Suppose G/Z(G) is cyclic. Write Z = Z(G). Let xZ be a generator for G/Z. Pick elements a and b in G.

Then aZ = xⁿZ and bZ = x^mZ for some integers n and m.

Thus, a = xⁿz₁ and b = x^mz₂ for some elements z₁ and z₂ in Z. So

ab = (xⁿz₁)(x^mz₂)
= xⁿ(x^mz₁)z₂
= x^m((xⁿz₁)z₂)
= (x^mz₂)(xⁿz₁)
= ba.

Hence G is abelian.

Note that G/Z(G) is never a nontrivial cyclic group. If G/Z(G) is cyclic then, as we have shown, G is abelian and so G = Z(G).