Problem:
Let $A=\begin{bmatrix}0 &-1\\1 &0\end{bmatrix}, B=\begin{bmatrix}0 &1\\-1 &1\end{bmatrix}\in GL(2,\Bbb Q)$, let $N=\langle \pm I\rangle$ and let $M=\langle A,B\rangle/N$
Prove that $M\cong SL(2,\Bbb Z)/N$.
Here is my proof.
Let $T=AB$, I claim that $SL(2,\Bbb Z)=\langle A,T\rangle$. If it is true, $SL(2,\Bbb Z)=\langle A,T\rangle=\langle A,B\rangle$; hence, $M=SL(2,\Bbb Z)/N$.
First, I observe that $$T^m\begin{bmatrix} a&b\\c&d\end{bmatrix}=\begin{bmatrix}a-mc&b-mc\\c&d\end{bmatrix}$$ and $$A\begin{bmatrix} a&b\\c&d\end{bmatrix}=\begin{bmatrix} -c&-d\\a&c\end{bmatrix}.$$ Pick any $M\in SL(2,\Bbb Z)$ of the form $\begin{bmatrix} a&b\\c&d\end{bmatrix}$. $a=qc+r$ for some $q$ and $0\leq r<|c|$; hence, the upper left entry of $T^qM$ is $r$. Now, the first column of $ST^qM$ is $[-c\quad r]^\top$ and $-c=q_1r+r_1$ for some $0\leq r_1<r$. Therefore, the upper left entry of $T^{q_1}ST^qM$ is $r_1$. This procedure can be done until the lower left entry of the matrix is 0. Since its determinant has to be 1, it must have the form $\begin{bmatrix} \pm1&m\\0&\pm\end{bmatrix}$, which is $T^m$ or $-T^{-m}$. Hence, we prove that $SL(2,\Bbb Z)=\langle A,T\rangle$.
My question:
Why did I actually "prove" that $M=SL(2,\Bbb Z)/N$? I am supposed to prove that they are isomorphic. Is my proof correct?
This problem is from Rotman's Advanced Modern Algebra 2nd Edition, Problem 1.86
Any help?