Subgroup is Normal Subgroup of Normalizer

Theorem
Let $G$ be a group.

A subgroup $H \le G$ is a normal subgroup of its normalizer:


 * $H \le G \implies H \lhd \map {N_G} H$

Proof
From Subgroup is Subgroup of Normalizer we have that $H \le \map {N_G} H$.

It remains to show that $H$ is normal in $\map {N_G} H$.

Let $a \in H$ and $b \in \map {N_G} H$.

By the definition of normalizer:
 * $b a b^{-1} \in H$

Thus $H$ is normal in $\map {N_G} H$.