Normalizer of Conjugate is Conjugate of Normalizer

Theorem
Let $G$ be a group.

Let $a \mathop \in G$.

Let $S$ be a subset of $G$.

Let $S^a$ denote the conjugate of $S$ by $a$.

Let $\map {N_G} S$ denote the normalizer of $S$ in $G$.

Then:
 * $\map {N_G} {S^a} = \paren {\map {N_G} S}^a$

That is, the normalizer of a conjugate is the conjugate of the normalizer:

Proof
From the definition of conjugate:
 * $S^a = \set {y \in G: \exists x \in S: y = a x a^{-1} } = a S a^{-1}$

From the definition of normalizer:
 * $\map {N_G} S = \set {x \in G: S^x = S}$

Thus:

Suppose that $x \in \map {N_G} S$.

It is to be shown that:
 * $a x a^{-1} \in \map {N_G} {S^a} = \map {N_G} {a S a^{-1} }$

To this end, compute:

Hence $a x a^{-1} \in \map {N_G} {S^a}$, and it follows that:


 * $z \in \paren {\map {N_G} S}^a \implies z \in \map {N_G} {S^a}$

Conversely, let $x \in \map {N_G} {S^a}$.

That is, let $x \in G$ such that $x a S a^{-1} x^{-1} = a S a^{-1}$.

Now if we can show that $a^{-1} x a \in \map {N_G} S$, then:


 * $x = a \left({a^{-1} x a}\right) a^{-1} \in \paren {\map {N_G} S}^a$

establishing the remaining inclusion.

Thus, we compute:

Combined with the above observation, this establishes that:


 * $z \in \map {N_G} {S^a} \implies z \in \paren {\map {N_G} S}^a$

Hence the result, by definition of set equality.