Sylow Conjugacy Lemma

Sylow Conjugacy Lemma: Let $G$ be a finite group , let $p$ be a prime, and let $P$ be a Sylow p-subgroup of $G$. If $Q\le G$ is a subgroup whose order is a power of $p$ (equivalently, $Q$ is a finite p-group ), then there exists $g\in G$ such that $Q\le gPg^{-1}$.

In particular, any two Sylow $p$-subgroups are conjugate in $G$ (take $Q$ to be another Sylow $p$-subgroup), a key input toward Sylow's second theorem .

Proof sketch: Consider the action of $Q$ on the set of left cosets $G/P$ by left multiplication. Counting fixed points modulo $p$ shows there is a fixed coset $gP$, which implies $g^{-1}Qg\le P$, i.e. $Q\le gPg^{-1}$.