Quotient of Commutator is Exactly Abelian

Proposition

Let $G$ be a finite group and let $N⊲G$ be a normal subgroup. Then $G/H$ is abelian if and only if $G^{\prime }⩽H$.

\begin{proof} Define the canonical map as $\pi :G\to G/H,g\mapsto \overline{g}$. If $G/H$ is abelian, then $\pi (ab{a}^{-1}{b}^{-1})=\overline{a}\overline{b}{\overline{a}}^{-1}{\overline{b}}^{-1}=\overline{e}$ and so $ab{a}^{-1}{b}^{-1}\in \ker \pi =H$. It yields that $G^{\prime }⩽H$. Conversely, if $ab{a}^{-1}{b}^{-1}\in H$ for any $a,b\in G$, then $\overline{ab{a}^{-1}{b}^{-1}}=\overline{e}$ and $\overline{a}\overline{b}=\overline{b}\overline{a}$. Therefore, we have $G/H$ is abelian. \end{proof}