Let $G$ be a finite group and $H$ be a subgroup of $G$. Suppose that for any prime power order element $x$ of $G$, there exists some element $g$ in $G$ such that $x^g$ is contained in $H$. Does it follow that $H=G$?
As Geoff Robinson pointed out in his comment, the answer is yes by a deep and difficult theorem in finite groups. It was proven in Fein; Kantor; Schacher: Relative Brauer groups. II., J. Reine Angew. Math. 328 (1981), 39–57. As far as I know, no simpler proof has been found yet (maybe Theo Johnson-Freyd has a better one ...?).