Let $G$ be a free group. Then $G/G^{(n)}$ ($G^{(n)}$ is the $n$th derived subgroup.) acts on $G^{(n)}/G^{(n+1)}$ by conjugation, which makes $G^{(n)}/G^{(n+1)}$ a $\mathbb{Z}[G/G^{(n)}]$-module. What can I say about this module? Namely, I wonder whether $G^{(n)}/G^{(n+1)}$ is a free $\mathbb{Z}[G/G^{(n)}]$-module. If I want to study about this subject, what would be a good reference?
$G/G^{(n)}$ is the free solvable group of class $n$. The module mentioned in your question is projective (a submodule of a free module of rank = rank of $G$).
You mean that it is $\mathbb{Z}[G\G^{(n)}]$-module? I have MKS's book but I don't know where to look at. –  user6569 Feb 24 '12 at 5:02
$\mathbb{Z}[G/G^{(n)}$-module. In MKS see Chapter 5. But the more recent papers I mentioned may help more. –  Mark Sapir Feb 24 '12 at 5:15