Normalizer of Conjugate is Conjugate of Normalizer

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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:

\(\ds \map {N_G} {S^a}\) \(=\) \(\ds \set {x \in G: \paren {S^a}^x = S^a}\)
\(\ds \) \(=\) \(\ds \set {x \in G: x a S a^{-1} x^{-1} = a S a^{-1} }\)


\(\ds \paren {\map {N_G} S}^a\) \(=\) \(\ds \set {x \in G: S^x = S}^a\)
\(\ds \) \(=\) \(\ds \set {x \in G: x S x^{-1} = S}^a\)
\(\ds \) \(=\) \(\ds \set {y \in G: \exists z \in \set {x \in G: x S x^{-1} = S}: y = a z a^{-1} }\)


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:

\(\ds \paren {a x a^{-1} } a S a^{-1} \paren {a x a^{-1} }^{-1}\) \(=\) \(\ds \paren {a x a^{-1} } a S a^{-1} \paren {a x^{-1} a^{-1} }\) Power of Conjugate equals Conjugate of Power
\(\ds \) \(=\) \(\ds a x S x^{-1} a^{-1}\)
\(\ds \) \(=\) \(\ds a S a^{-1}\) as $x \in \map {N_G} S$

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:

\(\ds \paren {a^{-1} x a} S \paren {a^{-1} x a}^{-1}\) \(=\) \(\ds a^{-1} x a S a^{-1} x^{-1} a\) Power of Conjugate equals Conjugate of Power
\(\ds \) \(=\) \(\ds a^{-1} a S a^{-1} a\) as $x \in \map {N_G} {S^a}$
\(\ds \) \(=\) \(\ds S\)

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.

$\blacksquare$


Sources

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