Commutator of Quotient Group Elements

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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

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

$\blacksquare$


Sources

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