Subgroup is Normal Subgroup of Normalizer

Theorem
Let $$G$$ be a group.

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

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

Proof

 * $$p \in H \implies H^p = H$$
 * $$q \in H \implies H^q = H$$

Let $$p, q \in H$$. Then:


 * $$p x p^{-1} \in H \iff x \in H$$
 * $$q x q^{-1} \in H \iff x \in H$$

So $$q^{-1} \left({q x q^{-1}}\right) q = x \implies q^{-1} \in H$$.

Now:

So $$H$$ is a subgroup of $$N_G \left({H}\right)$$.