Proof that a quotient is abelian iff the subgroup contains the commutator subgroup

Proof

We show that two cosets commute exactly when the commutator of their representatives lies in $N$.

1. Let $G$ be a group with identity $e$, and let $N$ be a normal subgroup of $G$. The quotient group $G/N$ has the operation $(aN)(bN) = (ab)N$.
2. We record a fact on left cosets: for $a, b \in G$, $aN = bN$ holds exactly when $a^{-1}b \in N$.
3. Proof of step 2. If $aN = bN$, then $b = be \in aN$, so $b = an$ with $n \in N$, and $a^{-1}b = n \in N$.
4. Conversely, let $n = a^{-1}b \in N$. Closure gives $nN \subseteq N$, and each $m \in N$ is $n(n^{-1}m) \in nN$. So $nN = N$, and $bN = (an)N = (aN)(nN) = aN$.
5. For $g, h \in G$, $(hg)(g^{-1}h^{-1}) = e$. By the uniqueness of inverses, $(hg)^{-1} = g^{-1}h^{-1}$.
6. Let $g, h \in G$. By step 1, $(gN)(hN) = (hN)(gN)$ holds exactly when $(hg)N = (gh)N$.
7. By steps 2 and 5, this holds exactly when $g^{-1}h^{-1}gh \in N$, that is, when the commutator $[g, h]$ lies in $N$.
8. By steps 6 and 7, $G/N$ is abelian exactly when $[g, h] \in N$ for all $g, h \in G$.
9. The commutator subgroup $[G, G]$ is the subgroup generated by the set $C$ of all commutators. So it is the intersection of all subgroups of $G$ that contain $C$.
10. If $C \subseteq N$, then $N$ is one of the subgroups in this intersection. So $[G, G] \subseteq N$.
11. If $[G, G] \subseteq N$, then $C \subseteq [G, G] \subseteq N$.
12. By steps 10 and 11, $C \subseteq N$ exactly when $[G, G] \subseteq N$. With step 8, $G/N$ is abelian if and only if $[G, G] \subseteq N$.