Gamma$ is vertex-transitive, then $3r\equiv k$ mod $2k$. Observe that under these conditions, $\Gamma$ is connected if and only if $r=1$ and thus making $k=3$, contradicting that $k\geq 9$. This shows that $r\neq s$ and view of Remark 38 then $r\neq-s$. ∎

Lemma 41**.**

Let $e$ be a $K$-edge of $T_{4}(k,r,s)$ and let $a$ be an integer greater than $4$. If $C$ is an cycle of length $a$ containing $e$ then there exists another cycle of length $a$, $C^{\prime}$, such that the intersection $C\cap C^{\prime}$ contains a $3$-path whose middle edge is $e$.

Proof.

Without loss of generality, we can assume $e=u_{0}u_{k}$. Define $H$ as the subgraph of $\Gamma$ induced by vertices at distance at most one from $e$, and let $C$ be a $k$-cycle through $e$ (see Figure 7.2). Observe that $C$ intersects $H$ in a $3$-path, $P$, whose middle edge is $e$. Now, let $\phi$ be the mapping that acts by multiplying the index of each vertex by $-1$; that is $\phi$ maps $u_{i}$ into $u_{-i}$, $v_{i}$ into $v_{-i}$, and $w_{i}$ to $w_{-i}$. Observe that $\phi$ is in fact an automorphism of $\Gamma$. Moreover, $\phi$ fixes each vertex and each edge of $H$. In particular, it fixes $P$. Therefore, $\phi(C)$ is a $k$-cycle through $e$ and $P$ lies in the intersection of $C$ and $\phi(C)$. To see that $\phi(C)$ is different from $C$, observe that $\phi$ interchanges the two vertices in each of the following sets: $\{v_{k+r},v_{k-r}\}$, $\{v_{r},v_{-r}\}$,$\{w_{k+s},w_{k-s}\}$ and $\{w_{s},w_{-s}\}$ (white vertices in Figure 7.2). It is clear that $C$ can visit at most one vertex in each of these four sets, and if $C$ visits one vertex in one of these sets, then $\phi(C)$ must visit the other. ∎

Lemma 42**.**

If $T_{4}(k,r,s)$ is vertex-transitive, then in $\mathbb{Z}_{2k}$ one of the equalities (A1)–(A4) below and one of the equalities (B1)–(B4) below hold:

Proof.

Suppose $T_{4}(k,r,s)$ is vertex-transitive. We will first show that one of the equalities (A1)–(A4) holds in $\mathbb{Z}_{2k}$. The rest of the lemma will follow from the fact that $T_{4}(k,r,s)\cong T_{4}(k,s,r)$. Since $T_{4}(k,r,s)$ is vertex-transitive, the edge-neighbourhood of any vertex in $U$ can be mapped by an automorphism to the edge-neighbourhood of any vertex in $V$. In particular, either there exists an automorphism mapping the $K$-edge $u_{0}u_{k}$ to the [math]-edge $u_{0}v_{0}$ or there is an automorphism mapping $u_{0}u_{k}$ to the $R$-edge $v_{0}v_{r}$. Therefore, the property described in Lemma 41 should also hold for $u_{0}v_{0}$ or for $v_{0}v_{r}$ (or possibly both). We will see what this means in terms of $k$, $r$ and $s$.

Suppose that the property described in Lemma 41 holds for $u_{0}v_{0}$. Observe that there is an $8$-cycle $C=(u_{0},v_{0},v_{-r},u_{-r},u_{k-r},v_{k-r},v_{k},u_{k},u_{0})$ through $u_{0}v_{0}$ (see Figure 7.3). Then there must be another $8$-cycle $C^{\prime}$ whose intersection with $C$ contains the $3$-path $(v_{-r},v_{0},u_{0},u_{k})$, but no $4$-path containing this $3$-path. This in turns implies that some vertex in $\{v_{-3r},u_{-2r}\}$ is adjacent to a vertex in $\{w_{k-s},w_{k+s}\}$. Since no vertex in $V$ is adjacent to a vertex in $W$, we have that either $u_{-2r}w_{k-s}\in E$ or $u_{-2r}w_{k+s}\in E$. This implies $2r\equiv k+s$ or $2r\equiv k-s$, and so (A1) or (A2) holds.

If, on the other hand, the property described in Lemma 41 holds for $v_{0}v_{r}$, then one vertex in $\{w_{r+s},w_{r-s}\}$ must be adjacent to a vertex in $\{w_{s},w_{-s}\}$ (see Figure 7.4), implying that one of the following expressions must be equal to [math] in $\mathbb{Z}_{2k}$: $r+3s$, $r+s$, $r-3s$, $r-s$. However, in view of Lemma 40, $r+s\neq 0$ and $r-s\neq 0$. Hence (A3) or (A4) holds. ∎

We are now ready to prove Theorem 39. Let $k\geq 9$ be an integer and suppose $T_{4}(k,r,s)$ is vertex-transitive. Then by Lemma 42, one of the equalities (A1)–(A4) and one of the equalities (B1)-(B4) must hold. If, for instance both (A1) and (B1) hold, then by adding them we see that $r=-s$, which contradicts Lemma 40. Similarly, we get that $r\pm s$ if both equalities in any of the following pair hold: (A1) and (B2), (A2) and (B2), (A3) and (B3). If (A1) and (B4), or (A2) and (B4) hold, then $k=r$, which contradicts the simplicity of $\Gamma$. If (A1) and (B3), (A2) and (B3), or (a3) and (B4) hold, then either $k=5$ or $\gcd(k,r,s)\neq 1$. It is readily seen that we get a contradiction in each of the $16$ possible cases that arise from Lemma 42. We conclude that a connected $T_{4}(k,r,s)$ cannot be vertex-transitive if $k\geq 9$.

Acknowledgements. The authors gratefully acknowledge support of the Slovenian Research Agency by financing the Research program P1-0294 and the second listed author Young Researcher scholarship.