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.

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Danial, for me this is the definition of the commutator subgroup, what's yours? –  Alex Youcis Sep 30 '13 at 14:56
@AlexYoucis The standard definition of the commutator subgroup would be the group generated by $\{xyx^{-1}y^{-1} \mid x, y \in G\}$. Note that $x$ and $y$ range over all of $G$, not only $X$. –  Dan Shved Sep 30 '13 at 14:59
@AlexYoucis For me it is not clear that $G'\subseteq K$, for example $xyzw(xy)^{-1}(zw)^{-1}\in G'$ but i cant prove that is in $K$ –  Ronald Sep 30 '13 at 15:01
The commutator subgroup of a free group of rank greater than 1 is not even finitely generated. For example, if I am remembering correctly, then for $F=\langle x,y \rangle$ free of rank 2, $F'$ is freely generated by the commutators $[x^m,y^n]$ with $m,n \ne 0$. –  Derek Holt Sep 30 '13 at 16:09

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.

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Yes, $\langle K^G \rangle = G'$. Inclusion $\langle K^G \rangle \subseteq G'$ is obvious. In the other direction, $x\langle K^G \rangle$ commutes with $y\langle K^G \rangle$ in the factorgroup $G / \langle K^G \rangle$ because $xyx^{-1}y^{-1} \in K$, so the factorgroup $G / \langle K^G \rangle$ is abelian, therefore $G' \subseteq \langle K^G \rangle$. –  Dan Shved Oct 2 '13 at 18:47
The same argument works in general, not only when $X$ consists of two elements. –  Dan Shved Oct 2 '13 at 18:49

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.

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