\lefteqn{b'_z(f_s^{(z-1,z)}(u_1,\dots,u_{m_z}),f_s^{(z-1,z)}(v_1,\dots,v_{m_z}))}\\
&=&b'_z ((f_{z}^{\mu_1+\mu_{r_{z-1}}}(u_1),\dots,
f_{z}^{\mu_{r_{z-1}-1}+\mu_{r_{z-1}}}(u_{r_{z-1}-1}),\\
&&\quad f_{z}^{\mu_{r_{z-1}}}(u_{r_{z-1}}),\dots,
f_{z}^{\mu_{r_{z-1}}}(u_{m_{z}})),\\
&&\qquad
(f_{z}^{\mu_1+\mu_{r_{z-1}}}(v_1),\dots,
f_{z}^{\mu_{r_{z-1}-1}+\mu_{r_{z-1}}}(v_{r_{z-1}-1}),\\
&&\quad f_{z}^{\mu_{r_{z-1}}}(v_{r_{z-1}}),\dots,
f_{z}^{\mu_{r_{z-1}}}(v_{m_{z}})))\\
&=&\sum_{i=1}^{r_z-1}b_z(u_i,v_i)e_i^{(z)}+\sum_{j=1}^{m_z}b_z(u_j,v_j)e_{r_z}^{(z)}\\
&=&b'_z((u_1,\dots, u_{m_z}),(v_1,\dots, v_{m_z})),\end{aligned}$$ where the second equality holds because $f_z\in
\mathrm{O}(V_z,b_z)$. Thus the result follows.

Let $T=T_1\times \cdots\times T_n$ and $S=S_1\times\cdots\times S_n$ be the direct products of the left braces $T_z$’s and $S_z$’s respectively. 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 $(t_1,\dots ,t_n),(t'_1,\dots ,t'_n)\in T$. Let $f^{(z,z')}\colon S_z\longrightarrow {\operatorname{Aut}}(T_z,+)$ be the trivial homomorphism for all $z'\neq z+1$. Note that the homomorphisms $f^{(z,z')}$ satisfy (\[commute\]). 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 $$\begin{aligned}
\alpha_{(s_1,\dots,s_n)}(t_1,\dots ,t_n)&=&(f^{(1,1)}_{s_1}\cdots
f^{(n,1)}_{s_n}(t_1),\dots ,f^{(1,n)}_{s_1}\cdots
f^{(n,n)}_{s_n}(t_n))\\
&=&(f^{(n,1)}_{s_n}(t_1),f^{(1,2)}_{s_1}(t_2),\dots
,f^{(n-1,n)}_{s_{n-1}}(t_n)),\end{aligned}$$ for all $(s_1,\dots,s_n)\in S$ and $(t_1, \dots, t_n)\in T$. By Lemma \[orthof\], and Lemma \[bilinear\], we have that $\alpha_{(s_1,\dots,s_n)}\in \mathrm{O}(T,b)$, for all $(s_1,\dots,s_n)\in S$ and we can construct the asymmetric product $T\rtimes_{\circ} S$ of $T$ by $S$ via $b$ and $\alpha$.

For $z\in {\mathbb{Z}}/(n)$, let $A_z=\{ ((t_1,\dots ,t_n),(s_1,\dots
,s_n))\in T\rtimes_{\circ}S\mid t_{z'}=0 \text{ and }s_{z'}=0 \text{
for all }z'\neq z \}$. By Proposition \[construction1\], $A_z$ is a left ideal of $T\rtimes_{\circ}S$, in fact it is the Sylow $p_z$-subgroup of the additive group of $T\rtimes_{\circ}S$.

\[simple0\] The asymmetric product $T\rtimes_{\circ} S$ is a simple 