Quotient Group is Abelian iff All Commutators in Divisor/Proof 1

Proof
Let $x, y \in G$.

First we establish the following:

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

Necessary Condition
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.