Generator for $[G,G]$ given that $G = \left$. If $S'$ is a generating set for $G$, let $S=S' \cup \{s^{-1} \,| \, s\in S\}$, then is the set $[S,S] = \{[s,z] \,|\, s,z \in S\}$ a generating set for the commutator subgroup $[G,G]$? I want to believe that this is true and "almost" have a proof, whose weak point might be the thing that can be used to generate a counter example. Clearly, if the following were true, then the answer to the question at the beginning would be positive.

Claim
$[s_1 \dots s_n, z_1 \dots z_m] \in \left<[S,S] \right>$, for $s_1, \dots, s_n, z_1, \dots, z_m \in S$.

First, show that $[s_1, \dots, s_n, z] \in \left<[S,S]\right>$, where $s_1, \dots, s_n, z \in S$. To do this, use induction on $n$. The case for $n=1$ is trivial. For $n >1$, 
\begin{align*}
[s_1 \dots s_n, z] =& (s_1\dots s_n)^{-1} z^{-1} s_1 \dots s_n z \\
                =& s_n^{-1} (s_1 \dots s_{n-1})^{-1} z^{-1} s_1 \dots s_n z \\
                =& s_n^{-1} (s_1 \dots s_{n-1})^{-1} z^{-1} s_1 \dots s_{n-1} z s_n [s_n, z] \\
                =& s_n^{-1} [s_1 \dots s_{n-1}, z] s_n [s_n,z] \\
                =& [s_1 \dots s_{n-1}, z]^{s_n} [s_n,z]
\end{align*}
By induction hypothesis, $[s_1 \dots s_{n-1}, z], [s_n,z] \in \left<[S,S]\right>$. Not sure how to complete this here, so let's just move on for now and pretend that $[s_1 \dots s_{n-1}, z]^{s_n} \in \left<[S,S]\right>$ by some miracle.
Next, we show $[z, s_1 \dots s_n] \in \left<[S,S]\right>$ where $z = z_1 \dots z_m$, and $z_1, \dots, z_m, s_1, \dots, s_n \in S$. We proceed by induction on $n$. The case $n=1$ is done by the above. Note for $n>1$, everything follows the same format as above except we have the order reversed. Thus we are done.

Consider $G=F_n$, free group on $n\ge 2$ generators. Then the commutator subgroup $C<G$ is normal, nontrivial and of infinite index. Therefore, $C$ cannot be finitely generated. Here is the proof:
Let $N<F_n$ be a nontrivial normal subgroup of infinite index. Realize $G$ as the fundamental group of a finite graph $X$. Then $N$ is the fundamental group of an infinite regular cover of $X$. Such a cover has to be an graph with infinite first Betti number (once you get one circuit, you move it around using deck-transformations to obtain infinitely many disjoint circuits). Hence, $N$ has abelianization of infinite rank and, hence, cannot be finitely generated. 
A similar argument applies to fundamental groups of surfaces of genus $\ge 2$. 
