is simple if $B$ and $0$ are its only ideals. If $B$ is finite, then the Sylow subgroups $P_1,\dots, P_n$ of the additive group of $B$ are left ideals of $B$. Furthermore $\mathcal{S}=\{
P_1,\dots ,P_n\}$ is a Sylow system of the multiplicative group of $B$. We say that a left brace $B$ is trivial if $a+b=ab$ for all $a,b\in B$.

Consider a finite left brace $(B,+,\cdot)$ such that $(B,\cdot)$ is a metabelian $A$-group. Let $P_{1},\ldots, P_{n}$ be the Sylow subgroups of its additive group. Thus $\mathcal{S}=\{ P_1,\dots
,P_n\}$ is a Sylow system of $(B,\cdot)$. Let $M(B)$ denote its associated system normalizer, that is $M(B)=
\bigcap_{i=1}^{n}N_{B}(P_{i})$. It is known that the multiplicative group of $B$ is $B=[B,B]\rtimes M(B)$, a semidirect product and $M(B)$ is an abelian group, [@hall; @taunt]. Then $[B,B]=T_{1}\times \cdots \times T_{n}$, for subgroups $T_{i}$ of $(P_{i},\cdot)$, and $M(B)=S_{1}\times \cdots \times S_{n}$ for subgroups $S_{i}$ of $(P_{i},\cdot)$. Since $(P_i,\cdot)$ is abelian, $P_{i} = T_{i}\times S_{i}$.

A special case that will be exploited in this paper is when $(B,+,\cdot)$ is an asymmetric product $B=[B,B]\rtimes_{\circ} M(B)$ of two trivial left braces $[B,B], M(B)$ via a symmetric bilinear map $b: [B,B]\longrightarrow M(B)$ (on $([B,B],+)$ with values in $(M(B),+)$) and a homomorphism $$\alpha: 