How to write the commutator subgroup in terms of the generators of the group?

Question

Let $G=\langle\ S\ |\ R\ \rangle$ be a finitely presented group. The commutator subgroup of $G$ is the group generated by $\{[a,b]\ |\ a,b\in G\}$ and is denoted by $[G,G]$, where $[a,b]=aba^{-1}b^{-1}$.

1. Is it true that $[G,G]$ is generated by $\{[s,t]\ |\ s,t\in S\}$ ?

I need to show that any $[a,b]$ can be written as a word in $[s,t]$ but I am not sure if it can be done.

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The above is a generalized question which I asked myself when considering the particular case below -

Let $W=\langle\ S\ |\ R\ \rangle$ where, $$S=\{s_1,\cdots,s_{2n}\}$$ $$R=\left\{s_i^2 : 1\le i\le 2n\right\}\bigcup\left\{(s_is_j)^2:1\le i,j\le 2n,j\neq i+n\right\}$$

2. Then is $[W,W]$ is generated by $\{[s_i,s_{i+n}]\ |\ 1\le i\le n \}$ ?

Thank you.

Answer 1

The answer is no. If $G$ is a free group of rank greater than $1$ then $[G,G]$ is not even finitely generated.

If what you were asking were true, then for any $2$-generator group $[G,G]$ would be cyclic, which is clearly not the case.

It is true in your specific example. In the case $n=1$, $W$ is the infinite dihedral group with $[W,W]$ generated by $[s_1,s_2]$.

For general $n$, $W$ is the direct product of its subgroups $\langle s_i,s_{i+n}\rangle$, so $[W,W]$ is free abelian of rank $n$, with generators $[s_i,s_{i+n}]$.

Answer 2

No, 1. is not true. There are groups $G$ where the commutator subgroup $[G,G]$ may contain elements which are not commutators. One example are the free groups. For instance, in the free group of rank $4$, freely generated by $x$, $y$, $z$, and $w$, the element $[x,y][z,w]$ of the commutator subgroup cannot be written in the form $[a,b]$ for some $a,b$ in the group.