I'm in trouble on Dummit and Foote Abstract Algebra Ex. 6.2 13:
Let $P,Q$ be distinct Sylow p-group with maximal $|P \cap Q|$. Show $N_G(P \cap Q)$ cotains more than one Sylow p-group and each pair intersect in $P \cap Q$.
It's easy to showing intersecting part. Let $A,B$ distinct Sylow p-group of $N_G(P \cap Q)$. We can find Sylow p-group of $G$: $A',B'$ s.t. $A \subseteq A'$, $B \subseteq B'$, and cleary $A \cap B \subseteq A' \cap B'$, and $|A \cap B| \leq |P \cap Q|$ by maximal choose, and $P \cap Q \subseteq A \cap B$ because A and B are conjugate for some Sylow p-group of $N_G(P \cap Q)$, R which contains $P \cap Q$(because $P \cap Q$ is p-group) and since $P\cap Q$ is normal, $P\cap Q\leq A,B$. So $P \cap Q=A \cap B$.
By above, only left thing to show is $N_G(P\cap Q)=P \cap Q$ is impossible. (It's easy to observe uniqueness of its Sylow p-group and this is statement is equivalent.)
I think we have to use $|P\cap Q|$ is maximal to show above. How to show this? $N_G(P\cap Q)\neq P \cap Q$
Well, $N_G(P \cap Q)$ cannot be a $p$-group. Suppose that it is. Then we can find an $R \in Syl_p(G)$, such that $N_G(P \cap Q) \subseteq R$. Assume that $R \neq P$. Then by the normalizers-grow principle applied to the $p$-subgroup $N_P(P \cap Q)$ of $P$, we have (and for the first inclusion note that $P \cap Q \subsetneq P$) $$ P \cap Q \subsetneq N_P(P \cap Q) = N_G(P \cap Q) \cap P \subseteq R \cap P. $$ Since $|P \cap Q|$ is maximal, $|P \cap Q|=|R \cap P|$, which cannot be the case since the first inclusion is strict. So, $R=P$ and in a similar manner you can argue that $R=Q$, whence $P=Q$, a contradiction.
Extra: one can actually show that $N_P(P \cap Q)$ and $N_Q(P \cap Q)$ are different Sylow $p$-subgroups of $N_G(P \cap Q)$.