are distinct prime numbers and $m_1,m_2$ are positive integers, then the Sylow subgroups of the multiplicative group of $B$ are not normal in $(B,\cdot)$ (see [@B3 Proposition 6.1]). Hence, by the Sylow Theorems, $p_1$ should be a divisor of $p_2^t-1$, for some $1\leq t\leq m_2$, and $p_2$ should be a divisor of $p_1^s-1$, for some $1\leq s\leq m_1$. Hence if $k_1$ is the order of $p_1$ in $({\mathbb{Z}}/(p_2))^*$ and $k_2$ is the order of $p_2$ in $({\mathbb{Z}}/(p_1))^*$, then $k_i\leq m_i$, for $i=1,2$.

By [@BCJO Remark 5.3], if $p$ is an odd prime and $k$ is the order of $2$ in $({\mathbb{Z}}/(p))^*$, then any left brace of order $2^kp$ is not simple.

For the construction of simple left braces as explained in Section 3 we use elements of prime order $p$ in the orthogonal group over ${\mathbb{Z}}/(q)$, for a prime $q\neq p$. We now first recall what is needed about the orders of these orthogonal groups.

Let $p$ be an odd prime number. Following [@Wi], we denote by $GO_{2m+1}(p)$ the orthogonal group (unique up to isomorphism) of a $(2m+1)$-dimensional vector space over the field of $p$ elements with respect to a non-singular symmetric bilinear form (equivalently a non-degenerate quadratic form), for 