Asymmetric product of left braces and simplicity; new solutions of the Yang-Baxter equation

TL;DR

This paper advances the understanding of finite simple left braces by using asymmetric products to construct new classes and interpret existing ones, significantly contributing to the classification of solutions to the Yang-Baxter equation.

Contribution

It introduces a novel application of asymmetric products to construct and interpret finite simple left braces, expanding the known classes and providing new examples with solvable multiplicative groups.

Findings

Constructed new classes of simple left braces.

Interpreted all known constructions as asymmetric products.

Produced finite simple left braces with solvable multiplicative groups of arbitrary derived length.

Abstract

The problem of constructing all the non-degenerate involutive set-theoretic solutions of the Yang-Baxter equation recently has been reduced to the problem of describing all the left braces. In particular, the classification of all finite left braces is fundamental in order to describe all finite such solutions of the Yang-Baxter equation. In this paper we continue the study of finite simple left braces with the emphasis on the application of the asymmetric product of left braces in order to construct new classes of simple left braces. We do not only construct new classes but also we interpret all previously known constructions as asymmetric products. Moreover, a construction is given of finite simple left braces with a multiplicative group that is solvable of arbitrary derived length.

Equations

$$r_{12}{r}_{23}{r}_{12}=r_{23}{r}_{12}{r}_{23},$$

$$a\cdot (b+c)+a=a\cdot b+a\cdot c,$$

$$y^{-1}=-{\lambda }_{y^{-1}}(y)\in I\text{mbox}andxy={\lambda }_x(y)+x\in I,$$

$$-y={\lambda }_y(y^{-1})\in N\text{mbox}andx+y=x{\lambda }_{x^{-1}}(y)\in N,$$

$$B^{n+1}=B\ast B^n=\{\sum _{i=1}^m{a}_i\ast b_i\mid a_i\in B,b_i\in B^n,\text{mbox}foreveryi\},$$

$$B^{(n+1)}=B^{(n)}\ast B=\{\sum _{i=1}^m{a}_i\ast b_i\mid a_i\in B^{(n)},b_i\in B,\text{mbox}foreveryi\},$$

$$r(a,b)=({\lambda }_a(b),{\lambda }_{{\lambda }_a(b)}^{-1}(a)),$$

$$\alpha_{x}(y,z)=(y,z)\left(\begin{array}[]{cc}0&1\\ 1&1\end{array}\right)^{x},$$

$$((y_1,z_1),x_1)\ast ((y_2,z_2),x_2)$$

$$HN/N\cong H/(H\cap N).$$

$$s\cdot b(t_2,t_3)+b(t_1\cdot {\alpha }_s(t_2+t_3),t_1)=b(t_1\cdot {\alpha }_s(t_2),t_1\cdot {\alpha }_s(t_3))+s,$$

$$(t_1,s_1)+(t_2,s_2)=(t_1+t_2,s_1+s_2+b(t_1,t_2)),$$

$$(t_1,s_1)\cdot (t_2,s_2)=(t_1\cdot {\alpha }_{s_1}(t_2),s_1\cdot s_2),$$

$${\lambda }_{(t_1,s_1)}(t_2,s_2)$$

$$\mathrm{Soc}(T{⋊}_{\circ }S)=\{(t,s)\mid {\lambda }_s={\mathrm{id}}_S,{\lambda }_t\circ {\alpha }_s={\mathrm{id}}_T,b(t,t^{\prime })=0\text{mbox}forall{t}^{\prime }\in T\}.$$

$${\lambda }_s(b(t_2,t_3))=b({\lambda }_{t_1}{\alpha }_s(t_2),{\lambda }_{t_1}{\alpha }_s(t_3)),$$

$${\lambda }_s(b(t_2,t_3))$$

$$b(t_2,t_3)$$

$$({\lambda }_a-\mathrm{id})(b)\in I,$$

$$({\lambda }_{(t,0)}-\mathrm{id})(t^{\prime },0)$$

$$({\lambda }_{(t,s)}-\mathrm{id})(0,s^{\prime })$$

$$({\lambda }_{(0,y)}-\mathrm{id})(x,z)$$

$${\lambda }_{(x,y)}(x^{\prime },y^{\prime })$$

$$(x,y)(x^{\prime },y^{\prime })(x,y{)}^{-1}$$

$$G_n\wr (G_{n-1}\wr \cdots (G_2\wr G_1)\cdots ),$$

$${\gamma }_2(f,g)=({\hat{\gamma }}_1(f),{\gamma }_1(g)),$$

$${\hat{\gamma }}_1(f)(h)$$

$$E_m=\omega (K[D_{m-1}])K[G]⋊D_m,$$

$$((1-h)\alpha ,1)=(\alpha ,h^{-1}{)}^{-1}(0,h{)}^{-1}(\alpha ,h^{-1})(0,h),$$

$$(0,h^{-1}{g}^{-1}hg)=(0,h{)}^{-1}(0,g{)}^{-1}(0,h)(0,g).$$

$$\omega (K[G])⋊D_1\subseteq E_1.$$

Full text

2010 MSC: 16T25, 20E22, 20F16.

Keywords: Yang-Baxter equation, set-theoretic solution, brace, simple brace, asymmetric product.

1 Introduction

Rump [23] introduced braces to study a class of solutions of the Yang-Baxter equation, an important equation in mathematical physics (see [27, 21]). Recall that a set-theoretic solution of the Yang-Baxter equation is a map $r\colon X\times X\longrightarrow X\times X$, such that $X$ is a non-empty set and

[TABLE]

where $r_{ij}$ denotes the map $X\times X\times X\longrightarrow X\times X\times X$ acting as $r$ on the $(i,j)$ components and as the identity on the remaining component. The solution $r$ is involutive if $r^{2}=\operatorname{id}_{X^{2}}$, and it is non-degenerate if $r(x,y)=(f_{x}(y),g_{y}(x))$ and $f_{x}$ and $g_{y}$ are bijective maps from $X$ to $X$ , for all $x,y\in X$. A systematic study of non-degenerate, involutive set-theoretic solutions of the Yang-Baxter equation (solutions of the YBE for short) was initiated in [13] and [17]. Solutions of this type have received a lot of attention in recent years (see for example [4, 5, 9, 10, 11, 14, 15, 16, 18, 22, 23, 24, 25, 26]).

Every solution $(X,r)$ of the YBE has associated two important groups: its structure group $G(X,r)$ defined by the presentation $\langle x\in X\mid xy=f_{x}(y)g_{y}(x)\rangle$ (where $r(x,y)=(f_{x}(y),g_{y}(x))$), and its permutation group $\mathcal{G}(X,r)=\langle f_{x}\mid x\in X\rangle$, a subgroup of $\operatorname{Sym}_{X}$. There exists a homomorphism $\phi\colon G(X,r)\longrightarrow\mathcal{G}(X,r)$ such that $\phi(x)=f_{x}$, for all $x\in X$. Etingof, Schedler and Soloviev proved in [13] that if $(X,r)$ is a finite solution of the YBE, then $G(X,r)$ and $\mathcal{G}(X,r)$ are solvable groups.

Rump introduced braces to study this type of solutions [23], and noted that for every solution $(X,r)$ of the YBE, the groups $G(X,r)$ and $\mathcal{G}(X,r)$ have a natural structure of left braces such that $\phi\colon G(X,r)\longrightarrow\mathcal{G}(X,r)$ becomes a homomorphism of left braces. A left brace is a set $B$ with two operations, $+$ and $\cdot$, such that $(B,+)$ is an abelian group, $(B,\cdot)$ is a group and

[TABLE]

for all $a,b,c\in B$.

In [4] a method is given to construct explicitly, given a left brace $B$, all the solutions $(X,r)$ of the YBE such that $\mathcal{G}(X,r)\cong B$ as left braces. Therefore, the problem of constructing all the finite solutions of the YBE is reduced to describing all the finite left braces. In particular, as a consequence of [4, Theorem 3.1], every finite left brace is isomorphic to the left brace $\mathcal{G}(X,r)$, for some finite solution $(X,r)$ of the YBE. Therefore the multiplicative group of every finite left brace is solvable. The finite solvable groups isomorphic to the multiplicative group of a left brace are called IYB groups [12]. Not all the finite solvable groups are IYB groups. In fact, there exist finite $p$-groups ($p$ a prime) that are not of this type (see [2]). On the other hand, there are some known classes of finite solvable groups that are IYB groups, (see [7, 12]). It is an open problem whether every finite nilpotent group of class three is an IYB group.

Every left brace $B$ has a left action $\lambda\colon(B,\cdot)\longrightarrow\operatorname{Aut}(B,+)$, defined by $\lambda(b)=\lambda_{b}$ and $\lambda_{b}(a)=b\cdot a-b$, for all $a,b\in B$ (see [11, Lemma 1]). It is called the lambda map of the left brace $B$. Note that, via the lambda map, the additive structure of $B$ is determined by the multiplicative structure of $B$ and vice-versa. It is known and easy to check that, in every left brace $B$, $0=1$, i. e. the neutral elements of the groups $(B,+)$ and $(B,\cdot)$ coincide. A left ideal of a left brace $B$ is a subgroup $I$ of $(B,+)$ such that $\lambda_{b}(a)\in I$, for all $b\in B$ and all $a\in I$. Note that, in this case, for $x,y\in I$, we have that

[TABLE]

thus $I$ also is a subgroup of $(B,\cdot)$. An ideal of a left brace $B$ is a normal subgroup $N$ of $(B,\cdot)$ such that $\lambda_{b}(a)\in N$, for all $b\in B$ and all $a\in N$. Note that that, in this case, if $x,y\in N$, then

[TABLE]

thus $N$ also is a subgroup of $(B,+)$, and hence it is a left ideal of $B$. A non-zero left brace $B$ is simple if $\{0\}$ and $B$ are the only ideals of $B$. We say that a left brace $B$ is trivial if $a+b=a\cdot b$, for all $a,b\in B$.

In order to classify all finite left braces, it is natural to start with classifying the finite simple left braces. To build a foundation of the theory of finite simple braces it is crucial to find various methods of constructions of such structures and to understand their common features. In this paper, we continue a systematic approach to this fundamental problem, initiated in [6]. It is known that every simple left brace of prime power order $p^{n}$ is a trivial brace of cardinality $p$, [23, Corollary on page 166]. Until recently, these were the only known examples of finite simple left braces. The first finite nontrivial simple left braces have been constructed in [3, Theorem 6.3 and Section 7]; the additive groups of which are isomorphic to $\mathbb{Z}/(p_{1})\times(\mathbb{Z}/(p_{2}))^{k(p_{1}-1)+1}$, where $p_{1},p_{2}$ are primes such that $p_{2}\mid p_{1}-1$ and $k$ is a positive integer. In [6], using the iterated matched product construction, we built new families of finite simple left braces of order $p_{1}^{n_{1}}\cdots p_{s}^{n_{s}}$, where $\{p_{1},\dots,p_{s}\}$ is any finite set of primes of cardinality greater that $1$ and $n_{1},\dots,n_{s}$ are positive integers depending on the different primes $p_{1},\dots,p_{s}$. It is remarkable that the derived length of the multiplicative group of all the previously known simple left braces is at most $3$.

There are some known restrictions on the possible cardinalities of finite simple left braces (see [24]), but it is not at all clear which cardinalities may occur.

In the study of finite simple groups the classification of all minimal simple groups obtained by Thompson was of great significance (see [19, page 444]). Recall that a minimal simple group is a finite non-commutative simple group all of whose proper subgroups are solvable. Thus it seems natural to ask for the meaning of a minimal non-trivial finite simple left brace. This should be a non-trivial finite simple left brace such that all the factors of the composition series of every proper subbrace are trivial. In Section 2, we introduce the concept of solvable left brace in a similar way as it is done in group theory. Using this concept, a minimal non-trivial finite simple left brace is a non-trivial simple left brace all of whose proper subbraces are solvable. We expect that the classification of such finite simple left braces will be important in the further study of finite simple left braces.

In Sections 3 and 4, we recall the two main tools of the approach presented in this paper: the asymmetric product of left braces and the wreath product of left braces, respectively. We also prove some technical results useful for the construction of new simple left braces.

In Section 5, using the asymmetric product, we construct new classes of simple left braces. Furthermore, we construct first examples of finite simple left braces with a multiplicative group that is solvable of arbitrary derived length. This is accomplished via the asymmetric product of special left braces, constructed using the wreath product, and trivial braces.

In Section 6, we interpret all previously known constructions of finite left braces as asymmetric products.

2 Solvable braces

As motivated in the introduction, in this section we define solvable left braces, a class of left braces including all the left nilpotent left braces and all the right nilpotent left braces, introduced in [9]. These concepts have some similarity with the corresponding concepts in group theory.

Let $B$ be a left brace. Consider the operation $*$ of $B$ defined by $a*b:=a\cdot b-a-b=(\lambda_{a}-\operatorname{id})(b)$, for all $a,b\in B$. Rump in [23] introduced two series of subbraces of $B$. One consists of left ideals $B^{n}$, defined inductively as follows: $B^{1}=B$ and

[TABLE]

for all positive integers $n$. Another one consisting of ideals $B^{(n)}$, defined as follows. $B^{(1)}=B$ and

[TABLE]

for all positive integers $n$. Note that $B^{(2)}=B^{2}$, but in general $B^{(3)}$ and $B^{3}$ are different because $*$ is not associative in general.

We say that a left brace $B$ is left nilpotent if there exists a positive integer $n$ such that $B^{n}=0$. A left brace $B$ is right nilpotent if there exists a positive integer such that $B^{(n)}=0$.

Remark 2.1

Smoktunowicz in [25, Theorem 1] proved that a finite left brace $B$ is left nilpotent if and only if its multiplicative group $(B,\cdot)$ is nilpotent. In particular, every left brace of order a power of a prime is left nilpotent.

On the other hand, a left brace $B$ is right nilpotent if and only if the solution of the YBE associated with $B$ is a multipermutation solution (see [9, Proposition 6]). Recall, if $B$ is a left brace, its associated solution of the YBE is $(B,r)$, where

[TABLE]

for all $a,b\in B$. In particular, every non-zero right nilpotent left brace $B$ has non-zero socle, that is $\operatorname{Soc}(B)=\{a\in B\mid a\cdot b=a+b\mbox{ for all }b\in B\}\neq\{0\}$.

In [1] it is proved that there exist two left braces $B_{1}$ and $B_{2}$ of order $8$ such that $\operatorname{Soc}(B_{i})=\{0\}$. Therefore $B_{1}$ and $B_{2}$ are left nilpotent but not right nilpotent.

Consider the trivial left braces $K=\mathbb{Z}/(2)\times\mathbb{Z}/(2)$ and $\mathbb{Z}/(3)$. Let $\alpha\colon\mathbb{Z}/(3)\longrightarrow\operatorname{Aut}(K,+)$ be the action defined by $\alpha(x)=\alpha_{x}$ and

[TABLE]

for all $x\in\mathbb{Z}/(3)$ and all $y,z\in\mathbb{Z}/(2)$. Then the semidirect product $B_{3}=K\rtimes\mathbb{Z}/(3)$ is a left brace which is not left nilpotent. Now we have

[TABLE]

for all $y_{i},z_{i}\in\mathbb{Z}/(2)$ and all $x_{i}\in\mathbb{Z}/(3)$. Hence $B_{3}^{(2)}=K\times\{0\}$ and, since $((y_{1},z_{1}),0)*((y_{2},z_{2}),x_{2})=((0,0),0)$, we get that $B_{3}^{(3)}=\{0\}$. Therefore $B_{3}$ is right nilpotent.

Recall that a trivial brace is a left brace $B$ such that any $a,b\in B$ satisfy $a\cdot b=a+b$. We also say that $B$ has trivial structure. Thus $B$ is a trivial brace if and only if $B^{2}=\{0\}$.

Let $B$ be a left brace. It was proven in [23, page 161] that $B^{2}$ is always an ideal of $B$. Observe that $B/B^{2}$ has trivial brace structure because $a\cdot b-a-b=a*b\in B^{2}$. Moreover, if $I$ is an ideal of $B$, $B/I$ has trivial structure if and only if $B^{2}\subseteq I$.

In particular, note that $[(B,\cdot),(B,\cdot)]\leq B^{2}$ because $(B/B^{2},\cdot)$ is abelian.

Definition 2.2

A solvable brace is any left brace $B$ which has a series $0=B_{0}\subseteq B_{1}\subseteq\dots\subseteq B_{m}=B$ such that $B_{i}$ is an ideal of $B_{i+1}$, and such that $B_{i+1}/B_{i}$ is a trivial brace for any $i\in\{0,1,\dots,m-1\}$.

Using the same arguments as in group theory, one can prove the following results.

Proposition 2.3

(Second Isomorphism Theorem).* Let $B$ be a left brace, let $H$ be a sub-brace of $B$, and let $N$ be an ideal of $B$. Then, $HN$ is a sub-brace of $B$. Moreover,*

[TABLE]

Proposition 2.4

1. (a)

Define $d_{1}(B)=B^{2}$, and $d_{i+1}(B)=d_{i}(B)^{2}$ for every positive integer $i$. Then, $B$ is a solvable brace if and only if $d_{k}(B)=0$ for some $k$. 2. (b)

Let $B$ be a left brace, and let $I$ be an ideal of $B$. If $I$ and $B/I$ are solvable braces, then $B$ is also solvable. 3. (c)

Any sub-brace, and any quotient of a solvable brace is solvable.

The third property means that there are neither simple sub-braces nor simple quotients in a solvable brace, except for the trivial simple braces $\mathbb{Z}/(p)$, where $p$ is a prime.

Remark 2.5

Note that for every left brace $B$, $d_{n}(B)\subseteq B^{n+1}\cap B^{(n+1)}$. Hence every left nilpotent left brace is solvable, and every right nilpotent left brace is solvable. We have seen in Remark 2.1 that there exists a finite left nilpotent left brace $B_{1}$ which is not right nilpotent, and there exists a finite right nilpotent left brace $B_{3}$ which is not left nilpotent. Hence the left brace $B_{1}\times B_{3}$ is neither left nilpotent nor right nilpotent. But clearly $B_{1}\times B_{2}$ is solvable.

On the other extreme end, one can introduce the following class of braces.

Definition 2.6

A left brace $B$ is called perfect if it satisfies $B^{2}=B$.

For instance, every non-trivial simple left brace is perfect. The following example shows that the converse is not true. So, this class should play an important role in the approach towards a description of all finite left simple braces, and should deserve a further study.

Example 2.7

Take $B$ to be the simple left brace of order $24$ defined in [3]. Since the multiplicative group of $B$ is isomorphic to $S_{4}$, we can define a morphism $\alpha:(B,\cdot)\rightarrow\operatorname{Aut}(\mathbb{Z}/(3))$, given by the sign morphism of $S_{4}$. Then, the semidirect product of braces $\mathbb{Z}/(3)\rtimes B$ with respect to $\alpha$, where $\mathbb{Z}/(3)$ is considered as a trivial brace, is a perfect brace which is not simple.

3 Asymmetric product

In this section we recall a construction of left braces introduced by Catino, Colazzo and Stefanelli [8], the asymmetric product of two left braces. Furthermore we give a result that can be used to construct new simple left braces using asymmetric products.

Let $S$ and $T$ be two (additive) abelian groups. Recall that a (normalized) symmetric $2$-cocyle on $T$ with values in $S$ is a map $b\colon T\times T\longrightarrow S$ such that

1. (i)

$b(0,0)=0$; 2. (ii)

$b(t_{1},t_{2})=b(t_{2},t_{1})$; 3. (iii)

$b(t_{1}+t_{2},t_{3})+b(t_{1},t_{2})=b(t_{1},t_{2}+t_{3})+b(t_{2},t_{3})$, for all $t_{1},t_{2},t_{3}\in T$.

As a consequence, we get that $b(t,0)=b(0,t)=0$, for all $t\in T$.

Theorem 3.1

([8, Theorem 3]). Let $T$ and $S$ be two left braces. Let $b\colon T\times T\longrightarrow S$ be a symmetric $2$-cocycle on $(T,+)$ with values in $(S,+)$, and let $\alpha\colon(S,\cdot)\longrightarrow\operatorname{Aut}(T,+,\cdot)$ be a homomorphism of groups such that

[TABLE]

where $\alpha_{s}=\alpha(s)$, for all $s\in S$ and $t_{1},t_{2},t_{3}\in T$. Then the addition and multiplication over $T\times S$ given by

[TABLE]

[TABLE]

define a structure of left brace on $T\times S$. We call this left brace the asymmetric product of $T$ by $S$ (via $b$ and $\alpha$) and denote it by $T\rtimes_{\circ}S$.

Note that the lambda map of $T\rtimes_{\circ}S$ is defined by

[TABLE]

and its socle is

[TABLE]

Moreover, the subset $T\times\{0\}$ is a normal subgroup of $(T\rtimes_{\circ}S,\cdot)$, and $\{0\}\times S$ is a left ideal of $T\rtimes_{\circ}S$.

A particular case of this theorem is when we assume that $b$ is a symmetric bilinear form. In this case, conditions (i)-(iii) are automatic, and condition (1) becomes

[TABLE]

which is equivalent to the two conditions:

[TABLE]

for all $s\in S$ and $t_{1},t_{2},t_{3}\in T$.

To study simplicity of left braces we will often make use of the following result without any further explicit reference.

Lemma 3.2

([6, Lemma 2.5]). Let $B$ be a left brace. Let $I$ be an ideal of $B$. Then

[TABLE]

for all $a\in I$ and all $b\in B$.

Theorem 3.3

Let $S$ be a simple non-trivial left brace. Let $T$ be a trivial brace and let $T\rtimes_{\circ}S$ be an asymmetric product of $T$ by $S$ via $b\colon T\times T\longrightarrow S$ and $\alpha\colon(S,\cdot)\longrightarrow\operatorname{Aut}(T,+)$. Suppose that for every $t\in T\setminus\{0\}$ there exists $t^{\prime}\in T$ such that $b(t,t^{\prime})\neq 0$. Then $T\rtimes_{\circ}S$ is simple if and only if $\sum_{s\in S}\operatorname{Im}(\alpha_{s}-\operatorname{id})=T$.

Proof. Let $I$ be a non-zero ideal of $T\rtimes_{\circ}S$. Let $(t,s)\in I$ be a nonzero element. Suppose that $s=0$. Then $t\neq 0$. There exists $t^{\prime}\in T$ such that $b(t,t^{\prime})\neq 0$, and we have

[TABLE]

Hence we may assume that $s\neq 0$. Since $S$ is a simple non-trivial left brace, $\operatorname{Soc}(S)=\{0\}$. So there exists $s^{\prime}\in S$ such that $\lambda_{s}(s^{\prime})\neq s^{\prime}$. Note that

[TABLE]

Since $S$ is a simple left brace, we easily get that $\{(0,y)\mid y\in S\}\subseteq I$. Let $x\in T$ and $y,z\in S$. We have

[TABLE]

and thus $((\alpha_{y}-\operatorname{id})(x),0)\in I$, because $\{(0,y)\mid y\in S\}\subseteq I$.

Let $T^{\prime}=\sum_{y\in S}\operatorname{Im}(\alpha_{y}-\operatorname{id})$. We have that $T^{\prime}\times S\subseteq I$. It is clear that $T^{\prime}\times S$ is a subgroup of the additive group of $T\rtimes_{\circ}S$. Let $(x,y)\in T\rtimes_{\circ}S$ and $(x^{\prime},y^{\prime})\in T^{\prime}\times S$. Note that

[TABLE]

and $\alpha_{y}\circ(\alpha_{z}-\operatorname{id})=(\alpha_{yzy^{-1}}-\operatorname{id})\circ\alpha_{y}$. Therefore $\lambda_{(x,y)}(x^{\prime},y^{\prime})\in T^{\prime}\times S$. Furthermore

[TABLE]

Hence $T^{\prime}\times S$ is an ideal of $T\rtimes_{\circ}S$. Thus $T\rtimes_{\circ}S$ is simple if and only if $T^{\prime}=T$, and the result follows.

4 Wreath products of left braces

In this section we recall the definition of the wreath product of left braces (see [11, Corollary 1]) and we study how to construct automorphisms of wreath products of left braces. The aim is to build new classes of left braces via the iterated wreath product construction.

Let $G_{1}$ and $G_{2}$ be left braces. Recall that the wreath product $G_{2}\wr G_{1}$ of the left braces $G_{2}$ and $G_{1}$ is a left brace which is the semidirect product of left braces $H_{2}\rtimes G_{1}$, where $H_{2}=\{f\colon G_{1}\longrightarrow G_{2}\mid|\{g\in G_{1}\mid f(g)\neq 1\}|<\infty\}$ is a left brace with the operations $(f_{1}\cdot f_{2})(g)=f_{1}(g)\cdot f_{2}(g)$ and $(f_{1}+f_{2})(g)=f_{1}(g)+f_{2}(g)$, for all $f_{1},f_{2}\in H_{2}$ and $g\in G_{1}$, and the action of $(G_{1},\cdot)$ on $H_{2}$ is given by the homomorphism $\sigma\colon(G_{1},\cdot)\longrightarrow\operatorname{Aut}(H_{2},+,\cdot)$ defined by $\sigma(g)(f)(x)=f(g^{-1}x)$, for all $g,x\in G_{1}$ and $f\in H_{2}$. We will denote $\sigma(g)(f)$ simply by $g(f)$.

As in group theory (see [20, Hilfssatz I.15.7(b)]), one can extend every automorphism of the left brace $G_{1}$ to an automorphism of $G_{2}\wr G_{1}$ in a natural way.

Proposition 4.1

Let $G_{1}$ and $G_{2}$ be two left braces. Let $\alpha_{1}\in\operatorname{Aut}(G_{1},+,\cdot)$. Let $H_{2}=\{f\colon G_{1}\longrightarrow G_{2}\mid|\{g\in G_{1}\mid f(g)\neq 1\}|<\infty\}$. Let $\widehat{\alpha}_{1}\colon H_{2}\longrightarrow H_{2}$ be the map defined by $\widehat{\alpha}_{1}(f)(g)=f(\alpha_{1}^{-1}(g))$, for all $f\in H_{2}$ and all $g\in G_{1}$. Then the map $\alpha_{2}\colon G_{2}\wr G_{1}\longrightarrow G_{2}\wr G_{1}$ defined by $\alpha_{2}(f,g)=(\widehat{\alpha}_{1}(f),\alpha_{1}(g))$, for all $f\in H_{2}$ and all $g\in G_{1}$, is an automorphism of left braces.

Proof. The proof is similar to the proof of [20, Hilfssatz I.15.7(b)] and we leave it to the reader.

In the situation described above, we say that $\alpha_{2}$ is the automorphism of $G_{2}\wr G_{1}$ induced by the automorphism $\alpha_{1}$ of $G_{1}$. Thus, given left braces $G_{1},G_{2},\dots,G_{n}$, every automorphism $\alpha_{1}$ of the left brace $G_{1}$, induces an automorphism $\alpha_{n}$ of the left brace

[TABLE]

where $\alpha_{i}$ is the automorphism of $G_{i}\wr(G_{i-1}\wr\cdots(G_{2}\wr G_{1})\cdots)$ induced by the automorphism $\alpha_{i-1}$ of $G_{i-1}\wr(G_{i-2}\wr\cdots(G_{2}\wr G_{1})\cdots)$ (as in Proposition 4.1), for all $n\geq i\geq 2$.

Proposition 4.2

Let $G_{1}$ and $G_{2}$ be two left braces. Let $\alpha_{1},\beta_{1}\in\operatorname{Aut}(G_{1},+,\cdot)$. Let $\alpha_{2},\beta_{2}$ be the automorphisms of the left brace $G_{2}\wr G_{1}$ induced by $\alpha_{1},\beta_{1}$, respectively. Then the automorphism of $G_{2}\wr G_{1}$ induced by $\alpha_{1}\circ\beta_{1}$ is $\alpha_{2}\circ\beta_{2}$.

Proof. Let $\gamma_{2}$ be the automorphism of $G_{2}\wr G_{1}$ induced by $\gamma_{1}=\alpha_{1}\circ\beta_{1}$. Let $H_{2}=\{f\colon G_{1}\longrightarrow G_{2}\mid|\{g\in G_{1}\mid f(g)\neq 1\}|<\infty\}$. Let $\widehat{\gamma}_{1}\colon H_{2}\longrightarrow H_{2}$ be the map defined by $\widehat{\gamma}_{1}(f)(g)=f(\gamma_{1}^{-1}(g))$, for all $f\in H_{2}$ and all $g\in G_{1}$. We have that

[TABLE]

for all $f\in H_{2}$ and all $g\in G_{1}$. Note that

[TABLE]

for all $f\in H_{2}$ and all $h\in G_{1}$. Hence $\widehat{\gamma}_{1}=\widehat{\alpha}_{1}\circ\widehat{\beta}_{1}$. Now it is clear that $\gamma_{2}=\alpha_{2}\circ\beta_{2}$ and the result follows.

Lemma 4.3

Let $G$ be a finite group and let $K$ be a field of characteristic $p$. Suppose that $p$ is not a divisor of the order of $G$. Consider the semidirect product $K[G]\rtimes G$ with respect to the action of $G$ on the additive group of $K[G]$ by left multiplication. Let $D_{0}=G$ and, for $m>0$, let $D_{m}=[D_{m-1},D_{m-1}]$. Let $E_{0}=K[G]\rtimes G$ and, for $m>0$, let $E_{m}=[E_{m-1},E_{m-1}]$. Then

[TABLE]

for all $m>0$. (Here $\omega(K[G])$ denotes the augmentation ideal of the group algebra $K[G]$.) In particular, the derived length of $E_{0}$ is $d+1$, where $d$ is the derived length of $G$.

Proof. We shall prove the result by induction on $m$. For $m=1$, we have that

[TABLE]

for all $h\in G$ and $\alpha\in K[G]$, and

[TABLE]

Hence

[TABLE]

On the other hand

[TABLE]

for all $\alpha,\beta\in K[G]$ and all $h,g\in G$. Hence $E_{1}\subseteq\omega(K[G])\rtimes D_{1}$. Therefore $E_{1}=\omega(K[G])\rtimes D_{1}$.

Suppose that $m>1$ and that (5) holds for $m-1$. Let $\alpha\in\omega(K[D_{m-2}])K[G]$. We have

[TABLE]

for all $h\in D_{m-1}$. Thus

[TABLE]

Since $p$ is not a divisor of $|G|$, the group algebras $K[D_{j}]$ are semisimple by Maschke’s Theorem, for all non-negative integers $j$. Hence

[TABLE]

and consequently

[TABLE]

if $m>1$. Hence

[TABLE]

Since

[TABLE]

for all $\alpha,\beta\in\omega(K[D_{m-2}])K[G]$ and all $h,g\in D_{m-1}$, we also have that

[TABLE]

and therefore $E_{m}=\omega(K[D_{m-1}])K[G]\rtimes D_{m}$. Thus the result follows.

5 New simple left braces

In [3], Bachiller gave the first examples of non-trivial finite simple left braces. They are of order $p_{1}^{n_{1}}p_{2}$ for any pair of primes $p_{1},p_{2}$ such that $p_{1}$ divides $p_{2}-1$, and $n_{1}$ is a positive integer that depends on $p_{2}$. In [6], for every $s>1$ and every set $\{p_{1},\dots,p_{s}\}$ of $s$ distinct primes, the authors constructed examples of finite simple left braces of order $p_{1}^{n_{1}}\cdots p_{s}^{n_{s}}$ for some positive integers $n_{1},\dots,n_{s}$ depending on the $p_{i}$’s.

In this section we generalize the construction of simple left braces of [3]. Moreover, using the iterated wreath product construction, we give examples of finite simple left braces with multiplicative group of arbitrary derived length. Note that the multiplicative groups of all the previously known finite simple left braces [3, 6] have derived length at most $3$.

5.1 Finite simple left braces with multiplicative group of arbitrary derived length

Let $R$ be a ring. We will consider the trivial brace associated to $R$, which is $(R,+,+)$. Let $G$ be a left brace. Note that the trivial left brace

[TABLE]

can be identified with the trivial left brace associated to the group ring $R[G]$ by

[TABLE]

for all $f\in H$. We know that the wreath product of left braces $R\wr G$ is the semidirect product $H\rtimes G$ with respect to the action defined by $g(f)(x)=f(g^{-1}x)$ for all $x,g\in G$ and $f\in H$. Note that with the above identification this action is

[TABLE]

so it coincides with the left multiplication by $g\in G$ in the group ring $R[G]$. Thus $R\wr G$ can be identified with the semidirect product $R[G]\rtimes G$ with respect to the action given by left multiplication.

Let $n$ be an integer greater than $1$. Let $p_{1},\dots,p_{n}$ be distinct prime numbers. Consider the trivial braces associated to the fields $\mathbb{F}_{p_{i}}$ of order $p_{i}$, for $i=1,\dots,n$. Sometimes we will need to use the ring product of $\mathbb{F}_{p_{i}}$. In order to avoid confusion, we always denote this product by a dot.

We shall define recursively two series of left braces $G_{j}$ and $\overline{G}_{j}$, for $j=1,\ldots,n$, in such a way that $\overline{G}_{j}$ is defined as a quotient of $G_{j}$, and that $G_{j+1}$ is defined as a wreath product. First of all, let $G_{1}=\overline{G}_{1}=\mathbb{F}_{p_{1}}$. By recurrence we define the $G_{j}$’s, for $2\leq j\leq n$, as $G_{j}=\mathbb{F}_{p_{j}}\wr\overline{G}_{j-1}$. For this to make sense, we must define next $\overline{G}_{j}$ as a quotient of $G_{j}$. Let $H_{j}=\{f\colon\overline{G}_{j-1}\longrightarrow\mathbb{F}_{p_{j}}\}$ be the set of all the functions from $\overline{G}_{j-1}$ to $\mathbb{F}_{p_{j}}$. Since $\mathbb{F}_{p_{j}}$ is a trivial brace, so is $H_{j}$ (for the pointwise addition of functions). Let $I_{j}=\{f\colon\overline{G}_{j-1}\longrightarrow\mathbb{F}_{p_{j}}\mid f\text{ is a constant map}\}$. Clearly $I_{j}$ is an ideal of $H_{j}$. By the definition of wreath product, we know that $G_{j}=H_{j}\rtimes\overline{G}_{j-1}$. Let

[TABLE]

Note that $N_{j}$ is a central subgroup of the multiplicative group of the left brace $G_{j}$. Indeed, for $(f_{1},0)\in N_{j}$ and $(f_{2},g)\in G_{j}$, we have

[TABLE]

where the second equality holds because $f_{1}$ is a constant map. Furthermore, since

[TABLE]

we have that $N_{j}\subseteq\operatorname{Soc}(G_{j})$. Thus $N_{j}$ is an ideal of the left brace $G_{j}$. We define $\overline{G}_{j}=G_{j}/N_{j}$. Let $\overline{H}_{j}=H_{j}/I_{j}$. It is clear that $\overline{G}_{j}=\overline{H}_{j}\rtimes\overline{G}_{j-1}$. Hence

[TABLE]

and, therefore, every element in $\overline{G}_{n}$ is of the form

[TABLE]

where $\overline{h}_{j}\in\overline{H}_{j}$ denotes the class of $h_{j}\in H_{j}$ modulo $I_{j}$, and $c\in\mathbb{F}_{p_{1}}$.

Note that $|\overline{G}_{1}|=|\mathbb{F}_{p_{1}}|=p_{1}$ and, for $2\leq j\leq n$,

[TABLE]

Thus $|\overline{G}_{j}|=p_{1}p_{2}^{m_{2}}\cdots p_{j}^{m_{j}}$, for some positive integers $m_{2},\dots,m_{j}$.

Now we introduce some actions and some bilinear forms in order to apply the asymmetric product construction to the braces $\overline{G}_{j}$. Suppose that $p_{j}$ is a divisor of $p_{i}-1$ for all $2\leq j\leq n$ and all $1\leq i<j$. The existence of such primes is a consequence of Dirichlet’s Theorem on arithmetic progressions. Let $A$ be the trivial brace associated to the ring $\mathbb{F}_{p_{2}}\times\dots\times\mathbb{F}_{p_{n}}$. Then $a=(1,\dots,1)$ is a generator of the trivial brace $A$. Besides the brace structure, $A$ is a ring, and we will use this structure sometimes. Again we will denote this ring product as a dot.

Since $p_{j}$ is a divisor of $p_{1}-1$, for all $2\leq j\leq n$, there exists $\gamma\in(\mathbb{F}_{p_{1}})^{*}$ of order $p_{2}\cdots p_{n}$. Let $\alpha_{1}(ia)$ be the automorphism of the left brace $\mathbb{F}_{p_{1}}$ defined by $\alpha_{1}(ia)(z)=\gamma^{i}\cdot z$, for all $z\in\mathbb{F}_{p_{1}}$, and $i\in\mathbb{Z}$. Since the additive order of $a$ is $p_{2}\cdots p_{n}$, $\alpha_{1}(ia)$ is well-defined. By Proposition 4.1, $\alpha_{1}(ia)$ induces an automorphism $\alpha_{2}(ia)$ of the left brace $G_{2}$. Recall that

[TABLE]

and

[TABLE]

It is clear that $\alpha_{2}(ia)(N_{2})=N_{2}$. Thus $\alpha_{2}(ia)$ induces an automorphism $\alpha^{\prime}_{2}(ia)$ of $\overline{G}_{2}$. By Proposition 4.2, $\alpha_{2}(ia)=\alpha_{2}(a)^{i}$ and we also have $\alpha^{\prime}_{2}(ia)=\alpha^{\prime}_{2}(a)^{i}$. Hence, the map $\alpha^{\prime}_{2}\colon A\longrightarrow\operatorname{Aut}(\overline{G}_{2},+,\cdot)$ is a morphism of groups. By induction one can see that $\alpha_{1}(ia)$ induces an automorphism $\alpha^{\prime}_{j}(ia)$ of the left brace $\overline{G}_{j}$, for all $2\leq j\leq n$, and the map $\alpha^{\prime}_{j}\colon A\longrightarrow\operatorname{Aut}(\overline{G}_{j},+,\cdot)$ is a morphism of groups. Note that

[TABLE]

for all $\overline{h}_{j}\in\overline{H}_{j}$ and all $\overline{g}_{j-1}\in\overline{G}_{j-1}$, where

[TABLE]

for all $\overline{g}\in\overline{G}_{j-1}$.

Next we are going to define some bilinear forms. For $f\in H_{j}$ put $\varepsilon_{j}(f)=\sum_{g\in\overline{G}_{j-1}}f(g)$. Let $b_{j}\colon H_{j}\times H_{j}\longrightarrow\mathbb{F}_{p_{j}}$ be the symmetric bilinear form defined by

[TABLE]

for all $f_{1},f_{2}\in H_{j}$, where $f_{1}\bullet f_{2}\in H_{j}$ is defined by $(f_{1}\bullet f_{2})(x)=f_{1}(x)\cdot f_{2}(x)$, for all $x\in\overline{G}_{j-1}$. Recall that $\varepsilon_{j}(f_{1})\cdot\varepsilon_{j}(f_{2})$ denotes the ring product of $\mathbb{F}_{p_{j}}$.

Let $\mathds{1}\colon\overline{G}_{j-1}\longrightarrow\mathbb{F}_{p_{j}}$ be the map defined by $\mathds{1}(g)=1$, for all $g\in\overline{G}_{j-1}$. Note that

[TABLE]

because $p_{j}$ is a divisor of $|\overline{G}_{j-1}|-1$. Thus $b_{j}(I_{j},H_{j})=0$ and then $b_{j}$ induces a symmetric bilinear form $b^{\prime}_{j}\colon\overline{H}_{j}\times\overline{H}_{j}\longrightarrow\mathbb{F}_{p_{j}}$. It is easy to see that $b^{\prime}_{j}$ is non-degenerate. We define $b\colon\overline{G}_{n}\times\overline{G}_{n}\longrightarrow A$ by

[TABLE]

for all $\overline{h}_{j},\overline{h^{\prime}}_{j}\in\overline{H}_{j}$, $2\leq j\leq n$, and all $z_{1},z^{\prime}_{1}\in\mathbb{F}_{p_{1}}$. It is easy to see that $b$ is bilinear.

In order to apply Theorem 3.1, we need to check that the bilinear form $b$ satisfies conditions (3) and (4). As a first step, we will check the analogous property for $b_{j}$.

Lemma 5.1

Let $h_{1},h_{2}\in H_{j}$ and $g\in\overline{G}_{j-1}$. Then

[TABLE]

(recall that $a=(1,\dots,1)$ is a generator of the trivial brace $A$) and

[TABLE]

where $g(h)\colon\overline{G}_{j-1}\longrightarrow\mathbb{F}_{p_{j}}$ is the map defined by $(g(h))(g^{\prime})=h(g^{-1}g^{\prime})$, for all $g^{\prime}\in\overline{G}_{j-1}$ and $h\in H_{j}$.

Proof. The right side of the first equality is by definition

[TABLE]

The action of $\widehat{\alpha}_{j-1}(a)$ over a map $f$ is a permutation of its images. Thus $\varepsilon_{j}(f)$, which is the sum of all the images of $f$, is invariant under this action. Therefore,

[TABLE]

The proof of the second equality is analogous, because the action of $g\in\overline{G}_{j-1}$ over a map is also a permutation of its images.

From the previous lemma, we are going to deduce the desired result for $b$.

Lemma 5.2

Let $g_{1},g_{2},x\in\overline{G}_{n}$. Then

[TABLE]

and

[TABLE]

(Recall that $a=(1,\dots,1)$ is a generator of the trivial brace $A$.)

Proof. The first equality is a direct consequence of Lemma 5.1, and the definition of $b$ in terms of all the $b^{\prime}_{j}$, plus the definition of $\alpha^{\prime}_{n}$ in terms of all the $\widehat{\alpha}_{j}$.

For the second equality, recall that $x$ is of the form

[TABLE]

for some $\overline{h}_{j}\in\overline{H}_{j}$ and $z\in\mathbb{F}_{p_{1}}$. We use the following notation: let $x_{1}=z$ and $x_{j}=(\overline{h}_{j},\overline{h}_{j-1},\dots,\overline{h}_{2},z)$, for $j=2,\dots,n-1$. Then, if $g_{k}$, $k=1,2$, is of the form

[TABLE]

the corresponding lambda map is equal to

[TABLE]

where $x_{j-1}(h_{k,j})\colon\overline{G}_{j-1}\longrightarrow\mathbb{F}_{p_{j}}$ is the map defined by

[TABLE]

for all $g\in\overline{G}_{j-1}$, for $j=2,\dots,n$. By Lemma 5.1,

[TABLE]

Hence

[TABLE]

Thus, the result is proved.

Since $A$ is a trivial brace, by Lemma 5.2 and Theorem 3.1, we can construct the asymmetric product $\overline{G}_{n}\rtimes_{\circ}A$ (via $b$ and $\alpha^{\prime}_{n}$).

Recall from (2) that, for $(g,c),(g^{\prime},c^{\prime})\in\overline{G}_{n}\rtimes_{\circ}A$,

[TABLE]

Consider the semidirect product of left braces $\mathbb{F}_{p_{j}}[\overline{G}_{j-1}]\rtimes\overline{G}_{j-1}$ with respect to the action of $\overline{G}_{j-1}$ on the additive group of the group ring $\mathbb{F}_{p_{j}}[\overline{G}_{j-1}]$ (considered as a trivial brace) by left multiplication. As mentioned above, we may identify $\mathbb{F}_{p_{j}}\wr\overline{G}_{j-1}$ with $\mathbb{F}_{p_{j}}[\overline{G}_{j-1}]\rtimes\overline{G}_{j-1}$.

Theorem 5.3

The left brace $\overline{G}_{n}\rtimes_{\circ}A$ is simple and the derived length of its multiplicative group is $n+1$.

Proof. In this proof we shall identify $H_{j}$ with $\mathbb{F}_{p_{j}}[\overline{G}_{j-1}]$ (via $f=\sum_{g\in\overline{G}_{j-1}}f(g)g$, for all $f\in H_{j}$). Thus $H_{j}=\mathbb{F}_{p_{j}}[\overline{G}_{j-1}]$, $I_{j}=\{\sum_{g\in\overline{G}_{j-1}}zg\mid z\in\mathbb{F}_{p_{j}}\}$. As usual, we denote by $\overline{h}$ the natural image of $h\in H_{j}$ in $\overline{H}_{j}=H_{j}/I_{j}$.

Note also that, for $j=2,\dots,n$, $H_{j}$, $\overline{H}_{j}$ and $A$ are trivial braces. In order to avoid confusion, we only use the sum operation for these trivial braces. We will only use the multiplication of the trivial brace $G_{1}$ for elements in the group ring $H_{2}$. Thus the product in $H_{j}$ will mean the product in the corresponding group ring.

We shall prove the simplicity of the left brace $\overline{G}_{n}\rtimes_{\circ}A$ by induction on $n$. For $n=2$, let $I$ be a nonzero ideal of $\overline{G}_{2}\rtimes_{\circ}A$. Let $(\overline{h}_{2},z_{1},c)\in I$ be a nonzero element, where $\overline{h}_{2}\in\overline{H}_{2}$, $z_{1}\in G_{1}$ and $c\in A$ . We consider three mutually exclusive cases.

Case 1. Suppose that $\overline{h}_{2}\neq 0$. Then $p_{1}(\overline{h}_{2},z_{1},c)=(p_{1}\overline{h}_{2},0,c^{\prime})\in I$, for some $c^{\prime}\in A$ and $p_{1}\overline{h}_{2}\neq 0$, because $\overline{H}_{2}$ has order a power of $p_{2}$. Thus we may assume that $(\overline{h}_{2},0,c)\in I$ and $\overline{h}_{2}\neq 0$. Since $b^{\prime}_{2}$ is non-degenerate, there exists $\overline{h}\in\overline{H}_{2}$ such that $b^{\prime}_{2}(\overline{h}_{2},\overline{h})\neq 0$. Hence

[TABLE]

Since $A$ has order $p_{2}$ and $b^{\prime}(\overline{h}_{2},\overline{h})\in A\setminus\{0\}$, we have that $\{0\}\times\{0\}\times A\subseteq I$. In particular $(0,0,a)\in I$, where $a=1$ is the fixed generator of $A$. Thus

[TABLE]

for all $z\in G_{1}$. Since $\gamma-1$ is a nonzero element of $\mathbb{F}_{p_{1}}$, we have $\{0\}\times G_{1}\times\{0\}\subseteq I$. Hence $\{0\}\times G_{1}\times A\subseteq I$. Now we have

[TABLE]

for all $\overline{h}\in\overline{H}_{2}$ and $z\in G_{1}$. Hence

[TABLE]

for all $h\in H_{2}$. Therefore $\pi_{2}(\omega(\mathbb{F}_{p_{2}}[G_{1}]))\times G_{1}\times A\subseteq I$, where $\pi_{2}\colon H_{2}\longrightarrow\overline{H}_{2}$ is the natural map. Note that

[TABLE]

Therefore $\overline{H}_{2}\times G_{1}\times A\subseteq I$. This implies that $\overline{G}_{2}\rtimes_{\circ}A=I$ in this case.

Case 2. Suppose that $\overline{h}_{2}=0$ and $z_{1}\neq 0$. Now $p_{2}(0,z_{1},c)=(0,p_{2}z_{1},0)\in I$ and $p_{2}z_{1}\neq 0$ because $G_{1}$ has order $p_{1}$. Now

[TABLE]

for all $\overline{h}\in\overline{H}_{2}$. In particular, for $h=1$ in the group ring $H_{2}$, we have $(\overline{z_{1}-1},0,-b^{\prime}_{2}(z_{1}-1,1))\in I$. Note $z_{1}-1\in H_{2}\setminus I_{2}$. Therefore $\overline{z_{1}-1}\neq 0$ in $\overline{H}_{2}$. So, by Case 1, $\overline{G}_{2}\rtimes_{\circ}A=I$, as desired.

Case 3. Suppose that $(h_{2},z_{1})=0$ and $c\neq 0$. Then $(0,0,a)\in I$ and, as in the proof of Case 1, we again get that $\overline{G}_{2}\rtimes_{\circ}A=I$.

Therefore for $n=2$ the left brace $\overline{G}_{2}\rtimes_{\circ}A$ is simple.

Suppose now that $n>2$ and assume that the result holds for $n-1$. Let $A_{n-1}=\mathbb{F}_{p_{2}}\times\dots\times\mathbb{F}_{p_{n-1}}$. Note that $\{0\}\times\overline{G}_{n-1}\times A_{n-1}\times\{0\}$ is a subbrace of $\overline{G}_{n}\rtimes_{\circ}A$ isomorphic to $\overline{G}_{n-1}\rtimes_{\circ}A_{n-1}$ which is simple by the induction hypothesis. Let $I$ be a nonzero ideal of $\overline{G}_{n}\rtimes_{\circ}A$ and let $(\overline{h}_{n},\dots,\overline{h}_{2},z_{1},c)$ be a nonzero element of $I$, where $\overline{h}_{j}\in\overline{H}_{j}$, $z_{1}\in G_{1}$ and $c\in A$. We again consider three mutually exclusive cases.

Case i. Suppose that $\overline{h}_{n}\neq 0$. Then

[TABLE]

and $p_{1}\cdots p_{n-1}\overline{h}_{n}\neq 0$, because $\overline{H}_{n}$ has order a power of $p_{n}$. Thus, without loss of generality, we may assume that $(\overline{h}_{n},0,\dots,0,0,c)\in I$ and $\overline{h}_{n}\neq 0$. Since $b^{\prime}_{n}$ is non-degenerate, there exists $\overline{h}\in\overline{H}_{n}$ such that $b^{\prime}_{n}(\overline{h}_{n},\overline{h})\neq 0$. Hence

[TABLE]

In particular $(0,\dots,0,(0,\dots,0,1))\in I$. Thus

[TABLE]

for all $z\in G_{1}$. Since $\alpha_{1}(0,\dots,0,1)\neq\operatorname{id}$ and the subbrace $\{0\}\times\overline{G}_{n-1}\times A_{n-1}\times\{0\}$ is simple, we have that $\{0\}\times\overline{G}_{n-1}\times A_{n-1}\times\{0\}\subseteq I$. Moreover, $\{0\}\times\overline{G}_{n-1}\times A\subseteq I$ because $(0,\dots,0,(0,\dots,0,1))\in I$. Now we have

[TABLE]

for all $\overline{h}\in\overline{H}_{n}$ and $g\in\overline{G}_{n-1}$, and then $\pi_{n}(\omega(\mathbb{F}_{p_{n}}[\overline{G}_{n-1}]))\times\overline{G}_{n-1}\times A\subseteq I$, where $\pi_{n}\colon H_{n}\longrightarrow\overline{H}_{n}$ is the natural map. Note that

[TABLE]

Therefore $\overline{G}_{n}\rtimes_{\circ}A=I$ in this case.

Case ii. Suppose that $\overline{h}_{n}=0$ and $(\overline{h}_{n-1},\dots,\overline{h}_{2},z_{1})\neq 0$. Since the subbrace $\{0\}\times\overline{G}_{n-1}\times A_{n-1}\times\{0\}$ is simple, we have that $\{0\}\times\overline{G}_{n-1}\times A_{n-1}\times\{0\}\subseteq I$. Hence, as in the proof of the last part of Case i, $\overline{G}_{n}\rtimes_{\circ}A=I$.

Case iii. Suppose that $(\overline{h}_{n},\dots,\overline{h}_{2},z_{1})=0$ and $c\neq 0$. In this case

[TABLE]

for all $z\in\mathbb{F}_{p_{1}}$. Since $\alpha_{1}(c)\neq\operatorname{id}$, by Case ii, $\overline{G}_{n}\rtimes_{\circ}A=I$, again as desired.

Therefore $\overline{G}_{n}\rtimes_{\circ}A$ is simple.

In the remainder of the proof we show that the derived length of the multiplicative group of the left brace $\overline{G}_{n}\rtimes_{\circ}A$ is $n+1$. Recall that this group is the semidirect product $\overline{G}_{n}\rtimes A$ (via the action $\alpha^{\prime}_{n}$) of the multiplicative groups of the left braces $\overline{G}_{n}$ and $A$ (recall that $A$ is the trivial brace associated to the ring $\mathbb{F}_{p_{2}}\times\dots\times\mathbb{F}_{p_{n}}$, and therefore the multiplicative group of the trivial brace $A$ coincides with its additive group). First we shall prove by induction on $n$ that

[TABLE]

Suppose that $n=1$. Note that if $z_{1}\in G_{1}$, then

[TABLE]

Hence $[G_{1}\rtimes A,G_{1}\rtimes A]=G_{1}\times\{0\}$. Suppose that $n\geq 2$ and that

[TABLE]

Let $g_{n-1}\in\overline{G}_{n-1}$ and $\overline{h_{n}}\in\overline{H}_{n}$. Since the left brace $H_{n}$ is trivial, we will use the additive notation on $H_{n}$. Then

[TABLE]

Thus

[TABLE]

By (6), which is also true for $n=2$, we obtain that

[TABLE]

Since

[TABLE]

we get that

[TABLE]

and thus

[TABLE]

as claimed. Hence, it is enough to show that the derived length of the multiplicative group $\overline{G}_{n}$ is $n$. It is clear that the derived length of $\overline{G}_{2}$ is $2$. Suppose that $n>2$ and that $\overline{G}_{n-1}$ has derived length $n-1$. Recall that $G_{n}=\mathbb{F}_{p_{n}}[\overline{G}_{n-1}]\rtimes\overline{G}_{n-1}$. Hence, by Lemma 4.3, the derived length of $G_{n}$ is $n$. Note that $N_{n}=I_{n}\times\{1\}$ is a central subgroup of $G_{n}=H_{n}\rtimes\overline{G}_{n-1}$, and by (6)

[TABLE]

Furthermore, it is easy to see that

[TABLE]

Since

[TABLE]

and

[TABLE]

the derived length of $\overline{G}_{n}$ also is $n$ and the result follows.

5.2 A generalized construction of simple braces

In this subsection we present a broad class of simple left braces, extending the construction of [3]. Let $A=\mathbb{Z}/(l_{1})\times\dots\times\mathbb{Z}/(l_{s})$ be a finite abelian group and let $p$ be a prime such that $p\mid q-1$ for every prime divisor $q$ of the order of $A$. For every $i=1,\dots,s$, let

$b_{i}$ be a non-degenerate symmetric bilinear form over $(\mathbb{Z}/(p))^{n_{i}}$, 2. 2.

$f_{i}$ and $c_{i}$ be elements of order $p$ and $l_{i}$ respectively in the orthogonal group of $b_{i}$, 3. 3.

$\gamma_{i}\in\mathbb{Z}/(l_{i})$ be an invertible element of order $p$.

Assume that

• (i)

$f_{i}c_{i}=c_{i}^{\gamma_{i}}f_{i}$,

• (ii)

$\gamma_{i}-1$ is invertible in $\mathbb{Z}/(l_{i})$.

Put $n=n_{1}+\dots+n_{s}$. Let $T=(\mathbb{Z}/(p))^{n}\rtimes A$ be the semidirect product of the trivial braces $(\mathbb{Z}/(p))^{n}$ and $A$ via the action

[TABLE]

defined by $\beta_{(a_{1},\dots,a_{s})}(u_{1},\dots,u_{s})=(c_{1}^{a_{1}}(u_{1}),\dots,c_{s}^{a_{s}}(u_{s}))$, where $\beta(a_{1},\dots,a_{s})=\beta_{(a_{1},\dots,a_{s})}$, for all $a_{i}\in\mathbb{Z}/(l_{i})$ and all $u_{i}\in(\mathbb{Z}/(p))^{n_{i}}$.

Lemma 5.4

The map $\alpha\colon(\mathbb{Z}/(p),+)\longrightarrow\operatorname{Aut}(T,+,\cdot)$ defined by

[TABLE]

where $\alpha(\mu)=\alpha_{\mu}$, for all $\mu\in\mathbb{Z}/(p)$, all $a_{i}\in\mathbb{Z}/(l_{i})$ and all $u_{i}\in(\mathbb{Z}/(p))^{n_{i}}$, is a homomorphism of groups.

Proof. It is clear that $\alpha_{\mu}\in\operatorname{Aut}(T,+)$. Because $f_{i}c_{i}=c_{i}^{\gamma_{i}}f_{i}$ for all $i=1,\dots,s$, it is straightforward to verify that $\alpha_{\mu}$ also is a multiplicative group homomorphism, and thus a brace homomorphism. Hence, $\alpha$ is a homomorphism.

Let $b\colon T\times T\longrightarrow\mathbb{Z}/(p),$ defined by

[TABLE]

for all $u_{i}\in(\mathbb{Z}/(p))^{n_{i}}$ and all $a_{i}\in\mathbb{Z}/(l_{i})$. Since each $b_{i}$ is a symmetric bilinear form, it is clear that $b$ is a symmetric $2$-cocyle on $T$ with values in $\mathbb{Z}/(p)$.

We shall see that $\alpha$ and $b$ satisfy condition (1). In this case this is equivalent to verify conditions (3) and (4), and for this we need to check:

[TABLE]

[TABLE]

for all $t_{1},t_{2},t_{3}\in T$ and all $\mu\in\mathbb{Z}/(p)$. That these equalities hold follows from the definitions of $\alpha$, $\lambda$ and $b$, and from the fact that $c_{i}$ and $f_{i}$ are in the orthogonal group of $b_{i}$. Hence, by Theorem 3.1, we get the asymmetric product $T\rtimes_{\circ}\mathbb{Z}/(p)$ of $T$ by $\mathbb{Z}/(p)$ via $b$ and $\alpha$.

Let $B=T\rtimes_{\circ}\mathbb{Z}/(p)$. Recall that the addition in $B$ is given by

[TABLE]

for all $u_{i},v_{i}\in(\mathbb{Z}/(p))^{n_{i}}$, all $a_{i},a^{\prime}_{i}\in\mathbb{Z}/(l_{i})$ and all $\mu,\mu^{\prime}\in\mathbb{Z}/(p)$. The lambda map of $B$ is given by

[TABLE]

where $a_{i},a^{\prime}_{i}\in\mathbb{Z}/(l_{i})$ and $u_{i},v_{i}\in(\mathbb{Z}/(p))^{n_{i}}$, $\mu,\mu^{\prime}\in\mathbb{Z}/(p)$.

The following technical remark is of use for the subsequent proofs.

Remark 5.5

For $i=1,\dots,s$, let $G_{i}$ be the subgroup of $\operatorname{Aut}((\mathbb{Z}/(p))^{n_{i}})$ generated by $c_{i}$ and $f_{i}$. Then there exists a natural homomorphism of $\mathbb{Z}/(p)$-algebras $\varphi_{i}\colon\mathbb{Z}/(p)[G_{i}]\longrightarrow\operatorname{End}((\mathbb{Z}/(p))^{n_{i}})$ such that $\varphi_{i}(c_{i})=c_{i}$ and $\varphi_{i}(f_{i})=f_{i}$. Since

[TABLE]

it is clear that for every $u_{i}\in(\mathbb{Z}/(p))^{n_{i}}$ and every $g\in\varphi(\omega(\mathbb{Z}/(p)[G_{i}]))$, $g(u_{i})\in\operatorname{Im}(c_{i}-\operatorname{id})+\operatorname{Im}(f_{i}-\operatorname{id})$. Furthermore, if $v_{i}\in\operatorname{Im}(c_{i}-\operatorname{id})+\operatorname{Im}(f_{i}-\operatorname{id})$, then $f_{i}(v_{i}),c_{i}(v_{i})\in\operatorname{Im}(c_{i}-\operatorname{id})+\operatorname{Im}(f_{i}-\operatorname{id})$.**

Lemma 5.6

Let

[TABLE]

Then $I^{\prime}$ is an ideal of $B$.

Proof. Clearly $I^{\prime}$ is a subgroup of the additive group of $B$.

Let $(u_{1},\dots,u_{s},a_{1},\dots,a_{s},\mu)\in B$ and $(w_{1},\dots,w_{s},a^{\prime}_{1},\dots,a^{\prime}_{s},\mu^{\prime})\in I^{\prime}$. We have

[TABLE]

Thus, by Remark 5.5, $\lambda_{(u_{1},\dots,u_{s},a_{1},\dots,a_{s},\mu)}(w_{1},\dots,w_{s},a^{\prime}_{1},\dots,a^{\prime}_{s},\mu^{\prime})\in I^{\prime}$. Finally

[TABLE]

by Remark 5.5. Thus, $I^{\prime}$ is a normal subgroup of the multiplicative group of $B$, and the result follows.

Theorem 5.7

The left brace $B=T\rtimes_{\circ}\mathbb{Z}/(p)$ is simple if and only if $\operatorname{Im}(c_{i}-\operatorname{id})+\operatorname{Im}(f_{i}-\operatorname{id})=(\mathbb{Z}/(p))^{n_{i}}$, for every $i=1,\dots,s$. In particular, a sufficient condition to be simple is that $c_{i}-\operatorname{id}$ is invertible for every $i=1,\dots,s$.

Proof. It is enough to prove that any non-zero ideal of $B$ contains the ideal $I^{\prime}$ defined in the previous lemma. So let $I$ be a nonzero ideal of $B$. Let $(u_{1},\dots,u_{s},a_{1},\dots,a_{s},\mu)$ be a nonzero element of $I$. We consider two cases.

Case 1: suppose that some $a_{i}$ is nonzero. Then

[TABLE]

for some $\mu^{\prime}\in\mathbb{Z}/(p)$, and $pa_{i}\neq 0$ because the order of $A$ is not a multiple of the prime $p$. Now

[TABLE]

and $p^{2}a_{i}\neq 0$. Thus we may assume that $(0,\dots,0,a_{1},\dots,a_{s},0)\in I$ and $a_{i}\neq 0$. Let $v_{i}\in(\mathbb{Z}/(p))^{n_{i}}$ and for $j\neq i$, let $v_{j}=0\in(\mathbb{Z}/(p))^{n_{j}}$. Then

[TABLE]

Since $c_{i}$ has order $l_{i}$ and $a_{i}\neq 0$, there exists $v_{i}\in(\mathbb{Z}/(p))^{n_{i}}$ such that $(c_{i}^{a_{i}}-\operatorname{id})(v_{i})\neq 0$. Now

[TABLE]

Thus we may assume that $(0,\dots,0,u_{i},0,\dots,0,0,\dots,0,0)\in I$ for some nonzero element $u_{i}\in(\mathbb{Z}/(p))^{n_{i}}$. Since $b_{i}$ is non-degenerate, there exists $w_{i}\in(\mathbb{Z}/(p))^{n_{i}}$ such that $b_{i}(w_{i},u_{i})\neq 0$. We have that

[TABLE]

Hence $\{(0,\dots,0,0,\dots,0,\mu^{\prime})\mid\mu^{\prime}\in\mathbb{Z}/(p)\}\subseteq I$. Note that for arbitrary $a_{1}^{\prime},\ldots,a_{n}^{\prime}$,

[TABLE]

Since, by assumption, $\gamma_{j}-1$ is invertible in $\mathbb{Z}/(l_{j})$, we have that

[TABLE]

Hence

[TABLE]

Now we have that

[TABLE]

for all $v_{j}\in(\mathbb{Z}/(p))^{n_{j}}$, all $a^{\prime}_{j}\in\mathbb{Z}/(l_{i})$ and all $\mu^{\prime}\in\mathbb{Z}/(p)$. Hence

[TABLE]

as desired.

Case 2: suppose that $a_{j}=0$ for all $j$. Thus $(u_{1},\dots,u_{s},0,\dots,0,\mu)\in I$ is a nonzero element. If $\mu\neq 0$, then

[TABLE]

Therefore, by Case 1, $I^{\prime}\subseteq I$. If $\mu=0$, then some $u_{i}\neq 0$ and

[TABLE]

Since $b_{i}$ is non-degenerate, there exists $w_{i}\in(\mathbb{Z}/(p))^{n_{i}}$ such that $b_{i}(w_{i},u_{i})\neq 0$. Therefore, by the first part of Case 2, $I^{\prime}\subseteq I$. Thus the result follows.

5.3 Concrete construction of simple left braces

In this section we give concrete realisations of simple braces as constructed in Theorem 5.7.

Let $m$ be a positive integer and let $p,p_{1},p_{2},\dots,p_{m}$ be different primes such that $p\mid p_{j}-1$, for all $j=1,\dots,m$. Let $s$ be a positive integer and let $A=\mathbb{Z}/(l_{1})\times\dots\times\mathbb{Z}/(l_{s})$, where each $l_{i}$ is an integer such that $l_{i}>1$ and

[TABLE]

for some nonnegative integers $r_{i,j}$.

Consider the natural isomorphism of rings

[TABLE]

Since $p\mid p_{j}-1$, for all $j=1,\dots,m$, if $r_{i,j}>0$, then there exist invertible elements $\gamma_{i,j}\in\mathbb{Z}/(p_{j}^{r_{i,j}})$ of order $p$. If $r_{i,j}=0$, let $\gamma_{i,j}=0$. Let $\gamma_{i}\in\mathbb{Z}/(l_{i})$ be the element corresponding to $(\gamma_{i,1},\dots,\gamma_{i,m})$ under the natural isomorphism. Then, $\gamma_{i}$ is an invertible element of order $p$ and $\gamma_{i}-1$ corresponds to the element $(\gamma_{i,1}-1,\dots,\gamma_{i,m}-1)$. Note that if $r_{i,j}>0$ then $(1+p_{j})^{p_{j}^{r_{i,j}-1}}=1$ in $\mathbb{Z}/(p_{j}^{r_{i,j}})$. Hence, in this case, $\gamma_{i,j}-1\notin p_{j}\mathbb{Z}/(p_{j}^{r_{i,j}})$, and consequently, $\gamma_{i,j}-1$ is invertible in $\mathbb{Z}/(p_{j}^{r_{i,j}})$. Therefore, $\gamma_{i}-1$ is invertible in $\mathbb{Z}/(l_{i})$.

For $i=1,\dots,s$, define the ring $R_{i}=\mathbb{F}_{p}[x]/(x^{l_{i}-1}+\cdots+x+1)$, and denote $\xi_{i}:=\overline{x}\in R_{i}$. Let $c_{i}\in\operatorname{Aut}(R_{i},+)$ be defined by

[TABLE]

for all $r_{i}\in R_{i}$. Let $f_{i}\in\operatorname{Aut}(R_{i},+)$ be defined by

[TABLE]

for all $r_{i}(\xi_{i})\in R_{i}$. Note that

[TABLE]

for all $r_{i}(\xi_{i})\in R_{i}$. Hence $f_{i}c_{i}=c_{i}^{\gamma_{i}}f_{i}$.

Note that

[TABLE]

Since $p\mid p_{j}-1$, for all $j=1,\dots,m$, we have that $p\mid l_{i}-1$, for all $i$. Since the characteristic polynomial of $c_{i}$ is $x^{l_{i}-1}+\cdots+x+1\in\mathbb{F}_{p}[x]$ and $p\mid l_{i}-1$, $c_{i}-\operatorname{id}$ is invertible.

Let $b_{i}\colon R_{i}\times R_{i}\longrightarrow\mathbb{F}_{p}$ be the unique bilinear form such that $b_{i}(\xi_{i}^{j},\xi_{i}^{k})=1-\delta_{j,k}$, for all $j,k\in\{0,1,\dots,l_{i}-2\}$. Note that $(1,\xi_{i},\dots,\xi_{i}^{l_{i}-2})$ is a basis of the $\mathbb{F}_{p}$-vector space $R_{i}$. The matrix of $b_{i}$ with respect to this basis is

[TABLE]

Since $p\mid l_{i}-1$, the determinant of this matrix is $(l_{i}-2)(-1)^{l_{i}-2}=(-1)^{l_{i}-1}\in\mathbb{F}_{p}$. Hence $b_{i}$ is non-degenerate. Furthermore

[TABLE]

and

[TABLE]

Therefore $f_{i}$ and $c_{i}$ are in the orthogonal group of $b_{i}$.

Let $T=(R_{1}\times\dots\times R_{s})\rtimes A$ be the semidirect product of the trivial braces $R_{1}\times\dots\times R_{s}$ and $A$ via the action

[TABLE]

defined by $\beta_{(a_{1},\dots a_{s})}(u_{1},\dots,u_{s})=(c_{1}^{a_{1}}(u_{1}),\dots,c_{s}^{a_{s}}(u_{s}))$, where $\beta(a_{1},\dots,a_{s})=\beta_{(a_{1},\dots,a_{s})}$, for all $a_{i}\in\mathbb{Z}/(l_{i})$ and all $u_{i}\in R_{i}$.

We define the map $\alpha\colon(\mathbb{Z}/(p),+)\longrightarrow\operatorname{Aut}(T,+,\cdot)$ by

[TABLE]

where $\alpha(\mu)=\alpha_{\mu}$, for all $\mu\in\mathbb{F}_{p}$, all $a_{i}\in\mathbb{Z}/(l_{i})$ and all $u_{i}\in R_{i}$.

Let $b\colon T\times T\longrightarrow\mathbb{F}_{p}$ be the symmetric bilinear form defined by

[TABLE]

for all $a_{i}\in\mathbb{Z}/(l_{i})$ and all $u_{i}\in R_{i}$.

Since $c_{i}-\operatorname{id}$ is invertible, Theorem 5.7 yields that the the asymmetric product $T\rtimes_{\circ}\mathbb{F}_{p}$ of $T$ by $\mathbb{F}_{p}$, via $b$ and $\alpha$, is a simple left brace.

Remark 5.8

In our approach we exploited the bilinear form with matrix $B=(b_{ij})$, where $b_{ii}=0$ and $b_{ij}=1$ for $i\neq j$. Every permutation matrix is in the orthogonal group of this bilinear form. In the concrete constructions we show how to construct an element $f$ in the orthogonal group that has the desired properties. However, one can give a general reason why such elements exist.

Assume that $q$ and $p$ are prime numbers such that $p\mid(q-1)$. Let $c\in S_{n}$ be an element of the symmetric group $S_{n}$ for some $n$. Assume that $c$ has order $q$. Assume also that $\gamma\in\mathbb{Z}_{q}^{*}$ has order $p$. We claim that $c^{\gamma}=fcf^{-1}$ for a permutation $f\in S_{n}$ of order $p$.

Notice that $c^{\gamma}$ also has order $q$ (since $p,q$ are relatively prime) and in fact $c,c^{\gamma}$ are products of the same number of disjoint cycles of length $q$. It is then enough to consider the case where $c=(1,2,\ldots,q)$, a cycle of length $q$, and $n=q$. In particular, there exists $g\in S_{q}$ such that $gcg^{-1}=c^{\gamma}$. By the hypothesis, $\gamma^{p}=qr+1$ for some nonnegative integer $r$. Then

[TABLE]

It is easy to see that the centralizer $C_{S_{q}}(c)$ of $c$ in the symmetric group $S_{q}$ is equal to the cyclic group $\langle c\rangle$. Therefore $g^{p}\in\langle c\rangle$. In particular, $g^{pq}=1$. Since $p\mid(q-1)$, it follows that $q=pk+1$ for some $k$ and then $g^{q}=g^{pk+1}\in gC_{S_{q}}(c)$. Therefore, $g^{q}cg^{-q}=gcg^{-1}=c^{\gamma}$.

Notice that $g^{q}\neq 1$, as otherwise $c^{\gamma}=c$, a contradiction, because $c$ has order $q$ and $\gamma\in\mathbb{Z}_{q}^{*}$ has order $p$. Therefore, $g^{q}\neq 1$ and then $g^{q}$ is a permutation of order $p$. Hence, we may put $f=g^{q}$.**

6 Other simple left braces as asymmetric product of left

braces

Our aim in this section is to show that all previously known nontrivial left simple braces (see [6, Theorems 3.1 and 3.6]) can be interpreted as asymmetric products of braces. This sheds a new light on these classes of examples and raises the question of a further application of the asymmetric product construction in the context of the challenging program of finding new (simple) left braces.

Let $H(p^{r},n,Q,f)$ be the left brace described in [6, Theorem 3.1], where $p$ is a prime number, $r,n$ are positive integers, $Q$ is a quadratic form over $(\mathbb{Z}/(p^{r}))^{n}$ (considered as a free module over the ring $\mathbb{Z}/(p^{r})$) and $f$ is an element in the orthogonal group of $Q$ of order $p^{r^{\prime}}$ for some $0\leq r^{\prime}\leq r$.

Recall that the additive group of $H(p^{r},n,Q,f)$ is the additive abelian group $(\mathbb{Z}/(p^{r}))^{n+1}$. The elements of $(\mathbb{Z}/(p^{r}))^{n+1}$ will be written in the form $(\vec{x},\mu)$, with $\vec{x}\in(\mathbb{Z}/(p^{r}))^{n}$ and $\mu\in\mathbb{Z}/(p^{r})$. The lambda map of $H(p^{r},n,Q,f)$ is defined by

[TABLE]

for $(\vec{x},\mu),(\vec{y},\mu^{\prime})\in(\mathbb{Z}/(p^{r}))^{n+1}$, where $q(\vec{x},\mu)=\mu-Q(\vec{x})$, and $b$ is the bilinear form $b(\vec{x},\vec{y})=Q(\vec{x}+\vec{y})-Q(\vec{x})-Q(\vec{y})$ associated to $Q$.

Let $T$ be the trivial brace $(\mathbb{Z}/(p^{r}))^{n}$ and let $S$ be the trivial brace $\mathbb{Z}/(p^{r})$. Let $\alpha\colon(S,\cdot)\longrightarrow\operatorname{Aut}(T,+,\cdot)$ be the map defined by $\alpha(\mu)=\alpha_{\mu}$ and

[TABLE]

for all $\mu\in S$ and $\vec{x}\in T$. It is clear that $\alpha$ is a homomorphism of groups. We will check that $\alpha$ and $-b$ satisfy condition (1). Since $-b$ is a symmetric bilinear form, this condition is equivalent to $-b(\vec{x},\vec{y})=-b(\alpha_{\mu}(\vec{x}),\alpha_{\mu}(\vec{y}))$, which is true because $f$ is in the orthogonal group of $Q$. Let $H^{\prime}(p^{r},n,-b,f)=T\rtimes_{\circ}S$ be the asymmetric product of $T$ by $S$ via $-b$ and $\alpha$.

Proposition 6.1

The map $\varphi\colon H(p^{r},n,Q,f)\longrightarrow H^{\prime}(p^{r},n,-b,f)$ defined by $\varphi(\vec{x},\mu)=(\vec{x},\mu-Q(\vec{x}))$ is an isomorphism of left braces.

Proof. Let $\vec{x},\vec{y}\in T$ and $\mu,\mu^{\prime}\in S$. We have

[TABLE]

and

[TABLE]

Thus the result follows.

Consider now the left brace $H_{1}\bowtie\dots\bowtie H_{s}$ described in [6, Theorem 3.6]. Recall that $H_{i}=H(p_{i}^{r_{i}},n_{i},Q_{i},f_{i})$, $s>1$, $\;p_{1},p_{2},\ldots,p_{s}$ are different prime numbers, $p_{1},p_{2},\dots,p_{s-1}$ are odd, $Q_{i}$ is a non-degenerate quadratic form over $(\mathbb{Z}/(p_{i}^{r_{i}}))^{n_{i}}$, $f_{i}$ is an element of order $p_{i}^{r^{\prime}_{i}}$ in the orthogonal group determined by $Q_{i}$, for some $0\leq r^{\prime}_{i}\leq r_{i}$, and the iterated matched product of left ideals is constructed with the maps

[TABLE]

with

[TABLE]

for $1\leq k<s$,

[TABLE]

and $\alpha^{(j,i)}_{(\vec{x}_{j},\mu_{j})}=\operatorname{id}_{H_{i}}$ otherwise, where, for $1\leq i<s$, $c_{i}$ is an element of order $p_{i+1}^{r_{i+1}}$ in the orthogonal group determined by $Q_{i}$, $c_{s}$ is an element of order $p_{1}^{r_{1}}$ of $\operatorname{Aut}((\mathbb{Z}/(p_{s}^{r_{s}}))^{n_{s}})$ and $v_{s}\in(\mathbb{Z}/(p_{s}^{r_{s}}))^{n_{s}}$, such that

[TABLE]

and

[TABLE]

for $1\leq i\leq s$. The lambda map of $H_{1}\bowtie\dots\bowtie H_{s}$ is defined by

[TABLE]

for all $(\vec{x}_{i},\mu_{i}),(\vec{y}_{i},\mu^{\prime}_{i})\in H_{i}$.

By [6, Theorem 3.6] the left brace $H_{1}\bowtie\dots\bowtie H_{s}$ is simple if and only if $c_{i}-\operatorname{id}$ is an automorphism for all $1\leq i\leq s$.

Let $T^{\prime}$ be the trivial brace $(\mathbb{Z}/(p_{1}^{r_{1}}))^{n_{1}}\times\dots\times(\mathbb{Z}/(p_{s}^{r_{s}}))^{n_{s}}$ and let $S^{\prime}$ be the trivial brace $\mathbb{Z}/(p_{1}^{r_{1}})\times\dots\times\mathbb{Z}/(p_{s}^{r_{s}})$. Let $\alpha^{\prime}\colon(S^{\prime},\cdot)\longrightarrow\operatorname{Aut}(T^{\prime},+,\cdot)$ be the map defined by $\alpha^{\prime}(\mu_{1},\dots,\mu_{s})=\alpha^{\prime}_{(\mu_{1},\dots,\mu_{s})}$ and

[TABLE]

for all $(\mu_{1},\dots,\mu_{s})\in S^{\prime}$ and $(\vec{x}_{1},\dots,\vec{x}_{s})\in T^{\prime}$. Let $b^{\prime}\colon T^{\prime}\times T^{\prime}\longrightarrow S^{\prime}$ be the map defined by

[TABLE]

for all $(\vec{x}_{1},\dots,\vec{x}_{s}),(\vec{y}_{1},\dots,\vec{y}_{s})\in T^{\prime}$. We will check that $\alpha^{\prime}$ and $b^{\prime}$ satisfy condition (1). Since $b^{\prime}$ is a symmetric bilinear form this is equivalent to

[TABLE]

which is true because, for $1\leq i<s$, $c_{i}$ and $f_{i}$ are in the orthogonal group of $Q_{i}$, $f_{s}$ is in the orthogonal group of $Q_{s}$ and

[TABLE]

Let $B=T^{\prime}\rtimes_{\circ}S^{\prime}$ be the asymmetric product of $T^{\prime}$ by $S^{\prime}$ via $b^{\prime}$ and $\alpha^{\prime}$.

Theorem 6.2

The map $\varphi^{\prime}\colon H_{1}\bowtie\dots\bowtie H_{s}\longrightarrow B$ defined by

[TABLE]

is an isomorphism of left braces.

Proof. The proof is straightforward using the definitions of the sum and the multiplication in both left braces and the formula (10).

Acknowledgments

The two first-named authors were partially supported by the grant MINECO MTM2014-53644-P. The third author is supported in part by Onderzoeksraad of Vrije Universiteit Brussel and Fonds voor Wetenschappelijk Onderzoek (Belgium). The fourth author is supported by a National Science Centre grant (Poland).