integers such that $m_z \geq\max\{
r_{z},r_{z-1}\}$ for all $z\in {\mathbb{Z}}/(n)$. Let $V_z$ be a ${\mathbb{Z}}/(p_z)$-vector space and let $b_z$ be a non-singular symmetric bilinear form over $V_z$. Assume also that there exist elements $f_z
\in O(V_z, b_z)$ (the associated orthogonal group) of order $p_{z-1}$.

For each $z\in {\mathbb{Z}}/(n)$, consider the trivial left braces $T_z=V_z^{m_z}$ and $S_z=({\mathbb{Z}}/(p_z))^{r_z}$. Note that $S_z$ also is a ${\mathbb{Z}}/(p_z)$-vector space. The standard basis elements of $({\mathbb{Z}}/(p_z))^{r_z}$ we denote by $e_{1}^{(z)}, \dots ,
e_{r_{z}}^{(z)}$. Let $b'_z:T_z\times T_z\longrightarrow S_z$ be the symmetric bilinear map defined by $$b'_z((u_1,\dots,u_{m_z}),(v_1,\dots,v_{m_z}))=\sum_{i=1}^{r_z-1}b_z(u_i,v_i)e_{i}^{(z)}+\sum_{j=1}^{m_z}b_z(u_j,v_j)e_{r_z}^{(z)},$$ for all $(u_1,\dots,u_{m_z}),(v_1,\dots,v_{m_z})\in T_z$ (we agree that the first sum is zero if $r_z=1$). Let $f^{(z-1,z)}:S_{z-1}\longrightarrow {\operatorname{Aut}}(T_z,+)$ be the map defined by $f^{(z-1,z)}(s)=f_s^{(z-1,z)}$ and $$\begin{aligned}
\lefteqn{f_s^{(z-1,z)}(u_1,\dots,u_{m_z})}\\
&=&(f_{z}^{\mu_1+\mu_{r_{z-1}}}(u_1),\dots,
f_{z}^{\mu_{r_{z-1}-1}+\mu_{r_{z-1}}}(u_{r_{z-1}-1}),\\
&&\quad f_{z}^{\mu_{r_{z-1}}}(u_{r_{z-1}}),\dots,
f_{z}^{\mu_{r_{z-1}}}(u_{m_{z}})),\end{aligned}$$ for all $s=\sum_{i=1}^{r_{z-1}}\mu_ie_{i}^{(z-1)}\in S_{z-1}$ and $(u_1,\dots,u_{m_z})\in T_z$ (we agree that if $r_{z-1}=1$, then $f_s^{(z-1,z)}(u_1,\dots,u_{m_z})=(f_{z}^{\mu_{1}}(u_{1}),\dots,
f_{z}^{\mu_{1}}(u_{m_{z}}))$).

\[orthof\] For each $z\in {\mathbb{Z}}/(n)$, the map $f^{(z-1,z)}:S_{z-1}\longrightarrow
{\operatorname{Aut}}(T_z,+)$ is a homomorphism of multiplicative groups and $f_s^{(z-1,z)}\in \mathrm{O}(T_{z},b'_z)$, for all $s\in S_{z-1}$.

Recall that $S_{z-1}$ is a trivial left brace. Thus the addition and the multiplication of $S_{z-1}$ coincide. Hence it is clear that $f^{(z-1,z)}:S_{z-1}\longrightarrow {\operatorname{Aut}}(T_z,+)$ is a homomorphism of multiplicative groups. Let $s=\sum_{i=1}^{r_{z-1}}\mu_ie_{i}^{(z-1)}\in S_{z-1}$. By assumption, $f_z \in O(V_z, b_z)$. Hence we have $$\begin{aligned}
