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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 $Z(P)\cap Z(P^g)\neq \{e\}$?

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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.

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+1. The same reasoning seems to work for any group $G$ of order $p^nq$ with $p < q$ prime, that has nonabelian Sylow $p$-subgroups, no normal Sylow subgroup, and $p \nmid |Z(G)|$ –  zcn Jul 9 at 8:17
    
The copy of the Klein four group $V_4$ that is a normal subgroup of $S_4$ is, of course, the intersection of all three Sylow $2$-subgroups. –  Jyrki Lahtonen Jul 9 at 8:58

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