Determining skew left braces of size np

TL;DR

This paper characterizes skew left braces of size np where p is an odd prime and n meets specific conditions, providing a classification, an algorithm for enumeration, and confirming a conjecture for size 12p.

Contribution

It introduces a new structural decomposition for skew left braces of size np and develops an algorithm to enumerate all such braces, confirming a conjecture for size 12p.

Findings

Classified skew left braces of size np under certain conditions.

Developed an algorithm to enumerate all skew left braces of size np.

Proved the conjecture for skew left braces of size 12p with p ≥ 7.

Abstract

We define the twofold semidirect product of two skew left braces, in which both the additive and multiplicative groups are semidirect products of the corresponding groups of the given skew left braces. We consider an odd prime $p$ and an integer $n$ satisfying $p\nmid n$, $p\nmid|\mathrm{Aut}(E)|$ for every group $E$ of order $n$ and such that each group of order $np$ has a unique $p$-Sylow subgroup. Under these conditions, we prove that any skew left brace of size $np$ is either a twofold semidirect product of the trivial brace of size $p$ and a skew left brace of size $n$ or a companion skew left brace of that one. We develop an algorithm to obtain all skew left braces of size $np$ from the skew left braces of size $n$ and provide a formula to count them. We use this result to describe all skew left braces of size $12p$ for $p\geq 7$, which proves a conjecture of V.G. Bardakov, M.V. Neshchadim and M.K. Yadav.

Equations

$$a\circ (b\cdot c)=a\circ b\cdot a^{-1}\cdot a\circ c,$$

$$s(12p)=\left\{\begin{array}[]{ll}324&\text{\ if }p\equiv 11\pmod{12},\\ 410&\text{\ if }p\equiv 5\pmod{12},\\ 606&\text{\ if }p\equiv 7\pmod{12},\\ 782&\text{\ if }p\equiv 1\pmod{12}.\end{array}\right.$$

$$(a,b)\cdot (a^{\prime },b^{\prime })=(a\cdot a^{\prime },b\cdot b^{\prime })(a,b)\circ (a^{\prime },b^{\prime })=(a\circ a^{\prime },b\circ b^{\prime })$$

$$(a,b)\cdot (a^{\prime },b^{\prime })=(a\cdot a^{\prime },b\cdot b^{\prime })$$

$$(a,b)\circ (a^{\prime },b^{\prime })=(a\circ \tau (b)(a^{\prime }),b\circ b^{\prime })$$

$$\begin{array}[]{llll}\lambda_{a}:&A&\rightarrow&A\\ &b&\mapsto&a^{-1}\cdot a\circ b\end{array}$$

$$\begin{array}[]{llll}\lambda_{a}^{-1}:&A&\rightarrow&A\\ &b&\mapsto&\overline{a}\circ(a\cdot b)\end{array},$$

$$\sigma :(B,\cdot )\to \mathrm{Aut}(A,\cdot ,\circ ),\tau :(B,\circ )\to \mathrm{Aut}(A,\cdot ,\circ )$$

$$(a,b)\cdot (a^{\prime },b^{\prime })=(a\cdot \sigma (b)(a^{\prime }),b\cdot b^{\prime })$$

$$(a,b)\circ (a^{\prime },b^{\prime })=(a\circ \tau (b)(a^{\prime }),b\circ b^{\prime })$$

$$\sigma (b_1\circ b_2\cdot b_1^{-1}){\lambda }_a={\lambda }_a\tau (b_1)\sigma (b_2)\tau (\overline{b_1}).$$

$$(a_1,b_1)\circ \left((a_2,b_2)\cdot (a_3,b_3)\right)=(a_1,b_1)\circ (a_2,b_2)\cdot (a_1,b_1{)}^{-1}\cdot (a_1,b_1)\circ (a_3,b_3)$$

$$\begin{array}[]{lll}(a_{1},b_{1})\circ\left((a_{2},b_{2})\cdot(a_{3},b_{3})\right)&=&(a_{1},b_{1})\circ(a_{2}\cdot\sigma(b_{2})(a_{3}),b_{2}\cdot b_{3})\\ &=&(a_{1}\circ\tau(b_{1})(a_{2}\cdot\sigma(b_{2})(a_{3})),b_{1}\circ(b_{2}\cdot b_{3}))\end{array}$$

$$\begin{array}[]{l}(a_{1},b_{1})\circ(a_{2},b_{2})\cdot(a_{1},b_{1})^{-1}\cdot(a_{1},b_{1})\circ(a_{3},b_{3})\\ =(a_{1}\circ\tau(b_{1})(a_{2}),b_{1}\circ b_{2})\cdot(\sigma(b_{1}^{-1})(a_{1}^{-1}),b_{1}^{-1})\cdot(a_{1}\circ\tau(b_{1})(a_{3}),b_{1}\circ b_{3})\\ =(a_{1}\circ\tau(b_{1})(a_{2})\cdot\sigma(b_{1}\circ b_{2})((\sigma(b_{1}^{-1})(a_{1}^{-1})),b_{1}\circ b_{2}\cdot b_{1}^{-1})\cdot(a_{1}\circ\tau(b_{1})(a_{3}),b_{1}\circ b_{3})\\ =(a_{1}\circ\tau(b_{1})(a_{2})\cdot\sigma(b_{1}\circ b_{2}\cdot b_{1}^{-1})(a_{1}^{-1})\cdot\sigma(b_{1}\circ b_{2}\cdot b_{1}^{-1})(a_{1}\circ\tau(b_{1})(a_{3})),b_{1}\circ b_{2}\cdot b_{1}^{-1}\cdot b_{1}\circ b_{3})\end{array}$$

$$\begin{array}[]{lll}a_{1}\circ\tau(b_{1})(a_{2}\cdot\sigma(b_{2})(a_{3}))&=&a_{1}\circ\left(\tau(b_{1})(a_{2})\cdot\tau(b_{1})(\sigma(b_{2})(a_{3})\right)\\ &=&a_{1}\circ\tau(b_{1})(a_{2})\cdot a_{1}^{-1}\cdot a_{1}\circ\tau(b_{1})(\sigma(b_{2})(a_{3}))\\ &=&a_{1}\circ\tau(b_{1})(a_{2})\cdot\lambda_{a_{1}}(\tau(b_{1})(\sigma(b_{2})(a_{3})))\end{array}$$

$$\begin{array}[]{l}a_{1}\circ\tau(b_{1})(a_{2})\cdot\sigma(b_{1}\circ b_{2}\cdot b_{1}^{-1})(a_{1}^{-1})\cdot\sigma(b_{1}\circ b_{2}\cdot b_{1}^{-1})(a_{1}\circ\tau(b_{1})(a_{3}))\\ =a_{1}\circ\tau(b_{1})(a_{2})\cdot\sigma(b_{1}\circ b_{2}\cdot b_{1}^{-1})(a_{1}^{-1}\cdot a_{1}\circ\tau(b_{1})(a_{3}))\\ =a_{1}\circ\tau(b_{1})(a_{2})\cdot\sigma(b_{1}\circ b_{2}\cdot b_{1}^{-1})(\lambda_{a_{1}}(\tau(b_{1})(a_{3}))).\end{array}$$

$${\lambda }_{a_1}(\tau (b_1)(\sigma (b_2)(a_3)))=\sigma (b_1\circ b_2\cdot b_1^{-1})({\lambda }_{a_1}(\tau (b_1)(a_3)))$$

$$\begin{array}[]{rcl}(a,b)\cdot(a^{\prime},b^{\prime})&=&(a+\sigma(b)a^{\prime},bb^{\prime})\\ (a,b)\circ(a^{\prime},b^{\prime})&=&(a+\tau(b)a^{\prime},b\circ b^{\prime})\end{array}.$$

$$(a,b)\circ \left((a^{\prime },b^{\prime })\cdot (a^{\prime \prime },b^{\prime \prime })\right)=(a,b)\circ (a^{\prime },b^{\prime })\cdot (a,b{)}^{-1}\cdot (a,b)\circ (a^{\prime \prime },b^{\prime \prime }).$$

$$\begin{array}[]{rcl}(a,b)\circ\left((a^{\prime},b^{\prime})\cdot(a^{\prime\prime},b^{\prime\prime})\right)&=&(a,b)\circ\left(a^{\prime}+\sigma(b^{\prime})a^{\prime\prime},b^{\prime}\cdot b^{\prime\prime}\right)\\ &=&\left(a+\tau(b)(a^{\prime}+\sigma(b^{\prime})a^{\prime\prime}),b\circ(b^{\prime}\cdot b^{\prime\prime})\right)\end{array}$$

$$\begin{array}[]{l}(a,b)\circ(a^{\prime},b^{\prime})\cdot(a,b)^{-1}\cdot(a,b)\circ(a^{\prime\prime},b^{\prime\prime})\\ =(a+\tau(b)a^{\prime},b\circ b^{\prime})\cdot(-\sigma(b^{-1})a,b^{-1})\cdot(a+\tau(b)a^{\prime\prime},b\circ b^{\prime\prime})\\ =(a+\tau(b)a^{\prime}-\sigma(b\circ b^{\prime})\sigma(b^{-1})a,(b\circ b^{\prime})\cdot b^{-1})\cdot(a+\tau(b)a^{\prime\prime},b\circ b^{\prime\prime})\\ =(a+\tau(b)a^{\prime}-\sigma(b\circ b^{\prime})\sigma(b^{-1})a+\sigma((b\circ b^{\prime})\cdot b^{-1})(a+\tau(b)a^{\prime\prime}),(b\circ b^{\prime})\cdot b^{-1}\cdot(b\circ b^{\prime\prime})).\end{array}$$

$$a+\tau (b)(a^{\prime }+\sigma (b^{\prime })a^{\prime \prime })=a+\tau (b)a^{\prime }+\tau (b)\sigma (b^{\prime })a^{\prime \prime }$$

$$\begin{array}[]{l}a+\tau(b)a^{\prime}-\sigma(b\circ b^{\prime})\sigma(b^{-1})a+\sigma((b\circ b^{\prime})\cdot b^{-1})(a+\tau(b)a^{\prime\prime})\\ =a+\tau(b)a^{\prime}-\sigma((b\circ b^{\prime})\cdot b^{-1})a+\sigma((b\circ b^{\prime})\cdot b^{-1})a+\sigma((b\circ b^{\prime})\cdot b^{-1})\tau(b)a^{\prime\prime}\\ =a+\tau(b)a^{\prime}+\sigma((b\circ b^{\prime})\cdot b^{-1})\tau(b)a^{\prime\prime}\end{array}$$

$$\tau (b)\sigma (b^{\prime })a^{\prime \prime }=\sigma ((b\circ b^{\prime })\cdot b^{-1})\tau (b)a^{\prime \prime }$$

$$\sigma (b^{\prime })=\sigma ((b\circ b^{\prime })\cdot b^{-1})=\sigma (b\circ b^{\prime })\sigma (b^{-1})$$

$$\sigma (b\circ b^{\prime })=\sigma (b)\sigma (b^{\prime }),$$

$$\begin{array}[]{lcll}{\pi_{1}}_{|G}:&G&\rightarrow&A\\ &(a,f)&\mapsto&a\end{array}$$

$$\{(a,{\lambda }_a):a\in A\}$$

$$a\circ b=a\cdot f(b),$$

$${\lambda }_x(y)=x^{-1}\cdot (x\circ y)=(-\sigma (a^{-1})m,a^{-1})\cdot (m+\tau (a)n,a\circ b)=(\sigma (a^{-1})\tau (a)n,{\lambda }_a(b)),$$

Full text

(1) Departament de Matemàtiques i Informàtica, Universitat de Barcelona, Gran Via de les Corts Catalanes 585, 08007, Barcelona (Spain)

(2) Charles University, Faculty of Mathematics and Physics, Department of Algebra, Sokolovska 83, 18600 Praha 8, Czech Republic

(3) Departament de Matemàtiques, Universitat Politècnica de Catalunya, Edifici Omega, Jordi Girona, 1-3, 08034, Barcelona (Spain)

Keywords: Left braces, Holomorphs, regular subgroups, Hopf Galois structures.

2020MSC: Primary: 16T05, 16T25, 12F10; Secondary: 20B35, 20B05, 20D20, 20D45.

††footnotetext: The first author was supported by grant PID2019-107297GB-I00 (Ministerio de Ciencia, Innovación y Universidades). The second author was supported by Czech Science Foundation, grant 21-00420M, and by Charles University Research Centre program UNCE/SCI/022. Third and fourth authors are supported by 2021 SGR 01468. Email addresses: teresa.crespo@ub.edu, daniel.gil-munoz@mff.cuni.cz, ana.rio@upc.edu, montse.vela@upc.edu

1 Introduction

In [7] Rump introduced an algebraic structure called brace to study set-theoretic solutions of the Yang-Baxter equation. Guarnieri and Vendramin in [6] gave the more general concept of skew brace. A skew (left) brace is a set $B$ with two operations $\cdot$ and $\circ$ such that $(B,\cdot)$ and $(B,\circ)$ are groups and the brace relation is satisfied, namely,

[TABLE]

for all $a,b,c\in B$, where $a^{-1}$ denotes the inverse of $a$ with respect to $\cdot$. We call $(B,\cdot$) the additive group and $(B,\circ)$ the multiplicative group of the skew left brace. We say that $B$ is a classical brace if its additive group is abelian.

Let $B_{1}$ and $B_{2}$ be skew left braces. A map $f:B_{1}\to B_{2}$ is said to be a skew brace morphism if $f(a\cdot b)=f(a)\cdot f(b)$ and $f(a\circ b)=f(a)\circ f(b)$ for all $a,b\in B_{1}$. If $f$ is bijective, we say that $f$ is an isomorphism. In that case we say that the skew braces $B_{1}$ and $B_{2}$ are isomorphic.

In [6] Guarnieri and Vendramin proved that given a group $N$, there is a bijective correspondence between isomorphism classes of skew left braces with additive group $N$ and conjugacy classes of regular subgroups of $\mathrm{Hol}(N)$.

In [3], we classified classical left braces of size $8p$, for $p$ a prime number, $p\neq 3,7$. In [4], we gave a method to determine all braces of size $np$ from the braces of size $n$ and certain classes of morphisms from the multiplicative group of these braces of size $n$ to $\mathbf{Z}_{p}^{*}$. For a prime number $p\geq 7$, we applied the obtained results to classify classical left braces of size $12p$. In particular we proved the conjectures in [1] on the numbers $b(8p),b(12p)$ of isomorphism classes of classical left braces of size $8p$ and $12p$, respectively.

Our aim in this paper is to generalize the method obtained in [4] to skew nonclassical left braces of size $np$, where $p$ is a prime number and $n$ an integer not divisible by $p$ and such that every group of order $np$ has a normal subgroup of order $p$. We shall apply our method to classify skew left braces of size $12p$, for $p$ a prime number, $p\geq 7$. In particular we will prove the following conjecture in [1] on the number $s(12p)$ of isomorphism classes of skew left braces of size $12p$, $p\geq 7$.

[TABLE]

We note that $s(24)=855,s(36)=400$ and $s(60)=418$ (see [9]).

2 Skew braces of size $np$

We recall the definition of direct and semidirect product of braces as defined in [2] and [8].

Let $A$ and $B$ be skew left braces. Then $A\times B$ together with

[TABLE]

is a skew left brace called the direct product of the braces $A$ and $B$.

Now, let $\tau:(B,\circ)\to\operatorname{Aut}(A,\cdot,\circ)$ be a group homomorphism. Consider in $A\times B$ the additive structure of the direct product $(A,\cdot)\times(B,\cdot)$

[TABLE]

and the multiplicative structure of the semidirect product $(A,\circ)\rtimes_{\tau}(B,\circ)$

[TABLE]

Then, we obtain a skew left brace, which is called the semidirect product of the skew left braces $A$ and $B$ via $\tau$.

We observe in particular that the additive group of the semidirect product of two left braces is the direct product of the corresponding additive groups. In [4] we proved that a classical brace of size $np$, where $p$ is a prime number and $n$ an integer not divisible by $p$ and such that every group of order $np$ has a normal subgroup of order $p$, is a direct or semidirect product of the trivial brace of size $p$ and a brace of size $n$. In this case, the additive group was a direct product of $C_{p}$ and an abelian group of order $n$. However we are now interested in skew braces of size $np$, for which both additive and multiplicative groups may be strictly semidirect products of $C_{p}$ and a group of order $n$. We want then to determine the relationship between these skew braces of order $np$ and skew braces of order $n$. To this end we shall define the double semidirect product of two skew braces.

Let us recall first that for a skew left brace $(A,\cdot,\circ)$ and each $a\in A$, the map

[TABLE]

is bijective with inverse

[TABLE]

where $\overline{a}$ denotes the inverse of $a$ with respect to $\circ$. It follows that $a\circ b=a\cdot\lambda_{a}(b),a\cdot b=a\circ\lambda_{a}^{-1}(b)$.

Proposition 1**.**

Let $A$ and $B$ be skew left braces and let

[TABLE]

be morphisms of groups. Consider in $A\times B$ the additive structure of the semidirect product $(A,\cdot)\rtimes_{\sigma}(B,\cdot)$

[TABLE]

and the multiplicative structure of the semidirect product $(A,\circ)\rtimes_{\tau}(B,\circ)$

[TABLE]

We assume that the following equality holds for any $a\in A$ and all $b_{1},b_{2}\in B$

[TABLE]

Then $(A\times B,\cdot,\circ)$ is a skew brace.

Proof.

We check the brace condition for $(A\times B,\cdot,\circ)$, i.e., the equality

[TABLE]

for $(a_{1},b_{1}),(a_{2},b_{2}),(a_{3},b_{3})\in A\times B$. The term on the left is

[TABLE]

and the term on the right is

[TABLE]

Since $B$ is a brace, the two second components are equal. Now, the first component of the term on the left is

[TABLE]

where the second equality follows from the brace condition for $A$. The first component of the term on the right is

[TABLE]

The equality of the first components is then equivalent to

[TABLE]

for all $a_{3}\in A$, hence to equality (2). ∎

Definition 2**.**

We call double semidirect product of the skew braces $A$ and $B$ via $\sigma$ and $\tau$ the skew brace $(A\times B,\cdot,\circ)$ in Proposition 1.

Remarks 3**.**

If $\sigma$ is the trivial morphism, then (2) is satisfied and we recover the notion of semidirect product of skew braces. 2. 2)

If $A$ is a trivial brace and $\operatorname{Aut}A$ is abelian, then (2) is equivalent to $\sigma(b_{1}\circ b_{2})=\sigma(b_{1})\sigma(b_{2})$ for all $b_{1},b_{2}\in B$, i.e. to $\sigma$ being a group morphism with respect to $\circ$.

Turning back to skew braces of size $np$, we establish the following hypothesis on $n$ and $p$ which will be maintained all along the paper.

Hypothesis**.**

Let $p$ be an odd prime and $n$ an integer such that $p$ does not divide $n$ and each group of order $np$ has a normal subgroup of order $p$.

In particular, we may consider $n=12$ and $p\geq 7$, which satisfy this hypothesis.

Proposition 4**.**

Any skew brace of size $np$ is a double semidirect product of the trivial brace of size $p$ and a skew brace of size $n$.

Proof.

Let $(B_{np},\cdot,\circ)$ be a left brace of size $np$ with additive group $N$ and multiplicative group $G$. Then, by the Schur-Zassenhaus theorem, $N=\mathbf{Z}_{p}\rtimes_{\sigma}E$ with $(E,\cdot)$ a group of order $n$ and $\sigma:E\to\operatorname{Aut}(\mathbf{Z}_{p})$ a group morphism and $G=\mathbf{Z}_{p}\rtimes_{\tau}F$ with $(F,\circ)$ a group of order $n$ and $\tau:F\to\operatorname{Aut}(\mathbf{Z}_{p})$ a group morphism. We note that $\operatorname{Aut}(\mathbf{Z}_{p})$ is the group of automorphisms of the trivial brace of size $p$, which is the unique brace of size $p$. We write $\mathbf{Z}_{p}$ additively and identify $\operatorname{Aut}(\mathbf{Z}_{p})$ with $\mathbf{Z}_{p}^{*}$. As a set, $B_{np}$ is the cartesian product of $\mathbf{Z}_{p}$ and a set $B_{n}$ with $n$ elements endowed with two operations $\cdot$ and $\circ$ such that $(B_{n},\cdot)$ and $(B_{n},\circ)$ are groups isomorphic to $(E,\cdot)$ and $(F,\circ)$, respectively.

For $(a,b),(a^{\prime},b^{\prime})\in B_{np}$, we have then

[TABLE]

We write the brace relation for $B_{np}$

[TABLE]

The lefthand side is

[TABLE]

and the righthand side is

[TABLE]

The equality of the second components gives a brace $B_{n}$ of size $n$ with additive group $E$ and multiplicative group $F$. The first component in the lefthand side is

[TABLE]

and the first component in the righthand side is

[TABLE]

The equality of the first components is then equivalent to

[TABLE]

for all $a^{\prime\prime}\in\mathbf{Z}_{p}$, hence to

[TABLE]

which is equivalent to

[TABLE]

for all $b,b^{\prime}\in B_{n}$, hence to $\sigma$ being also a morphism with respect to $\circ$.

We have then obtained that $B_{np}$ is the double semidirect product of the trivial brace $\mathbf{Z}_{p}$ and the skew brace $B_{n}$. ∎

We recall now precisely the correspondence between braces and regular subgroups of the holomorph given by Guarnieri and Vendramin. We note that, if $G$ is a regular subgroup of $\mathrm{Hol}(A,\cdot)$, then

[TABLE]

is bijective.

Proposition 5** ([6] th. 4.2).**

Let $A$ be a skew left brace. Then

[TABLE]

is a regular subgroup of $\mathrm{Hol}(A,\cdot)$, isomorphic to $(A,\circ)$.

Conversely, if $(A,\cdot)$ is a group and $G$ is a regular subgroup of $\mathrm{Hol}(A,\cdot)$, then $A$ is a skew left brace with $(A,\circ)\simeq G$, where

[TABLE]

and $({\pi_{1}}_{|G})^{-1}(a)=(a,f)\in G$.

These assignments give a bijective correspondence between the set of isomorphism classes of skew left braces $(A,\cdot,\circ)$ and the set of conjugation classes of regular subgroups of $\mathrm{Hol}(A,\cdot)$.

Using this correspondence we can see how braces of size $np$ give braces of size $n$ and how to obtain braces of size $np$ from the trivial brace of size of size $p$ and braces of size $n$.

Let $(B_{np},\cdot,\circ)$ be a brace of size $np$ with additive group $N=\mathbf{Z}_{p}\rtimes_{\sigma}E$, as above, and multiplicative group $G=\mathbf{Z}_{p}\rtimes_{\tau}F$. By applying Proposition 5, we obtain that $\{(x,\lambda_{x})\,:\,x\in N\}$ is a regular subgroup of $\mathrm{Hol}(N)$, isomorphic to $G$. Now, for $x=(m,a),y=(n,b)\in N$, we have

[TABLE]

hence $\pi_{2}\lambda_{x}=\lambda_{a}\pi_{2}$, where $\pi_{2}:N=\mathbf{Z}_{p}\rtimes_{\sigma}E\rightarrow E$ is the projection on the second component.

Let us see these regular groups explicitly in $\mathrm{Hol}(N)$. Following [5], the automorphisms of $N$ can be described as matrices

[TABLE]

with $k\in\mathbf{Z}_{p}^{*}=\operatorname{Aut}(\mathbf{Z}_{p})$, $\lambda\in\operatorname{Aut}(E)$, such that $\sigma\lambda=\sigma$ and $\gamma:E\to\mathbf{Z}_{p}$ a $1$-cocycle. The action on $N$ is given by

[TABLE]

Under our hypothesis, Schur Zassenhaus condition on subgroups of order $|E|$ being conjugates of $E$ by elements in $\mathbf{Z}_{p}$ leaves us with the coboundaries

[TABLE]

For a trivial $\sigma$, we get the direct product $\operatorname{Aut}(\mathbf{Z}_{p})\times\operatorname{Aut}(E)$ and for a non trivial $\sigma$ the group $\operatorname{Aut}(N)$ has order $p(p-1)s$ with $s=\#\{\lambda\in\operatorname{Aut}(E):\sigma\lambda=\sigma\}$. This subgroup is the stabilizer $\Sigma_{\sigma}$ of $\sigma$ under the action

[TABLE]

The orbit of $\sigma$ gives the $\sigma^{\prime}$ such that $\mathbf{Z}_{p}\rtimes_{\sigma^{\prime}}E\simeq N$.

All together, if we write

[TABLE]

then

[TABLE]

with

[TABLE]

where $\rtimes$ indicates operation in $\mathbf{Z}_{p}\rtimes_{\sigma}E$. For a brace $B_{np}$ as before, this is the $\cdot$ operation. The set $\{(0,a)\mid a\in E\}$ is stable under this operation. For the multiplicative group $G=\{(x,\lambda_{x})\mid\ x\in N\}$, if $x=(m,a)$ we have

[TABLE]

The set $\{(a,\lambda_{a})\mid a\in E\}$ is a group isomorphic to $F$, and a regular subgroup of $\mathrm{Hol}(E)$. This describes the subbbrace $B_{n}$ of $B_{np}$ with additive structure isomorphic to $E$ and multiplicative structure isomorphic to $F$. Notice that in the description of automorphisms of $N$ we got the condition $\sigma\in\mathrm{Hom}(B_{n},\mathbf{Z}_{p}^{*})$, namely $\sigma\lambda_{a}=\sigma$ for all $a\in E$.

From the above, we have also obtained a way to construct a brace of size $np$ from a brace of size $n$. If $E$ is a group of order $n$, $\sigma\in\mathrm{Hom}(E,\mathbf{Z}_{p}^{*})$ and $F$ is a regular subgroup of $\mathrm{Hol}(E)$ such that $\pi_{2}(F)$ is inside the stabilizer of $\sigma$ by the action of $\operatorname{Aut}(E)$, then for any $\tau\in\mathrm{Hom}(F,\mathbf{Z}_{p}^{*})$

[TABLE]

is a regular subgroup of $\mathrm{Hol}(\mathbf{Z}_{p}\rtimes_{\sigma}E)$ isomorphic to $\mathbf{Z}_{p}\rtimes_{\tau}F.$

But we can also consider the opposite group operation on $N$, so that elements in $N$ are $(\sigma(a)m,a)$, with $m\in\mathbf{Z}_{p}$ and $a\in E$. Then,

[TABLE]

is a regular embedding which restricted to $E$ provides an embedding of $F$:

[TABLE]

[TABLE]

Using again [6] we get the regular subgroup

[TABLE]

isomorphic to $\mathbf{Z}_{p}\rtimes_{\tau}F$.

Proposition 6**.**

Let $p$ be a prime and $n$ a positive integer not divisible by $p$ and such that every group of order $np$ has a normal subgroup of order $p$. Let $B_{n}$ a brace of order $n$ with additive structure $(E,\cdot)$ and multiplicative structure $(F,\circ)$. For every $\sigma\in\mathrm{Hom}(B_{n},\mathbf{Z}_{p}^{*})$ and every $\tau\in\mathrm{Hom}(F,\mathbf{Z}_{p}^{*})$,

[TABLE]

[TABLE]

are regular subgroups of $\mathrm{Hol}(\mathbf{Z}_{p}\rtimes_{\sigma}E)$ isomorphic to $\mathbf{Z}_{p}\rtimes_{\tau}F$.

We have $G(1,\tau)=G(1,\tau)^{\prime}$ for any $\tau$. For a nontrivial $\sigma$, $G(\sigma,\tau)$ and $G(\sigma,\tau)^{\prime}$ are not conjugate in $\mathrm{Hol}(N)$.

Any regular subgroup of $\mathrm{Hol}(\mathbf{Z}_{p}\rtimes_{\sigma}N)$ isomorphic to $\mathbf{Z}_{p}\rtimes_{\tau}F$ is conjugate to $G(\sigma,\tau)$ or $G(\sigma,\tau)^{\prime}$ by an element of $\operatorname{Aut}(N)$.

When $\sigma=1$, the coboundaries are also trivial, and both subgroups are equal. For a nontrivial $\sigma$, since $\mathbf{Z}_{p}^{*}$ is abelian, the first coefficient of the matrix is invariant under conjugation, and this shows that the groups are not conjugate in $\mathrm{Hol}(N)$ for any $\tau$.

Let us prove the last statement. If $G\subseteq\mathrm{Hol}(N)$ is a regular subgroup isomorphic to $\mathbf{Z}_{p}\rtimes_{\tau}F$, then via $\mathbf{Z}_{p}\rtimes_{\tau}F\simeq G\hookrightarrow\mathrm{Hol}(N)$, using [6] and $\mathrm{Hom}(\mathbf{Z}_{p},\mathbf{Z}_{p}^{*})=1$, we get a semiregular embedding of $\mathbf{Z}_{p}$ in $\mathrm{Hol}(N)$,

[TABLE]

so that $f\in\operatorname{Aut}(\mathbf{Z}_{p})$ and $h\in\mathrm{Hom}(\mathbf{Z}_{p},\mathbf{Z}_{p})$, and the embedding is

[TABLE]

for some $k\in\mathbf{Z}_{p}^{*},\ l\in\mathbf{Z}_{p}$. It is easy to check that conjugation by elements in $\operatorname{Aut}(N)$ splits this family into two classes represented by $\nu_{1,0}$ and $\nu_{1,l}$ for any $l\in\mathbf{Z}_{p}^{*}$, and that conjugation is given by an element in $\operatorname{Aut}(\mathbf{Z}_{p})$.

On the other hand, we get a semiregular embedding of $F$ in $\mathrm{Hol}(N)$. By [6] and using $\mathrm{Hom}(E,\mathbf{Z}_{p})=1$, it has to be

[TABLE]

with $\alpha\in\mathrm{Hom}(B_{n},\mathbf{Z}_{p}^{*})$. The conjugacy class by elements of $\operatorname{Aut}(N)$ of the image of $\Psi_{\alpha}$ is

[TABLE]

so that $\Psi_{\alpha}(F)$ and $\Psi_{\alpha^{\prime}}(F)$ are not conjugate when $\alpha\neq\alpha^{\prime}$. Notice that $\{(\lambda_{M}(x),\lambda_{M}\lambda_{a}\lambda_{M}^{-1})\mid(a,\lambda_{a})\in F\}=\Phi_{\lambda_{M}}(F).$ It is equal to $F$ if and only if $\lambda_{M}$ is a brace automorphism. Notice also that the image is invariant under conjugation by $\operatorname{Aut}(\mathbf{Z}_{p})$. This gives the following result.

Proposition 7**.**

Let $F,F^{\prime}$ be regular subgroups of $\mathrm{Hol}(E)$ giving multiplicative structures for a brace of order $n$. If $\Psi_{\alpha}(F)$ and $\Psi_{\alpha}(F^{\prime})$ are conjugate in $\mathrm{Hol}(N)$ then they are conjugate in $\mathrm{Hol}(E)$.

Back to the proof of the previous proposition, up to conjugation by elements in $\operatorname{Aut}(\mathbf{Z}_{p})$, the group $G$ has to be the semidirect product of $\Psi_{\alpha}(F)$ and either $\nu_{1,0}(\mathbf{Z}_{p})$ or $\nu_{1,l}(\mathbf{Z}_{p})$. An easy computation shows that the only possibilities are those corresponding to $G(\sigma,\tau)$ and $G(\sigma,\tau)^{\prime}$, namely $\nu_{1,0}$ and $\alpha(a)=\tau(a,\lambda_{a})\sigma(a)^{-1}$ or $\nu_{1,-1}$ and $\alpha(a)=\tau(a,\lambda_{a})$. Indeed,

[TABLE]

gives the image of the product $(m+\tau(a,\lambda_{a})n,(a\lambda_{a}(b),\lambda_{a}\lambda_{b}))$ if and only if $\alpha(a)=\tau(a,\lambda_{a})\sigma(a)^{-1}$, and

[TABLE]

[TABLE]

gives the image of the product if and only if $l=-1$ and $\alpha(a)=\tau(a,\lambda_{a})$.

In order to count different brace structures, we have to determine conjugation orbits in the families $\{G(\sigma,\tau)\}_{\sigma,\tau},$ and $\{G(\sigma,\tau)^{\prime}\}_{\sigma,\tau}$ or equivalently, isomorphism classes of braces structures. Therefore, we should take $\sigma$ up to brace automorphisms of $(B_{n},\cdot,\circ)$ and $\tau$ up to brace automorphisms of $(B_{n},\circ)$.

2.1 Algorithm

Our aim is to determine all skew braces of size $np$ from skew braces of size $n$. Therefore we need a precomputation

Step 0

Determine isomorphism classes of groups $E$ of order $n$ and the number of braces of size $n$ of each type $(E,F)$, identifying $F$ with a representative for a conjugacy class of regular subgroups of $\mathrm{Hol}(E)$.

For every $E$ compute the stabilisers $\Sigma_{\sigma}$ for the action

[TABLE]

namely, $\Sigma_{\sigma}=\{g\in\operatorname{Aut}(E):\ \sigma g=\sigma\}$.

Next, we consider as an input a pair $(E,F)$, namely a brace $B_{n}$. The steps to count all the braces of size $np$ having additive group isomorphic to $\mathbf{Z}_{p}\rtimes E$ and multiplicative group isomorphic to $\mathbf{Z}_{p}\rtimes F$ are the following:

Step 1

Determine $\mathrm{Hom}(B_{n},\mathbf{Z}_{p}^{*})=\{\sigma\in\mathrm{Hom}(E,\mathbf{Z}_{p}^{*}):\ \pi_{2}(F)\subseteq\Sigma_{\sigma}\ \}$

Step 2

Determine $\operatorname{Aut}(B_{n})=\{g\in\operatorname{Aut}(E):\ \Phi_{g}(F)=F\ \}$

Step 3

Compute the number of orbits of the action

[TABLE]

Notice that the orbit of $\sigma=1$, which corresponds to the direct product, has a single element. The remaining orbits will give rise to two different braces.

The number of additive structures for $F$ is the number of orbits. Since orbits preserve order and isomorphism class of kernel, we can keep track of the number of structures according to these parameters.

Step 4

For each $\sigma$ in a system of representatives of the above orbits, compute orbits of the action

[TABLE]

Notice that $\operatorname{Aut}(B_{n})\cap\Sigma_{\sigma}$ is a subgroup of the normaliser of $\pi_{2}(F)$ in $\Sigma_{\sigma}$.

The number of multiplicative structures for $F$ is twice the number of orbits except for $\sigma$ trivial, when we get a single one. Since orbits preserve order and isomorphism class of kernel, we can keep track of the number of structures according to these parameters.

2.2 Number of braces of size $np$

Taking into account the description given in terms of orbits, using the Burnside lemma we can give formulas for the number of different braces of size $np$ in terms of the number of braces of size $n$.

Theorem 8**.**

Let $p$ be a prime and $n$ a positive integer not divisible by $p$ and such that every group of order $np$ has a normal subgroup of order $p$. For braces $B_{np}$ of size $np$ the number of additive structures is

[TABLE]

where $B_{n}$ runs over the braces of size $n$. The total amount of braces of size $np$ is

[TABLE]

where $H=\mathrm{Hom}((B_{n},\circ),\mathbf{Z}_{p}^{*})$, $\sigma\in\mathrm{Hom}(B_{n},\mathbf{Z}_{p}^{*})$ runs over a system of representatives for additive structures and $A_{\sigma}=\operatorname{Aut}(B_{n})\cap\mathrm{Stab}_{\operatorname{Aut}(B_{n},\cdot)}(\sigma)$. The case $\sigma=1$, where $A_{1}=\operatorname{Aut}(B_{n})$, corresponds to the semidirect product of $B_{n}$ and the trivial brace of order $p$. The remaining cases are double semidirect products of these two braces.

Equivalently, the number of additive structures can be expressed as

[TABLE]

and the total amount as

[TABLE]

where $\operatorname{Fix}$ denotes the number of fixed points.

3 Groups of order 12 and skew braces of size 12

There are five groups of order 12, up to isomorphism, two abelian ones $C_{12}$ and $C_{6}\times C_{2}$ and three non-abelian ones, the alternating group $A_{4}$, the dihedral group $D_{2\cdot 6}$ and the dicyclic group $\operatorname{Dic}_{12}$. By computation with Magma, we obtain that the number of skew left braces with additive group $E$ and multiplicative group $F$ is as shown in the following table.

[TABLE]

We determine now for each group $E$ of order 12, its automorphism group $\operatorname{Aut}E$, the group $\mathrm{Hom}(E,\mathbf{Z}_{p}^{*})$ and stabilizers of the corresponding group action.

Since $\mathbf{Z}_{p}^{*}=\langle\zeta\rangle$ is a cyclic group of order $p-1$, for each $d\mid p-1$ it has a unique subgroup $\mu_{d}=\langle\zeta^{\frac{p-1}{d}}\rangle$ of order $d$. We write $\zeta_{d}$ for $\zeta^{\frac{p-1}{d}}$. In order to have elements in $\mathrm{Hom}(E,\mathbf{Z}_{p}^{*})$ with image $\mu_{d}$ it is necessary that $d\mid\gcd(\exp(E),p-1)$, where $\exp(E)$ is the exponent of the group $E$.

3.1 $E=C_{12}$

Let $E=\langle c\rangle\simeq C_{12}$. Then, $\operatorname{Aut}E=\langle g_{5}\rangle\times\langle g_{7}\rangle\simeq\mathbf{Z}_{12}^{*}\simeq\mathbf{Z}_{2}\times\mathbf{Z}_{2}$ with $g_{i}(c)=c^{i}$. If $D=\gcd(12,p-1)$, then

[TABLE]

If $|\operatorname{Ker}\sigma|=k$, then the order of $\sigma$ in $\mathrm{Hom}(E,\mathbf{Z}_{p}^{*})$ is $d=\frac{12}{k}$. The Euler totient function gives the number $\varphi(d)$ of $\sigma$ having order $d$. Such a sigma can be identified as a pair $(d,j)$ meaning $\sigma(c)=\zeta_{d}^{j}$ for $j$ relatively prime to $d$. Elements of the same order are in the same orbit under the action of $\operatorname{Aut}(E)$

[TABLE]

and the size of the kernel parametrizes the orbits. As for stabilisers, since

[TABLE]

we have $\Sigma_{\sigma}=\{g_{i}\in\operatorname{Aut}(E):i\equiv 1\bmod d\},$ which identifies as subgroup of $\mathbf{Z}_{12}^{*}=\{1,5,7,11\}$. In the following table we compile all the information:

[TABLE]

The number of isomorphism classes of semidirect products $\mathbf{Z}_{p}\rtimes C_{12}$ is equal to the number of divisors of $\gcd(12,p-1)$, namely $\sigma_{0}(D)$, and can be labelled by the size $k$ of $\operatorname{Ker}\sigma$ (size 12 means direct product):

[TABLE]

3.2 $E=C_{6}\times C_{2}$

Let us write $E=\langle a\rangle\times\langle b\rangle$, with $a$ of order 6, $b$ of order 2. The automorphism group is $\operatorname{Aut}(C_{6}\times C_{2})\simeq D_{2\cdot 6}=\langle g_{1},\ g_{2}\rangle$, where

[TABLE]

$g_{1}$ has order 2, $g_{2}$ has order 6 and $g_{1}g_{2}g_{1}=g_{2}^{-1}$.

Since the exponent of $E$ is $6$, elements $\sigma\in\mathrm{Hom}(E,\mathbf{Z}_{p}^{*})$ have order dividing $6$. If $D=\gcd(6,p-1)$, we have

[TABLE]

For each $d\mid D$, if we write $(j\bmod d,i\bmod 2)$ for

[TABLE]

we get the following orbits for the action of $\operatorname{Aut}(E)$

[TABLE]

and again the order determines the orbit. Therefore, the size of kernel parameterise isomorphism classes of semidirect products.

[TABLE]

Finally, let us compute stabilizers for the first element in each orbit and take the corresponding conjugates. If $\sigma(b)=1$, then $\sigma(g_{1}(a))=\sigma(ab)=\sigma(a),$ $\sigma(g_{1}(b)=\sigma(b)$ and we have $g_{1}\in\mathrm{Stab}_{\sigma}$. Since $\langle g_{1}\rangle$ has index 6, this is the stabilizer for $d=6$. For the other values, the (cyclic) orbits above show the power of $g_{2}$ that we have to take: $g_{2}^{3}$ for $d=2$ and $g_{2}^{2}$ for $d=3$.

[TABLE]

3.3 $E=A_{4}$

Let $E$ be the alternating group $A_{4}$, and take generators $(1,2,3)$ and $(1,2)(3,4)$. We have $\operatorname{Aut}A_{4}\simeq S_{4}$, the isomorphism given by restriction to $A_{4}$ of inner automorphisms of $S_{4}$. Therefore the action is

[TABLE]

for any $s\in S_{4}$.

Since the exponent of $E$ is $3$, nontrivial elements $\sigma\in\mathrm{Hom}(E,\mathbf{Z}_{p}^{*})$ have order $3$, which forces them to be trivial on $(1,2)(3,4)$ and every element of order 2, namely, $\operatorname{Ker}\sigma\simeq V_{4}$. For $p\not\equiv 1\bmod 3$, $\mathrm{Hom}(E,\mathbf{Z}_{p}^{*})=1$ and for $p\equiv 1\bmod 3$

[TABLE]

For any $s\in S_{4}$ such that $(1,2,3)(s(1),s(2),s(3))$ has order 1 or 2, for example $s=(1,2)$, we have $\sigma_{1}=\sigma_{2}\phi_{s}$ so that they are in the same orbit for the action of $\operatorname{Aut}(E)$. The stabiliser is a normal subgroup of $S_{4}$ of index $2$, namely $\Sigma_{\sigma_{1}}=\Sigma_{\sigma_{2}}=A_{4}$.

[TABLE]

3.4 $E=D_{2\cdot 6}$

Let us write $E=\langle r,s\mid r^{6}=\operatorname{Id},\ s^{2}=\operatorname{Id},\ srs=r^{5}\rangle$. We note that $\operatorname{Aut}E$ is generated by

[TABLE]

$g_{1}$ has order 2, $g_{2}$ has order 6 and $g_{1}g_{2}g_{1}=g_{2}^{-1}$, hence $\operatorname{Aut}D_{2\cdot 6}\simeq D_{2\cdot 6}$.

Since $E$ has exponent $6$ and the unique normal subgroups with cyclic kernel are those of índex 2, we have that for every $p$,

[TABLE]

$(1,-1)$ has cyclic kernel $\langle r\rangle$ while $(-1,-1)$ and $(-1,1)$ have kernel isomorphic to dihedral group $D_{2\cdot 3}$. Therefore, they give at have at least two orbits under the action of $\operatorname{Aut}(E)$. Since $(-1,-1)g_{2}(r)=(-1,-1)(r)=-1$ and $(-1,-1)g_{2}(s)=(-1,-1)(rs)=1$, namely $(-1,-1)g_{2}=(-1,1)$, we have three orbits for the action, which means three isomorphism classes of semidirect (or direct) products $\mathbf{Z}_{p}\rtimes E$, parameterised by the isomorphism class of the kernel. For $\sigma=(1,-1)$, we denote $\mathbb{Z}_{p}\rtimes_{\sigma}E=\mathbb{Z}_{p}\rtimes_{6}^{c}E$ (cyclic kernel), while for $\sigma=(-1,-1)$ we write $\mathbb{Z}_{p}\rtimes_{\sigma}E=\mathbb{Z}_{p}\rtimes_{6}^{d}E$ (dihedral kernel).

As for stabilisers, the size is determined by the length of the orbit and we have

[TABLE]

Therefore, the action of $\operatorname{Aut}(E)$ can be described as follows:

[TABLE]

3.5 $E=\operatorname{Dic}_{12}$

Let us write $E=\langle x,y\mid x^{3}=1,y^{4}=1,yxy^{-1}=x^{2}\rangle$. The group $\operatorname{Aut}E$ is generated by

[TABLE]

$g_{1}$ has order 2, $g_{2}$ has order 6 and $g_{1}g_{2}g_{1}=g_{2}^{-1}$, hence $\operatorname{Aut}\operatorname{Dic}_{12}\simeq D_{2\cdot 6}$.

The exponent of the dicyclic group is $6$ (the element $xy$ has order $6$) and the normal subgroups with cyclic quotient are $\langle x^{2}\rangle$ and the derived subgroup $\langle x\rangle$. Therefore, $\sigma(x)=1$ for all $\sigma\in\mathrm{Hom}(E,\mathbf{Z}_{p}^{*})$ and the image of $y$ determines $\sigma$. If $D=\gcd(4,p-1)$, we have

[TABLE]

Again, we can identify the elements by a pair $(d,j)$. For $d=2$ we have a single element, and for $d=4$, since $g_{1}(y)=y^{-1}$, we have $(4,1)$ and $(4,3)$ in the same orbit. In this case, the stabiliser is a normal subgroup of $\operatorname{Aut}(E)$ of index $2$. We have

[TABLE]

and $(4,1)g_{2}^{3}=(4,1)g_{2}=(4,3).$ Therefore, the action goes as follows:

[TABLE]

4 Skew braces of size $12p$

Once we have done the precomputations for all groups $E$, we bring in the multiplicative group $F$ of the brace $B_{12}$. In all the sub-cases we follow the procedure described at the end of Section 2 and the notations in Section 3 for $E$, $\operatorname{Aut}(E)$ and $\mathrm{Hom}(E,\mathbf{Z}_{p}^{*})$.

4.1 $F=C_{12}$

According to the table in Section 3 and Proposition 4, for a brace of size 12 with multiplicative group $C_{12}$, the additive group can be any group except for $A_{4}$. In all the sub-cases the group $\mathrm{Hom}(F,\mathbf{Z}_{p}^{*})$ is as in 3.1 so that $\tau$ identifies as a pair $(d,j)$ meaning that the given generator of $F$ has image $\zeta_{d}^{j}.$

Case $E=C_{12}$

Let $E=\langle c\rangle\simeq C_{12}$. There is a unique conjugacy class of regular subgroups of $\mathrm{Hol}(E)$ isomorphic to $C_{12}$, given by the normal subgroup

[TABLE]

We have stabilisers $\Sigma_{\sigma}=\{g_{i}\in\operatorname{Aut}(E):i\equiv 1\bmod\mbox{ ord}(\sigma)\}.$ Since $\pi_{2}(F)=\{\operatorname{Id}\}$ we have $\mathrm{Hom}(B_{12},\mathbf{Z}_{p}^{*})=\mathrm{Hom}(E,\mathbf{Z}_{p}^{*})$. Since $F$ is a normal subgroup of $\mathrm{Hol}(E)$, we have $\Phi_{g}(F)=F$ for all $g\in\operatorname{Aut}(E)$ and $\operatorname{Aut}(B_{12})=\operatorname{Aut}(E).$ Therefore, the orbits of the action of $\operatorname{Aut}(B_{12})$ on $\mathrm{Hom}(B_{12},\mathbf{Z}_{p}^{*})$ are those computed in 3.1 and we have one orbit for each size of kernel. The number of addditive structures is the number of divisors $\sigma_{0}(\gcd(12,p-1))$.

Finally we have to compute orbits of the action of $\Sigma_{\sigma}$ on $\mathrm{Hom}(F,\mathbf{Z}_{p}^{*})$. Since $\Phi_{g_{i}}(c,\operatorname{Id})=(g_{i}(c),g_{i}\operatorname{Id}g_{i}^{-1})=(c^{i},\operatorname{Id})$ we have $(d,j)$ and $(d,j^{\prime})$ in the same orbit if and only if $j\equiv ij^{\prime}\bmod d$ for some $i\in\Sigma_{\sigma}$. Therefore, we have the following representatives for the orbits

[TABLE]

If we count the number of classes according to the size $\frac{12}{d}$ of $\operatorname{Ker}\tau$, we obtain

[TABLE]

For $k=12$, the trivial $\sigma$, we have one structure for each $\tau$, otherwise we have two multiplicative structures for each additive one.

Proposition 9**.**

Let $p$ be a prime number, $p\geq 7$. The number of braces with additive group $N:=\mathbf{Z}_{p}\rtimes_{k}C_{12}$ and multiplicative group $G:=\mathbf{Z}_{p}\rtimes_{k^{\prime}}C_{12}$ is as shown in the following table, where we need $p\equiv 1\pmod{\frac{12}{k}}$ for a kernel of size $k$ to occur.

[TABLE]

Case $E=C_{6}\times C_{2}$

We write $E=\langle a\rangle\times\langle b\rangle$. There is one brace with additive group $C_{6}\times C_{2}$ and multiplicative group $C_{12}$, hence one conjugation class of regular subgroups of $\mathrm{Hol}(E)$ isomorphic to $C_{12}$. We may consider as representative the following regular subgroup

[TABLE]

where $(a,g_{1})^{m}$ is $(a^{m}b^{\frac{m}{2}},\operatorname{Id})$ if $m$ is even and $(a^{m}b^{\frac{m-1}{2}},g_{1})$ if $m$ is odd.

Since $\pi_{2}(F)=\langle g_{1}\rangle$, computation of stabilisers $\Sigma_{\sigma}$ in 3.2 gives that the $\sigma$’s to consider are $(1\bmod d,0\bmod 2)$ for every $d|6$, meaning $\sigma(a)=\zeta_{d}$, $\sigma(b)=1$. Since there is only one for each order, they give a set of representatives for the action of $\operatorname{Aut}(B_{12})$. Therefore, we have 4 possible additive structures.

We consider now $\tau\in\mathrm{Hom}(F,\mathbf{Z}_{p}^{*})$ and the action by elements $\Phi_{g}$ such that $\Phi_{g}(F)=F$. We need $gg_{1}g^{-1}=g_{1}$, that is, $g$ belongs to the centraliser of $g_{1}$, which is the Klein group $\langle g_{2}^{3},g_{1}\rangle$. We have

[TABLE]

For $k=12,6$ we have all these automorphisms in $\Sigma_{\sigma}$, therefore, $(d,j)$ and $(d,j^{\prime})$ are conjugate. If $k=4,2$, the intersection between $\Sigma_{\sigma}$ and the centralizer is $\langle g_{1}\rangle$ and the orbit consists just in $(d,j)$ and $(d,7j\bmod d)$. Therefore, we have the following representatives for the orbits:

[TABLE]

Proposition 10**.**

Let $p$ be a prime number, $p\geq 7$. The number of braces with additive group $N:=\mathbf{Z}_{p}\rtimes(C_{6}\times C_{2})$ and multiplicative group $G:=\mathbf{Z}_{p}\rtimes C_{12}$ is as shown in the following table, where we need $p\equiv 1\pmod{\frac{12}{k}}$ for a kernel of size $k$ to occur.

[TABLE]

Case $E=D_{2\cdot 6}$

We write $E=\langle r,s\rangle$. There are two braces with additive group $D_{2\cdot 6}$ and multiplicative group $C_{12}$, hence two conjugation classes of regular subgroups of $\mathrm{Hol}(D_{2\cdot 6})$ isomorphic to $C_{12}$. We may consider the following regular subgroups of $\mathrm{Hol}(E)$ isomorphic to $C_{12}$,

[TABLE]

where $c_{1}^{m}$ is $(r^{6-m/2},g_{2}^{m})$ for $m$ even and $(r^{\frac{k-1}{2}}s,g_{2}^{m})$ for $m$ odd, and

[TABLE]

where $c_{2}^{m}$ is $(r^{m/2},\operatorname{Id})$ for $m$ even and $(r^{\frac{k+1}{2}}s,g_{2}g_{1})$ for $m$ odd. The groups $\pi_{2}(F_{1})=\langle g_{2}\rangle\simeq C_{6}$ and $\pi_{2}(F_{2})=\langle g_{2}g_{1}\rangle\simeq C_{2}$ are not conjugate in $\operatorname{Aut}(E)$, hence $F_{1}$ and $F_{2}$ are not conjugate in $\mathrm{Hol}(E)$. These groups show also that the unique nontrivial element in $\mathrm{Hom}(F,\mathbf{Z}_{p}^{*})$ to consider is $\sigma=(1,-1)$, meaning $\sigma(r)=1,\sigma(s)=-1$, whose stabiliser is $\operatorname{Aut}(E)$. Therefore, we have 2 additive structures for each $F$. The nontrivial one has cyclic kernel and we denoted it by $\mathbf{Z}_{p}\rtimes_{6}^{c}D_{2\cdot 6}$.

Regarding orbits on $\mathrm{Hom}(F,\mathbf{Z}_{p}^{*})$, since $\pi_{2}(F_{1})=\langle g_{2}\rangle$ is a normal subgroup, we can take any $g$ in $\operatorname{Aut}(E)$ to satisfy $g\pi_{2}(F_{1})g^{-1}=\pi_{2}(F_{1})$. But

[TABLE]

and

[TABLE]

Therefore, we are left with the automorphisms in $\langle g_{2}^{3},g_{2}^{2}g_{1}\rangle\simeq V_{4}$ which happens to be the normaliser of $\pi_{2}(F_{2})=\langle g_{2}g_{1}\rangle$. For automorphisms of this group we have

[TABLE]

[TABLE]

Therefore, in $\mathrm{Hom}(F,\mathbf{Z}_{p}^{*})$ we have just orbit for each $d$.

Proposition 11**.**

Let $p$ be a prime number, $p\geq 7$. The number of braces with additive group $N:=\mathbf{Z}_{p}\rtimes D_{2.6}$ and multiplicative group $G:=\mathbf{Z}_{p}\rtimes C_{12}$ is as shown in the following table, where we need $p\equiv 1\pmod{\frac{12}{k}}$ for a kernel of size $k$ to occur.

[TABLE]

Case $E=\operatorname{Dic}_{12}$

We write $E=\langle\ x,y\ \rangle$. There are two braces with additive group $\operatorname{Dic}_{12}$ and multiplicative group $C_{12}$, hence two conjugation classes of regular subgroups of $\mathrm{Hol}(\operatorname{Dic}_{12})$ isomorphic to $C_{12}$. We may consider representatives

[TABLE]

with $\pi_{2}(F_{1})=\langle g_{2}g_{1}\rangle\simeq C_{2}$ and $\pi_{2}(F_{2})=\langle g_{2}^{2}\rangle\simeq C_{3}$. Since $\pi_{2}(F_{1}),\pi_{2}(F_{2})\subseteq\langle g_{2}^{2},g_{1}g_{2}\rangle$, we have $\pi_{2}(F_{1}),\pi_{2}(F_{2})\subseteq\Sigma_{\sigma}$ for all $\sigma\in\mathrm{Hom}(E,\mathbf{Z}_{p}^{*})$.

In order to compute brace automorphisms we consider first the normalizer of $\pi_{2}(F_{1})$ in $\operatorname{Aut}(E)$, which is $\langle g_{2}^{3},g_{2}g_{1}\rangle$, and the normalizer of $\pi_{2}(F_{2})$, which is $\langle g_{2}\rangle$. Since

[TABLE]

[TABLE]

we have one orbit for each order and 3 possible additive structures for each $F$. On the other hand, we have

[TABLE]

and this shows that in both cases the automorphism group of the brace is the maximal subgroup $\langle g_{2}^{3},g_{2}g_{1}\rangle\simeq V_{4}$.

For $\sigma=(1,1)$ or $(2,1)$, we have $\operatorname{Aut}(B_{12})\cap\Sigma_{\sigma}=\operatorname{Aut}(B_{12})$ and $\operatorname{Aut}(B_{12})$ acts on $\mathrm{Hom}(F,\mathbf{Z}_{p}^{*})=\{(d,j)\}$ giving a single orbit for each order $d$. But, for $\sigma=(4,1)$, we have $\operatorname{Aut}(B_{12})\cap\Sigma_{\sigma}=\langle g_{2}g_{1}\rangle$. Then, $\tau=(d,j)$ and $\tau^{\prime}=(d,j^{\prime})$ are in the same orbit if, and only if, $j\equiv 5j^{\prime}\bmod d$, and we have orbits represented by $(1,1),(2,1),(3,1),(4,1),(4,3),(6,1),(12,1),(12,7).$

Proposition 12**.**

Let $p$ be a prime number, $p\geq 7$. The number of braces with additive group $N:=\mathbf{Z}_{p}\rtimes\operatorname{Dic}_{12}$ and multiplicative group $G:=\mathbf{Z}_{p}\rtimes C_{12}$ is as shown in the following table, where we need where we need $p\equiv 1\pmod{\frac{12}{k}}$ for a kernel of size $k$ to occur.

[TABLE]

4.2 $F=C_{6}\times C_{2}$

According to the table in Section 3 and Proposition 4, for a brace of size 12 with multiplicative group $C_{12}$, the additive group can be any group. In all the sub-cases the group $\mathrm{Hom}(F,\mathbf{Z}_{p}^{*})$ is as in 3.2 so that $\tau$ identifies as a pair $(j\bmod d,i\bmod 2)$ meaning $\tau(a)=\zeta_{d}^{j},\ \tau(b)=(-1)^{i}$ for $F=\langle a\rangle\times\langle b\rangle$.

Case $E=C_{12}$

There is one brace with additive group $C_{12}$ and multiplicative group $C_{6}\times C_{2}$, hence one conjugation class of regular subgroups of $\mathrm{Hol}(C_{12})$ isomorphic to $C_{6}\times C_{2}$. We may consider the following representative,

[TABLE]

where $a$ has order 6, $b$ has order 2. Since $\pi_{2}(F)=\langle g_{7}\rangle$ we cannot consider $\sigma$ of order 4 or 12. Since $F$ is a normal subgroup of $\mathrm{Hol}(E)$ all automorphisms of $E$ are brace automorphisms and we have a single orbit for each order. Therefore, we have 4 possible additive structures.

Regarding the action on $\mathrm{Hom}(F,\mathbf{Z}_{p}^{*})$, since

[TABLE]

we have the following orbits, acording to the order of $\tau$:

[TABLE]

In the last two cases, when $\Sigma_{\sigma}=\langle g_{7}\rangle$ some points get disconnected and we have 2 and 4 orbits, respectively.

Proposition 13**.**

Let $p$ be a prime number, $p\geq 7$. The number of braces with additive group $N=\mathbf{Z}_{p}\rtimes C_{12}$ and multiplicative group $G=\mathbf{Z}_{p}\rtimes(C_{6}\times C_{2})$ is as shown in the following table, where we need $p\equiv 1\pmod{\frac{12}{k}}$ for a kernel of size $k$ to occur.

[TABLE]

Case $E=C_{6}\times C_{2}$

Let $E=\langle a\rangle\times\langle b\rangle$. There is one brace with additive group $C_{6}\times C_{2}$ and multiplicative group $C_{6}\times C_{2}$, hence one conjugation class of regular subgroups of $\mathrm{Hol}(C_{6}\times C_{2})$ isomorphic to $C_{6}\times C_{2}$, which has to be that of the normal subgroup

[TABLE]

Clearly, we have a single orbit for each order of $\sigma\in\mathrm{Hom}(E,\mathbf{Z}_{p}^{*})$ and 4 possible additive structures.

Regarding the action on $\tau\in\mathrm{Hom}(F,\mathbf{Z}_{p}^{*})$, for $\sigma=1$ we have one orbit for each order and for $\tau=1$ the orbit has a single point. According to the stabilisers and orbits computed in 3.2, in the remaining cases we have

[TABLE]

Proposition 14**.**

Let $p$ be a prime number, $p\geq 7$. The number of braces with additive group $N=\mathbf{Z}_{p}\rtimes(C_{6}\times C_{2})$ and multiplicative group $G=\mathbf{Z}_{p}\rtimes(C_{6}\times C_{2})$ is as shown in the following table, where we need where we need $p\equiv 1\pmod{\frac{12}{k}}$ for a kernel of size $k$ to occur.

[TABLE]

Case $E=A_{4}$

There are two braces with additive group $A_{4}$ and multiplicative group $C_{6}\times C_{2}$, hence two conjugation classes of regular subgroups of $\mathrm{Hol}(A_{4})$ isomorphic to $C_{6}\times C_{2}$. We may consider the following representatives,

[TABLE]

where $a_{1},a_{2}$ and have order 6 and $b_{1},b_{2}$ have order 2. We have $\pi_{2}(F_{1})=\langle\phi_{(1,2,4)}\rangle\simeq C_{3}$ and $\pi_{2}(F_{1})\simeq V_{4}$. Therefore, on one hand we see that the groups are not conjugate in $\mathrm{Hol}(A_{4})$ and one the other hand that we can consider all $\sigma\in\mathrm{Hom}(E,\mathbf{Z}_{p}^{*}).$

Since $\pi_{2}(F_{1})=\langle\phi_{(1,2,4)}\rangle$, in order to find brace automorphisms we need $\phi_{s}$ in its normaliser, namely $s\in\langle(2,4),(1,2,4)\rangle\simeq S_{3}$. We have

[TABLE]

Since $\phi_{(2,4)}(1,2,3)=(1,4,3)=(1,3,2)(1,4)(2,3)$, we have that for $F_{1}$ it is enough to consider $\sigma_{1}$. The normaliser of $\pi_{2}(F_{1})=V_{4}$ is all $S_{4}$ and we find

[TABLE]

Since $\phi_{(1,2)}(1,2,3)=(1,3,2)(1,4)$ we are also left with $\sigma_{1}$ for $F_{2}$. We have two additive structures for each $F$.

We complete the description of the group of brace automorphisms, since

[TABLE]

shows that it is not $S_{4}$ and contains the maximal group $\langle(1,2),(1,2,3)\rangle\simeq S_{3}$. All together,

[TABLE]

The intersection with $\Sigma_{\sigma_{1}}$ is

[TABLE]

To get the multiplicative structures we need the orbits of $(s,\tau)\longrightarrow\tau\Phi_{\phi_{s}}$. We have the same action on generators $\phi_{c}(a)=ab$, $\phi_{c}(b)=a^{3}$, so that

[TABLE]

and we obtain the following orbits:

[TABLE]

Proposition 15**.**

Let $p$ be a prime number, $p\geq 7$. The number of braces with additive group $N:=\mathbf{Z}_{p}\rtimes A_{4}$ and multiplicative group $G:=\mathbf{Z}_{p}\rtimes(C_{6}\times C_{2})$ is as shown in the following table, where we need $p\equiv 1\pmod{\frac{12}{k}}$ for a kernel of size $k$ to occur.

[TABLE]

Case $E=D_{2\cdot 6}$

We write $E=\langle r,s\rangle$. There are two braces with additive group $D_{2\cdot 6}$ and multiplicative group $C_{6}\times C_{2}$, hence two conjugation classes of regular subgroups of $\mathrm{Hol}(D_{2\cdot 6})$ isomorphic to $C_{6}\times C_{2}$. We may consider the following representatives:

[TABLE]

where $a_{1},a_{2}$ have order 6, and $b_{1},b_{2}$ have order 2. We have $\pi_{2}(F_{2})=\langle g_{1}\rangle\simeq C_{2}$ and $\pi_{2}(F_{1})=\langle g_{2}^{4}\rangle\simeq C_{3}$. This shows that they are not conjugate and that they are subgroups of all the stabilisers $\Sigma_{\sigma}$. Moreover, since $g_{2}^{3}$ normalizes both $F_{1}$ and $F_{2}$ we have only one orbit in order $6$ with dihedral kernel and we have 3 additive structures.

Since the normaliser of $\langle g_{1}\rangle$ is $\langle g_{2}^{3},g_{1}\rangle$ and

[TABLE]

we have that for $F_{1}$ the group of brace automorphisms is $\langle g_{2}^{3},g_{1}\rangle\simeq V_{4}$. On the other hand,

[TABLE]

shows that for $F_{2}$ the group of brace automorphisms is also the maximal subgroup $\langle g_{2}^{3},g_{1}\rangle$. Since the action of generators is the same, there is no difference in computation of orbits for $F_{1}$ or $F_{2}$. We obtain

[TABLE]

For $\sigma$ of order $6$ with dihedral kernel $g_{2}^{3}\not\in\Sigma_{\sigma}$ and in first and last case orbits split.

Proposition 16**.**

Let $p$ be a prime number, $p\geq 7$. The number of braces with additive group $N:=\mathbf{Z}_{p}\rtimes D_{2.6}$ and multiplicative group $G:=\mathbf{Z}_{p}\rtimes(C_{6}\times C_{2})$ is as shown in the following table, where we need $p\equiv 1\pmod{\frac{12}{k}}$ for a kernel of size $k$ to occur.

[TABLE]

Case $E=\operatorname{Dic}_{12}$

We write $E=\langle x,y\rangle$. There are two braces with additive group $\operatorname{Dic}_{12}$ and multiplicative group $C_{6}\times C_{2}$, hence two conjugation classes of regular subgroups of $\mathrm{Hol}(\operatorname{Dic}_{12})$ isomorphic to $C_{6}\times C_{2}$. We may consider the following representatives

[TABLE]

where $a_{1},a_{2}$ have order 6, $b_{1},b_{2}$ have order 2. We note that since the second component of $a_{2}$ is $\operatorname{Id}$ and the second components of $a_{1}$ and $a_{1}^{-1}$ are equal to $g_{1}$, they are not conjugated. We have $\pi_{2}(F_{1})=\pi_{2}(F_{2})=\langle g_{1}\rangle$ and therefore we cannot consider $\sigma$ order $4$ and then we have $\Sigma_{\sigma}=\operatorname{Aut}(E)$. We need to determine brace automorphisms in order to determine the orbits of the action on $\mathrm{Hom}(F,\mathbf{Z}_{p}^{*})$

The normaliser of $\langle g_{1}\rangle$ in $\operatorname{Aut}(E)$ is $\langle g_{1},g_{2}^{3}\rangle$ and we have

[TABLE]

This group of brace automorphism gives raise to the following orbits

[TABLE]

and both groups give raise to the same number of multiplicative structures.

Proposition 17**.**

Let $p$ be a prime number, $p\geq 7$. The number of braces with additive group $N=\mathbf{Z}_{p}\rtimes\operatorname{Dic}_{12}$ and multiplicative group $G=\mathbf{Z}_{p}\rtimes(C_{6}\times C_{2})$ is as shown in the following table, where we need $p\equiv 1\pmod{\frac{12}{k}}$ for a kernel of size $k$ to occur.

[TABLE]

4.3 $F=A_{4}$

According to the table in Section 3 and Proposition 4, for a brace with multiplicative group a semidirect product of $\mathbf{Z}_{p}$ and $A_{4}$, the additive group is either a semidirect product of $\mathbf{Z}_{p}$ and $C_{6}\times C_{2}$ or a semidirect product of $\mathbf{Z}_{p}$ and $A_{4}$. In both sub-cases the group $\mathrm{Hom}(F,\mathbf{Z}_{p}^{*})$ is as in 3.3 so that its elements are $\tau_{j}$, with $j=0,1,2$, defined by $\tau_{j}(1,2,3)=\zeta_{3}^{j}$ and $\tau_{j}$ trivial on elements of order 2.

Case $E=C_{6}\times C_{2}$

There is one brace with additive group $E=\langle a\rangle\times\langle b\rangle\simeq C_{6}\times C_{2}$ and multiplicative group $A_{4}$, hence one conjugation class of regular subgroups of $\mathrm{Hol}(C_{6}\times C_{2})$ isomorphic to $A_{4}$. We may consider representative

[TABLE]

Indeed, $(a,g_{2}^{2})$ has order 3, $(b,\operatorname{Id})$ has order 2, and $(a,g_{2}^{2})(b,\operatorname{Id})=(a^{4},g_{2}^{2})$ which has order 3. The Klein subgroup is $\langle(b,\operatorname{Id}),(a^{3},\operatorname{Id})\rangle$.

We have $\pi_{2}(F)=<g_{2}^{2}>$ and therefore it is a subgroup of $\Sigma_{\sigma}$ only when $\sigma$ has order $1$ or $3$. For $\sigma=(1,0)$ of order $3$, we have $(1,0)g_{1}g_{2}=(1,0)g_{2}=(2,0)$. Since

[TABLE]

there is one orbit for each order and two different additive structures.

Let us check if elements of $\Sigma_{\sigma}=\langle g_{2}^{2},g_{1}\rangle$ are brace automorphisms. Since

[TABLE]

the action on $\mathrm{Hom}(F,\mathbf{Z}_{p}^{*})$ is just given by the identity and orbits consist on single elements. For a non trivial $\sigma$, $\tau_{1}$ and $\tau_{2}$ give different multiplicative structures.

Proposition 18**.**

Let $p$ be a prime number, $p\geq 7$. The number of braces with additive group $N:=\mathbf{Z}_{p}\rtimes(C_{6}\times C_{2})$ and multiplicative group $G:=\mathbf{Z}_{p}\rtimes A_{4}$ is as shown in the following table, where we need $p\equiv 1\pmod{\frac{12}{k}}$ for a kernel of size $k$ to occur.

[TABLE]

Case $E=A_{4}$

There are four braces with additive group $A_{4}$ and multiplicative group $A_{4}$, hence four conjugation classes of regular subgroups of $\mathrm{Hol}(A_{4})$ isomorphic to $A_{4}$. If we write $\rho=(1,2,3)$ and $\mu=(1,2)(3,4)$, we may consider the following representatives

[TABLE]

We have $\pi_{2}(F_{1})={\operatorname{Id}}$, $\pi_{2}(F_{2})=\langle\phi_{\rho}\rangle\simeq C_{3}$, and $\pi_{2}(F_{3})=\pi_{2}(F_{4})\simeq A_{4}$. But they are not conjugate, since in $F_{3}$ all elements of order $3$ are of the form $(x,\phi_{x^{-1}})$. On the other hand, we see that $\pi_{2}(F_{i})\subseteq\Sigma_{\sigma}$ for all $i=1..4$ and all $\sigma\in\mathrm{Hom}(E,\mathbf{Z}_{p}^{*})$.

Since

[TABLE]

and $((1,2)(3,4),\operatorname{Id})$ and $((1,2)(3,4),\phi_{(1,2)(3,4)}))$ are invariant under $\Phi_{\phi_{(1,2)}}$ we have $\Phi_{2}(F_{i})=F_{i}$ for all $i$ and elements of order 3 of $\mathrm{Hom}(E,\mathbf{Z}_{p})^{*}$ are in the same orbit. We have two possible additive structures for each $F_{i}$.

As for morphisms $\tau\in\mathrm{Hom}(F,\mathbf{Z}_{p}^{*})$, since $\Sigma_{\sigma}$ stabilizes them, the action give orbits of a single elememt, so that we have one orbit in order 1 and two orbits in order 3.

Proposition 19**.**

Let $p$ be a prime number, $p\geq 7$. The number of braces with additive group $N=\mathbf{Z}_{p}\rtimes A_{4}$ and multiplicative group $G=\mathbf{Z}_{p}\rtimes A_{4}$ is as shown in the following table, where we need $p\equiv 1\pmod{3}$ for the group $\mathbf{Z}_{p}\rtimes_{4}A_{4}$ to occur.

[TABLE]

4.4 $F=D_{2\cdot 6}$

Let us assume that $F$ is isomorphic to the dihedral group $D_{2\cdot 6}$. From the table in Section 3 we know that for a brace of size $12$ with multiplicative group $F$, the additive group can be in any isomorphism class of groups of order $12$ except $A_{4}$. Moreover, in all the sub-cases the group $\mathrm{Hom}(F,\mathbf{Z}_{p}^{*})$ is as in 3.4, so $\tau$ identifies with any of the pairs $(\pm 1,\pm 1)$. Let us call $\tau_{1}=(1,-1)$, $\tau_{2}=(-1,-1)$ and $\tau_{3}=(-1,1)$, so that $\tau_{1}$ has kernel isomorphic to $C_{6}$ and $\tau_{2}$, $\tau_{3}$ have kernel isomorphic to $D_{2\cdot 3}$.

Case $E=C_{12}$

There are two braces with additive group $C_{12}$ and multiplicative group $D_{2\cdot 6}$, hence two conjugation classes of regular subgroups of $\mathrm{Hol}(C_{12})$ isomorphic to $D_{2\cdot 6}$. These are represented by

[TABLE]

respectively. Here $a_{1}$, $a_{2}$ have order $6$ and $b_{1}$, $b_{2}$ have order $2$. Since $\pi_{2}(F_{1})=\langle g_{11}\rangle$ and $\pi_{2}(F_{2})=\mathrm{Aut}(E)$, in both cases we have $\pi_{2}(F_{i})\subseteq\Sigma_{\sigma}$ only for $\sigma=(d,1)$, $d=1,2$. On the other hand, conjugating $a_{1}$ and $b_{1}$ by the three nontrivial automorphisms of $C_{12}$, we obtain

[TABLE]

Hence, $\Phi_{g}(F_{1})=F_{1}$ for every $g$ and $\mathrm{Aut}(B_{12})=\mathrm{Aut}(E)$. As for $F_{2}$, the only $g\in\mathrm{Aut}(E)$ such that $\Phi_{g}(F_{2})=F_{2}$ is $g=g_{5}$, so $\mathrm{Aut}(B_{12})=\langle g_{5}\rangle$. In any case, each morphism $\sigma=(1,1),(2,1)$ give rise to two orbits (the ones at the first two rows at the first table in 3.1). For each of $F_{1}$ and $F_{2}$, the first morphism gives an additive structure, while the second one gives two additive structures.

Now, we consider the morphisms $\tau\colon F\longrightarrow\mathbb{Z}_{p}^{*}$. For $F_{1}$, we have that $\tau_{2}\Phi_{g_{7}}=\tau_{3}$ with $g_{7}\in\Sigma_{\sigma}=\mathrm{Aut}(E)$. If we now take $F_{2}$, we need to consider the action of $\langle g_{5}\rangle$ on $\mathrm{Hom}(F,\mathbb{Z}_{p}^{*})$, and we see that $\tau_{2}\Phi_{g_{5}}\neq\tau_{3}$. Therefore, for each $\sigma$, the number of orbits of the action of $\mathrm{Aut}(B_{12})\cap\Sigma_{\sigma}$ over $\mathrm{Hom}(F,\mathbf{Z}_{p}^{*})$ is as follows:

[TABLE]

Proposition 20**.**

Let $p$ be a prime number, $p\geq 7$. The number of braces with additive group $N:=\mathbf{Z}_{p}\rtimes C_{12}$ and multiplicative group $G:=\mathbf{Z}_{p}\rtimes D_{2\cdot 6}$ is as shown in the following table.

[TABLE]

Case $E=C_{6}\times C_{2}$

There is one brace with additive group $C_{6}\times C_{2}$ and multiplicative group $D_{2\cdot 6}$, and hence one conjugacy class of regular subgroups of $\mathrm{Hol}(C_{6}\times C_{2})$ isomorphic to $D_{2\cdot 6}$. This is represented by

[TABLE]

Here $r$ has order $6$, $s$ has order $2$ and $srs=r^{-1}$.

Since $\pi_{2}(F)=\langle g_{2}^{3}\rangle$, we have that $\pi_{2}(F)=\Sigma_{\sigma}$ only for $\sigma=(0,0)$ and $\sigma=(1,0),(0,1),(1,1)$ (corresponding to the first two rows of the final table at 3.2). On the other hand, it is easily checked that $\Phi_{g}(F)=F$ if and only if $g\in\langle g_{2}^{3},g_{1}g_{2}^{2}\rangle$. None of these automorphisms satisfy that $(0,1)=(1,0)g$ or $(0,1)=(1,1)g$, while $(1,0)=(1,1)g_{1}g_{2}^{2}$. Hence, we obtain two orbits (other than the trivial one, given by $(0,0)$).

As for the morphisms $\tau\colon F\longrightarrow\mathbb{Z}_{p}^{*}$, we have that $\tau_{2}=\tau_{3}\Phi_{g}$ if and only if $g\in\{g_{1}g_{2}^{2},g_{1}g_{2}^{5}\}$, with $g_{1}g_{2}^{5}\in\Sigma_{(0,1)}$, while for $\sigma=(1,0),(1,1)$, no $g\in\mathrm{Aut}(B_{12})$ satisfies that $\tau_{2}\Phi_{g}=\tau_{3}$. We collect the number of orbits for the action corresponding to each $\sigma$ in the following table.

[TABLE]

Proposition 21**.**

Let $p$ be a prime number, $p\geq 7$. The number of braces with additive group $N=\mathbf{Z}_{p}\rtimes C_{6}\times C_{2}$ and multiplicative group $G=\mathbf{Z}_{p}\rtimes D_{2\cdot 6}$ is as shown in the following table.

[TABLE]

Case $E=D_{2\cdot 6}$

There are four braces with additive group $D_{2\cdot 6}$ and multiplicative group $D_{2\cdot 6}$, and hence four conjugacy classes of regular subgroups of $\mathrm{Hol}(D_{2\cdot 6})$ isomorphic to $D_{2\cdot 6}$. These are represented by

[TABLE]

[TABLE]

[TABLE]

[TABLE]

For $j\in\{1,2,3,4\}$, $\varphi_{j}$ has order $6$, $\psi_{j}$ has order $2$, and $\psi_{j}\varphi_{j}\psi_{j}^{-1}=\varphi_{j}^{-1}$.

Let us call $\sigma_{1}=(1,-1)$, $\sigma_{2}=(-1,-1)$ and $\sigma_{3}=(-1,1)$. We see immediately that $\pi_{2}(F_{1})=\{\mathrm{Id}\}$, $\pi_{2}(F_{2})=\langle g_{1}\rangle$ and $\pi_{2}(F_{3})=\pi_{2}(F_{4})=\langle g_{1},g_{2}^{2}\rangle$. All of these are contained in $\Sigma_{\sigma_{i}}$ for $i\in\{1,2,3\}$, so $\mathrm{Aut}(B_{12})=\{1,\sigma_{1},\sigma_{2},\sigma_{3}\}$. On the other hand, we find that:

• •

For $j=1,3$, $\Phi_{g}(F_{j})=F_{j}$ for every $g\in\mathrm{Aut}(E)$.

• •

$\Phi_{g}(F_{2})=F_{2}$ if and only if $g\in\langle g_{1}\rangle$.

• •

$\Phi_{g}(F_{4})=F_{4}$ if and only if $g\in\langle g_{1}g_{2}^{2}\rangle$.

Now, we have that $\sigma_{3}=\sigma_{2}g_{1}g_{2}^{i}$ if and only if $i$ is odd, but this is a brace automorphism only for $F_{1}$ and $F_{3}$. Thus, other than the trivial one, there are two orbits for $F_{1}$ and $F_{3}$ and three orbits for $F_{2}$ and $F_{4}$.

We examine the classes of morphisms $\tau\colon F_{i}\longrightarrow\mathbb{Z}_{p}^{*}$. For $F_{1}$, we have that $\tau_{3}\Phi_{g_{1}g_{2}^{i}}=\tau_{2}$ if and only if $i$ is odd. Hence, $\tau_{2}\simeq\tau_{3}$ under $\mathrm{Aut}(E)$ but define two different classes under $\Sigma_{\sigma_{2}}=\Sigma_{\sigma_{3}}$. For $F_{2}$, since $\tau_{3}\Phi_{g_{1}}\neq\tau_{2}$, we have that $\tau_{2}\not\simeq\tau_{3}$. For $F_{3}$, we have that $\tau_{3}\Phi_{g_{2}^{i}}=\tau_{2}$ if and only if $i$ is odd, so we obtain the same conclusions as with $F_{1}$. Finally, for $F_{4}$ we have that $\tau_{3}\Phi_{g_{1}g_{2}^{2}}\neq\tau_{2}$, so $\tau_{2}\not\simeq\tau_{3}$.

We collect the numbers of orbits corresponding to each $\sigma$ in the following tables.

[TABLE]

[TABLE]

$F_{1},F_{3}\qquad\qquad\qquad\qquad\qquad\qquad F_{2},F_{4}$

Proposition 22**.**

Let $p$ be a prime number, $p\geq 7$. The number of braces with additive group $N=\mathbf{Z}_{p}\rtimes D_{2\cdot 6}$ and multiplicative group $G=\mathbf{Z}_{p}\rtimes D_{2\cdot 6}$ is as shown in the following table.

[TABLE]

Case $E=\operatorname{Dic}_{12}$

There are four braces with additive group $\operatorname{Dic}_{12}$ and multiplicative group $D_{2\cdot 6}$, and hence four conjugacy classes of regular subgroups of $\mathrm{Hol}(\operatorname{Dic}_{12})$ isomorphic to $D_{2\cdot 6}$. These are represented by

[TABLE]

We have that $\pi_{2}(F_{1})=\langle g_{1},g_{2}^{3}\rangle$, $\pi_{2}(F_{2})=\mathrm{Aut}(E)$, $\pi_{2}(F_{3})=\langle g_{2}^{3}\rangle$ and $\pi_{2}(F_{4})=\langle g_{1},g_{2}^{2}\rangle$. Hence, for each $j=1,2,3,4$, $\pi_{2}(F_{j})\subseteq\Sigma_{\sigma}$ only for $\sigma=(d,1)$ with $d=1,2$, in which case $\Sigma_{\sigma}=\mathrm{Aut}(E)$. On the other hand, we find that

[TABLE]

and for $j\in\{3,4\}$, $\Phi_{g}(F_{j})=F_{j}$ for every $g\in\mathrm{Aut}(E)$.

Let us examine the classes of morphisms $\tau\colon F_{j}\longrightarrow\mathbb{Z}_{p}^{*}$ for each $j$. For $F_{1}$ we have that $\tau_{2}(s_{1})\neq\tau_{3}\Phi_{g_{1}g_{2}^{3}}(s_{1})$, so $\tau_{2}\not\simeq\tau_{3}$. For $F_{2}$, $\tau_{2}(s_{2})\neq\tau_{3}\Phi_{g_{1}g_{2}^{5}}(s_{2})$, whence $\tau_{2}\not\simeq\tau_{3}$ neither. For $F_{3}$, we see that $\tau_{2}=\tau_{3}\Phi_{g_{1}g_{2}^{2}}$ with $g_{1}g_{2}^{2}$, hence $\tau_{2}\simeq\tau_{3}$. Finally, for $F_{4}$, we see that $\tau_{2}=\tau_{3}\Phi_{g_{1}}$, so $\tau_{2}\simeq\tau_{3}$. Then, the orbits for each $\sigma$ are as in the following table.

[TABLE]

Proposition 23**.**

Let $p$ be a prime number, $p\geq 7$. The number of braces with additive group $N=\mathbf{Z}_{p}\rtimes Dic_{12}$ and multiplicative group $G=\mathbf{Z}_{p}\rtimes D_{2\cdot 6}$ is as shown in the following table.

[TABLE]

4.5 $F=\operatorname{Dic}_{12}$

We take for $F$ the dicyclic group of order 12. According to the table in Section 3 and Proposition 4, for a brace of size 12 with multiplicative group $F=\operatorname{Dic}_{12}$, the additive group can be any group of order 12. In all the sub-cases the group $\mathrm{Hom}(F,\mathbf{Z}_{p}^{*})$ is as in 3.5 so that $\tau$ identifies as a pair $(1,1),(2,1),(4,1)$ or $(4,3)$.

Case $E=C_{12}$

Let $E=\langle c\rangle\simeq C_{12}$. There is one brace with additive group $C_{12}$ and multiplicative group $\operatorname{Dic}_{12}$, hence one conjugation class of regular subgroups of $\mathrm{Hol}(C_{12})$ isomorphic to $\operatorname{Dic}_{12}$. We may consider representative

[TABLE]

where $x$ has order 3, $y$ has order 4 and $y\,x=x^{2}\,y$.

Since $\pi_{2}(F)=\langle g_{5}\rangle$ we have $\pi_{2}(F)\subseteq\Sigma_{\sigma}$ only for $\sigma=(d,j)$ with $d=1,2,4$. Since $F$ is a normal subgroup of $\mathrm{Hol}(E)$, we have $\Phi_{g}(F)=F,\,\forall g\in\operatorname{Aut}(E)$. Since $\sigma=(4,1)$ and $\sigma^{\prime}=(4,3)$ are in the same orbit, the action of $\operatorname{Aut}(B_{12})$ gives one orbit for each order. All together, we have 3 additive structures for $F$ if $p\equiv 1\pmod{4}$ and 2 otherwise.

For $\sigma$ of order $1$ or two, $\Sigma_{\sigma}=\operatorname{Aut}(E)=\operatorname{Aut}(B_{12})$ and the action on $\mathrm{Hom}(F,\mathbf{Z}_{p}^{*})$ gives one orbit for each order. For $\sigma$ of order $4$, to consider the action of $\operatorname{Aut}(B_{12})\cap\Sigma_{\sigma}$ we just have to check the action of $g_{5}$ on $y$. Since

[TABLE]

$\tau=(4,1)$ and $\tau^{\prime}=(4,3)$ are not in the same orbit and this gives two multiplicative structures with kernel of order 3 for $\sigma$.

Proposition 24**.**

Let $p$ be a prime number, $p\geq 7$. The number of braces with additive group $N=\mathbf{Z}_{p}\rtimes C_{12}$ and multiplicative group $G=\mathbf{Z}_{p}\rtimes\operatorname{Dic}_{12}$ is as shown in the following table, where we need $p\equiv 1\pmod{\frac{12}{k}}$ for a kernel of size $k$ to occur.

[TABLE]

Case $E=C_{6}\times C_{2}$

We write $E=\langle a\rangle\times\langle b\rangle$. There is one brace with additive group $C_{6}\times C_{2}$ and multiplicative group $\operatorname{Dic}_{12}$, hence one conjugation class of regular subgroups of $\mathrm{Hol}(C_{6}\times C_{2})$ isomorphic to $\operatorname{Dic}_{12}$ for which we may consider as representative the following subgroup

[TABLE]

Since $\pi_{2}(F)=\langle g_{1}g_{2}\rangle$ there is only one nontrivial $\sigma$ to consider, namely that of order 2 identified as $(1,1)$ and defined by $\sigma(a)=\sigma(b)=\sigma_{2}=-1$, which has stabiliser $\langle g_{2}^{3},g_{2}^{4}g_{1}\rangle\simeq V_{4}$. Since we only have $1$ and $\sigma$, it is clear that they are fixed points for the action of $\operatorname{Aut}(B_{12)}$ and we have two orbits, namely 2 additive structures for $F$.

Since $F$ is normal in $\mathrm{Hol}(E)$, the elements of $\langle g_{2}^{3},g_{2}^{4}g_{1}\rangle$ are brace automorphisms and to determine its action on $\mathrm{Hom}(F,\mathbf{Z}_{p}^{*})$, we just have to check the action on $y$. From $\Phi_{g_{1}g_{2}}(y)=y^{3}$, we obtain that $\tau=(4,3)=(4,1)\Phi_{g_{1}g_{2}}$ and there is just one orbit in order 4.

Proposition 25**.**

Let $p$ be a prime number, $p\geq 7$. The number of braces with additive group $N=\mathbf{Z}_{p}\rtimes(C_{6}\times C_{2})$ and multiplicative group $G=\mathbf{Z}_{p}\rtimes\operatorname{Dic}_{12}$ is as shown in the following table, where we need $p\equiv 1\pmod{4}$ for the group $\mathbf{Z}_{p}\rtimes_{3}\operatorname{Dic}_{12}$ to occur.

[TABLE]

Case $E=A_{4}$

There are two braces with additive group $A_{4}$ and multiplicative group $\operatorname{Dic}_{12}$, hence two conjugation classes of regular subgroups of $\mathrm{Hol}(A_{4})$ isomorphic to $\operatorname{Dic}_{12}$. We may consider the following representatives

[TABLE]

where $x_{1},x_{2}$ have order 3, $y_{1},y_{2}$ have order 4 and $y_{i}x_{i}=x_{i}^{2}y_{i}$ for $i=1,2$. We have $\pi_{2}(F_{1})=\langle\phi_{(1,2,3)}\phi_{(1,2)}\rangle\simeq S_{3}$ and $\pi_{2}(F_{1})=\langle\phi_{(1,3,2,4)}\rangle$ which shows that $F_{1},F_{2}$ are not conjugate in $\mathrm{Hol}(A_{4})$ and that their projections are not subgroups of $A_{4}$, so that we obtain only one additive structure for each $F_{1},F_{2}$, corresponding to the direct product.

Since $\Sigma_{1}=\operatorname{Aut}(E)$ in order to determine multiplicative structures, we have one orbit for order 1 and one for order 2, and we only need to check if $\tau=(4,1)$ and $\tau^{\prime}=(4,3)$ differ in a brace automorphism. We have

[TABLE]

[TABLE]

So, for $F_{1}$ and $F_{2}$ we obtain one class of morphisms $\tau$ with kernel of order 6 and, if $p\equiv 1\pmod{4}$, there is another class of morphisms $\tau$ with kernel of order 3.

Proposition 26**.**

Let $p$ be a prime number, $p\geq 7$. The number of braces with additive group $N=\mathbf{Z}_{p}\rtimes A_{4}$ and multiplicative group $G=\mathbf{Z}_{p}\rtimes\operatorname{Dic}_{12}$ is as shown in the following table, where we need $p\equiv 1\pmod{4}$ for the group $\mathbf{Z}_{p}\rtimes_{3}\operatorname{Dic}_{12}$ to occur.

[TABLE]

Case $E=D_{2\cdot 6}$

There are two braces with additive group $D_{2\cdot 6}$ and multiplicative group $\operatorname{Dic}_{12}$, hence two conjugation classes of regular subgroups of $\mathrm{Hol}(D_{2\cdot 6})$ isomorphic to $\operatorname{Dic}_{12}$. We take representatives

[TABLE]

where $x_{1},x_{2}$ have order 3, $y_{1},y_{2}$ have order 4 and $y_{i}x_{i}=x_{i}^{2}y_{i}$ for $i=1,2$.

Since $\pi_{2}(F_{1})=\langle g_{2}^{3}\rangle$ and $\pi_{2}(F_{2})=\langle g_{2}^{2},g_{2}g_{1}\rangle$, we see that the groups are not conjugate and that the only nontrivial $\sigma$ to consider is that of order $2$ with cyclic kernel. So, the number of additive structures for each $F_{i}$ is $2$.

Again, to determine multiplicative structures it is enough to check if $(4,3)$ and $(4,1)$ differ in a brace automorphism. We have

[TABLE]

[TABLE]

So, for each $F_{1},F_{2}$ we obtain one orbit for each order.

Proposition 27**.**

Let $p$ be a prime number, $p\geq 7$. The number of braces with additive group $N=\mathbf{Z}_{p}\rtimes D_{2\cdot 6}$ and multiplicative group $G=\mathbf{Z}_{p}\rtimes\operatorname{Dic}_{12}$ is as shown in the following table, where we need $p\equiv 1\pmod{4}$ for the group $\mathbf{Z}_{p}\rtimes_{3}\operatorname{Dic}_{12}$ to occur.

[TABLE]

Case $E=\operatorname{Dic}_{12}$

There are two braces with additive group $\operatorname{Dic}_{12}$ and multiplicative group $\operatorname{Dic}_{12}$, hence two conjugation classes of regular subgroups of $\mathrm{Hol}(\operatorname{Dic}_{12})$ isomorphic to $\operatorname{Dic}_{12}$. We may consider the following representatives

[TABLE]

where $x_{1},x_{2}$ have order 3, $y_{1},y_{2}$ have order 4 and $y_{i}x_{i}=x_{i}^{2}y_{i}$ for $i=1,2$. Since $\pi_{2}(F_{1})=\{\operatorname{Id}\}$ and $\pi_{2}(F_{2})=\langle g_{2}^{2},g_{2}g_{1}\rangle=\langle g_{2}^{2},g_{1}g_{2}\rangle$ we see that groups are not conjugate and that we can consider all possible $\sigma$. We have

[TABLE]

[TABLE]

and $(4,3)=(4,1)g_{1}\mbox{ in }\mathrm{Hom}(E,\mathbf{Z}_{p}^{*}).$ Then, the number of additive structures for each $F_{i}$ is $3$ if $p\equiv 1\pmod{4}$ and $2$ otherwise.

For $\sigma=(2,1)$, since $\Sigma_{\sigma}=\operatorname{Aut}(E)$, the above computation shows that $(4,3)=(4,1)\Phi_{g_{1}}$ and we have just one orbit of order 4. But for $\sigma$ order 4, since $\Sigma_{\sigma}$ is also the stabiliser of the $\tau$’s of order 4, we have fixed points and 2 orbits.

Proposition 28**.**

Let $p$ be a prime number, $p\geq 7$. The number of braces with additive group $N:=\mathbf{Z}_{p}\rtimes\operatorname{Dic}_{12}$ and multiplicative group $G:=\mathbf{Z}_{p}\rtimes\operatorname{Dic}_{12}$ is as shown in the following table, where we need $p\equiv 1\pmod{\frac{12}{k}}$ for a kernel of size $k$ to occur.

[TABLE]

4.6 Total numbers

For a prime number $p\geq 7$ we compile in the following tables the total number of skew left braces of size $12p$.

The additive group is a semidirect product $\mathbf{Z}_{p}\rtimes E$ and the multiplicative group is a semidirect product $\mathbf{Z}_{p}\rtimes F$. In the first column we have the possible $E$’s and in the first row the possible $F$’s.

• •

If $p\equiv 11\pmod{12}$

[TABLE]

• •

If $p\equiv 5\pmod{12}$

[TABLE]

• •

If $p\equiv 7\pmod{12}$

[TABLE]

• •

If $p\equiv 1\pmod{12}$

[TABLE]

With the results summarized in the above tables, the validity of conjecture (1) is then established.