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Let $G$ be a finite group, $H$ a normal subgroup of $G$ and $P$ a Sylow $p$-subgroup of $H$. Let $N_G(P)$ be the normalizer of $P$ in $G$. Show that $G=N_G(P)H$.

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See the remark of Jack - this is called the Frattini argument and follows from the fact that Sylow $p$-subgroups are conjugated: let $g \in G$, since $H$ is normal, $H^g=g^{-1}Hg=H$ and hence $P^g \subseteq H$. So $P^g \in Syl_p(H)$ and by the Sylow theory in $H$, one can find an $h \in H$ with $P^g=P^h$. It follows that $gh^{-1} \in N_G(P)$, so $g \in HN_G(P)$.

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