Commutator of Quotient Group Elements

Theorem
Let $G$ be a group.

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

Let $\sqbrk {x, y}$ denote the commutator of $x, y \in G$:
 * $\sqbrk {x, y} = x^{-1} y^{-1} x y$

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

where $x N$ and $y N$ are left cosets of $N$, and so elements of the quotient group $G / N$ of $G$ by $N$.