---
tag: RDK4W7Z3A1
kind: proof
name: Proof that a quotient is abelian iff the subgroup contains the commutator subgroup
msc: [20F12]
proves: 0MS8QGA6S5
uses: [0YYWCNTPWR, 0W77CVBCHQ, 05FD317V2W, 0D9NDGE9KP, 06RGT4V6Y6, 08SJ5K2CBG, 0RG9SPQSWM, 071DDFVG4J]
sources:
  - url: https://en.wikipedia.org/wiki/Commutator_subgroup
    locator: Abelianization
---

## 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|0W77CVBCHQ]] of $G$. The [[quotient group|0YYWCNTPWR]] $G/N$ has the operation $(aN)(bN) = (ab)N$.
2. We record a fact on [[left cosets|05FD317V2W]]: 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|0D9NDGE9KP]], $(hg)^{-1} = g^{-1}h^{-1}$.
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|06RGT4V6Y6]] $[g, h]$ lies in $N$.
8. By steps 6 and 7, $G/N$ is [[abelian|071DDFVG4J]] exactly when $[g, h] \in N$ for all $g, h \in G$.
9. The [[commutator subgroup|08SJ5K2CBG]] $[G, G]$ is the [[subgroup generated by|0RG9SPQSWM]] 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 $[G, G] \subseteq N$.

## Idea

The cosets $gN$ and $hN$ commute exactly when $gh$ and $hg$ differ by an element of $N$, and the difference is the commutator $[g, h]$. So the quotient is abelian when $N$ contains all commutators. Since $N$ is a subgroup, this is the same as containing the subgroup they generate.