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If $H$ is a subgroup of a group $ G$. Then the normalizer of $H$ is defined as,

$N_G(H)=${$g\in G: gHg^{-1}=H$}.

However, $gHg^{-1}=H$ if and only if $gH=Hg$, which implies that $H$ is a normal subgroup of $G$ (since left and right cosets of $H$ are identical). Is this observation correct? Please someone help me out. Thanks in advance.

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    $\begingroup$ Yes correct, $N_G(H)=G$ if and only if $H \unlhd G$. Observe that always $H \unlhd N_G(H)$, in fact the normalizer is the largest subgroup of $G$ containing $H$ in which $H$ is normal. $\endgroup$ – Nicky Hekster Sep 11 at 11:27
  • $\begingroup$ @Nicky Hekster Thanks for a fruitful comment. $\endgroup$ – gete Sep 11 at 12:32
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You've just noticed that a subgroup is a normal subgroup of its normalizer.

This is, in a sense, the whole idea of the normalizer. In fact, as @Nicky Hekster observes, the normalizer is the largest such subgroup.

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