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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?

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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 $P \cap Q=O_p(N_G(P \cap Q))$.

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  • $\begingroup$ In last paragraph, did you mean that any $p$-subgroup of $N_G(P\cap Q)$ contains $P\cap Q$, since for $H$ a Sylow $p$-subgroup of $N_G(P\cap Q)$, then $P\cap Q\leq xHx^{-1},\;x\in N_G(P\cap Q)$, which implies $x(P\cap Q)x^{-1}=P\cap Q\leq H$? $\endgroup$ – Xie Jul 17 '19 at 9:58
  • $\begingroup$ Yes exactly. In general, a normal $p$-subgroup is contained in every Sylow $p$-subgroup. $\endgroup$ – Nicky Hekster Jul 17 '19 at 15:47

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