Lang Sylow Theorem

Question

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 $|P|$?

Answer 1

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

Answer 2

Note that $HP=\{hp|h\in H p\in P\}$ is a subgroup provided $H\subseteq N(P)$. This set clearly has power dividing $|H||P|$.

Answer 3

if $H$ and $P$ are Sylow subgroups then $H \cap P$ also has order a power of $p$. since the order of $H$ is also a power of $P$ then $(HP:P)=(H:H\cap P)$ is a power of $p$.