Solution
= Solution
\b[Can be false.] Use $G=S_3$ and the <normal subgroup> $H=A_3$ from the root solution. The <subgroup> $H$ is cyclic of order three, hence an <abelian group>, and the <quotient group> has the two <cosets> $A_3$ and $(12)A_3$, so is cyclic of order two and abelian. Yet $(12)(23)=(123)$ whereas $(23)(12)=(132)$, proving that $S_3$ is not abelian. An abelian kernel and an abelian quotient do not force all elements of the original group to commute.