Show that there exist infinitely many matrices $\gamma \in \Gamma_1$ with $\gamma x_1 = x_2$ I am trying to solve the following exercise:
For $x \in \bar{\mathbb{Q}} := \mathbb{Q} \, \cup \, \{ \infty \}$ and $\gamma = 
  \left[ {\begin{array}{cc}
   a & b \\
   c & d \\
  \end{array} } \right] \in \Gamma_1 \ $where $\Gamma_1 = SL_2(\mathbb{Z}),$ we define $\gamma x \in \bar{\mathbb{Q}}$ by
$$ \gamma x = \left\{
\begin{array}{ll}
\frac{a}{c}, & \textrm{if} \, x = \infty \\
\infty, & \textrm{if} \, x = -\frac{d}{c} \\
\frac{ax+b}{cx+d}, & \textrm{else}
\end{array}
\right. $$
Let $x_1,x_2 \in \bar{\mathbb{Q}}.$ Show that there exists infinitely many matrices $\gamma \in \Gamma_1$ with $\gamma x_1 = x_2$
My approach would be to look at all the cases and see what comes out of it. I just don’t know how to draw backs to the matrix and why there are so many of them. Could someone give me an approach? Or how to solve such tasks in general? Thank you in advance!
 A: Hint Remember that you must ensure that $a d - b c = 1$. Start by finding infinitely many such matrices $A$ such that $\gamma_A x_1 = 0$. Then find infinitely many such matrices $B$ such that $\gamma_B 0 = x_2$. Then use the matrices $B A$.
Note that $\gamma_B 0 = x_2$ is easy to solve because it reduces to $\gamma_{B}^{-1} x_2 = 0$, so that there is only one problem.
First problem
Let us assume that $x = p/q \in \mathbb{Q}$ with $p$ and $q$ coprimes. Let $\gamma=\pmatrix{a& b\cr c& d}$ be a matrix such that $\gamma x=0$. Note that $a$ and $b$ are coprime because $a d - b c= 1$, it follows easily that $p/q = -b/a$ and finally $(a, b) = \pm (q, -p)$. Let $(u, v)$ be a Bezout pair such that $u p + v q = 1$. The condition on the determinant implies that for some $k\in \mathbb{Z}$ one has $c = u - k q$ and $d = v + k p$, finally
\begin{equation}
\gamma = \pm \begin{pmatrix}q & -p\cr u- k q& v + kp\end{pmatrix}
\qquad k\in\mathbb{Z}
\end{equation}
Second problem With the same assumptions, let us look for a matrix $\gamma$ such that $\gamma 0 = x$. Then one has $\gamma^{-1}x = \gamma^{-1}\gamma 0 = 0$ and $\gamma^{-1}$ is a matrix of the previous form. It follows that here
\begin{equation}
\gamma = \pm \begin{pmatrix}v+k p & p\cr -u+ k q& q\end{pmatrix}
\qquad k\in\mathbb{Z}
\end{equation}
Initial problem
Assuming that $x_1,x_2\in \mathbb{Q}$ take two matrices $\gamma_1$
and $\gamma_2$ having the previous forms (for $x_1=p_1/q_1$ and $x_2=p_2/q_2$). The matrices $\gamma = \gamma_2\gamma_1$ solve the initial problem. There are infinitely many of them by letting $k_1$ or $k_2$ vary.
It remains to handle the special cases where one or both $x_i$s can be equal $\infty$.
