## The centralizer and normalizer of a group center is the group itself

Let $G$ be a group. Prove that $C_G(Z(G)) = G$ and deduce that $N_G(Z(G)) = G$.

[With a simplification pointed out by Jer.]

First we show that $C_G(Z(G)) = G$.

$(\subseteq)$ is clear. $(\supseteq)$ Suppose $g \in G$. Then by definition, for all $a \in Z(G)$, we have $ga = ag$. That is, for all $a \in Z(G)$, we have $a = gag^{-1}$. Thus $g \in C_G(Z(G))$.

Since $C_G(Z(G)) \leq N_G(Z(G))$, we have $N_G(Z(G)) = G$.