If \( G \) is a group and \( N \) is a subgroup of \( G \) generated by commutators ( \( x y x^{-1} y^{-1} \) ), then \( N \) is a normal subgroup and \( G / H \) is abelian iff \( N \le H \). Here \( N \) is called a commutator subgroup.
For a proof, the only non-routine step is to show that \( g x_1 x_2 \ldots x_n g^{-1} \in N \) for an arbitrary \( g \in G \) and commutators \( x_1, x_2, \ldots, x_n \). Since $$ g x_1 x_2 \ldots x_n g^{-1} = (g x_1 g^{-1}) (g x_2 g^{-1}) \ldots (g x_n g^{-1}), $$ it suffices to show that \( g x y x^{-1} y^{-1} g^{-1} \in N \) for \( x, y, g \in G \).
But $$\begin{aligned}
g x y x^{-1} y^{-1} g^{-1} &= g x ( g^{-1} x^{-1} x g ) y x^{-1} y^{-1} g^{-1} \\
&= (g x g^{-1} x^{-1}) x (g y) x^{-1} (y^{-1} g^{-1}) \\
&= (g x g^{-1} x^{-1}) (x (g y) x^{-1} (g y)^{-1})
\end{aligned}$$ is a product of commutators, so must be in \( N \).