**Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 2.2 Exercise 2.2.2**

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

Solution: 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 $N_G(Z(G)) = G$.