Finite index subgroup of Fuchsian group of the first type is also Fuchsian of the first type Just trying to see why the statement in the title is true. For $\Gamma$ a Fuchsian group of the first type on the complex upper half plane $\mathbb{H}$, a subgroup $\Gamma'$ of finite index is also Fuschian of the first type. A Fuchsian group on $\mathbb{H}$ is a discrete subgroup of PSL$_2(\mathbb{R})$. To say that $\Gamma$ is of the first type means that every $x\in\mathbb{R}\cup\{\infty\}$ is the limit of an orbit $\Gamma z$ for some $z\in\mathbb{H}$ (using the topology of the Riemann sphere $\mathbb{C}\cup\{\infty\}$). So for every $x\in\mathbb{R}\cup\{\infty\}$ we can take a sequence $\gamma_i\in\Gamma,i\in\mathbb{N}$ and some $z\in\mathbb{H}$ so that $\lim_{i\rightarrow\infty}\gamma_iz=x.$ In Topics In Classical Automorphic Forms it is said that $\Gamma'$ is also Fuschian of the first type. I can't see an obvious subsequence of $\gamma_i$ contained in $\Gamma'$ that would work, but perhaps there is one, or I'm looking at it from the wrong angle.
 A: Here's a proof. I'm going to represent the hyperbolic plane $\mathbb H$ using the Poincaré disc model $\mathbb D$ instead of the upper half plane model. So with $\mathbb H = \mathbb D$ we also have $\partial\mathbb H = S^1$. Convergence in $\mathbb H \cup \partial \mathbb H = \mathbb D \cup S^1 = \overline{\mathbb D}$ is therefore governed by the Euclidean metric $d_{\mathbb E}$, so it is easier to talk about convergence to boundary points using $\mathbb D$ than it is using the upper half plane model.
Now I'll state a geometric fact that one can use to resolve this problem.

Lemma: For any $z \in \mathbb H$ there exists $r > 0$ such that for any $\gamma \in \Gamma$ there exists $\delta \in \Gamma'$ such that $d_{\mathbb H}(\gamma \cdot z,\delta \cdot z) < r$.

In this statement, the distance function $d_{\mathbb H}$ refers to distance in $\mathbb H$. In words, this lemma says that the entire $\Gamma$-orbit of $z$ is contained in the $r$ neighborhood of the $\Gamma'$ orbit of $z$. This lemma is true for any group acting by isometries on any metric space; I'll leave the proof as an exercise for you to ponder (hint: first pick right coset representatives).
To see how this proves what you want, consider a point $x \in \partial\mathbb H = S^1$. Using that $\Gamma$ is of the first type, choose $z \in \mathbb H$ and a sequence $\gamma_i \in \Gamma$ such that $\lim_{i \to \infty} \gamma_i \cdot x = z$. Applying the lemma, choose $r > 0$ and a sequence $\delta_i \in \Gamma'$ such that $d(\gamma_i \cdot z, \delta_i \cdot z) < r$.
We want to prove that $\lim_{i \to \infty} \delta_i \cdot z = x$. To do this, we use a fact about the Poincaré disc model $\mathbb D$: as the center of a hyperbolic ball of radius $r$ approaches $S^1$, the Euclidean diameter of that ball approaches zero. So, knowing that $\lim_{i \to \infty} \gamma_i \cdot z = x \in S^1$, and that $d_{\mathbb H}(\gamma_i \cdot z,\delta_i \cdot z) < r$, it follows that $\lim_{i \to \infty} d_{\mathbb E}(\gamma_i \cdot z,\delta_i \cdot z ) = 0$. Therefore,
$$\lim_{i \to \infty} \delta_i \cdot z = \lim_{i \to \infty} \gamma_i \cdot z = x
$$
