Proving that if $G/Z(G)$ is cyclic, then $G$ is abelian 
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?

Here's part of the proof that $G$ is abelian. Hopefully this will get you started...
Let $Z(G)=Z$. If $G/Z$ is cyclic, then it has a generator, say $G/Z = \langle gZ \rangle$. This means that for each coset $xZ$ there exists some $i \in \mathbb{Z}$ such that $xZ=(gZ)^i=g^iZ$.
Suppose that $x,y \in G$. Consider $x \in xZ=g^iZ$ so that $x=g^iz$ for some $z\in Z$. 
Represent $y$ in a similar manner and consider $xy$ and $yx$. Why are they equal?
Edit: !!!Spoiler alert!!! :) Here's the rest of the story.
$yZ \in G/Z = \langle gZ \rangle$ so that $yZ=(gZ)^j=g^jZ$ for some $j \in \mathbb{Z}$.
Therefore, $y \in yZ=g^jZ$ so that $y=g^jz_0$ for some $z_0 \in Z$. 
Finally, $xy=g^izg^jz_0=g^ig^jzz_0=g^{i+j}zz_0=g^{j+i}zz_0=g^jg^izz_0=g^jz_0g^iz=yx$
The second equality follows because $z$ is in the center and thus commutes with everything. Then we're just messing with powers of $g$ (which commute with themselves). The next to last equality follows because $z_0$ is in the center and thus commutes with everything.
