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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 ?
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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.

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  • $\begingroup$ Thanks Lee, very nice explanation! But is there any such strategy to find the general form of the group and the commutator? $\endgroup$ – Ronald Dec 4 '15 at 4:19
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    $\begingroup$ Perhaps you might be able to formulate a new question along those lines. But you should specify which particular groups you are asking about, and how those groups are given (as you did in this question). Otherwise, the answer to your question will be a simple "no". $\endgroup$ – Lee Mosher Dec 4 '15 at 13:54
  • $\begingroup$ At least I like to see the way of thinking and steps from a professional. In the example above, I think I can deal with it but not in the professional way. If you like to include this work here, it would be great! and if you like I can post another question.. thanks a gain $\endgroup$ – Ronald Dec 4 '15 at 14:56

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