Quotient Group is Abelian iff All Commutators in Divisor

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 :


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

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

Proof
Let $x, y \in G$.

Then:

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.