The intersection of a normal subgroup and Sylow $p$-subgroup Let $G$ be a group and $P\in Syl_p(G)$, $H$ is normal in $G$. I want to show that $P\cap H\in Syl_p(H)$. 
So I let $P_0\in Syl_p(H)$. $P\cap H$ is a $p$ subgroup of $H$, so by Sylow 2nd Theorem, $P\cap H \leq P_0$. 
And by Sylow's 2nd and 3rd theorem, I get that there exists $g\in G$ such that $P_0 \leq gPg^{-1}$. 
I think I want to prove that $P_0 \leq P\cap H$ next in order to conclude that $P_0=P\cap H$ but got stuck at this part.
 A: Hint: $PH$ is a subgroup when $H$ is a normal subgroup. Apply the formula 
$$|PH| = \frac{|P||H|}{|P \cap H|}$$
A: You're almost there.
So, you have chosen a $P_0\in Syl_p(H)$ such that $P\cap H\le P_0$. Then, we have $P_0\le gPg^{-1}$. Also, $P_0\le H$, so, as $gHg^{-1}=H$, it means
$$g^{-1}P_0\,g\le P\cap H\,.$$
Assuming everything is finite, by calculating sizes, we are ready, as $|g^{-1}P_0\,g|=|P_0|$ and both are included in $H$, so $|P\cap H|=|P_0|$ also follows.
A: Let's prove by contradiction.
Suppose the converse, which is $P \cap H $ is not a Sylow $p$-subgroup of $H$. Then by Sylow theorems, there must exist a Sylow $p$-subgroup $P'$ of $G$ and $q\in G$, such that
$$ P'=qPq^{-1}\ \text{and}\ P\cap H<P'\cap H .$$
Since $H$ is normal in $G$, we have: $$P'\cap H=(qPq^{-1})\cap H=q(P\cap H)q^{-1}<q(P'\cap H)q^{-1}$$
but $|P'\cap H|=|q(P'\cap H)q^{-1}|$. Contradiction!

Addendum:
For non-normal subgroup $H$, the statement may not be valid. For example, take $G=S_3$ the symmetric group on three letters. Let $H=\langle (12)\rangle$ and $P=\langle (13)\rangle$ a Sylow $2$-subgroup of $S_3$, then clearly $P\cap H=(1) $ is not a Sylow $2$-subgroup of $H$.
