Regular $t$-balanced Cayley maps on split metacyclic $2$-groups

TL;DR

This paper classifies all regular t-balanced Cayley maps on a specific class of split metacyclic 2-groups, advancing understanding of symmetric embeddings of Cayley graphs on these groups.

Contribution

It provides a complete classification of regular t-balanced Cayley maps for split metacyclic 2-groups, a previously uncharacterized class.

Findings

Complete classification achieved for the specified groups.

Identification of conditions for the existence of such Cayley maps.

Enhanced understanding of symmetric embeddings on split metacyclic 2-groups.

Abstract

A regular $t$-balanced Cayley map on a group $\Gamma$ is an embedding of a Cayley graph on $\Gamma$ into a surface with certain special symmetric properties. We completely classify regular $t$-balanced Cayley maps for a class of split metacyclic $2$-groups.

Equations

$$\rho ({\omega }^{-1})=({\rho }^t(\omega ){)}^{-1}\text{for all }\omega \in \Omega ;$$

$$\overline{\phi }(\overline{\eta }\overline{\mu })=\overline{\phi (\eta \mu )}=\overline{\phi (\eta ){\phi }^{\pi (\eta )}(\mu )}=\overline{\phi }(\overline{\eta }){\overline{\phi }}^{\overline{\pi }(\overline{\eta })}(\overline{\mu }).$$

$$\Lambda (n,m;r)=⟨\alpha ,\beta \mid {\alpha }^n={\beta }^m=1,\beta \alpha {\beta }^{-1}={\alpha }^r⟩,$$

$$\max \{2,a-b\}\le c\le a-3\text{and}b\ne c.$$

$${\beta }^y{\alpha }^x$$

$$({\alpha }^{x_1}{\beta }^{y_1})({\alpha }^{x_2}{\beta }^{y_2})$$

$$({\alpha }^x{\beta }^y{)}^u$$

$$[{\alpha }^{x_1}{\beta }^{y_1},{\alpha }^{x_2}{\beta }^{y_2}]$$

$${\Lambda }^{\mathrm{ab}}:=\Lambda /[\Lambda ,\Lambda ]\cong {\mathbb{Z}}_{(r-1,n)}\times {\mathbb{Z}}_m.$$

$${\kappa }_1(u,v)=u,{\kappa }_2(u,v)=v,{\kappa }_3(u,v)=u+v.$$

$$2$$

$$y_2$$

$${\sigma }_{x_2,y_2}^{x_1,y_1}({\alpha }^u{\beta }^v)={\alpha }^{x_1[u{]}_{r^{y_1}}+r^{y_{1}u}{x}_2[v{]}_{r^{y_2}}}{\beta }^{y_{1}u+y_{2}v}.$$

$$h_j=p_1[x_j{]}_{r^{q_1}}+r^{q_1{x}_j}{p}_2[y_j{]}_{r^{q_2}},j=1,2.$$

$$[x_j{]}_{r^{q_1}}$$

$$r^{q_1{x}_j}{p}_2$$

$$h_1{\equiv }^{(a)}{p}_1{x}_1(1+2^{c-1}{q}_1(x_1-1))+p_2{y}_1,h_2{\equiv }^{(a)}{p}_1{x}_2+p_2{y}_2.$$

$${\sigma }_{p_2,q_2}^{p_1,q_1}\circ {\sigma }_{x_2,y_2}^{x_1,y_1}={\sigma }_{p_1{x}_2+p_2{y}_2,q_1{x}_2+q_2{y}_2}^{p_1{x}_1(1+2^{c-1}{q}_1(x_1-1))+p_2{y}_1,q_1{x}_1+q_2{y}_1}.$$

$$|{\alpha }^x{\beta }^y|=2^{\max \{a-\Vert x\Vert ,b-\Vert y\Vert \}}.$$

$${\omega }_{\ell +ti}={\omega }_i^{-1},i=1,\dots ,d.$$

$${\Delta }_{+}$$

$${\omega }_i{\omega }_d^{-1}$$

$${\omega }_d^{-2}={\omega }_{\ell }{\omega }_d^{-1}={\eta }_{\ell }\cdots {\eta }_1.$$

$${\phi }_{+}({\eta }_j)={\eta }_{j+1}.$$

$$x_1^2+x_2{y}_1{\equiv }^{(c-1)}1,x_1+y_2{\equiv }^{\prime }{y}_2^2+x_2{y}_1-1{\equiv }^{\prime }0.$$

$${\phi }_{+}({\alpha }^2)=({\alpha }^2{)}^{x_1}(\alpha \beta {)}^{y_1},{\phi }_{+}(\alpha \beta )=({\alpha }^2{)}^{x_2}(\alpha \beta {)}^{y_2}.$$

$$1=\varphi_{+}\big{(}(\alpha^{2})^{2^{a-1}}\big{)}=((\alpha^{2})^{x_{1}}(\alpha\beta)^{y_{1}})^{2^{a-1}}=(\alpha^{2x_{1}+[y_{1}]_{1+2^{c}}}\beta^{y_{1}})^{2^{a-1}}$$

$$\Xi=\big{\langle}\alpha^{2^{c}},\beta^{2^{c}}\big{\rangle}=\big{\langle}\alpha^{2^{c}},(\alpha\beta)^{2^{c}}\big{\rangle}$$

$$\Vert y_1\Vert \ge b-c-2,\Vert x_2\Vert \ge a-b+1,\Vert y_2-1\Vert \ge a-c-2.$$

$$\overline{{\eta }_1}$$

$$\overline{{\eta }_1}+\overline{{\eta }_2}$$

$$c\ge b-2-\Vert y_1{u}_1+(y_2+1)v_1\Vert =b-3;$$

$$\overline{{\alpha }^2}$$

$$\overline{\beta }$$

Full text

Keywords: regular Cayley map; $t$-balanced; split metacyclic $2$-group; reduction.

MSC2020: 05C25, 05C10, 20B25.

1 Introduction

Suppose $\Gamma$ is a finite group and $\Omega$ is a generating set of $\Gamma$ such that $\omega^{-1}\in\Omega$ whenever $\omega\in\Omega$, and the identity $1\notin\Omega$. The Cayley graph ${\rm Cay}(\Gamma,\Omega)$ is the graph having the vertex set $\Gamma$ and the arc set $\Gamma\times\Omega$, where for $\eta\in\Gamma,\omega\in\Omega$, the arc from $\eta$ to $\eta\omega$ is denoted as $(\eta,\omega)$.

A cyclic permutation $\rho$ on $\Omega$ canonically induces a permutation on the arc set via $(\eta,\omega)\mapsto(\eta,\rho(\omega))$, and this equips each vertex $\eta$ with a “cyclic order”, which means a cyclic permutation on the set of arcs emanating from $\eta$. This determines an embedding of ${\rm Cay}(\Gamma,\Omega)$ into a closed oriented surface, which is characterized by the property that each connected component of the complement of the Cayley graph is a disk. Such an embedding is called a Cayley map and denoted by $\mathcal{CM}(\Gamma,\Omega,\rho)$. An isomorphism of Cayley maps $\mathcal{CM}(\Gamma,\Omega,\rho)\to\mathcal{CM}(\Gamma^{\prime},\Omega^{\prime},\rho^{\prime})$ is by definition an isomorphism ${\rm Cay}(\Gamma,\Omega)\to{\rm Cay}(\Gamma^{\prime},\Omega^{\prime})$ which can be extended to an orientation-preserving homeomorphism between their embedding surfaces.

A Cayley map is called regular if its automorphism group acts regularly on the arc set, i.e., for any two arcs, there exists an automorphism sending one arc to the other. It was shown in [10] that $\mathcal{CM}(\Gamma,\Omega,\rho)$ is regular if and only if there exist a skew-morphism which is a bijective function $\varphi:\Gamma\to\Gamma$, and a power function $\pi:\Gamma\to\{1,\ldots,\#\Omega\}$ (where $\#\Omega$ is the cardinality of $\Omega$), such that $\varphi|_{\Omega}=\rho$, $\varphi(1)=1$ and $\varphi(\eta\mu)=\varphi(\eta)\varphi^{\pi(\eta)}(\mu)$ for all $\eta,\mu\in\Gamma.$

Let $d=\#\Omega$, and $t$ be an integer with $t^{2}\equiv 1\pmod{d}$. A regular Cayley map $\mathcal{CM}(\Gamma,\Omega,\rho)$ is called $t$-balanced if

[TABLE]

in particular, it is called balanced if $t\equiv 1\pmod{d}$ and anti-balanced if $t\equiv-1\pmod{d}$. It is the residue modulo $d$ rather than $t$ itself, that plays a key role. From now on we assume $0<t<d$, and abbreviate “regular $t$-balanced Cayley map” to “RBCMt”.

Recall some facts on RBCMt from [1] Proposition 1.2.

Proposition 1.1**.**

(a)* A Cayley map $\mathcal{CM}(\Gamma,\Omega,\rho)$ is a RBCM1 if and only if $\rho$ can be extended to an automorphism of $\Gamma$.*

(b)* Suppose $t>1$. A Cayley map $\mathcal{CM}(\Gamma,\Omega,\rho)$ is a RBCMt if and only if $\rho$ can be extended to a skew-morphism of $\Gamma$, $\pi(\omega)=t$ for all $\omega\in\Omega$ and $\pi(\eta)\in\{1,t\}$ for all $\eta\in\Gamma$.*

(c)* When the conditions in (b) are satisfied, $\Gamma_{+}:=\{\eta\in\Gamma\colon\pi(\eta)=1\}$ is a subgroup of index $2$, consisting of elements which are products of an even number of generators, $\varphi(\Gamma_{+})=\Gamma_{+}$, and $\varphi_{+}:=\varphi|_{\Gamma_{+}}$ is an automorphism.*

By (1), there is an involution $\iota$ on $\{1,\ldots,d\}$ with $\omega_{i}^{-1}=\omega_{\iota(i)}$ and $\iota(i+1)\equiv\iota(i)+t\pmod{d}$ for all $i$. Let $\ell=\iota(d)$, then $\iota(i)\equiv\ell+ti\pmod{d}$, and the condition $\iota^{2}={\rm id}$ is equivalent to $(t+1)\ell\equiv 0\pmod{d}$, which together with $t^{2}\equiv 1\pmod{d}$ implies $(t-1,d)\mid 2\ell$. We say that the RBCMt has type I or type II if $(t-1,d)\nmid\ell$ or $(t-1,d)\mid\ell$, respectively.

Remark 1.2**.**

Observe that $(t-1,d)\mid\ell$ if and only if $\Omega$ contains an element of order 2, so RBCMt’s of different type cannot be isomorphic. On the other hand, according to Lemma 2.4 of [11], two RBCMt’s of the same type $\mathcal{CM}(\Gamma_{j},\Omega_{j},\rho_{j}),j=1,2$ are isomorphic if and only if there exists an isomorphism $\sigma:\Gamma_{1}\to\Gamma_{2}$ such that $\sigma(\Omega_{1})=\Omega_{2}$ and $\sigma\circ\rho_{1}=\rho_{2}\circ\sigma$.

When $\mathcal{CM}(\Gamma,\{\omega_{1},\ldots,\omega_{d}\},\rho)$ has type I (resp. type II), by re-indexing the $\omega_{i}$’s if necessary, we may assume $\ell=(t-1,d)/2$ (resp. $\ell=(t-1,d)$).

So far, people have completely classified RBCMt’s for the following classes of groups: dihedral groups (Kwak, Kwon and Feng [11], 2006), dicyclic groups (Kwak and Oh [12], 2008), semi-dihedral groups (Oh [14], 2009), cyclic groups (Kwon [13], 2013). In 2017 the first author [1] reduced the classification of RBCMt’s on abelian groups to a problem about polynomial rings, and gave a complete classification for RBCMt’s on abelian $2$-groups. In 2018 Yuan, Wang and Qu [15] classified RBCM1’s for the so-called minimal nonabelian metacyclic groups. For results on more general regular Cayley maps, see [2, 4, 6, 7, 9].

It is still challenging to study regular Cayley maps on nonabelian groups. We propose a “reduction method”, through which known results about RBCMt’s on simpler groups may be applicable. A key ingredient is the following observation.

Lemma 1.3**.**

Let $\mathcal{CM}(\Gamma,\Omega,\rho)$ be a RBCMt with skew-morphism $\varphi$. Suppose $\Xi$ is a normal subgroup of $\Gamma$ which is contained in $\Gamma_{+}$ and invariant under $\varphi_{+}$. Let $\overline{\Gamma}=\Gamma/\Xi$, and let $\overline{\Omega}$ denote the image of $\Omega$ under the quotient map $\Gamma\twoheadrightarrow\overline{\Gamma}$. Then $\rho$ induces a permutation $\overline{\rho}$ on $\overline{\Omega}$ and gives rise to a RBCMt $\mathcal{CM}(\overline{\Gamma},\overline{\Omega},\overline{\rho})$. Furthermore, if $\mathcal{CM}(\Gamma,\Omega,\rho)$ has type II, then so does $\mathcal{CM}(\overline{\Gamma},\overline{\Omega},\overline{\rho})$.

Proof.

For $\eta\in\Gamma$, let $\overline{\eta}$ denote its image under the quotient map $\Gamma\twoheadrightarrow\overline{\Gamma}$.

The map $\overline{\varphi}:\overline{\Gamma}\to\overline{\Gamma}$, $\overline{\eta}\mapsto\overline{\varphi(\eta)}$ is well-defined, as $\varphi(\xi\eta)=\varphi(\xi)\varphi(\eta)$ for any $\xi\in\Xi$. Let $\pi$ be the power function of $\mathcal{CM}(\Gamma,\Omega,\rho)$. It induces a function $\overline{\pi}:\overline{\Gamma}\twoheadrightarrow\{1,t\}$ in an obvious way. For all $\eta,\mu$, we have

[TABLE]

So $\rho=\varphi|_{\Omega}$ induces a permutation $\overline{\rho}$ on $\overline{\Omega}$, building $\mathcal{CM}(\overline{\Gamma},\overline{\Omega},\overline{\rho})$ into a RBCMt.

The assertion about type follows from the first sentence of Remark 1.2. ∎

The idea is, to understand a RBCMt $\mathcal{M}$ on $\Gamma$, we take a suitable subgroup $\Xi$, investigate the quotient RBCMt $\overline{\mathcal{M}}$ on $\Gamma/\Xi$, and use knowledge on $\overline{\mathcal{M}}$ to extract information about $\mathcal{M}$ as much as possible.

In this paper, we apply the reduction method to classify RBCMt’s for a class of split metacyclic 2-groups.

A general split metacyclic group can be presented as

[TABLE]

for some positive integers $n,m,r$ such that $r^{m}\equiv 1\pmod{n}$; see [8] Page 2. We focus on $\Lambda(2^{a},2^{b};1+2^{c})$, with

[TABLE]

These groups constitute a major part of split metacyclic $2$-groups of Class A, as introduced on [5] Page 2. The artificial restriction (3) is imposed for simplicity, so that the paper has a clear structure and a moderate length; if $b=c$ is allowed, then some annoying subtleties will arise, but nothing interesting will happen.

The main result is Theorem 3.10. As shown in [15], any metacyclic $p$-group for odd prime $p$ does not admit a RBCM1; (by Proposition 1.1, it does not admit a RBCMt for $t>1$). On the contrary, we shall see that the metacyclic $2$-group $\Lambda(2^{a},2^{b};1+2^{c})$ admits a rich family of RBCMt’s, consisting of $2^{a-c-1}$ isomorphism classes. To some extent, we can say that the richness and complexity of RBCMt’s on metacyclic groups are concentrated on metacyclic $2$-groups.

Section 2 presents a preliminary on metacyclic groups. Section 3 comprises the main steps of classifying RBCMt’s. First, we combine Lemma 1.3 and the previous work [1] on RBCMt’s on abelian $2$-groups to deduce several constraints on RBCMt’s on metacyclic $2$-groups, stated as Lemma 3.2. Second, based on the work [3] on automorphisms of metacyclic groups, we show that each RBCMt can be “normalized”, in the sense that it is isomorphic to one with the property that $\varphi_{+}$ and $\omega_{d}$ are in certain special forms. Third, we solve a system of congruence equations which characterize conditions for given data to determine a normalized RBCMt. Finally we state the classification as Theorem 3.10.

Notation 1.4**.**

For positive integers $u,s$, let $[u]_{s}=1+s+\cdots+s^{u-1}$; let $[0]_{s}=0$.

For $u\neq 0$, let $\|u\|$ denote the largest $k$ with $2^{k}\mid u$; set $\|0\|=+\infty$.

For an element $\theta$ of a finite group, let $|\theta|$ denote its order.

Let $\mathbb{Z}_{n}=\mathbb{Z}/n\mathbb{Z}$, which is a quotient ring of $\mathbb{Z}$.

For an abelian $2$-group $\Gamma$, let ${\rm rk}(\Gamma)$ denote its rank.

Given a normal subgroup $\Xi\vartriangleleft\Gamma$, the image of $\eta\in\Gamma$ under the quotient $\Gamma\twoheadrightarrow\Gamma/\Xi$ is usually denoted by $\overline{\eta}$, (but for $u\in\mathbb{Z}$, its image under $\mathbb{Z}\twoheadrightarrow\mathbb{Z}_{n}$ is still denoted by $u$), and if an automorphism $\phi$ of $\Gamma$ satisfies $\phi(\Xi)=\Xi$, then its induced automorphism on $\Gamma/\Xi$ is denoted by $\overline{\phi}$.

A RBCMt $\mathcal{CM}(\Gamma,\Omega,\rho)$ is shorten as $\mathcal{CM}(\Gamma,\Omega)$ if $\Omega$ can be written as $\{\omega_{1},\ldots,\omega_{d}\}$ and $\rho(\omega_{i})=\omega_{i+1}$. The subscript in $\omega_{i}$ is always understood as modulo $d$. Let ${\rm Aut}^{+}(\Gamma)=\{\tau\in{\rm Aut}(\Gamma)\colon\tau(\Gamma_{+})=\Gamma_{+}\}$.

Since various congruences modulo powers of $2$ will appear in the computations, to simplify the writing we use $A\equiv^{(k)}B$ to indicate $A\equiv B\pmod{2^{k}}$. Furthermore, abbreviate $A\equiv^{(a-1)}B$ to $A\equiv B$, and $A\equiv^{(b)}B$ to $A\equiv^{\prime}B$.

2 Preliminary on metacyclic groups

A general element of $\Lambda=\Lambda(n,m;r)$ can be written as $\alpha^{x}\beta^{y}$. By (2) we have

[TABLE]

Here $r^{y}$ is understood as $r^{y-m[y/m]}$ if $y<0$, $[u]_{r^{y}}$ is understood as $[u-n[u/n]]_{r^{y}}$ if $u<0$, and the commutator $[\eta,\mu]=\eta\mu\eta^{-1}\mu^{-1}$. Consequently, the commutator subgroup is generated by $\langle\alpha^{r-1}\rangle$, hence the abelianization

[TABLE]

Lemma 2.1**.**

There are three index $2$ subgroups of $\Lambda=\Lambda(n,m;r)$, namely, $\langle\alpha^{2},\beta\rangle$, $\langle\alpha,\beta^{2}\rangle$ and $\langle\alpha^{2},\alpha\beta\rangle$.

Proof.

Each homomorphism $\Lambda\to\mathbb{Z}_{2}$ factors through $\Lambda^{{\rm ab}}$, and there are exactly three epimorphisms $\kappa_{j}:\Lambda^{{\rm ab}}\cong\mathbb{Z}_{(r-1,n)}\times\mathbb{Z}_{m}\twoheadrightarrow\mathbb{Z}_{2},j=1,2,3$, given by

[TABLE]

Let $\widetilde{\kappa}_{j}$ denote the composite of the quotient $\Lambda\twoheadrightarrow\Lambda^{{\rm ab}}$ and $\kappa_{j}$. It is easy to see that $\ker\widetilde{\kappa}_{1}=\langle\alpha^{2},\beta\rangle$, $\ker\widetilde{\kappa}_{2}=\langle\alpha,\beta^{2}\rangle$, $\ker\widetilde{\kappa}_{3}=\langle\alpha^{2},\alpha\beta\rangle$. ∎

The following is a special case of [3] Theorem 2.9, in which, $\Lambda_{1}=\{2\}$, $\Lambda_{2}=\Lambda^{\prime}=\emptyset$, $a_{2}=c_{2}=a$, $b_{2}=b$, $d_{2}=c$, $t=2^{a}$, $m=2^{b}$, $m_{0}=1$.

If there is an automorphism $\sigma$ of $\Lambda(n,m;r)$ sending $\alpha$ and $\beta$ to $\alpha^{x_{1}}\beta^{y_{1}}$ and $\alpha^{x_{2}}\beta^{y_{2}}$, respectively, then we denote such automorphism $\sigma$ by $\sigma^{x_{1},y_{1}}_{x_{2},y_{2}}$ in this paper.

Lemma 2.2**.**

Suppose $\|r-1\|=c\geq 2$. Each automorphism of $\Lambda(2^{a},2^{b};r)$ is given by $\sigma^{x_{1},y_{1}}_{x_{2},y_{2}}:\alpha\mapsto\alpha^{x_{1}}\beta^{y_{1}},\beta\mapsto\alpha^{x_{2}}\beta^{y_{2}}$ for some integers $x_{1},y_{1},x_{2},y_{2}$ with

[TABLE]

Actually, any $x_{1},y_{1},x_{2},y_{2}$ satisfying these define an automorphism.

Acting on general elements,

[TABLE]

Given $\sigma^{x_{1},y_{1}}_{x_{2},y_{2}}$ and $\sigma^{p_{1},q_{1}}_{p_{2},q_{2}}$, the composite $\sigma^{p_{1},q_{1}}_{p_{2},q_{2}}\circ\sigma^{x_{1},y_{1}}_{x_{2},y_{2}}$ sends $\alpha$ to $\alpha^{h_{1}}\beta^{q_{1}x_{1}+q_{2}y_{1}}$ and sends $\beta$ to $\alpha^{h_{2}}\beta^{q_{1}x_{2}+q_{2}y_{2}}$, with

[TABLE]

Let $r=1+2^{c}$. Since $2(\|q_{1}\|+c)\geq b+c\geq a$, we have $r^{q_{1}u}\equiv^{(a)}1+2^{c}q_{1}u$, so

[TABLE]

Suppose $c>b$ which will hold in the next section. Then $\|x_{2}\|>a-c$, $\|p_{2}\|>a-c$, so that $2^{c-1}x_{2}\equiv^{(a)}0$, and $p_{2}[y_{j}]_{r^{q_{2}}}\equiv^{(c)}p_{2}y_{j}$, implying

[TABLE]

Thus

[TABLE]

3 Classifying regular $t$-balanced Cayley maps for a class of split metacyclic 2-groups

Let $\Delta=\Lambda(2^{a},2^{b};1+2^{c})$ for $(a,b,c)$ satisfying (3). In particular, $b\geq 3$, $c\geq 2$.

By [3] Lemma 2.1, $\|[u]_{(1+2^{c})^{y}}\|=\|u\|$. Then by (4),

[TABLE]

Suppose $\mathcal{CM}(\Delta,\{\omega_{1},\ldots,\omega_{d}\})$ is a RBCMt with skew-morphism $\varphi$. As in Remark 1.2, we may further assume $\ell\in\{(t-1,d)/2,(t-1,d)\}$, so

[TABLE]

Let $\eta_{j}=\omega_{j}\omega_{j-1}^{-1}=\omega_{j}\omega_{\ell+t(j-1)}$. Then

[TABLE]

in particular,

[TABLE]

Moreover, $\varphi(\omega_{j}\omega_{\ell+t(j-1)})=\omega_{j+1}\varphi^{t}(\omega_{\ell+t(j-1)})=\omega_{j+1}\omega_{\ell+tj}$, i.e.,

[TABLE]

3.1 Constraints

Lemma 3.1**.**

Suppose $\mathcal{CM}(\Gamma,\{\mu_{1},\ldots,\mu_{m}\})$ is a RBCMt with skew-morphism $\psi$, and $\Gamma$ is an abelian 2-group such that ${\rm rk}(\Gamma)={\rm rk}(\Gamma_{+})=2$. Then

1. (i)

there exists an isomorphism $\Gamma_{+}\cong\mathbb{Z}_{2^{k^{\prime}}}\times\mathbb{Z}_{2^{k}}$ for some $k^{\prime}\geq k\geq 1$, sending $\theta_{1}$ to $(1,0)$ and $\theta_{2}$ to $(-1,1)$, where $\theta_{j}=\mu_{j}-\mu_{j-1}$; 2. (ii)

$\mathcal{CM}(\Gamma,\{\mu_{1},\ldots,\mu_{m}\})$* has type I, and $m=2^{k+1}\mid t+1$;* 3. (iii)

$\psi_{+}^{2}={\rm id}$.

Proof.

RBCMt’s on abelian $2$-groups were completely classified in [1] Section 4.2, Corollary 4.3 and Corollary 4.7; obviously ${\rm rk}(\Gamma)={\rm rk}(\Gamma_{+})=2$ only occurs in the last case of Section 4.2. The conditions (i)–(iii) can be easily verified. ∎

Lemma 3.2**.**

For our RBCMt $\mathcal{CM}(\Delta,\{\omega_{1},\ldots,\omega_{d}\})$, the following holds:

1. (i)

it has type I, with $\Delta_{+}=\langle\alpha^{2},\beta\rangle\cong\Lambda(2^{a-1},2^{b};1+2^{c})$; 2. (ii)

$c>b$, and $\|t+1\|>b$; 3. (iii)

$\varphi_{+}=\sigma^{x_{1},y_{1}}_{x_{2},y_{2}}$* for some $x_{1},y_{1},x_{2},y_{2}$ with $2\nmid y_{1}$ and*

[TABLE]

Remark 3.3**.**

As a consequence of (ii), $c\geq 4$.

Be careful: here $\varphi_{+}=\sigma^{x_{1},y_{1}}_{x_{2},y_{2}}$ means that it sends $\alpha^{2}$ to $\alpha^{2x_{1}}\beta^{y_{1}}$ and sends $\beta$ to $\alpha^{2x_{2}}\beta^{y_{2}}$.

Proof.

The proof consists of three parts.

Assume $\Delta_{+}=\langle\alpha^{2},\alpha\beta\rangle$, $\eta_{j}=\alpha^{u_{j}}\beta^{v_{j}}$ ($j=1,\ldots,d$), and

[TABLE]

Since $|(\alpha^{2})^{x_{2}}(\alpha\beta)^{y_{2}}|=|\varphi_{+}(\alpha\beta)|=|\alpha\beta|$, by (6) we have $2\nmid y_{2}$; from

[TABLE]

we see $2\mid y_{1}$. From (10) we see that all the $v_{j}$’s have the same parity, which, by (8), must be odd. By (9), $2\mid v_{1}+\cdots+v_{\ell}$, so $\ell$ is even.

On the other hand, one can verify that the subgroup

[TABLE]

is normal in $\Delta$ and invariant under $\varphi_{+}$. By Lemma 1.3 there is a quotient RBCMt $\overline{\mathcal{M}}$ on $\Delta/\Xi\cong\mathbb{Z}_{2^{c}}\times\mathbb{Z}_{2^{c}}$. Clearly ${\rm rk}(\Delta/\Xi)={\rm rk}(\Delta_{+}/\Xi)=2$, hence by Lemma 3.1, $\overline{\mathcal{M}}$ has type I, and $4\mid t+1$. By Lemma 1.3, our RBCMt has type I. Hence $\ell=(t-1,d)/2$, contradicting $2\mid\ell$. 2. 2.

Assume $\Delta_{+}=\langle\alpha,\beta^{2}\rangle$, $\eta_{j}=\alpha^{u_{j}}\beta^{2v_{j}}$ ($j=1,\ldots,d$) and $\varphi_{+}=\sigma^{x_{1},y_{1}}_{x_{2},y_{2}}$. Since $\Delta_{+}\cong\Lambda(2^{a},2^{b-1};(1+2^{c})^{2})$, by Lemma 2.2 we have

[TABLE]

The subgroup \Xi^{\prime}=\big{\langle}\alpha^{2^{c}},\beta^{2^{b-1}}\big{\rangle} is normal in $\Delta$ and invariant under $\varphi_{+}$, with $\Delta/\Xi^{\prime}\cong\mathbb{Z}_{2^{c}}\times\mathbb{Z}_{2^{b-1}}$ and $\Delta_{+}/\Xi^{\prime}\cong\mathbb{Z}_{2^{c}}\times\mathbb{Z}_{2^{b-2}}$. In $\Delta_{+}/\Xi^{\prime}$,

[TABLE]

By Lemma 1.3 and Lemma 3.1, $4\mid t+1$ and $\ell=(t-1,d)/2$, so that $2\nmid\ell$.

If $2\mid x_{2}$, then $2\nmid x_{1}$, so that $u_{j}\equiv u_{1}\pmod{2}$; by (8), $2\nmid u_{1}$, hence $\eta_{\ell}\cdots\eta_{1}=\alpha^{u^{\prime}}\beta^{v^{\prime}}$ for some odd $u^{\prime}$, but this contradicts (9). Hence $2\nmid x_{2}$, and consequently $b-1\geq a\geq c+3$.

By Lemma 3.1 (i), $|\overline{\eta_{1}}|=2^{b-2}$ and $|\overline{\eta_{1}}+\overline{\eta_{2}}|=2^{c}$. Hence $2\nmid v_{1}$, and

[TABLE]

the inequality comes from (6), and the equality relies on (11) which implies $\|y_{1}\|\geq b-c-2\geq 2>\|y_{2}+1\|=1$. This contradicts $b-1\geq c+3$. 3. 3.

Therefore by Lemma 2.1, $\Delta_{+}=\langle\alpha^{2},\beta\rangle\cong\Lambda(2^{a-1},2^{b};1+2^{c})$. Suppose $\varphi_{+}=\sigma^{x_{1},y_{1}}_{x_{2},y_{2}}$, and $\eta_{j}=\alpha^{2u_{j}}\beta^{v_{j}}$.

The subgroup $\langle\alpha^{2^{c}}\rangle$ is normal in $\Delta$ and invariant under $\varphi_{+}$. Applying Lemma 3.1 to the quotient RBCMt on $\Delta/\langle\alpha^{2^{c}}\rangle\cong\mathbb{Z}_{2^{c}}\times\mathbb{Z}_{2^{b}}$, we obtain $4\mid t+1$ and $\ell=(t-1,d)/2$. So $2\nmid\ell$.

If $2\mid y_{1}$, then $v_{j}\equiv v_{1}\pmod{2}$ for all $j$; by (8), $2\nmid v_{1}$, and since $2\nmid\ell$, we have $\eta_{\ell}\cdots\eta_{1}=\alpha^{2u^{\prime}}\beta^{v^{\prime}}$ for some odd $v^{\prime}$, contradicting (9). Hence $2\nmid y_{1}$, and then by Lemma 2.2, $b\leq c+\|y_{1}\|=c$. Since $b\neq c$, actually $c>b$.

By Lemma 3.1 (iii), $\overline{\varphi_{+}}^{2}={\rm id}$. The expression for $\overline{\varphi_{+}}^{2}$ is

[TABLE]

Thus, $x_{1}^{2}+x_{2}y_{1}\equiv^{(c-1)}1$ and $x_{1}+y_{2}\equiv^{\prime}y_{2}^{2}+x_{2}y_{1}-1\equiv^{\prime}0$.

∎

3.2 Normalization

Lemma 3.4**.**

Let $\sigma^{z_{1},w_{1}}_{z_{2},w_{2}}\in{\rm Aut}(\Delta_{+})$. There exists $\tau\in{\rm Aut}^{+}(\Delta)$ with $\tau_{+}=\sigma^{z_{1},w_{1}}_{z_{2},w_{2}}$ if and only if $2\mid w_{1}$ and $\|w_{2}-1\|\geq a-c$.

Proof.

If $\tau=\sigma^{p_{1},q_{1}}_{p_{2},q_{2}}\in{\rm Aut}^{+}(\Delta)$, then by Lemma 2.2, $2\mid p_{2}$ and $\|q_{2}-1\|\geq a-c$. As an automorphism of $\Delta_{+}$, $\tau_{+}(\alpha^{2})=\alpha^{p_{1}(2+2^{c}q_{1})}\beta^{2q_{1}}$, $\tau_{+}(\beta)=\alpha^{p_{2}}\beta^{q_{2}}$, hence

[TABLE]

So $2\mid w_{1}$ and $\|w_{2}-1\|\geq a-c$ are necessary for there to exist $\tau\in{\rm Aut}^{+}(\Delta)$ with $\tau_{+}=\sigma^{z_{1},w_{1}}_{z_{2},w_{2}}$.

Conversely, suppose $2\mid w_{1}$ and $\|w_{2}-1\|\geq a-c$. Put

[TABLE]

It is clear that $\tau\in{\rm Aut}^{+}(\Delta)$ with $\tau_{+}=\sigma^{z_{1},w_{1}}_{z_{2},w_{2}}$. ∎

Lemma 3.5**.**

Suppose $h$ is odd, $e>2$, and $s^{2}\equiv^{(e)}h$. There exists a sequence $\{\tilde{s}_{k}\}_{k=2}^{\infty}$ such that $\tilde{s}_{k}^{2}\equiv^{(k(e-1))}h$ and $\tilde{s}_{k+1}\equiv^{(k(e-1)-1)}\tilde{s}_{k}$ for each $k$. Consequently, for each $\tilde{e}>e$, there exists $\tilde{s}$ such that $\tilde{s}^{2}\equiv^{(\tilde{e})}h$ and $\tilde{s}\equiv^{(e-1)}s$.

Proof.

We construct $\tilde{s}_{k}$ recursively. Since $s$ is odd, we may take $a_{1}\in\mathbb{Z}$ such that $h\equiv^{(2(e-1))}s^{2}+2^{e}sa_{1}$. Set $\tilde{s}_{2}=s+2^{e-1}a_{1}$. Then clearly $\tilde{s}_{2}^{2}\equiv^{(2(e-1))}h$ and $\tilde{s}_{2}\equiv^{(e-1)}s$.

Assume $k\geq 2$ and $\tilde{s}_{k}$ has been obtained. Take $a_{k}\in\mathbb{Z}$ with

[TABLE]

and set

[TABLE]

Then $\tilde{s}_{k+1}\equiv^{(k(e-1)-1)}\tilde{s}_{k}$ and $\tilde{s}_{k+1}^{2}\equiv^{((k+1)(e-1))}h$, due to $2k(e-1)-2\geq(k+1)(e-1)$. ∎

Lemma 3.6**.**

There exists $\tau_{1}\in{\rm Aut}^{+}(\Delta)$ such that $(\tau_{1}\varphi\tau_{1}^{-1})_{+}=\sigma^{z,1}_{0,-z}$ for some $z\equiv^{(c-2)}-1$.

Proof.

We are going to find $u_{1},v_{1},u_{2},v_{2},z,w$ satisfying the following:

[TABLE]

In view of (5), these will ensure

[TABLE]

Take $\gamma$ with $(1+2^{c-1}(y_{1}-1))\gamma\equiv 1$, and let

[TABLE]

Remember that $c\geq a-b>a-c\geq 3$ and $\|x_{2}\|\geq a-b$. Then $f(x_{1})\equiv^{(a-b)}0$. By Lemma 3.5, there exists $z$ with $f(z)\equiv 0$ and $z\equiv^{(a-b-1)}x_{1}$. Note that $\|z-y_{2}\|=\|x_{1}-y_{2}\|=1$.

Let $u_{1}=y_{1}$, $v_{1}=0$, $v_{2}=1$, $w=y_{2}-u_{2}$, and $u_{2}=(1+2^{c-1}(y_{1}-1))z-x_{1}.$ Then $f(z)\equiv 0$ is equivalent to

[TABLE]

It is easy to verify that (12)–(15) all hold. Now it holds that

[TABLE]

by Lemma 3.4, $\sigma^{u_{1},v_{1}}_{u_{2},v_{2}}=(\tau_{1})_{+}$ for some $\tau_{1}\in{\rm Aut}^{+}(\Delta)$.

Consider the automorphism of $\Delta_{+}/\langle\alpha^{2^{c}}\rangle$ induced by $\tau_{1}\varphi_{+}\tau_{1}^{-1}$. Similarly as the final part of the proof of Lemma 3.2, we have $z^{2}\equiv^{(c-1)}0$ and $z+w\equiv^{\prime}w^{2}-1\equiv^{\prime}0$. Thus $(\tau_{1}\varphi\tau_{1}^{-1})_{+}=\sigma^{z,1}_{0,-z}$. Note that $w\equiv y_{2}\equiv 1\pmod{4}$, implying $\|z+1\|\geq c-2$. ∎

Lemma 3.7**.**

Suppose $x\equiv^{(c-2)}z\equiv^{(c-2)}-1$. If $\tau\in{\rm Aut}^{+}(\Delta)$ with $\tau_{+}=\sigma^{p_{1},q_{1}}_{p_{2},q_{2}}$, then $\tau_{+}\circ\sigma^{z,1}_{0,-z}\circ\tau_{+}^{-1}=\sigma^{x,1}_{0,-x}$ is equivalent to

[TABLE]

In particular, $\tau_{+}\circ\sigma^{z,1}_{0,-z}\circ\tau_{+}^{-1}=\sigma^{z,1}_{0,-z}$ if and only if $p_{2}\equiv 0$ and $p_{1}-q_{2}\equiv^{\prime}2zq_{1}.$

Proof.

By Lemma 3.4, $2\mid q_{1}$ and $\|q_{2}-1\|\geq a-c$.

By (5), $\tau_{+}\circ\sigma^{z,1}_{0,-z}=\sigma^{x,1}_{0,-x}\circ\tau_{+}$ is equivalent to

[TABLE]

Since $x\equiv^{(c-2)}z\equiv^{(c-2)}-1$, we have $\|x+z\|=1$, hence by (19), $\|p_{2}\|\geq a-2$. Then (20) implies $x\equiv^{(a-2)}z$, and consequently by (18), $p_{1}-q_{2}\equiv^{\prime}2zq_{1}$. Now it holds that $c-1\geq b\geq a-c$, and one has

[TABLE]

which together with (17) implies

[TABLE]

Since p_{2}\equiv p_{2}\cdot p_{1}\big{(}1+2^{c-1}(z-1)q_{1}\big{)}, we have $x\equiv z+p_{2}$.

Conversely, assuming (16), it is rather easy to deduce (17)–(20). ∎

Lemma 3.8**.**

(i)* There exists $\tau_{2}\in{\rm Aut}^{+}(\Delta)$ such that $(\tau_{2})_{+}\circ\sigma^{z,1}_{0,-z}=\sigma^{z,1}_{0,-z}\circ(\tau_{2})_{+}$ and $(\tau_{2}\tau_{1})(\omega_{d})=\alpha^{\tilde{u}}\beta$ for some odd $\tilde{u}$.*

(ii)* For any $\tilde{u}^{\prime}$ with $\tilde{u}^{\prime}\equiv^{(a-c)}\tilde{u}$, there exists $\tau\in{\rm Aut}^{+}(\Delta)$ such that $\tau_{+}\circ\sigma^{z,1}_{0,-z}=\sigma^{z,1}_{0,-z}\circ\tau_{+}$ and $\tau(\alpha^{\tilde{u}}\beta)=\alpha^{\tilde{u}^{\prime}}\beta$.*

Proof.

(i) Suppose $\tau_{1}(\omega_{d})=\alpha^{u_{0}}\beta^{v_{0}}$. Note that $u_{0}$ is odd: otherwise it is impossible for $\tau_{1}(\eta_{1}),\ldots,\tau_{1}(\eta_{d}),\tau_{1}(\omega_{d})$ to generate $\Delta$.

Take $y$ with $yu_{0}\equiv^{\prime}1-v_{0}$. Let $p=1+4zy$, and let

[TABLE]

Let $\tau_{2}=\sigma^{(1-2^{c-1}y)p,y}_{0,1}$, so that $(\tau_{2})_{+}=\sigma^{p,2y}_{0,1}$. Then $(\tau_{2}\tau_{1})(\omega_{d})=\alpha^{\tilde{u}}\beta$, and by Lemma 3.7, $(\tau_{2})_{+}\circ\sigma^{z,1}_{0,-z}=\sigma^{z,1}_{0,-z}\circ(\tau_{2})_{+}$.

(ii) Take $y$ with $y\tilde{u}\equiv-2^{a-c}\overline{q}$, with $\overline{q}$ to be determined. Let $p^{\prime}=1+2^{a-c}\overline{q}+4zy$. Consider

[TABLE]

Obviously, we can find $\overline{q}$ such that $u(\overline{q})\equiv^{(a)}\tilde{u}^{\prime}$.

Let $\tau=\sigma^{p^{\prime},0}_{0,1+2^{a-c}\overline{q}}$. Now $\tau_{+}=\sigma^{p^{\prime},0}_{0,1+2^{a-c}\overline{q}}$ commutes with $\sigma^{z,1}_{0,-z}$ and $\tau(\alpha^{\tilde{u}}\beta)=\alpha^{\tilde{u}^{\prime}}\beta$. ∎

Concluding from the above lemmas, up to isomorphism we may just assume $\varphi_{+}=\sigma^{z,1}_{0,-z}$ for a unique $z$ with $0\leq z<2^{a-2}$ and $z\equiv^{(c-2)}-1$, and $\omega_{d}=\alpha^{\tilde{u}}\beta$ such that $\tilde{u}$ is an odd number whose residue modulo $2^{a-c}$ is unique.

3.3 Expressing necessary and sufficient conditions in terms of congruence equations

Remember that for each $k$,

[TABLE]

Implied by $z\equiv^{(c-2)}-1$,

[TABLE]

Suppose $\eta_{i}=\alpha^{2u_{i}}\beta^{v_{i}}$. Then $\omega_{i}\omega_{d}^{-1}=\eta_{i}\cdots\eta_{1}=\alpha^{2f_{i}}\beta^{g_{i}}$, where

[TABLE]

So $\omega_{i}=\alpha^{2f_{i}+(1+2^{c}g_{i})\tilde{u}}\beta^{g_{i}+1}.$

The condition (7) is equivalent to

[TABLE]

Also the condition $1=\omega_{d}\omega_{d}^{-1}=\alpha^{2f_{d}}\beta^{g_{d}}$ implies that

[TABLE]

From (5) and $z^{2}\equiv^{(c-1)}1$ we see $\varphi_{+}^{2}=\sigma^{s,0}_{0,1}$, with $s=z^{2}+2^{c-1}(z-1)$. Hence

[TABLE]

Clearly, $u_{2}\equiv u_{1}\pmod{2}$. It follows from (8) that the $u_{i}$’s are all odd.

Put

[TABLE]

Since

[TABLE]

we have

[TABLE]

Obviously, $s\equiv^{(c-1)}1$, so that for each $n$,

[TABLE]

Now (22), (23), (27) imply

[TABLE]

Lemma 3.9**.**

Let $h=(\ell-1)/2$. The conditions (24)–(26)* hold if and only if*

[TABLE]

Proof.

Let $e=(t+1)/2$. Let (25)i=2k stand for (25) when $i=2k$, and so forth.

The condition (25)i=2k reads

[TABLE]

which, due to $\|t+1\|\geq b+1$, is equivalent to (32). Conversely, if (32) is satisfied, then (25)i=2k+1 holds, too:

[TABLE]

In virtue of $2^{c}\overline{u}\equiv 0$ and (21), the condition (24)i=2k reads

[TABLE]

Then the difference between (24)i=2k+2 and (24)i=2k is equal to

[TABLE]

where (21), (31) have been used.

Setting $k=0$ in (37), we obtain

[TABLE]

Clearly, (24)i=2k holds for all $k$ if and only if (38) and (39) hold. With (29) referred to, (38), (39) are equivalent to

[TABLE]

Note that the second equation is equivalent to (34) and forces $\|s-1\|=c-1$. Hence $\|z+1\|=c-2$, and by (28), $\|\overline{u}\|=c-2$.

The condition (24)i=2k+1 reads

[TABLE]

So the difference between (24)i=2k+3 and (24)i=2k+1 equals

[TABLE]

which can be implied by (40), assuming (38).

Setting $k=0$ in (41), we obtain

[TABLE]

it combined with (40) implies $e\overline{u}\equiv 0$, which is equivalent to (35).

By (32), (34), $2^{c-1}v_{1}\equiv h(1-s)-2h(z+1)^{2}-2^{c}$, and hence (40) becomes (33).

Finally, (26) holds if and only if

[TABLE]

which are equivalent to (36), as is easy to verify. ∎

Now since $\|z+1\|=c-2$ and $0\leq z<2^{a-2}$, we may write

[TABLE]

By (34), using $(2x-1)^{2}\equiv 1\pmod{4}$ and $c-1\geq b\geq a-c$, we obtain

[TABLE]

By (28), $\|\overline{u}\|=c-2$. By (33), $\tilde{u}\equiv^{(c-1)}h\overline{u}-u_{1}\equiv^{(c-1)}2^{c-2}h-u_{1}$, so that

[TABLE]

According to the conclusion in the end of Section 3.2 we may just set

[TABLE]

By (42), (43),

[TABLE]

Hence by (28),

[TABLE]

Using

[TABLE]

we convert (33) into

[TABLE]

implying

[TABLE]

So (44) becomes

[TABLE]

Finally, (45) implies

[TABLE]

Hence by (30) and (32),

[TABLE]

where the meanings of fractions are self-evident. So

[TABLE]

3.4 The result

Recall

[TABLE]

For each $x$ with $1\leq x\leq 2^{a-c-1}$, let

[TABLE]

and put $\mathcal{M}(x)=\mathcal{CM}(\Delta,\{\omega_{1},\ldots,\omega_{d}\})$ with $\omega_{i}=\alpha^{2f_{i}+(1+2^{c}g_{i})\tilde{u}}\beta^{g_{i}+1}.$

Theorem 3.10**.**

If $\Delta$ admits $d$-valent RBCMt’s, then necessarily $\|d\|,\|t+1\|\geq\max\{a-c+2,b+1\}$ and $c>b$. When these hold, each $d$-valent RBCMt on $\Delta$ has type I and is isomorphic to $\mathcal{M}(x)$ for a unique $x$ with $1\leq x\leq 2^{a-c-1}$.