## In a finite group, conjugation permutes sub-Sylow subgroups

Let $G$ be a finite group and $p$ a prime. Prove that if $H \leq G$ and $Q \in \mathsf{Syl}_p(H)$, then $gQg^{-1} \in \mathsf{Syl}_p(gHg^{-1})$ for all $g \in G$.

Note that $gQg^{-1} \leq gHg^{-1}$. Moreover, because conjugation by $g$ is an automorphism, $|gQg^{-1}| = |Q|$ and $|gHg^{-1}| = |H|$. Hence $[gHg^{-1} : gQg^{-1}] = |gHg^{-1}|/|gQg^{-1}| = |H|/|Q| = [H : Q]$ is not divisible by $p$, so that $gQg^{-1}$ is a Sylow $p$-subgroup of $gHg^{-1}$.