Normalizer of Center is Group

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: