Powerful multiplicative groups do not force right nilpotence in finite braces

Abstract

For every odd prime $p$, we construct a finite left brace $A_p$ of order $p^{2p+1}$ whose additive group is elementary abelian and whose multiplicative group $G_p$ is a powerful $p$-group, but such that $A_p$ is not right nilpotent. More precisely, $G_p^{\prime }=G_p^p\cong C_p^2$, $\mathrm{cl}(G_p)=2$, $\exp (G_p)=p^2$, and $\mathrm{Soc}(A_p)=0$. Thus powerfulness does not force right nilpotence even for multiplicative groups of class two with derived subgroup of order $p^2$. The obstruction is explicit: $A_p$ contains a three-dimensional trivial ideal $T$ satisfying $T\ast A_p=T$, whereas its left series is the ideal-power filtration of a nilpotent commutative algebra. Both $T$ and $A_p/T$ are right nilpotent, so right nilpotence of finite left braces is not closed under extensions. The construction is uniform and arises from a finite local commutative algebra, a square-zero derivation, and an invariant character, yielding a regular affine subgroup as the kernel of a character on a semidirect product. It disproves the Shalev--Smoktunowicz conjecture in every odd characteristic and yields finite irretractable involutive set-theoretic solutions of the Yang--Baxter equation whose permutation groups are powerful $p$-groups of class two.