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

## Theorem

Let $G$ be a group.

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

Let $G / N$ be the quotient group of $G$ by $N$.

Then the quotient group $G / N$ is abelian if and only if:

$\forall x, y \in G: \sqbrk {x, y} \in N$

where $\sqbrk {x, y}$ denotes the commutator of $x$ and $y$.

## Proof

Let $x, y \in G$.

Then:

 $\text {(1)}: \quad$ $\ds \paren {x N} \paren {y N}$ $=$ $\ds \paren {y N} \paren {x N}$ $\ds \leadstoandfrom \ \$ $\ds \sqbrk {x N, y N}$ $=$ $\ds N$ Definition of Commutator of Group Elements $\ds \leadstoandfrom \ \$ $\ds \sqbrk {x, y} N$ $=$ $\ds N$ Commutator of Quotient Group Elements

Let $G / N$ be abelian.

Then by definition:

$\forall x, y \in G: \paren {x N} \paren {y N} = \paren {y N} \paren {x N}$

and it follows from $(1)$ that:

$\forall x, y \in G: \sqbrk {x, y} \in N$

Conversely, let:

$\forall x, y \in G: \sqbrk {x, y} \in N$

Again it follows from $(1)$ that:

$\forall x, y \in G: \paren {x N} \paren {y N} = \paren {y N} \paren {x N}$

That is, that $G / N$ is abelian.

$\blacksquare$