# Intersections of $p$-Sylow subgroups of group of order $p^2q$

Let $G$ be a group of order $p^2q$ ($p$ and $q$ are distinct primes) and $P_1$ and $P_2$ be distinct $p$-Sylow subgroups of $G$. Then if $Z(G) = \{e\}$, $P_1\cap P_2 = \{e\}$

How do I go about proving the above? Using Lagrange's theorem, I have that $|P_1\cap P_2|\in\{1,p\}$ (excluding $p^2$ as the subgroups are distinct), so it only remains to exclude $p$. But if $|P_1\cap P_2| = p$, then the intersection is a cyclic subgroup of $G$. How does this lead to a contradiction? I cannot see how.

Here is a sketch of a proof, with details missing (for you to work out): note that $P_1$ and $P_2$ are abelian and that, together, they generate $G$, so their intersection must be central.
• They can't generate $G$ :s – José Alejandro Aburto Araneda Sep 8 '17 at 1:57