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 $G$.

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

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:

$$ $$ $$ $$ $$