= Solution
{c}
For a <nonabelian group>, <7/b/iv> and <6/a/iii> imply $|G|=|G/A|\,|A|\geq4|A|$. Substituting $|A|\leq|G|/4$ into <7/c/i> yields
$$
\boxed{m(G)\leq\frac12\left(|G|+\frac{|G|}{4}\right)=\frac58|G|.}
$$
The 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, $\pm1$ 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 $\pm1,\pm i,\pm j,\pm k$ changes only its sign. Hence $m(G)=5=(5/8)|G|$.
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