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I am considering finitely generated subgroup $G=\langle A,B,C\rangle$ of $GL_2(\mathbb{Q})$ such that $A,B,C$ all have the upper triangular form $$A=\begin{pmatrix}a_1 & a_2 \\ 0 & a_3\end{pmatrix},\;B=\begin{pmatrix}b_1 & b_2 \\ 0 & b_3\end{pmatrix},\;C=\begin{pmatrix}c_1 & c_2 \\ 0 & c_3\end{pmatrix}.$$

Are there known conditions on the parameters $a_i,b_i,c_i$ such that $G$ is isomorphic to the free group $F_3$? Inversely, many conditions can be established, such that this is not the case, by simple matrix multiplication to obtain, e.g., $A^2BC^{-1}=Id_2$. But I do not manage do find the correct relations to make the group free.

I look in particular at the example $$A=\begin{pmatrix}4 & 1 \\ 0 & 1\end{pmatrix},\;B=\begin{pmatrix}2 & -1 \\ 0 & 3\end{pmatrix},\;C=\begin{pmatrix}4 & -1 \\ 0 & 3\end{pmatrix}.$$ So the question is in this case: Is the group generated by these three examples isomorphic to $F_3$.

Jfischer
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    Hint: The group of upper triangular matrices is solvable. – Moishe Kohan May 01 '24 at 13:08
  • The matrices in your example are not in $GL_2(\mathbb{Z})$ btw. – Daniel Arreola May 01 '24 at 13:13
  • @DanielArreola Why? They are invertible $2\times 2$ matrices if I am not completely mistaken. But the generated group is not a subgroup is $GL_2(\mathbb{Z})$ that is true. $A$ is not invertible as an integer matrix... Sorry, have to change the question then... – Jfischer May 01 '24 at 13:15
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    @Jfischer Their inverses are not integer matrices. – Daniel Arreola May 01 '24 at 13:16
  • @MoisheKohan I had to change the underlying field to $\mathbb{Q}$, the upper triangular matrices should still be solvable, right? – Jfischer May 01 '24 at 13:20
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    You can use use any commutative ring you like and also you can use matrices of any size. https://math.stackexchange.com/questions/1753556/why-is-the-group-of-unit-upper-triangular-matrices-solvable – Moishe Kohan May 01 '24 at 13:24

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