# 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 $N$.

## Proof

 $\displaystyle \sqbrk {x N, y N}$ $=$ $\displaystyle \paren {x N}^{-1} \paren {y N}^{-1} \paren {x N} \paren {y N}$ Definition of Commutator of Group Elements $\displaystyle$ $=$ $\displaystyle \paren {x^{-1} N} \paren {y^{-1} N} \paren {x N} \paren {y N}$ Quotient Group is Group: inverse of $x N$ is $x^{-1} N$ $\displaystyle$ $=$ $\displaystyle \paren {x^{-1} y^{-1} N} \paren {x y N}$ Definition of Coset Product $\displaystyle$ $=$ $\displaystyle x^{-1} y^{-1} x y N$ Definition of Coset Product $\displaystyle$ $=$ $\displaystyle \sqbrk {x, y} N$ Definition of Commutator of Group Elements

$\blacksquare$