Normalizer of Center is Group

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Theorem

Let $G$ be a group.

Let $\map Z G$ be the center of $G$.

Let $x \in G$.

Let $\map {N_G} x$ be the normalizer of $x$ in $G$.


Then:

$\map Z G = \set {x \in G: \map {N_G} x = G}$


That is, the center of a group $G$ is the set of elements $x$ of $G$ such that the normalizer of $x$ is the whole of $G$.


Thus:

$x \in \map Z G \iff \map {N_G} x = G$

and so:

$\index G {\map {N_G} x} = 1$

where $\index G {\map {N_G} x}$ is the index of $\map {N_G} x$ in $G$.


Proof

$\map {N_G} x$ is the normalizer of the set $\set x$.

Thus:


\(\ds \map {N_G} x\) \(=\) \(\ds G\)
\(\ds \leadstoandfrom \ \ \) \(\ds \forall a \in G: \, \) \(\ds \set x^a\) \(=\) \(\ds \set x\) Definition of Normalizer
\(\ds \leadstoandfrom \ \ \) \(\ds \forall a \in G: \, \) \(\ds a x a^{-1}\) \(=\) \(\ds x\) Definition of Conjugate of Group Element
\(\ds \leadstoandfrom \ \ \) \(\ds \forall a \in G: \, \) \(\ds a x\) \(=\) \(\ds x a\)
\(\ds \leadstoandfrom \ \ \) \(\ds x\) \(\in\) \(\ds \map Z G\) Definition of Center of Group

$\blacksquare$


Sources

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