5.11 (1976)

Let $G$ be a finite group and suppose that there exists a non-empty proper subset $\pi$ of the set of all primes dividing $|G|$ such that the centralizer of every non-trivial $\pi$-element is a $\pi$-subgroup. Does it follow that $G$ contains a subgroup $U$ such that $U^g \cap U = 1$ or $U$ for every $g \in G$, and the centralizer of every non-trivial element of $U$ is contained in $U$?