I am looking for a correct proof of this statement: If $G$ is a group such that $G/Z(G)$ is cyclic, then $G$ is commutative.
Proof: $G/Z(G)$ is isomorphic to $\operatorname{Inn}(G)$ and is cyclic, and then for every $a$ and $b$ in $G$ the inner isomorphisms $\gamma_a$ and $\gamma_b$ satisfy $\gamma_a \gamma_b = \gamma_{ab} = \gamma_{ba} = \gamma_b \gamma_a$, and therefore for every $a,b \in G$, $ab = ba$.
Is that proof complete, or am I missing something? Thanks a lot for the help.