Normalizer of Conjugate

Theorem

The normalizer of a conjugate is the conjugate of the normalizer:

$S \subseteq G \implies N_G \left({S^a}\right) = \left({N_G \left({S}\right)}\right)^a$

Proof

From the definition of conjugate, $S^a = \left\{{y \in G: \exists x \in S: y = a x a^{-1}}\right\} = a S a^{-1}$.

From the definition of normalizer, $N_G \left({S}\right) = \left\{{x \in G: S^x = S}\right\}$.

Thus:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle N_G \left({S^a}\right)\)	\(=\)	\(\displaystyle \left\{ {x \in G: \left({S^a}\right)^x = S^a}\right\}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left\{ {x \in G: x a S a^{-1} x^{-1} = a S a^{-1} }\right\}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \left({N_G \left({S}\right)}\right)^a\)	\(=\)	\(\displaystyle \left\{ {x \in G: S^x = S}\right\}^a\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left\{ {x \in G: x S x^{-1} = S}\right\}^a\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left\{ {y \in G: \exists z \in \left\{ {x \in G: x S x^{-1} = S}\right\}: y = a z a^{-1} }\right\}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

Sources

• Allan Clark: Elements of Abstract Algebra (1971)... (previous)... (next): $\S 48 \gamma$