You can prove this using algebraic topology$^{\dagger}$: in fact, it follows from an exercise in Hatcher's book (see this related math.SE question). Basically: use Reidemeister-Schreier. I have written out a "full-ish" proof - but if you are comfortable with covering spaces then this proof is ridiculously quick and elegant!

The covering-spaces proof that $F_2^{\prime}=[F(a, b), F(a, b)]\cong F_{\infty}$ is as follows: The free group $F(a, b)$ is the fundamental group of a bouquet of two circles:

Using $\widetilde{X}$ to denote the universal cover, by standard covering-space theory the derived subgroup $F_2^{\prime}=[F(a, b), F(a, b)]$ is isomorphic to the fundamental group of $\widetilde{X}/F_2^{\prime}$. Now, $\widetilde{X}$ is just the Cayley graph of $\langle a, b; -\rangle$,

while $\widetilde{X}/F_2^{\prime}$ is the Cayley graph of $\mathbb{Z}\times\mathbb{Z}$,

Then, $\pi_1(\widetilde{X}/F_2^{\prime})$ is the group with presentation $\langle V; R\cup T\rangle$ where $V$ is the set of $1$-cells (edges) of $\widetilde{X}/F_2^{\prime}$, $R$ is the set of $2$-cells of $\widetilde{X}/F_2^{\prime}$, and $T$ is a spanning tree for $\widetilde{X}/F_2^{\prime}$. Now, $\widetilde{X}/F_2^{\prime}$ has no $2$-cells, and taking any spanning tree we are still left with infinitely many edges (for example, take the tree $T=\bigcup_{y\in\mathbb{Z}}\{(x, y): x\in\mathbb{Z}\}\cup\{(0, x): x\in\mathbb{Z}\}$). Hence, we have an infinitely generated group with no relations, so $F_2^{\prime}=[F(a, b), F(a, b)]\cong F_{\infty}$ as required.

In fact, there is nothing special about $\mathbb{Z}\times\mathbb{Z}$ here:

**Thereom.** Let $G$ be an infinite group. If $\phi: F_n\rightarrow G$ is a surjection from the free group of rank $n$, $\infty>n>1$, to $G$ then $\ker(\phi)$ is either trivial or is not finitely generated.

*Proof*. Suppose $\ker(\phi)$ is non-trivial, and write $N:=\ker(\phi)$. Then, similar to the above, $\widetilde{X}/N$ is a graph; it is essentially the Cayley graph for $G$ given by the generating set $\operatorname{im}(\phi(X))\subset G$ (note the subtlety if there are $x_1, x_2\in X$ such that $\phi(x_1)=\phi(x_2)$). Now, as above, $N=\pi_1(\widetilde{X}/N)$ is the group with presentation $\langle V; R\cup T\rangle$ where $V$ is the set of $1$-cells (edges) of $\widetilde{X}/N$, $R$ is the set of $2$-cells of $\widetilde{X}/N$, and $T$ is a spanning tree for $\widetilde{X}/N$. As $\widetilde{X}/N$ is a graph it has no $2$-cells. It remains to show that for any spanning tree $T$ of $\widetilde{X}/N$ there are infinitely many edges $e$ in $(\widetilde{X}/N)\setminus T$. As $\ker(\phi)$ is non-trivial there exists a non-empty word $W\in F(X)$ such that $\phi(W)=1$. This corresponds to a loop in $\widetilde{X}/N$, and indeed at each vertex $v$ there exists such a loop $\mathcal{l}_{W, v}$ of length $d:=|W|$. Note that the tree $T$ cannot contain every edge of the loop $\mathcal{l}_{W, v}$. As $G$ is infinite there exists an infinite sequence of pairwise non-equal vertices $v_1, v_2, \ldots$ of $\widetilde{X}/N$ such that no two vertices $v_i, v_j$ are linked by a path of length $\leq2d$. Therefore, the loops $\mathcal{l}_{W, v_i}$ and $\mathcal{l}_{W, v_j}$ contain no edges in common. Hence, there are infinitely many edges $e$ in $(\widetilde{X}/N)\setminus T$. We therefore have an infinitely generated group with no relations, so $N$ is not finitely generated as required.

$^{\dagger}$ Serios recently reminded me of this proof in the comments to this fine math.SE answer.

Let $x=[a^2,b]=a^{-2}b^{-1}ab$ and $y=[a,b]=a^{-1}b^{-1}ab$.

Conisder all words of the form $uvw$, where $u,v,w \in \{x,x^{-1},y,y^{-1}\}$, and $v\ne u^{-1}$, $v \ne w^{-1}$. You will find that there is no such product $uvw$ in which the whole of $v$ cancels.

So the reduction of an reduced word $u_1u_2\cdots u_k$ with $u_i \in \{x,x^{-1},y,y^{-1}\}$ has length at least $k$ and so cannot reduce to the identity. So the subgroup generated by $x$ and $y$ is free.

The Nielsen proof that a subgroup of a free group is free involves an algorithmic method for reducing a generating set of a subgroup to a free generating set, so you could apply that method to examples like this.

## Best Answer

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.