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## Proof that if group $G/Z(G)$ is cyclic, then $G$ is commutative. [closed]

Hello. I finding for correct proof of this statement: If group $G/Z(G)$ is cyclic, then $G$ is commutative.

Proof: $G/Z(G)$ is isomorphic to $In(G)$ that is cyclic, and than for every inner isomorphism $\gamma_a$ is $\gamma_a \gamma_b = \gamma_{ab} = \gamma_{ba} = \gamma_b \gamma_a$ and therefore for every $a,b \in G$ is $ab = ba$.

Is that proof complete, or I miss something? Thanks a lot for help.

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Why "therefore"? – darij grinberg Nov 29 2010 at 8:48
Because $\gamma_a$ is isomorphism and in my opinion therefore $\gamma_a = \gamma_b$ implies $a = b$ - or i'm wrong? – tomas.lang Nov 29 2010 at 8:54
tomas, I think this question would fit in better at math.stackexchange.com. – Jonas Meyer Nov 29 2010 at 9:02
This is a standard exercise. Each element of $G$ has the form $a^n z$ where $a$ is fixed and $z\in Z(G)$... – Robin Chapman Nov 29 2010 at 9:10