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$H$ normal subgroup of a group $G$ with cardinality finite. $p$ prime number dividing $|H|$. $P$ a $p$-Sylow subgroup of $H$, how can I prove that then $G=HN_G(P)$ where $N_G(P)$ is the normalizer of $P$ in $G$?

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What have you tried? –  Alex Becker Jan 2 '12 at 6:04

1 Answer 1

up vote 7 down vote accepted

If $g\in G$, then $gHg^{-1}=H$, and so $gPg^{-1}\subseteq H$. Since $gPg^{-1}$ is a $p$-Sylow subgroup of $H$, by Sylow's Theorems we know that $gPg^{-1}$ is conjugate to $P$ in $H$. That is, there exists $h\in H$ such that $hPh^{-1} = gPg^{-1}$. Therefore, $g^{-1}hPh^{-1}g = P$, so $h^{-1}g\in N_G(P)$.

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