Normalizer is Subgroup

Theorem
Let $G$ be a group.

The normalizer of a subset $S \subseteq G$ is a subgroup of $G$.


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

Proof
Let $a, b \in N_G \left({S}\right)$.

Then:

Therefore $a b \in N_G \left({S}\right)$.

Now let $a \in N_G \left({S}\right)$:


 * $a \in N_G \left({S}\right) \implies S^{a^{-1}} = \left({S^a}\right)^{a^{-1}} = S^{a^{-1} a} = S$

Therefore $a^{-1} \in N_G \left({S}\right)$.

Thus, by the Two-Step Subgroup Test, $N_G \left({S}\right) \le G$.