One more normalizer q

**sfspitfire23** · October 31st 2009, 02:04 PM

If H is a subgroup of G, show that H is normal to the normalizer $(N(H))$

**Bruno J.** · October 31st 2009, 02:49 PM

$H$ 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.

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