4
$\begingroup$

Can't find a flaw in that proof:

Induction by the length of derived series.

Base: if $[G, G]=e$ then the group is abelian...

Assume that statement is true for n-1.

We have group $G$ with the derived series length $n$, and a subgroup $H$.

[G, G]=G' has derived series of length $n-1$, so H \cap G' is finitely generated.

H/(H\cap G') is a subgroup of finitely generated abelian subgroup G/G'. So $H$ is finitely generated.

  • 0
    @User1729: If he'd proven that $H\cap G'$ is finitely generated, then the finite generation of both $H/(H\cap G')$ and $H\cap G'$ would imply that $H$ is finitely generated: more generally, if $N$ and $G/N$ are finitely generated, then so is $G$: pick $g_1,\ldots,g_m$ whose images generated $G/N$, and $m_1,\ldots,m_k$ that generated $N$. Given $g\in G$, white $gN=g_{i_1}^{a_1}\cdots g_{i_{\ell}}^{a_{\ell}}N$, hence $g=g_{i_1}^{a_1}\cdots g_{i_{\ell}}^{a_{\ell}}n$ for some $n$, and now express $n$ in terms of $m_1,\ldots m_k$. The problem was elsewhere.2012-02-10

3 Answers 3

5

You are implicitly assuming that G' is finitely generated, but this need not be the case.

For example, consider the lamplighter group $\mathbb{Z}_2 \wr \mathbb{Z}$. This group is a semidirect product $\mathbb{Z}_2^\omega \rtimes \mathbb{Z}$, where $\mathbb{Z}_2^\omega$ is the direct sum of infinitely many copies of $\mathbb{Z}_2$, and $\mathbb{Z}$ acts on $\mathbb{Z}_2^\omega$ by translation. This group is clearly solvable, and it is has a standard two-element generating set. However, the commutator subgroup is a translation-invariant subgroup of $\mathbb{Z}_2^\omega$, and is therefore not finitely generated.

1

In order to apply the induction hypothesis to H\cap G', you need G' to be finitely generated. Did you prove that the derived subgroup of a finitely generated solvable group is necessarily finitely generated?

HINT: The commutator subgroup of the free group of rank $2$ is free of infinite rank; what happens if you mod out by [G',G']? You get the free metabelian group of rank $2$. Prove that its commutator subgroup is free abelian of infinite rank.

1

Another example is the Baumslag-Solitar group $G=\langle x,y\mid xyx^{-1}=y^2\rangle$ then the normal subgroup generated by $y$ is $A=Z[1/2]$ (dyadic rationals) an infinitely generated abelian group. In this case $A=G'$ the commutator subgroup.