Finite group having a $p$-subgroup has nontrivial center

Question

Let $G$ be a finite group, $H\le G$ a $p$-subgroup. I want to show that $Z(G)$ is non-trivial. I let $G$ act on $H$ by conjugation - or vice versa - but I every time it ended up with $$Z(G)\subseteq Z(H).$$ To add, by the class equation I can only say that $Z(H) \ge p$. How can I show that $Z(G)$ non-trivial? An appropriate action would be enough.

Answer 1

This is false. Consider $G = S_3$ and $H = \{(1),(123),(132)\}$. Then, $H \le G$ and $H$ is a 3-subgroup of $G$ but $Z(S_3) = \{(1)\}$.