A proof that BS(1,2) is not polycyclic

Question

I am looking for examples of finitely generated solvable groups that are not polycyclic. In Wikipedia Baumslag-Solitar group $BS(1,2)$ is an example. But how to prove this fact?

Answer 1

For a polycyclic group, every subgroup is finitely generated, in particular the commutator subgroup is finitely generated. Now assume that $BS(1,2)$ is polycyclic. Hence its derived subgroup is finitely generated. However, the derived subgroup is isomorphic to the group of $2$-adic rationals, i.e., the group of all rationals with denominators powers of $2$. This group is not finitely generated, a contradiction.

More generally, all Baumslag-Solitar groups $BS(1,m)$ for $m\ge 2$ are metabelian, but not polycyclic, because the commutator subgroup $\mathbb{Z}[\frac{1}{m}]$ is not finitely generated.

Answer 2

The Baumslag-Solitar group $BS(1, 2) = \mathbb{Z} \ltimes \mathbb{Z}[\frac{1}{2}]$ under the isomorphism $a \to (0, 1), b \to (1, 0)$, where the generator $g$ of the left term acts by $g.1 = 2$. But $\mathbb{Z}[\frac{1}{2}]\subset BS(1, 2)$ is clearly not finitely generated, so $BS(1, 2)$ is not polycyclic.