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There is a homomorphism $\langle x,y\rangle\to\langle x\rangle$ of free groups, sending $y$ to $1$. We can combine this with the other obvious homomorphism to get a surjective homomorphism $$ \langle x,y\rangle \to \langle x\rangle\times\langle y\rangle $$ The kernel is the commutator subgroup, and it is freely generated by the elements $[x^i,y^j]$ for $i,j\in\mathbb{Z}$ with $i,j\neq 0$.

We can define a homomorphism $$ \langle x,y,z\rangle \to \langle x,y\rangle \times \langle x,z\rangle \times \langle y,z\rangle $$ in a similar way, and let $K$ denote the kernel. In some notes of Vershinin I have seen this called the fat commutator subgroup. It is a subgroup of a free group and so must be free.

Question: Is there a known basis?

Let $X$ denote the smallest normal subgroup of $\langle x,y,z\rangle$ containing $x$, or equivalently the kernel of the map $\langle x,y,z\rangle \to \langle y,z\rangle$. Define $Y$ and $Z$ similarly, so $K=X\cap Y\cap Z$. Put $$ A = \{[u,[v,w]] \;:\; (u,v,w)\in X\times Y\times Z\}. $$ It is easy to see that $A\subseteq K$. There is a comment of Vershinin which might mean that $A$ generates $K$, or it might mean something a bit more complicated. In any case, the document that I found did not contain a proof or reference. Also, I think that $A$ is too big to generate $K$ freely.

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    $\begingroup$ You should be able to work this out fairly easily by just drawing the Cayley graph of the image group, finding a spanning tree, etc. Did you try it? I realize I'm being a bit lazy, because today is a busy day for me. Please don't read this comment as negative. $\endgroup$ – Benjamin Steinberg Oct 6 '15 at 13:21
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We can do this step by step, using the fact that any surjective map from a free group onto a free group splits. More precisely: a basis for the group $X$ is given by $\{wxw^{-1}\}_{w\in \langle y,z\rangle}$. We then calculate the image of this basis under the projection onto $\langle x,y\rangle$. We get that $wxw^{-1}$ is mapped onto $y^{s(w)}xy^{-s(w)}$ where $s(w)$ is the exponent sum of $y$ in $w$. The elements $\{y^ixy^{-i}\}_{i\in\mathbb{Z}}$ form a free basis of the normal closure of $x$ in $\langle x,y\rangle$. We thus have a map of free groups which maps the basis of the first group surjectively onto a basis of the second group. From this we can easily deduce that $\{wxw^{-1}y^{s(w)}x^{-1}y^{-s(w)}\}_{w\in\langle y,z\rangle - \langle y\rangle}$ is a basis for $X\cap Z$. We continue by calculating the image of this basis inside $Y$. We get that $wxw^{-1}y^{s(w)}x^{-1}y^{-s(w)}\mapsto z^{t(w)}xz^{-t(w)}x^{-1}$ where $t(w)$ is the exponent sum of $z$ in $w$. Since $\{z^ixz^{-i}x^{-1}\}_{i\in\mathbb{Z}-0}$ is a basis of a free subgroup of $Y$, we get that a basis of $X\cap Y\cap Z$ is given by $$\{wxw^{-1}y^{s(w)}x^{-1}y^{-s(w)}xz^{t(w)}x^{-1}z^{-t(w)}\}_{w\in \langle y,z\rangle - \langle y\rangle - \langle z \rangle}$$

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