## If a finite group has a normal Sylow p-subgroup, then all subgroups also have a normal Sylow p-subgroup

Let $P$ be a normal Sylow $p$-subgroup of a finite group $G$ and let $H \leq G$. Prove that $P \cap H$ is the unique Sylow $p$-subgroup of $H$.

Note that $P \cap H$ is normal in $H$, and $P \cap H \leq P$ is a $p$-subgroup.

Suppose now that $P \cap H$ is not a Sylow $p$-subgroup of $H$; then there exists an element $x$ of $p$-power order in $H$ which is not in $P$. However, as the unique Sylow $p$-subgroup of $G$, this is a contradiction because all elements of $p$-power order in $G$ are contained in $P$. Thus $P \cap H$ is a Sylow $p$-subgroup of $H$.

Another method to show that $P \cap H$ is a Sylow $p$-subgroup of $H$: $|H:P\cap H|=|HP:P|$ is a divisor of $|G:P|=|G:HP||HP:P|$, so $p$ does not divide $|H:P \cap H|$.