The generator set of the commutator subgroup of a free group 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. 
 A: 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.
A: I came across this question and I think it might be helpful to add a comment.
If we let $K$ be the least normal subgroup of $G$ containing $Y$ instead of the subgroup generated by $Y$, then $K$ is the commutator subgroup of $G$. 
It is clear that $K\subset [G,G]$ on one hand.
On the other hand, since we have that if $G/N$ is abelian for some normal subgroup $N$ then $N\supset[G,G]$, it suffices to show that $G/K$ is abelian. 
Note that $\{xN:x\in X\}$ generates $G/N$, it is enough to show that $xN\cdot yN=yN\cdot xN$ for any $x,y\in X$.  We have $xy=yx\cdot(x^{-1}y^{-1}xy)$, and $x^{-1}y^{-1}xy=(xy)^{-1}(xyx^{-1}y^{-1})(xy)\in K$. It follows that $xN\cdot yN=xyN=yxN=yN\cdot xN$ and therefore $G/N$ is abelian.
