The action of the Modular Group $PSL(2, \mathbb{Z})$ on $\mathbb{Q} \cup \infty$

Question

How to prove that the action of the Modular Group $PSL(2, \mathbb{Z})$ on $\mathbb{Q} \cup \infty$ is transitive, i.e., for every $q_1,q_2 \in \mathbb{Q} \cup \infty$, there is $g\in PSL(2, \mathbb{Z})$ such that $g(q_1)=q_2$.

Answer 1

To recall, the action of the modular group $\text{PSL}(2,\mathbb Z)$ on $\mathbb Q \cup \{\infty\}$ is defined for $g = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ (with determinant $ad-bc=1$) and for $q \in \mathbb Q \cup \{\infty\}$ by the formula $$g(q) = \frac{aq+b}{cq+d} $$

Step 1: First consider $q_1 = 0$ and any $q_2 = \frac{m}{n}$, with $\frac{m}{n}$ in lowest terms, of course. Use the fact that $\gcd(m,n)=1$ to find $a,b \in \mathbb Z$ such that $am+bn=1$.

Then let $g = \begin{pmatrix} b & m \\ -a & n \end{pmatrix} \in \text{SL}(2,\mathbb Z)$ (check that the determinant is, actually, $+1$, by using the equation $am+bn=1$).

Now compute: $$g(0) = \frac{b \cdot 0 + m}{-a \cdot 0 + n} = \frac{m}{n} $$

This works except in the very special case that $\frac{m}{n} = \frac{1}{0}$, but in that case just use $g = \begin{pmatrix}0 & 1 \\ -1 & 0 \end{pmatrix}$

Step 2: Next, consider arbitrary $q_1,q_2$.

Use Step 1 to choose $g_1 \in \text{SL}(2,\mathbb Z)$ so that $g_1(0)=q_1$

Use Step 1 again to choose matrix $g_2 \in \text{SL}(2,\mathbb Z)$ so that $g_2(0)=q_2$.

It follows that $g_2 \circ g^{-1}_1(q_1) = g_2(0)=q_2$.

Step 3: Since the group operations on $\text{PSL}(2,\mathbb Z)$ are just matrix operations, you can then compute the matrix $g_2 \circ g^{-1}$ using ordinary matrix inversion and matrix multiplication.