# Solvable group which has a subgroup that is not finitely-generated

Let $a= \begin{pmatrix}1&0 \\ 1&1 \end{pmatrix}$ and $b= \begin{pmatrix}2&0 \\ 0&1 \end{pmatrix}$. If $G=\langle a,b\rangle \subseteq GL(2,\mathbb Q)$.

• How to prove that $G$ is solvable?
• What is the commutator subgroup of $G$? And why it can't be finitely generated ?

The entire lower diagonal subgroup $\begin{pmatrix} q & 0 \\ r & s \end{pmatrix}$ of $GL(2,\mathbb{Q})$ is solvable, because the map to the abelian group $(\mathbb{Q}-0) \oplus (\mathbb{Q}-0)$ defined by $\begin{pmatrix} q & 0 \\ r & s \end{pmatrix} \mapsto (q,s)$ has abelian kernel $\begin{pmatrix} 1 & 0 \\ r & 1 \end{pmatrix}$. So your group $G$ is solvable too, since it is a subgroup of a solvable group.
With a little work one can show that the general element of your group $G$ has the form $\begin{pmatrix} 2^k & 0 \\ m / 2^l & 1 \end{pmatrix}$ where $k,l$ range over the integers and $m$ ranges over the odd integers, and that its commutator subgroup is the group of all such matrices with $k=0$, namely $\begin{pmatrix} 1 & 0 \\ m / 2^l & 1 \end{pmatrix}$. This is isomorphic to the additive group of "dyadic rational numbers", meaning those rational numbers of the form $m / 2^l$. This group is not finitely generated, because if you take any finite subset $m_1 / 2^{l_1}$, …, $m_I / 2^{l_I}$, the group generated by that subset consists solely of rational numbers having a common denominator $2^L$ for some integer $L$, and so that subgroup cannot be the entire group of dyadic rationals.