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?


1 Answer 1


$G/G^{(n)}$ 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 can read about free solvable groups in Hanna Neumann's "Varieties of groups", and in Karras-Magnus-Solitar "Combinatorial group theory". Among newer papers, see for example Romanovskiĭ, N. S. Coproducts of rigid groups. Algebra Logika 49 (2010), no. 6, 803--818, 845--846, 848; translation in Algebra Logic 49 (2011), no. 6, 539–550 and references there. See also Lichtman, A. I. Automorphism groups of free soluble groups. J. Algebra 174 (1995), no. 1, 132–149. See also this paper, especially p. 8 where they define Magnus embeddings and Fox derivatives.

  • $\begingroup$ 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. $\endgroup$
    – hopflink
    Feb 24, 2012 at 5:02
  • $\begingroup$ $\mathbb{Z}[G/G^{(n)}$-module. In MKS see Chapter 5. But the more recent papers I mentioned may help more. $\endgroup$
    – user6976
    Feb 24, 2012 at 5:15

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