Normalizer of Center is Group

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Theorem

Let $G$ be a group.

Let $Z \left({G}\right)$ be the center of $G$.

Let $x \in G$.

Let $N_G \left({x}\right)$ be the normalizer of $x$ in $G$.

Let $\left[{G : N_G \left({x}\right)}\right]$ be the index of $N_G \left({x}\right)$ in $G$.


Then:

$Z \left({G}\right) = \left\{{x \in G: N_G \left({x}\right) = G}\right\}$


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 $x \in Z \left({G}\right) \iff N_G \left({x}\right) = G$, and so $\left[{G : N_G \left({x}\right)}\right] = 1$.


Proof

$N_G \left({x}\right)$ is the normalizer of the set $\left\{{x}\right\}$. Thus:


\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle N_G \left({x}\right)\) \(=\) \(\displaystyle G\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)                    
\(\displaystyle \) \(\displaystyle \iff\) \(\displaystyle \) \(\displaystyle \forall a \in G: \left\{ {x}\right\}^a\) \(=\) \(\displaystyle \left\{ {x}\right\}\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)          Definition of Normalizer          
\(\displaystyle \) \(\displaystyle \iff\) \(\displaystyle \) \(\displaystyle \forall a \in G: a x a^{-1}\) \(=\) \(\displaystyle x\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)          Definition of Conjugate          
\(\displaystyle \) \(\displaystyle \iff\) \(\displaystyle \) \(\displaystyle \forall a \in G: a x\) \(=\) \(\displaystyle x a\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)                    
\(\displaystyle \) \(\displaystyle \iff\) \(\displaystyle \) \(\displaystyle x\) \(\in\) \(\displaystyle Z \left({G}\right)\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)          Definition of Center of a Group          

$\blacksquare$


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