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 $Q = \langle A \rangle$ is soluble. Further, if $O_p(G) = 1$, we show that $p$ is odd, $F(Q)$ is a non-trivial $p'$-group and $Q/F(Q)$ is an elementary abelian $p$-group. We also provide examples which show this conclusion is best possible.