to the multiplicative group of a left brace. In fact, there exist finite $p$-groups that are not isomorphic to the multiplicative group of any left brace [@B2]. The groups isomorphic to the multiplicative group of a finite left brace are called involutive Yang-Baxter groups (or simply IYB groups) in [@CJR].

In [@BDG2 Corollary 4.3] it is stated that any finite solvable $A$-group is an IYB group. However, its proof is not correct, because it is based on the statement that every finite solvable $A$-group has a normal Sylow subgroup; this is not true, as for example the group $A_4\times S_3$ is a solvable $A$-group without Sylow normal subgroups.

First, we give a proof of [@BDG2 Corollary 4.3].

\[Atype-IYB\] Every finite solvable $A$-group is an IYB group.

Let $G$ be a finite solvable $A$-group. Let $G=L_0>L_1>\dots
>L_n=1$ be the derived series of $G$. We shall prove the result by induction on $n$. It is known that every finite abelian group is IYB (see for example [@CJR Corollary 3.10]). Suppose that $n>1$ and that every finite solvable $A$-group of derived length $n-1$ is an IYB group. By [@taunt (4.5)], $M_G(L_{n-2})$ is a complement in $G$ of $L_{n-1}$, i.e. $G\cong
L_{n-1}\rtimes M_G(L_{n-2})$. Clearly 