has order $p_{z-1}$ and $$\begin{aligned}
\lefteqn{\left(
\begin{array}{c|c}
0&I_{p_{z-1}-1}\\
\hline I_{p_{z-1}-1}&0
\end{array}\right)}\\
&=& \left(
\begin{array}{c|c}
C_z&0\\
\hline 0&(C_z^{-1})^t
\end{array}\right)^t\left(
\begin{array}{c|c}
0&I_{p_{z-1}-1}\\
\hline I_{p_{z-1}-1}&0
\end{array}\right)\left(
\begin{array}{c|c}
C_z&0\\
\hline 0&(C_z^{-1})^t
\end{array}\right)
.\end{aligned}$$ Hence $f_z\in \mathrm{O}(V_z,b_z)$ and $f_z-{\operatorname{id}}$ is bijective.

Let $n'_z =\max\{\dim(V_z),\dim(V_{z-1})\}$ and let $l_{z}
=n'_z (\dim(V_z)+1)$. Let $m_z\geq l_z$. Write $m_z
=m'_{z}\dim(V_z)+ r_z$ with $0< r_z \leq \dim(V_z)$ and a positive integer $m'_{z}$. Note that $m_z\geq l_z=n'_z(\dim(V_z)+1)=
n'_z\dim(V_z)+n'_z\geq n'_z\dim(V_z)+r_z$. Hence $m'_z\geq n'_z\geq
\max\{ r_z,r_{z-1}\}$. Let $T_z =V_{z}^{m'_{z}}$ and $S_z=({\mathbb{Z}}/(p_{z}))^{r_{z}}$. Let $T=T_1\times \cdots\times T_n$ and $S=S_1\times \cdots\times S_n$. Then, using Theorem \[simple0\] and Proposition \[construction1\], one can construct a simple left brace $T\rtimes_{\circ}S$ that has a metabelian multiplicative group with abelian Sylow subgroups, and $$|T\rtimes_{\circ}S|=p_1^{m'_1\dim(V_1)+r_1}p_2^{m'_2\dim(V_2)+r_2}\cdots p_n^{m'_n\dim(V_n)+r_n}=p_1^{m_1}p_2^{m_2}\cdots p_n^{m_n}.$$ Therefore the result follows.

\[dimension\][In the proof of the main result (Theorem \[MainResult\]), we have seen that the positive integers $l_1,\dots, l_n$ of the statement can be defined as $$l_z=\max\{\dim(V_z),\dim(V_{z-1})\}(\dim(V_z)+1),$$ for all $z\in{\mathbb{Z}}/(n)$, where $V_z$ is a finite dimensional ${\mathbb{Z}}/(p_z)$-vector space with a non-singular symmetric bilinear form $b_z$ and $f_z\in \mathrm{O}(V_z,b_z)$ of order $p_{z-1}$ such that $f_z-{\operatorname{id}}$ is bijective.]{}

Search for the best bounds {#bounds}
==========================

In this section we study the minimal possible orders of finite simple left braces with multiplicative $A$-group that can arise 