Squares of conjugacy classes and a variant on the Baer-Suzuki Theorem

For $p$ a prime, $G$ a finite group and $A$ a normal subset of elements of order $p$, we prove that if $A^2=\{ab\mid a,b\in A\}$ consists of $p$-elements then $Q=⟨A⟩$ is soluble. Further, if $O_p(G)=1$, we show that $p$ is odd, $F(Q)$ is a non-trivial $p^{\mathrm{\prime }}$-group and $Q\mathrm{/}F(Q)$ is an elementary abelian $p$-group. We also provide examples which show this conclusion is best possible.