$G/Z(G)$ is cyclic useful for proving groups abelian? It's a common exercise to prove in an abstract algebra book that if $G/Z(G)$ is cyclic then $G$ must be abelian. But I've always found the exercise strange because if $G$ is abelian then $Z(G)=G$ and the quotient is trivial.
Is there a specific example of this being a useful technique to proving a group is abelian? As it seems  you must know enough about a specific group $G/Z(G)$ to proves it's cyclic, but not enough to notice that it the trivial group, which would prove the commutativity of $G$ immediately.
 A: A very good example to serve its usefulness is 
Is it possible to have  a group $G$ such that 0$(G/Z(G))=91$?solve  this and get the beauty of the result
A: If $G$ is a group of order $p^2$, then $G$ is abelian. It follows immediately from the exercise. And then one knows $G \cong \mathbb{Z}/p^2 $ or $G \cong \mathbb{Z}/p \oplus\mathbb{Z}/p$ by the structure theorem.
If $G$ is a group of order $p^3$, then $G$ doesn't have to be abelian. But it is abelian if the center of $G$ has order $\neq p$. In other words, in order to classify (non-abelian) groups of order $p^3$, we may assume that the center has order $p$. This is the first step in their classification.
A: If $\overline{G}=G/Z(G)$ is cyclic, then, exist $\overline{x}$ such that $\overline{G} = <\overline{x}>$. (Note that, $G/Z(G) = <x>Z(G)$).
Thus for all $g \in G$, exist $i$ such that $\overline{g}=\overline{x}^i$. Consequentely, $g = x^iz$, where $z \in Z(G)$.
Then $g_1,g_2 \in G$, we have that $g_j = x^{i_j}z_j, j=1,2$, where $z_j \in Z(G)$. Now is very simple see that $g_1g_2 = g_2g_1$, ie, $G$ is abelian.
