subgroup of finitely generated solvable group is finitely generated (false proof) 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.
 A: 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.
A: 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.
A: 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.
