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?