I will prove this by the contrapositive:

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> If $G/Z(G)$ is cyclic then $G$ is abelian.
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Proof: We assume that $G/Z(G)$ is cyclic. This means it is generated by a left coset $(aZ(G))^n=e$ for some integer $n$. By defined operation $a^n Z(G)=e$.  

Let $x,y \in G$ and $z\_0, z\_1 \in Z(G)$, so this leads to $$a^n z\_0=x,\ a^m z\_1=y.$$.  

This implies: $$xy=a^n z\_0,\ a^m z\_1 = a^{nm}(z\_0 z\_1)=a^m z\_1,\ a^n z\_0=yx.$$

This what I've been told is the write answer but I am wondering however, why is it necessary to let $x,y \in G$ and set our two elements $a^n z\_0$ and $a^m z\_1$ to elements $x, y \in G$? Also is the fact that $a^n Z(G)=e$ wrong?

Thanks in advance.