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In a finite group $G$, let $P \in Syl_p(N_G(P))$, i.e. $P$ is a Sylow $p$-subgroup in its normaliser, does this imply $P \in Syl_p(G)$, i.e. it is a Sylow $p$-subgroup in its entire group?

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Yes, by Sylow's Theorems there is some $S \in Syl_p(G)$, with $P=S \cap N_G(P)$. Assume $P \subsetneq S$. Then apply the normalizers grow principle in nilpotent groups: $P \subsetneq N_S(P)$. But $N_S(P)=S \cap N_G(P)=P$, a contradiction. Hence $P=S$ and is a Sylow $p$-subgroup of $G$.

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The answer is yes. Since $P$ is a $p$-group, it is contained in some Sylow $p$-subgroup $Q$ of $G$, and if $P$ were properly contained in $Q$, then by a standard fact about $p$-groups we would have that $P$ is a proper subgroup of $N_Q(P) = N_G(P) \cap Q$, so that $P$ is not a Sylow $p$-subgroup of $N_G(P)$, contrary to assumption.

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Hint: For any subgroup $H\le G$, $H$ is a subgroup of its own normalizer.

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  • $\begingroup$ Yes, and therefore $P \unlhd N_G(P)$, do not see how this helps, but is it a simple argument from thereon and I am just blind for it? $\endgroup$ – StefanH Mar 17 '15 at 17:34

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