Lang Sylow Theorem STATEMENT: Let $P$ be a $p$-Sylow subgroup of $G$ and H is a $p$-subgroup of G. Suppose first that $H$ is contained in the normalizer of $P$. We prove that $H\subseteq P$. Indeed, $HP$ is then a subgroup of the normalizer, and $P$ is normal in $HP$. But
$$(HP:P)=(H:H\cap P)$$
So if $HP\neq P$, then $HP$ has order a power of p, and the order is larger than $|P|$, contradicting the hypothesis that $P$ is a Sylow group. Hence $HP=P$ and $H\subseteq P$.
QUESTION: How does Lang get to the conclusion that $HP$ must have order a power of $p$ greater than $|P|$?

If $H$ normalizes $P$, then $HP$ is a subgroup and $|HP|=|H||P|/|H \cap P|$, whence $HP$ is a $p$-group. Since $P \subseteq HP$, $|P| \leq |HP|$.
