= Solution
Suppose every <group commutator> belongs to $H$. For $h\in H$ and $g\in G$, the particular <group commutator> $h^{-1}g^{-1}hg$ belongs to $H$. Multiplying on the left by $h$ gives $g^{-1}hg\in H$. Thus $g^{-1}Hg\subseteq H$ for every $g$; applying the same inclusion to $g^{-1}$ gives the reverse inclusion. \b[Therefore $H$ is a <normal subgroup>.]
We can now form the <quotient group>. Its <group commutators> are
$$
(aH)^{-1}(bH)^{-1}(aH)(bH)=(a^{-1}b^{-1}ab)H=H.
$$
A <group commutator> is the <identity element> exactly when the two elements commute. Hence \b[$G/H$ is an <abelian group>]. Equivalently, the hypothesis says that $H$ contains the <commutator subgroup>, so the quotient factors through the <abelianization>.
Back to article page
