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Possible Duplicate:
Proof that if group $G/Z(G)$ is cyclic, then $G$ is commutative
If $G/Z(G)$ is cyclic, then $G$ is abelian

If $G$ is a group and $Z(G)$ the center of $G$, show that if $G/Z(G)$ is cyclic, then $G$ is abelian.

This is what I have so far:

We know that all cyclic groups are abelian. This means $G/Z(G)$ is abelian. $Z(G)= \{z \in G \mid zx=xz \text{ for all } x \in G \}$. So $Z(G)$ is abelian.

Is it sufficient to say that since $G/Z(G)$ and $Z(G)$ are both abelian, $G$ must be abelian?

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    It's certainly not sufficient just to *say* it. You would have to *prove* it. Don't waste too much time trying to prove this, though, since it's false. For example, if $G$ is the [quaternion group](http://en.wikipedia.org/wiki/Quaternion_group), then $G/Z(G)$ and $Z(G)$ are both abelian, but $G$ is not. You really do need the assumption that $G/Z(G)$ is cyclic, not just abelian.2011-11-03
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    Oh, okay. How do I use the fact that G/Z(G) is cyclic to show that G is abelian?2011-11-03

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