Normalizer of Conjugate is Conjugate of Normalizer

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:

Suppose that $x \in N_G \left({S}\right)$.

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

To this end, compute:

Hence $a x a^{-1} \in N_G \left({S^a}\right)$, and it follows that:


 * $z \in N_G \left({S}\right)^a \implies z \in N_G \left({S^a}\right)$

Conversely, let $x \in N_G \left({S^a}\right)$.

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 $a^{-1} x a \in N_G \left({S}\right)$, then:


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

establishing the remaining inclusion.

Thus, we compute:

Combined with the above observation, this establishes that:


 * $z \in N_G \left({S^a}\right) \implies z \in N_G \left({S}\right)^a$

Hence the result, by definition of set equality.