Let $G$ be a finite group of order $n$. Let $H$ be a subgroup of $G$. Define the normalizer $N(H)$ of $H$, and prove that the number of distinct conjugates of $H$ is equal to the index of $N(H)$ in $G$. If $p$ is a prime dividing $n$, deduce that the number of Sylow $p$-subgroups of $G$ must divide $n$.

[You may assume the existence and conjugacy of Sylow subgroups.]

Prove that any group of order 72 must have either 1 or 4 Sylow 3-subgroups.