A question about Sylow subgroups

Question

Let $G$ be a finite group and $P\neq\{e\}$ be a Sylow $p$-subgroup of $G$ and $P^g\neq P$ be its conjugate in $G$. If we know that $P\cap P^g\neq \{e\}$, can we conclude that $Z(P)\cap Z(P^g)\neq \{e\}$?

Answer 1

I think $G=S_4$, $p=2$ is a counterexample. The Sylow $2$-subgroups are all isomorphic to the dihedral group $D_4$, and $Z(D_4)\cong C_2$. There are three Sylow $2$-subgroups - all of order eight. As $S_4$ has eight elements of order three, the union of the Sylow $2$-subgroups has sixteen elements. Hence they must intersect non-trivially. Yet their centers must intersect trivially. For if an element $g$ were centralized by two distinct Sylow $2$-subgroups, its centralizer would have to be all of $S_4$ ($D_4$ is a maximal subgroup). This contradicts the fact that $S_4$ itself has trivial center.