all $z_1, z_2,z \in {\mathbb{Z}}/(n)$, $s_1\in S_{z_1}$, $s_2\in S_{z_2}$ and $s\in S_z$. Let $T=T_1\times \cdots\times T_n$ and $S=S_1\times
\cdots\times S_n$ denote the direct products of the trivial left braces $T_z$’s and $S_z$’s respectively. Thus $T$ and $S$ are trivial left braces. Let $b\colon T\times T\longrightarrow S$ be the symmetric bilinear map defined by $$b((t_1,\dots ,t_n),(t'_1,\dots ,t'_n))=(b_1(t_1,t'_1),\dots ,b_n(t_n,t'_n)),$$ for all $t_z,t'_z\in T_z$. Let $\alpha\colon
(S,\cdot)\longrightarrow {\operatorname{Aut}}(T,+)$ be the map defined by $\alpha(s_1,\dots, s_n)=\alpha_{(s_1,\dots, s_n)}$ and $$\begin{aligned}
\alpha_{(s_1,\dots,s_n)}(t_1,\dots,t_n) &=&(f_{s_1}^{(1,1)}\cdots
f_{s_n}^{(n,1)}(t_1),\dots,f_{s_1}^{(1,n)}\cdots
f_{s_n}^{(n,n)}(t_n)),\end{aligned}$$ for all $(s_1,\dots,s_n)\in S$ and $(t_1,\dots,t_n)\in T$.

\[bilinear\] $\alpha$ is a homomorphism of groups and $\alpha_{(s_1,\dots,
s_n)}\in \mathrm{O}(T,b)$ for all $(s_1,\dots ,s_n)\in S$. Therefore we can construct the asymmetric product $T\rtimes_{\circ} S$ of $T$ by $S$ via $b$ and $\alpha$.

By (\[commute\]), it is clear that $\alpha$ is a homomorphism of groups. Let $(s_1,\dots,s_n)\in S$ and $(t_1,\dots
,t_n),(t'_1,\dots ,t'_n)\in T$. We have $$\begin{aligned}
\lefteqn{b(\alpha_{(s_1,\dots,s_n)}(t_1,\dots,t_n),\alpha_{(s_1,\dots,s_n)}(t'_1,\dots,t'_n))}\\
 &=&b((f_{s_1}^{(1,1)}\cdots
f_{s_n}^{(n,1)}(t_1),\dots,f_{s_1}^{(1,n)}\cdots
f_{s_n}^{(n,n)}(t_n)),\\
&&\quad (f_{s_1}^{(1,1)}\cdots
f_{s_n}^{(n,1)}(t'_1),\dots,f_{s_1}^{(1,n)}\cdots
f_{s_n}^{(n,n)}(t'_n)))\\
 &=&(b_1(f_{s_1}^{(1,1)}\cdots
f_{s_n}^{(n,1)}(t_1),f_{s_1}^{(1,1)}\cdots
f_{s_n}^{(n,1)}(t'_1)),\\
&&\quad \dots,b_n(f_{s_1}^{(1,n)}\cdots
f_{s_n}^{(n,n)}(t_n),f_{s_1}^{(1,n)}\cdots
f_{s_n}^{(n,n)}(t'_n)))\\
 &=&(b_1(t_1,t'_1),\dots,b_n(t_n,t'_n)).\end{aligned}$$ Hence $\alpha_{(s_1,\dots,s_n)}\in \mathrm{O}(T,b)$ for all $(s_1,\dots ,s_n)\in S$. Thus the result follows.

For $z\in {\mathbb{Z}}/(n)$, let $A_z=\{ ((t_1,\dots ,t_n),(s_1,\dots
,s_n))\in T\rtimes_{\circ}S\mid t_{z'}=0 \text{ and }s_{z'}=0 \text{
for all }z'\neq z \}$.

\[construction1\] Each $A_z$ is a left ideal of $T\rtimes_{\circ}S$ with abelian 