Verification of Proof that if $G$ is not abelian $G/Z(G)$ is not cyclic I will prove this by the contrapositive:

If $G/Z(G)$ is cyclic then $G$ is abelian.

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.
 A: Your proof is correct (except, as the comment points out, for the statement that $aZ(G)$ has finite order).
You need to start with two arbitrary elements $x, y\in G$ because you want to show that $G$ is abelian --- that is, that for any two elements of $G$, that $xy = yx$. So you start the proof by picking two arbitrary elements.
Next, the only way to make progress is to recognize that $x$ and $y$ are each in some $Z(G)$-coset; since the quotient is cyclic, that means that they are in the cosets $a^mZ(G)$ and $a^nZ(G)$. You aren't really setting $x$ and $y$ to those values; rather, you are recognizing the fact that they are in these cosets and using that to manipulate the values to prove what you want.
A: If $G/Z(G)$ is cyclic, then there is $g\in G$ such that, for every $a\in G$, there is an integer $n$ such that: $$aZ(G)=(gZ(G))^n=g^nZ(G)\iff g^{-n}a\in Z(G)$$ Therefore, for every $a,b\in G$, there are integers $n$ and $m$, and central elements $z$ and $z'$, such that: $$ab=g^nzg^mz'=g^ng^mzz'=g^mg^nz'z=g^mz'g^nz=ba$$ So, $G$ is Abelian (and hence the only cyclic quotient $G/Z(G)$ is the trivial one).
