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Tate $p$-nilpotent Theorem. If $P$ is a Sylow $p$-subgroup of $G$ and $N$ is a normal subgroup of $G$ such that $P \cap N \leq \Phi (P)$, then $N$ is $p$-nilpotent.

My question is the following: If $P \cap N \leq \Phi (P)$ for only one Sylow p-subgroup of $G$, is $N$ $p$-nilpotent? Remark: $G$ may have more than one Sylow for the prime $p$.

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    Consider $P^g\cap N^g \le \Phi(P)^g=P^g\cap N \le \Phi(P^g)$.2012-11-17

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That situation is not possible. Let $P$ be a Sylow $p$-subgroup such that $P \cap N \leqslant \Phi(P)$ and consider $Q\cap N$ for another Sylow $p$-subgroup $Q$. We have that there is a $g$ so that $P^g=Q$, and since $N$ is normal, $(P\cap N)^g=P^g\cap N^g=P^g \cap N=Q\cap N\leq \Phi(P)^g=\Phi(P^g)=\Phi(Q).$

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    So, this happens to all Sylow $p$-subgroups of $G$ or never happen. Thank you very much.2012-11-18