If $P\in{\rm Syl}_p(G)$, when is $P\cap H\in{\rm Syl}_p(H)$ Let $G$ be a finite group and let $P$ be a $p$-Sylow subgroup of $G$.
If $H\leq G$, I was wondering when $H\cap P$ is a $p$-Sylow subgroup of $H$.
Clearly, this is not true in general since it can happen that $P$ and $P'$ are $p$-Sylow subgroups of $G$ and that $P\cap P'\not=P'$ so is not a  $p$-Sylow subgroup of $P'$.
But what if $H$ is a normal subgroup? Then the previous counterexample does not work anymore, since if $P'$ is normal is the unique $p$-sylow subgroup of $G$.
I do not know if this is a general fact in Sylow's theory.
 A: If $H$ is normal then $H\cap P$ is a Sylow $p$-subgroup of $H$. To see this, let $Q$ be a Sylow $p$-subgroup of $H$. Extend $Q$ to a Sylow $p$-subgroup $P_1$ of $G$. There exists $g\in G$ such that $P_1^g=P$. Thus $Q^g\leq P$. But $Q^g$ is a Sylow $p$-subgroup of $H^g=H$, so $Q^g=P\cap H$.
A: A subgroup $S$ is said to be subnormal, if there exist subgroups $H_i$ of $G$ such that
$$S \lhd H_1 \lhd \dotsm \lhd H_r=G.$$
In this situation one writes $S \lhd \lhd G$. Subnormality in is an important topic on its own in (finite and infinite) group theory. Of course every normal subgroup is subnormal. The following holds. $G$ is assumed to be finite.
Proposition Let $S \lhd \lhd G$ and $P \in Syl_p(G)$, then $P \cap S \in Syl_p(G)$. Conversely, every Sylow $p$-subgroup $H \in Syl_p(S)$ arises in this way: $H=P \cap S$ for some $P \in Syl_p(G)$.
Proof Let us first prove the first assertion: if $S=G$ there is nothing to prove. So we can assume that $S$ is proper and we apply induction on $|G|$. Let $S \subseteq T \lt G$ with $T \lhd G$. Then $p$ does not divide $|T:P \cap T|=|PT:P|$. Hence $P \cap T \in Syl_p(T)$ and $P \cap S=(P \cap T) \cap S \in Syl_p(S)$ by induction.
Conversely, let $H \in Syl_p(S)$, again we can assume that $S \lt G$. Let $T \lhd G$ as in the previous paragraph. By induction we can find a $Q \in Syl_p(T)$ with $Q \cap S=H$. Since $T$ is normal, we can find a $P \in Syl_p(G)$ with $Q=P \cap T$. Hence, $H=Q \cap S=P \cap T \cap S=P\cap S$.$\square$ 
Remark A famous conjecture of O. Kegel and H. Wielandt is about the converse of the proposition. It was actually proved by Peter Kleidman $30$ years ago using the Classification of the Finite Simple Groups!
Theorem (P. Kleidman, 1991) If $S \subseteq G$, then for all primes $p$ and all Sylow $p$-subgroups $P$ of $G$, $P \cap S \in Syl_p(S)$ if and only if $S \lhd \lhd G$.
