Tate $p$-nilpotent theorem 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 $p$.

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, $$(P\cap N)^g=P^g\cap N^g=P^g \cap N=Q\cap N\leq \Phi(P)^g=\Phi(P^g)=\Phi(Q).$$
