Let $G$ be a group and let $D(G)=[G,G]$ be the commutator subgroup of $G$.
Let $N$ be a subgroup of $G$.
Prove that the subgroup $N$ is normal in $G$ and $G/N$ is an abelian group if and only if $N \supset D(G)$.

Recall that for any $a, b \in G$, the commutator of $a$ and $b$ is
\[ [a,b]=a^{-1}b^{-1}ab \in G.\]
The commutator subgroup $D(G)=[G,G]$ is a subgroup of $G$ generated by all commutators.
That is,
\[D(G)=[G,G]=\langle [a,b] \mid a,b \in G \rangle.\]

Proof.

$(\implies)$ Suppose that $N$ is a normal subgroup of $G$ and the quotient $G/N$ is an abelian group.
Then for any elements $a, b \in G$, we have
\begin{align*}
abN=aN\cdot bN=bN \cdot aN=baN.
\end{align*}
(Here we used the fact that $N$ is normal, hence $G/N$ is a group.)

From this, we obtain that
\[a^{-1}b^{-1}abN=N\]
and thus $[a,b]=a^{-1}b^{-1}ab\in N$.
Since any generator $[a,b]$ is in $N$, we have $D(G)\subset N$.

$(\impliedby)$ On the other hand, let us assume that $N \supset D(G)$.
We first show that $N$ is a normal subgroup of $G$.
For any $g \in G$, $x\in N$, we have
\begin{align*}
gxg^{-1}=gxg^{-1}x^{-1}x=[g^{-1},x^{-1}]\cdot x\in N
\end{align*}
since the commutator $[g^{-1},x^{-1}]\in D(G)\subset N$ and $x \in N$.
Thus $N$ is normal in $G$.

Now that $N$ is normal in $G$, the quotient $G/N$ is a group. We show that $G/N$ is an abelian group.
For any $a, b \in G$, we have
\begin{align*}
aN\cdot bN&=ab N\\
&=baa^{-1}b^{-1}abN\\
&=ba[a,b]N\\
&=baN \qquad \text{ since } [a,b] \in N\\
&=bN\cdot aN.
\end{align*}
Therefore the group operations of $G/N$ is commutative, and hence $G/H$ is abelian.

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Hint.
We use the following fact to prove the problem.
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