Past exam of the mathematics course of the University of Oxford / 2021 / Prelims / Mathematics I / 7 / b / iv

If $G$ is not Abelian, the contrapositive of ii says that $G/A$ is not a cyclic group. The quotient group is finite because $G$ is finite. By iii, every group with fewer than four elements is a cyclic group. HenceEquivalently, the center of a group satisfies $|A|\le |G|/4$, by the coset count in iii.