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.

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