Subgroup is Subgroup of Normalizer

Theorem

Let $G$ be a group.

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

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

Proof

Subset of Normalizer

First we show that $H$ is a subset of $N_G \left({H}\right)$.

This follows directly from 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 $x \in N_G \left({H}\right)$.

So $H \subseteq N_G \left({H}\right)$ as we wanted to show.

$\Box$

Subgroup of Normalizer

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

Thus $H$ is itself a group.

So by definition of subgroup:

$(1): \quad H \subseteq N_G \left({H}\right)$

$(2): \quad $ is a group

it follows that $H$ is a subgroup of $N_G \left({H}\right)$.

$\blacksquare$

Sources

• Allan Clark: Elements of Abstract Algebra (1971)... (previous)... (next): $\S 35 \gamma$
• Thomas A. Whitelaw: An Introduction to Abstract Algebra (1978)... (previous)... (next): Exercise $8.12 \ \text{(i)}$