$P$ is a $p$-Sylow subgroup of $G$ and $H$ is a subgroup of $G$. Show that $P\cap H$ is contained in a $p$-Sylow subgroup of $H$

Question

Let $G$ a group, let $H$ a subgroup of $G$ and let $P\in\operatorname{Syl}_p(G)$. Show that $P\cap H$ is a subset of some $U\in\operatorname{Syl}_p(H)$.

I know that \begin{align} |G| &=p^a\cdot nk,\\ |H| &=p^b\cdot n, \\ |P| &=p^a\end{align} for $b\leq a, (p,n),(p,k)=1$. The trivial case is $a=b$, but i don't know how to handle the case $b<a$.

Answer 1

Here is a little hint: what's the order of the elements of $P \cap H $ ? Then use Sylow's theorem to deduce that this is contain in a P-sylow of H.

Answer 2

Being the intersection of two subgroups, $P \cap H$ is a subgroup. Moreover, it is contained in $P$, so it has order $p^k$ for some $k$.

Let $Q$ be a maximal $p$-subgroup of $H$ that contains $P \cap H$. It has to exist since you can take $Q_0 = P \cap H$, and then search for a bigger $Q_1$. If you can't find it, then $Q=Q_0$. Otherwise, take $Q_1$ and search for a bigger $Q_2$, and so on. This process has to terminate, since $H$ is finite.

The key is that now, from Sylow's theorems, we know that every maximal $p$-subgroup of $H$ is a Sylow $p$-subgroup of $H$. So take $U=Q$ and you are done.