Let $P$, $Q$ be two Sylow p-subgroups of $G$, is it true that $N_P(Q)=Q\cap P$? Let $P$, $Q$ be two Sylow p-subgroups of $G$, is it true that $N_P(Q)=Q\cap P$? 
I have seen an explanation somewhere that any element $x$ outside of $Q$ cannot normalise $Q$, since otherwise $\langle Q,x\rangle$ will be a larger $p$-subgroup of $G$ than $Q$. However I don't really understand the above explanation, or know if it is correct.
Thanks for any help!
 A: Let $P$ be a $p$-subgroup of a group $G$ and let $S \in$ Syl$_p(G)$. Then of course $P \cap S \subseteq N_P(S)$. Conversely, let $x \in N_P (S)$. Then $x$ is a $p$-element and normalizes $S$. That means that $S\langle x \rangle$ is actually a subgroup of $G$, not just a set. But the order of this subgroup equals $|S\langle x \rangle|=\frac{|S||\langle x \rangle|}{|S \cap \langle x \rangle|}$, which is a power of $p$. But $S\subseteq S\langle x \rangle$, and since $S$ is a Sylow subgroup, hence a maximal  $p$-subgroup, we must have $S=S\langle x \rangle$. So $x \in S$. This shows that $N_P(S)=P \cap S$. In particular, if $S,T \in$ Syl$_p(G)$, then $N_S(T)=S \cap T=N_T(S)$. A nice symmetrical formula!
A: Another proof which I do not see very often. Let $P$ be a $p$-subgroup of $G$ and let $S \in Syl_p(G)$. Then of course $S \unlhd N_G(S)$ and hence $S$ is the unique Sylow $p$-subgroup of its normalizer. Obviously, $N_P(S) \subseteq N_G(S)$. But $N_P(S)$ is a $p$-subgroup of $N_G(S)$ and by Sylow theory it must be contained in some Sylow subgroup of $N_G(S)$. But $S$ is the only Sylow subgroup, hence $N_P(S) \subseteq S$. Again we conclude $N_P(S) \subseteq P \cap S$, yielding $N_P(S)=P \cap S$, since the other inclusion is trivial.
