Subgroup is Subgroup of Normalizer

Theorem

Let $G$ be a group.

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

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

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

$x \in H \implies x H = H$

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.

$\Box$

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 $\map {N_G} H$.

$\blacksquare$

1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $2$: Subgroups and Cosets: $\S 35 \gamma$
1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): Chapter $8$: Homomorphisms, Normal Subgroups and Quotient Groups: Exercise $12 \ \text{(i)}$
1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): Chapter $10$: The Orbit-Stabiliser Theorem: Example $10.10$