Subgroup is Normal Subgroup of Normalizer

Theorem
Let $$G$$ be a group.

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


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

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

This follows directly from Sundry Coset Results: Theorem 3:
 * $$x \in H \implies x H = H$$

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.

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

Thus $$H$$ is itself a group.

So by definition of subgroup:


 * 1) $$H \subseteq N_G \left({H}\right)$$;
 * 2) $$H$$ is a group,

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