Subgroup is Subgroup of Normalizer

Theorem
Let $G$ be a group.

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


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

Subset of Normalizer
First we show that $H$ is a subset of $\map {N_G} H$.

This follows directly from Left Coset Equals Subgroup iff Element in Subgroup:
 * $x \in H \implies x H = H$

As $x \in H \implies x^{-1} \in H$ it also follows that $x \in H \implies H x^{-1} = H$.

Thus:
 * $x \in H \implies x H x^{-1} = H^x = H$

and so:
 * $x \in \map {N_G} H$

So:
 * $H \subseteq \map {N_G} H$

as we wanted to show.

Subgroup of Normalizer
By hypothesis, $H$ is a subgroup of $G$.

Thus $H$ is itself a group.

So by definition of subgroup:


 * $(1): \quad H \subseteq \map {N_G} H$
 * $(2): \quad $ is a group

it follows that $H$ is a subgroup of $\map {N_G} H$.