# [Math] The generator set of the commutator subgroup of a free group

abstract-algebragroup-theory

Let $G$ be a free group generated by the set $X$. Let $Y=\{xyx^{-1}y^{-1}|x,y\in X\}$, and let $K$ be the subgroup generated by $Y$. How to show that $K$ is the commutator subgroup $G'$ of $G$?

It is clear that $K\subseteq G'$. I tried to use the universal mapping property of free groups to show $\supseteq$ but I failed.

Danial, it seems that $K$ is not (in general) the commutator subgroup of $G$.
For instance, if $X = \{x,y\}$ consists of just two elements, and $G$ is generated by $X$ as a free group, then $K = \langle xyx^{-1}y^{-1} \rangle$ is a cyclic group. The commutator of $G$ is much larger.