# One more normalizer q

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• October 31st 2009, 02:04 PM
sfspitfire23
One more normalizer q
If H is a subgroup of G, show that H is normal to the normalizer $(N(H))$
• October 31st 2009, 02:49 PM
Bruno J.
$H$ is clearly a subgroup of $N_G(H)$.

To see that $H \lhd N_G(H)$, take any $x \in N_G(H)$. By definition, $xHx^{-1}=H$. Since this is true for all $x \in N_G(H)$, the statement follows.