Normalizer is Subgroup

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Theorem

Let $G$ be a group.

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

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

Then:

$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle S^{ab}$	$=$	$\displaystyle \left({S^b}\right)^a$	$\displaystyle $	$\displaystyle $	$\displaystyle $	Conjugate of a Set by Product
$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle $	$=$	$\displaystyle S^a$	$\displaystyle $	$\displaystyle $	$\displaystyle $	Definition of a Normal Subgroup
$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle $	$=$	$\displaystyle S$	$\displaystyle $	$\displaystyle $	$\displaystyle $	Definition of a Normal Subgroup

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

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

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

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

$\blacksquare$

\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle S^{ab}\)	\(=\)	\(\displaystyle \left({S^b}\right)^a\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Conjugate of a Set by Product
\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle S^a\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Definition of a Normal Subgroup
\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle S\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Definition of a Normal Subgroup