Task I'm trying to figure out is:
Let P be a Sylow p-subgroup of $G$ and let $Q$ be any p-subgroup. Prove that $Q\cap P=Q\cap N_{G}(P)$.
I've seen some facts that might be useful, but nothing seems to come to mind.
I know $n_{(p)} = |G:N_{G}(P)|$, also $|Q| \le p^a$ where a is the maximal power of the prime, but i don't see how it helps.
Please keep in mind i only know the basics regarding Sylow theory when answering.
Best Answer
Here is a hands-down proof making use of Sylow Theory inside $N_G(P)$. Since $Q \cap N_G(P)$ is a $p$-subgroup of $N_G(P)$ it must be contained in some Sylow $p$-subgroup of $N_G(P)$. But $P \unlhd N_G(P)$, whence $P$ is the only Sylow $p$-subgroup of $N_G(P)$, implying $Q \cap N_G(P) \subseteq P$. So $Q \cap N_G(P) \subseteq P \cap Q$ and the reverse inclusion is trivial.