is an ideal of $B$, a contradiction. On the other hand, if $B_{1}\subseteq M(B )$ then $B_{2}\cdots B_{n}$ is a subgroup normalized by $B_{1}$. Hence $B_{2}\cdots B_{n}$ is a normal subgroup of $B$, hence an ideal of $B$, a contradiction again.]{}

The main new construction {#construction}
=========================

In this section, using the construction of left braces introduced in Section \[agroup\], we shall construct a new family of simple left braces, based on the existence of special elements of prime order $p$ in the orthogonal group $\mathrm{O}(V,b)$ of a non-singular symmetric bilinear form $b$ over a finite dimensional ${\mathbb{Z}}/(q)$-vector space $V$, for a prime $q\neq p$. Using this construction of simple left braces, we shall prove the main result of this paper, Theorem \[MainResult\].

To check the simplicity of a left brace the following lemma is useful.

\[ideal\] [@BCJO Lemma 2.5] If $I$ is an ideal of a left brace $B$, then $(\lambda_b-{\operatorname{id}})(a)\in I$, for all $a\in B$ and $b\in I$.

First, we introduce and fix some notation. Let $n>1$ be an integer. For $z\in {\mathbb{Z}}/(n)$, let $p_z$ be a prime number and let $r_{z}$ be a positive integer. We assume that $p_{z}\neq p_{z'}$ for $z\neq z'$.

Let $m_z$ be positive 