# Abelian Quotient Group

 It has been suggested that this page or section be merged into Quotient Group is Abelian iff All Commutators in Divisor. (Discuss)

## Theorem

Let $G$ be a group.

Let $H$ be a normal subgroup of $G$.

Let $G / H$ denote the quotient group of $G$ by $H$.

Then $G / H$ is abelian if and only if $H$ contains every element of $G$ of the form $a b a^{-1} b^{-1}$ where $a, b \in G$.

## Proof

Let $G / H$ be abelian.

Then:

 $\, \displaystyle \forall a H, b H \in G / H: \,$ $\displaystyle a H b H$ $=$ $\displaystyle b H a H$ $\displaystyle \leadsto \ \$ $\displaystyle a b H$ $=$ $\displaystyle b a H$ $\displaystyle \leadsto \ \$ $\displaystyle a b \paren {b a}^{-1}$ $\in$ $\displaystyle H$ $\displaystyle \leadsto \ \$ $\displaystyle a b a^{-1} b^{-1}$ $\in$ $\displaystyle H$

The argument reverses.

Suppose that $\forall a, b \in G: a b a^{-1} b^{-1} \in H$.

 $\, \displaystyle \forall a, b \in G: \,$ $\displaystyle a b a^{-1} b^{-1}$ $\in$ $\displaystyle H$ $\displaystyle \leadsto \ \$ $\displaystyle a b \paren {b a}^{-1}$ $\in$ $\displaystyle H$ $\displaystyle \leadsto \ \$ $\displaystyle a b H$ $=$ $\displaystyle b a H$ $\displaystyle \leadsto \ \$ $\displaystyle a H b H$ $=$ $\displaystyle b H a H$

$\blacksquare$