Is $PSL_2(\mathbb Z)$ a Fuchsian group of the first kind?

Question

We know that as a discrete subgroup of $PSL_2(\mathbb R)$, $PSL_2(\mathbb Z)$ is a Fuchsian group. But how to prove/disprove that it is of the first kind. i.e. if every point on the extended real line is its limit point of some orbit?

Answer 1

Hint: Prove that every rational point on the real line is fixed by a parabolic element of $PSL(2,{\mathbb Z})$. A sub-hint: Think first about stabilizers of nonzero elements of ${\mathbb Z}^2$ in $SL(2,{\mathbb Z})$.

Answer 2

Let's me add the following details to Moishe Kohan's answer and comments and please let me know if anything is wrong:

It is easy to see by Bezout theorem about coprime numbers that every rational point on $\mathbb R$ is $PSL(2,{\mathbb Z})$ equivalent to $\infty$. Since $\infty$ is a limit point of sequence $T^n z$ where $T:z\to z+1$. So $\mathbb Q \cup \infty$ is contained in the limit set of $PSL(2,\mathbb Z)$.

Now here is the harder part:

Any irrational point $x\in \mathbb R$ can be approximated by a squence of rational numbers $(r_n)$. Although $(r_n)$ don't belong to the orbit $PSL(2,\mathbb R).i$, for each $n$, there exists a sequence $(g_{nk}.i)$ converging to $r_n$ as $k \to \infty$. Now the diagonal sequence $(g_{nn}.i)$ converges to $x$ as $n\to \infty$.