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.
 A: As mentioned in the other answer, what you claim is not true and counterexamples can be found. However, if $S$ is a generating set for $G$, then you can prove that
$$\{[s,z]^g: s \in S, z \in S, g \in G\}$$
is a generating set for $[G,G]$.
A: This won't be true in general. Suppose $G$ is generated by a set $S$ with only two elements. Then $[S,S]$ contains only two non-identity elements, inverse to one another, and so $\langle[S,S]\rangle$ will be cyclic.
But, for example, the symmetric group $S_n$ is generated by a set of two elements, but its commutator subgroup $A_n$ is not cyclic if $n>3$.
A: 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 $\ge 2$. 
A: Another example of a finitely generated group $G = \langle S \rangle$ in which the commutator subgroup is not finitely generated is the wreath product of two infinite cyclic groups, which is 2-generated and metabelian.
A: A similar but true statement is that $[G,G]/[G,G,G]$ is generated by (the image of) $[S,S]$.
This is just a rephrasing of Mikko's answer. $[G,G] = \langle [g,h] : g,h \in G \rangle$ but $g$ and $h$ are products of elements of $S \cup S^{-1}$ and from $$\begin{align*}
[st,h] &=[s,h][s,h,t][t,h] \\
[g,st] &= [g,t][g,s][g,s,t] \\
[s^{-1},h] &= [h,s][h,s,s^{-1}] \\
[g,s^{-1}] &= [s,g][s,g,s^{-1}] \\
\end{align*}$$
we get that $[G,G]/[G,G,G] = \langle [s,t] [G,G,G] : s,t \in S \rangle$.
In other words, for nilpotent groups you can do approximately what you want, where “approximately” has the standard meaning of modulo terms of a central series (the lower central series).
