$N_G (P \cap Q) $ has more than one Sylow $p$-subgroup under some conditions. Let $p$ be a prime number.
Let $G$ be a group with more than one Sylow $p$-subgroup
Over all pairs  of distinct Sylow $p$-subgroups, let $P$ and $Q$ be chosen so that $|P \cap Q|$ is maximal.
I want to prove that $N_G(P\cap Q)$ has more than one Sylow $p$-subgroup and that any two distinct Sylow $p$-subgroups of $N_G(P\cap Q)$ intersect in the Subgroup $P\cap Q$.
Could you help me?
 A: We can assume $P \neq Q$, whence $P \cap Q \subsetneq P$. Then by the normalizer grow principle, we have $P \cap Q \subsetneq N_P(P \cap Q)$. Let $R$ be a Sylow $p$-subgroup of $N_G(P \cap Q)$ containing its $p$-subgroup $N_P(P \cap Q)$. Now $R \subseteq U$ for some $U\in Syl_p(G)$. Assume that $U \neq P$. Hence, $P \cap Q \subsetneq N_P(P \cap Q) \subseteq R \subseteq U$. But also $N_P(P \cap Q) \subseteq P$, so $P \cap Q \subsetneq N_P(P \cap Q) \subseteq P \cap U$, violating the maximality of $|P \cap Q|$. Hence $U =P$. We conclude that $R=P \cap N_G(P \cap Q)=N_P(P \cap Q)$. Similarly, we see that $S=Q \cap N_G(P \cap Q)=N_Q(P \cap Q)$ is a Sylow $p$-subgroup of $N_G(P \cap Q)$. Now if $R = S$, then $P \cap Q \subsetneq N_P(P \cap Q) \subseteq P \cap Q$, which is absurd. Finally, $R \cap S = P \cap N_G(P \cap Q) \cap Q=N_{P \cap Q}(P \cap Q)=P \cap Q$.
Since $P \cap Q$ is normal in $N_G(P \cap Q)$, a Sylow $p$-subgroup of $N_G(P \cap Q)$ contains $P \cap Q$. Because the order of $P \cap Q$ is maximal, this implies that any two distinct Sylow $p$- subgroups of $N_G(P \cap Q)$ intersect in exactly $P \cap Q$, since the above paragraph shows that a Sylow $p$-subgroup must be of the form $U \cap N_G(P \cap Q)$, for some $U \in Syl_p(G)$. So in fact $P \cap Q=O_p(N_G(P \cap Q))$.
