Solution
= Solution
{c}
If $G$ is not <Abelian>, the contrapositive of <7/b/ii> says that $G/A$ is not a <cyclic group>. The <quotient group> is finite because $G$ is finite. By <7/b/iii>, every <group> with fewer than four elements is a <cyclic group>. Hence
$$
\boxed{|G/A|\geq4.}
$$
Equivalently, the <center of a group> satisfies $|A|\leq|G|/4$, by the <coset> count in <6/a/iii>.