GZ-Theorem

Theorem

If $G/Z(G)$ is cyclic, then $G$ is abelian.

\begin{proof} Assume that $G/Z(G)=⟨xZ(G)⟩$. Then for all $g\in G$ there exists $n\in N,z\in Z(G)$ such that $x^{n}z=g$, where $xZ(G)$ generates $G/Z(G)$. It is easy to verify that $[x^n{z}_1,x^m{z}_2]=1$. \end{proof}