Proof blueprint - If $G/Z(G)$ cyclic then $G$ Abelian - Fraleigh p. 153 15.37 
(1.) Why didn't Fraleigh state the result in the direct form like in my title? Why state it with the negations and then prove the contrapositive? Isn't this extra unnecessary work? 
(2.) How do you envisage and envision to consider the cosets $aZ(G), bZ(G)$?
I know we presupposed $G/Z(G)$ is cyclic for the proof. 
(3.) If $G/Z(G)$ is cyclic then $G/Z(G)$ has one element. What is it?
(4.) What's the intuition?
 A: *

*I think that, although the proof is by contrapositive and therefore establishes the result of your title (and as a consequence the result you referred to in point (3)), Fraleigh stated the result as he did because it is really a result about non-Abelian groups (modulo their center they cannot possibly be cyclic). For Abelian groups one learns nothing new; it is obvious that modulo their center they are trivial.

*As for intuition, I think this could be done simpler. Given a group that is cyclic modulo its center, there is a representative $x$ of the generator of the quotient, and then the group is generated by $x$ and $Z(G)$ together. But that is a bunch of elements that all commute with each other, so the group must be Abelian (and in the end there is no need for $x$ after all).
A: 1) He didn't state the full contrapositive statement because it's strictly not needed. He has stated what he wants to show in the proof, he has said that he wants to proof the contrapositive, and he assumes the negation of the conclusion. At the end he has shown the negation of the hypothesis, and the theorem is proven.
2) You consider the cosets $aZ(G)$ and $bZ(G)$ because you want, in the end, to prove that $ab = ba$.
3) If a quotient group $G/H$ has one element, then $H = G$, and the one element in the quotient group is the (co)set of all elements in $G$.
