Normalizer is Subgroup

Theorem
Let $$G$$ be a group.

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

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

Proof

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

$$S^{ab} = \left({S^a}\right)^b = S^b = S$$

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


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

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

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

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