of $T\rtimes_{\circ}S$ is the semidirect product of two abelian groups via the action $\alpha$. Hence, if $\alpha$ is not trivial, this group is metabelian and not abelian.

Since the multiplicative group of each $A_z$ is abelian, the last assertion also follows.

[Assume that $(B,+,\cdot)$ is a finite simple left brace and $(B,\cdot)$ is a metabelian $A$-group. If one of the Sylow subgroups of $(B,\cdot)$ is cyclic, then $B$ is a cyclic (and trivial) left brace. Indeed, let $B_{1},\ldots , B_{n}$ be the Sylow subgroups of the additive group of $B$. Then $\mathcal{S}=\{ B_{1},\ldots ,
B_{n}\}$ is a Sylow system of $(B,\cdot)$. Say, $B_1$ is cyclic $p_1$-group, for some prime $p_1$, and suppose that $n>1$. Write the multiplicative group of $B$ as $B= [B,B] M(B)=[B,B]\rtimes
M(B)$. Let $S_1$ be the Sylow $p_1$-subgroup of $[B,B]$ and let $S_2$ be the Sylow $p_1$-subgroup of $M(B)$. Then $S_1S_2$ is an inner direct product because $(B,\cdot)$ is an $A$-group and it is a Sylow $p_1$-subgroup of $B$. Hence, $B_1$ being cyclic implies that $B_1\subseteq [B,B]$ or $B_1\subseteq M(B)$. If $B_{1}\subseteq
[B,B]$, then $B_{1}$ is a normal subgroup of $(B,\cdot)$ because $[B,B]$ is abelian and thus has a unique Sylow subgroup of order $|B_1|$. Hence, $B_{1}$ 