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 $G$.

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


Proof

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

Then:

\(\ds S^{a b}\) \(=\) \(\ds \paren {S^b}^a\) Conjugate of Set by Group Product
\(\ds \) \(=\) \(\ds S^a\) Definition of Normal Subgroup
\(\ds \) \(=\) \(\ds S\) 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, $\map {N_G} S \le G$.

$\blacksquare$


Sources

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