**The normalizer of a group center is the group itself**

We want to prove the following statement: N_G(Z(G)) = G.

**Proof.** “\subseteq“: This is by definition true since N_G(Z(G)) \subseteq G

“\supseteq“: Take g \in G. For all z \in Z(G) we have gz = zg. This implies that z = gzg^{-1} for all z \in Z(G). So we get that Z(G) = gZ(G)g^{-1}, which implies that g \in N_G(Z(G)). As g \in G was taken arbitrarily, we have that N_G(Z(G)) \supseteq G.