# Normalizer is Subgroup

## Theorem

Let $G$ be a group.

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

$S \subseteq G \implies \map {N_G} S \le G$

## Proof

Let $a, b \in \map {N_G} S$.

Then:

 $\displaystyle S^{a b}$ $=$ $\displaystyle \paren {S^b}^a$ Conjugate of Set by Group Product $\displaystyle$ $=$ $\displaystyle S^a$ Definition of Normal Subgroup $\displaystyle$ $=$ $\displaystyle S$ Definition of Normal Subgroup

Therefore $a b \in \map {N_G} S$.

Now let $a \in \map {N_G} S$:

$a \in \map {N_G} S \implies S^{a^{-1} } = \paren {S^a}^{a^{-1} } = S^{a^{-1} a} = S$

Therefore $a^{-1} \in \map {N_G} S$.

Thus, by the Two-Step Subgroup Test, $\map {N_G} S \le G$.

$\blacksquare$