Subgroup is Normal Subgroup of Normalizer

Theorem
Let $G$ be a group.

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


 * $H \le G \implies H \lhd N_G \left({H}\right)$

Proof
From Subgroup is Subgroup of Normalizer we have that $H \le N_G \left({H}\right)$.

It remains to show that $H$ is normal in $N_G \left({H}\right)$.

Let $a \in H$ and $b \in N_G \left({H}\right)$.

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

Thus $H$ is normal in $ N_G \left({H}\right)$.