Correction to “Faithful permutation representations of toroidal regular maps”

1 Introduction

In [1], we claim that we give a complete list of the possible degrees of a faithful transitive permutation representation of the groups of the toroidal regular maps. Indeed, the list given for type $\{4,4\}$ is complete; however, recently we were surprised with the existence of exceptional degrees for the map $\{3,6\}$. After, we struggled to find the reason why there were some degrees missing in our classification. The fact is that there is a gap in one proof, having consequences in two of our main theorems. Our goal is to fill in that gap.

In what follows, let G be the group of symmetries of a toroidal regular map of type $\{4,4\}$ or $\{3,6\}$, and suppose G is represented as a faithful transitive permutation representation group of degree n. Let T be the translation group generated by unitary independent translations of order s defining the groups of $\{4,4\}_\mathbf{{s}}$ or $\{3,6\}_\mathbf{{s}}$ for $\mathbf{s}\in \{(s,0),(s,s)\}$ (see [1]). We recall the following results.

Lemma 1.1

[1] [Lemma 1.1] If T is transitive, then $n=s^2$.

An immediate consequence of Lemma 1.1 is that if $n\ne s^2$, T is intransitive. Note that T is a normal subgroup of G that is a direct product of two cyclic groups of order s. Consider that $\alpha$ and $\beta$ are the actions of the generators of T on a block and let $K{:}{=}\langle \alpha ,\, \beta \rangle$. In [1], we consider the cases where K is a cyclic group or a direct product, but these cases do not cover all possibilities. In what follows, we complete all the proofs of [1] where this has any effect.

2 The size of a T-orbit

In what follows, o(x) denotes the order of x.

Lemma 2.1

If $B{:}{=}|K:\langle \alpha \rangle |$ and $C{:}{=}|K:\langle \beta \rangle |$, then the size of a T-orbit is $k=ds$ where $d=gcd(B,C)$.

Proof

Consider that $\alpha$ and $\beta$ are the actions of the generators of T on a block of size k. Then, $K{:}{=}\langle \alpha ,\, \beta \rangle$, $A{:}{=}o(\alpha )$, $B{:}{=}|K:\langle \alpha \rangle |$ and $C{:}{=}|K:\langle \beta \rangle |$. The order of K is AB and acts regularly on the block; hence, $k=AB$. As $\alpha$ and $\beta$ commute, we have the following.

$$\begin{aligned} K/ \langle \alpha \rangle= & {} \{\langle \alpha \rangle ,\, \langle \alpha \rangle \beta , \langle \alpha \rangle \beta ^2,\ldots , \langle \alpha \rangle \beta ^{B-1}\} \text{ and } \\ K/ \langle \beta \rangle= & {} \{\langle \beta \rangle ,\, \langle \beta \rangle \alpha , \langle \beta \rangle \alpha ^2,\ldots , \langle \beta \rangle \alpha ^{C-1}\}. \end{aligned}$$

Thus, B divides $o(\beta )$ and C divides $o(\alpha )=A$. Let $D{:}{=}A/C$. As $k=AB=o(\beta )C$, we have $o(\beta )=DB$. Now $s=lcm(o(\alpha ), o(\beta ))=lcm(CD,\,BD)=D\,lcm(C,B)$ and $k=AB=DCB=D\,lcm(C,B)\,gcd(C,B)=s\,gcd(C,B)$. To conclude the proof, consider $d= gcd(C,B)$. $\square$

The item (i) of the following lemma is a consequence of Lemma 2.1.

Lemma 2.2

[1] [Lemma 2.2] If $n\ne s^2$, then G is embedded into $S_k\wr S_m$ with $n=km$ $(m,\,k>1)$ and

• (i) $k=ab$ where $s=lcm(a,b)$ and,

• (ii) m is a divisor of $\frac{|G|}{s^2}$.

From now on, we use m for number of blocks (the T-orbits) and k size of a T-orbit.

Consider first the regular map $\{4,4\}_{(s,0)}$. In [1], the existence of a faithful transitive permutation representation of degree $n=mab$ was guarantied whenever $s=lcm(a,b)$ and m is a divisor of $\frac{|G|}{s^2}=4$. Moreover, we proved that the degrees of $\{4,4\}_{(s,s)}$ are the degrees of $\{4,4\}_{(s,0)}$ multiplied by 2. Consequently, there are no changes on our results for $\{4,4\}_{(s,s)}$. Anyway we take this opportunity to restate the classification of all possible degrees for the regular maps of type $\{4,4\}$.

Theorem 2.3

Let $s>1$. The degrees of a faithful transitive permutation representation of a toroidal regular map of type $\{4,4\}_\mathbf{{s}}$ are

• (1) 2ds, 4ds and 8ds for any divisor d of s, for $\mathbf{s}=(s,0)$;

• (2) 4ds, 8ds and 16ds for any divisor d of s, for $\mathbf{s}=(s,s)$.

Proof

Thus, this is a consequence of Theorems 4.2 and 4.4 of [1]. Note that if $k=ab$ and $s=lcm(a,b)$, then $k=ab=gcd(a,b)s$; thus, $k=ds$ for some divisor d of s. Conversely, if $k=ds$ with d being a divisor of s, then $k=ab$ for some integers a an b such that $s=lcm(a,b)$ . $\square$

For the map $\{3,6\}_{(s,0)}$, we did not find faithful transitive permutation representations of degrees 2ab and 4ab when both a and b are different from s. Indeed, the only degrees found, when T is intransitive, were $2s^2$, $4s^2$ and the indexes of the core-free subgroups given in the following proposition.

Proposition 2.4

[1] [Proposition 2.4] Let G be the group of $\{3,6\}_ {(s,0)}$ ($s\ge 2$).

• (1) $H=\langle u^a, v^b\rangle$ is core-free and $|G:H|=12ab$, where $s=lcm(a,b)$.

• (2) If $s=lcm(a,b)$ and $s\ne ab$ then $H=\langle u^a, v^b\rangle \rtimes \langle \rho _0\rho _2 \rangle$ is core-free and $|G:H|=6ab$.

• (3) If d is a divisor of s then $H=\langle u^d \rangle \rtimes \langle \rho _0,\rho _2 \rangle$ and $H'=\langle u^d \rangle \rtimes \langle \rho _0\rho _2 \rangle$ are core-free. Moreover, $|G:H|= 3ds$ and $|G:H'|= 6ds$.

Thus, for $m\in \{3,6,12\}$ there are no gaps to be filled. We only need to consider the case $m\in \{2,4\}$.

In Sect. 3, we consider $m\in \{2,4\}$ for the map $\{3,6\}_{(s,0)}$. In Sect. 4, we determine the core-free subgroups corresponding to the degrees found in the previous section. Finally, in Sect. 5, we give a classification of all possible degrees for $\{3,6\}$, including the degrees of $\{3,6\}_{(s,s)}$.

3 Cases $m=2$ and $m=4$

Let $\rho _0$, $\rho _1$ and $\rho _2$ be the distinguished generators of the group of automorphisms G of the map $\{3,6\}_{(s,0)}$. Consider that $T=\langle u,\, v\rangle$ with $u=\rho _0(\rho _1\rho _2)^2\rho _1$ and $v= u^{\rho _1}=(\rho _0\rho _1\rho _2)^2$, and let $t=u^{-1}v$. The actions of u, v and t on the block i are denoted by $u_i$, $v_i$ and $t_i$, respectively. In addition $K=\langle u_1,\, v_1\rangle$, that is, the action of T on the block 1.

Let us first prove a result to be used in both cases, $m=2$ and $m=4$.

Proposition 3.1

Let q be an odd number. The modular equation

$$\begin{aligned} x^2-x+1\equiv 0\ \text {mod}\ q \end{aligned}$$

has a solution if and only if all prime divisors p of q are such that $p\equiv 1 \ \text {mod}\ 6$.

Proof

Let $2^*$ be an inverse of 2 modulo q (which exists because q is odd). Then, the equation $x^2-x+1\equiv 0\ \text {mod}\ q$, which is equivalent to $(x-2^*)^2\equiv (-3)(2^*)^2\ \text {mod}\ q$, has a solution if and only if $-3$ is a quadratic residue modulo p for every prime divisor of q [2]. In addition, $-3$ is a quadratic residue modulo p for every prime divisor of q if and only if $p\equiv 1 \ \text {mod}\ 3$, or, as by assumption p is odd, $p\equiv 1 \ \text {mod}\ 6$. $\square$

Proposition 3.2

If $m=2$, then $k=sd$ where d is a divisor of s and all prime divisors p of s/d are such that $p\equiv 1\ \text {mod}\ 6$.

Proof

Let $m=2$. The following graphs represent all the possible block actions determined in [1].

In any of the three cases $o(u_1)=o(u_2)=o(v_1)=o(v_2)=o(t_1)=o(t_2)=s$. Let $d=|K:\langle u_1\rangle |=|K:\langle v_1\rangle |$, then $k=sd$. Assume $d\ne s$ since the degree $2s^2$ is already known to exist.

Case 1 Let $u_1^d=v_1^j$. Conjugating by $\rho _0$, we obtain $u_1^{-d}=t_1^j= u_1^{-j}v_1^{j}=u_1^{d-j}$; hence, $j\equiv 2d\ \text {mod}\ s$. Now conjugating the equation $u_1^d=v_1^{2d}$ by $\rho _2$, we obtain $u_2^d=t_2^{-2d}=u_2^{2d}v_2^{-2d}$; thus, $u_2^d=v_2^{2d}$, which gives $u^d=v^{2d}$, a contradiction.

Case 2 Let $u_1^d=v_1^j$. Conjugating by $\rho _2$, we obtain $u_1^d=t_1^{-j}=u_1^jv_1^{-j}=u_1^{j-d}$; hence, $j\equiv 2d\ \text {mod}\ s$. Now conjugating the equation $u_1^d=v_1^{2d}$ by $\rho _1\rho _0$, we obtain $t_1^d=u_1^{-2d}$; thus, $v_1^d=u_1^{-d}=v_1^{-2d}$. Hence, $3d\equiv 0\ \text {mod}\ s$, and therefore, $u_1^d=v_1^{-d}$ and, conjugating by $\rho _1$, $v_2^d=u_2^{-d}$. Consequently, $(u_1u_2)^d=(v_1v_2)^{-d}$, a contradiction.

Case 3 Let $u_1^d=v_1^j$. As $|K:\langle u_1\rangle |=d$, j must be a multiple of d, say $j=\alpha d$. From the equality $u_1^d=v_1^{\alpha d}$ and as $o(u_1)=o(v_1)$, we have $gcd(\alpha , s/d)=1$. Conjugating the equation $u_1^d=v_1^{\alpha d}$ by $\rho _0\rho _1$, we obtain $v_1^{-d}=t_1^{-\alpha d}=u_1^{\alpha d}v_1^{-\alpha d}$. Thus, $u_1^{\alpha d}=v_1^{\alpha d-d}$. Consequently $v_1^{\alpha ^2 d}=v_1^{\alpha d-d}$, which implies $d(\alpha ^2-\alpha +1)\equiv 0\ \text {mod}\ s$, or equivalently $\alpha ^2-\alpha +1\equiv 0\ \text {mod}\ (s/d)$. The rest follows from Proposition 3.1. $\square$

Proposition 3.3

If $m=4$, then $k=sd$ where d is a divisor of s and all prime divisors p of s/d are such that $p\equiv 1\ \text {mod}\ 6$.

Proof

Using GAP [3], it can be checked that there is only one possibility for the action of G given by the following graph.

Let $\Delta _i$ denote the block i, $i\in \{1,\ldots ,4\}$, as follows:

$$\begin{aligned} \Delta _2= & {} \Delta _1\rho _0=\Delta _1\rho _1,\,\Delta _3=\Delta _1\rho _2 \text{ and } \\ \Delta _4= & {} \Delta _3\rho _0=\Delta _3\rho _1 \end{aligned}$$

We have $o(u_i)=s$ for $i\in \{1,\,2,\,3,\,4\}$, and the same holds for $o(v_i)$ and $o(t_i)$.

Let $|K:\langle u_1\rangle |=|K:\langle v_1\rangle |=d$. Assume that $d \ne s$. Analogously to Proposition 3.2, we may write $u_1=v_1^{\alpha d}$ with $gcd(\alpha , s/d)=1$. Then, conjugating by $\rho _0\rho _1$, we get $v_1^{(\alpha -1)d}=v_1^{\alpha ^2 d}$, which implies $\alpha ^2 -\alpha +1\equiv 0\ \text {mod}\ (s/d)$. The rest follows from Proposition 3.1. $\square$

4 Core-free subgroups

The following proposition gives core-free subgroups that will guaranty the existence of the degrees obtained in Propositions 3.2 and 3.3.

Proposition 4.1

Let d be a divisor of s. Suppose that there exists $\alpha$, coprime with s/d, such that $\alpha ^2-\alpha +1\equiv 0\ \text {mod}\ (s/d)$. Then, $\langle (v^{-\alpha }u)^{d},\, \rho _0\rho _1\rangle$ and $\langle (v^{-\alpha }u)^{d},\, \rho _1\rho _2\rangle$ are core-free subgroups of G with indexes 4sd and 2sd, respectively.

Proof

Let us consider $H{:}{=}\langle (v^{-\alpha }u)^{d},\, \rho _0\rho _1\rangle$. We have $(v^{-\alpha }u)^{\rho _0\rho _1}=t^{\alpha }v^{-1}=v^{\alpha -1}u^{-\alpha }=v^{\alpha ^2}u^{-\alpha }=(v^{-\alpha }u)^{-\alpha }$. Hence, $H=\langle (v^{-\alpha }u)^{d}\rangle \rtimes \langle \rho _0\rho _1\rangle$. Furthermore, $|H|=3s/d$, and therefore, $|G:H|=4sd$.

Now let us prove that H is core-free. Suppose that $H\cap H^{\rho _0}$ is nontrivial. Then, there exist $i,\,i'\in \{0,\ldots ,\, s/d-1\}$ and $j,\,j'\in \{0,1,2\}$ such that $(v^{-\alpha }u)^{id}(\rho _0\rho _1)^j=(t^{-\alpha }u^{-1})^{i'd}(\rho _0\rho _1)^{j'}$. Clearly, this is only possible when $j=j'$, $id=i'd$ and $(\alpha -1)id \equiv id\ \text {mod}\ s$, or equivalently $\alpha ^2i\equiv i\ \text {mod}\ (s/d)$. As $\alpha ^2-\alpha +1\equiv 0\ \text {mod}\ (s/d)$, $\alpha$ is a cubic root of $-1$ modulo (s/d), $i\equiv -\alpha i \ \text {mod}\ (s/d)$. Consequently, $i\equiv -2i\ \text {mod}\ (s/d)$. As, by Proposition 3.1, 3 does not divide (s/d), we obtain $i=0$. Consequently, $H\cap H^{\rho _0}\le \langle \rho _0\rho _1\rangle$. Now suppose that $\langle \rho _0\rho _1\rangle \cap H^{\rho _2}$ is nontrivial. Then, either $(\rho _1\rho _2)^2$, $(\rho _0\rho _1\rho _2\rho _0)^2$, v or t belongs to $\langle t^{\alpha } u\rangle$, which is not possible.

For $H{:}{=}\langle (v^{-\alpha }u)^{d},\, \rho _1\rho _2\rangle$, the proof is analogous. $\square$

5 The degrees of the regular maps of type $\{3,6\}$

The results given previously have consequences on Theorems 5.6 and 5.7 of [1]. The correct statement of these results should be as follows:

Theorem 5.1

Let $s\ge 2$. The degrees of a faithful transitive permutation representation of a toroidal regular map of type $\{3,6\}_{(s,0)}$ are $n=s^2$ and the following.

• (1) 3ds, 6ds or 12ds for any divisor d of s,

• (2) 2ds and 4ds if and only if d is a divisor of s and all prime divisors of s/d are equal to $1 \ \text {mod}\ 6$.

Proof

This is a consequence of results included in the previous sections. $\square$

Let us now consider the map $\{3,6\}_{(s,s)}$. We first recall the following.

Corollary 5.2

[1] [Corollary 5.2] If n is a degree of $\{3,6\}_{(s,0)}$ (resp. $\{3,6\}_{(s,s)}$), then 3n is a degree of $\{3,6\}_{(s,s)}$ (resp. $\{3,6\}_{(3s,0)}$);

Theorem 5.3

Let $s\ge 2$. The degrees of a faithful transitive permutation representation of a toroidal regular map of type $\{3,6\}_{(s,s)}$ are $n=3s^2$ and the following.

• (1) 9ds, 18ds or 36ds for any divisor d of s,

• (2) 6ds and 12ds if and only if d is a divisor of s and all prime divisors of s/d are equal to $1 \ \text {mod}\ 6$.

Proof

By Corollary 5.2, all the degrees given on Theorem 5.1 multiplied by 3 are degrees for the map $\{3,6\}_{(s,s)}$. Moreover, the degrees of $\{3,6\}_{(s,s)}$ multiplied by 3 are degrees of $\{3,6\}_{(3s,0)}$. Hence, by Theorem 5.1, the set of degrees of $\{3,6\}_{(s,s)}$ must be contained in

$$\begin{aligned} \{3s^2,\,3\delta s,\,6\delta s,12\delta s\} \end{aligned}$$

with $\delta$ being any divisor of 3s, or in

$$\begin{aligned} \{2\delta s,\, 4\delta s\}. \end{aligned}$$

if and only if $\delta$ is a divisor of 3s and all prime divisors of $3s/\delta$ are equal to $1 \ \text {mod}\ 6$.

If all prime factors of $3s/\delta$ are $1 \ \text {mod}\ 6$, $\delta$ must be divisible by 3. Say $\delta =3d$, thus $\{2\delta s,\, 4\delta s\}=\{6d s,\, 12d s\}$ where d is a divisor of s and all prime divisors of s/d are equal to $1 \ \text {mod}\ 6$. These degrees are in (2).

Let us now prove that the degrees $3\delta s$, $6\delta s$ and $12\delta s$ correspond to the ones listed in (1). We need only to prove that $\delta$ is divisible by 3.

Consider now that G is the group of $\{3,6\}_{(3s,0)}$ and K is the action of T on block 1, where $T=\langle u,v\rangle$ is the translation group of order $(3s)^2$. The group of $\{3,6\}_{(s,s)}$ is a factorization of the group of $\{3,6\}_{(3s,0)}$ by $\langle (uv)^s\rangle$. A faithful transitive permutation representation of $\{3,6\}_{(3s,0)}$ on n points corresponds to a permutation representation of $\{3,6\}_{(s,s)}$ on n/3 triples of points of the form

$$\begin{aligned} \{x,x(uv)^s, x(uv)^{2s}\}. \end{aligned}$$

Let $B{:}{=}|K:\langle u_1\rangle |$ and $C{:}{=}|K:\langle v_1\rangle |$, where $u_1$ and $v_1$ are the action of u and v on block 1. By Lemma 2.1, $\delta =gcd(B,C)$.

Suppose that B divides s. Then, $(u_1v_1)^s=u_1^{is}$ for some integer $i\in \{0,1,2\}$. In this case, the triples of points are as follows:

$$\begin{aligned} \{x, xu_1^{is}, xu_1^{2is}\}, \end{aligned}$$

for x in block 1. But then both $u^s$ and $v^s$ fix these triples. By conjugation with $\rho _0$, $\rho _1$ and $\rho _2$, we get that the order of u and v in $\{3,6\}_{(s,s)}$ is at most s, a contradiction. Thus, B does not divide s. In particular, 3 must divide B. Analogously, 3 divides C, and therefore, $\delta =gcd(B,C)$ is divisible by 3, as wanted. $\square$