TITLE: How $H \cap P$ is a Sylow $p$-subgroup of $H$?
QUESTION [5 upvotes]: I recently came across the following problem :

True or false: If $P$ is a Sylow $p$-subgroup of a finite group $G$, then for any subgroup $H$ of $G$, $H \cap P$ is a Sylow $p$-subgroup of $H$.

The statement is false. For, take $G=S_3$ ,$P=\{e,(13)\}$  and $H=\{e,(12)\}$. Note that $P$ is the Sylow $2$-subgroup of $G$. Here $$H \cap P=\{e\}$$   but $\{e\}$ is not a Sylow $2$-Sylow subgroup of $H$.
Am I correct? What is the importance of this(true/false) statement if any ?Any help?

REPLY [2 votes]: Even a more general fact holds true: 

If $G$ is a finite group, $H \lhd \lhd G$ (subnormal) and $P \in Syl_p(G)$, then $H \cap P \in Syl_p(H)$.


Proof Since $H \lhd \lhd G$, we can find a subnormal series: $H=H_0 \lhd H_1 \cdots \lhd H_i \lhd \cdots \lhd G=H_r$. Since $H_{r-1} \lhd G$, $P \cap H_{r-1} \in Syl_p(H_{r-1})$. But $H_{r-2} \lhd H_{r-1}$, so $P \cap H_{r-1} \cap H_{r-2}=P \cap H_{r-2} \in Syl_p(H_{r-2})$. Now induction finishes the proof.