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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.

For a nonabelian group, iv and iii imply $|G|=|G/A||A|\ge 4|A|$. Substituting $|A|\le |G|/4$ into i yieldsThe constant is sharp: the quaternion group has eight elements and the five conjugacy classes $\{1\}$, $\{-1\}$, $\{i,-i\}$, $\{j,-j\}$, $\{k,-k\}$. To verify this directly, $\pm 1$ are central elements, conjugation by $j$ sends $i$ to $-i$, and similarly for $j,k$; its relations $i^2=j^2=k^2=-1$ and $ij=k=-ji$, $jk=i=-kj$, $ki=j=-ik$ show that conjugating $i,j,k$ by any of $\pm 1,\pm i,\pm j,\pm k$ changes only its sign. Hence $m(G)=5=(5/8)|G|$.