Let \(G\) be a group such that the quotient \(G/Z(G)\) is cyclic. Then \(G\) is abelian.
Proof: Let \(x, y \in G\).
Let \(gZ(G)\) be a generator of \(G/Z(G)\).
Then, \(xZ(G) = g^aZ(G)\) and \(yZ(G) = g^bZ(G)\) for some \(a, b \in \mathbb{Z}\).
Note that \(x \in xZ(G) = g^aZ(G)\) and thus, \(x = g^az\) for some \(z \in Z(G)\). Similarly, \(y = g^bz'\) for some \(z' \in Z(G)\).
Now, noting that \(z\) and \(z'\) commute with \(g\) and each other, we get that
\[xy = g^azg^bz' = g^{a+b}zz' = g^{b+a}z'z = g^bz'g^az = yx.\]