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.
Finite index subgroup of Fuchsian group of the first type is also Fuchsian of the first type
group-theoryhyperbolic-geometry
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The claim that Iwaniec makes is simply false; the mistake originates in Lemma 8 in:
C. L. Siegel, Discontinuous groups. Ann. of Math. (2) 44 (1943), 674--689.
The mistake in the proof is that Siegel incorrectly assumes that every Fuchsian group has a finitely-sided fundamental polygon. (I looked closely in the paper, Siegel's definition of a Fuchsian group only requires discreteness, no assumptions on fundamental polygons.)
The correct result is:
Theorem. The following are equivalent for a Fuchsian group $\Gamma< PSL(2,R)$ of the first kind:
$\Gamma$ is finitely generated.
$\Gamma$ is a lattice in $PSL(2,R)$, i.e. ${\mathbb H}^2/\Gamma$ has finite area.
One (equivalently every) fundamental polygon of $\Gamma$ has only finitely many sides.
See for instance Theorem 10.1.2 in
A.F.Beardon, "Geometry of Discrete Groups", Springer Verlag, 1983.
The example that Lee Mosher gave you is definitely valid although a proof that the uniformizing group $\Gamma$ is of the 1st kind would take some nontrivial effort involving extremal length considerations. An easier example is the following.
Take a torsion-free Fuchsian subgroup $\Gamma_1< PSL(2,R)$ such that ${\mathbb H}^2/\Gamma$ has finite area. Take a nontrivial normal subgroup $\Gamma_2< \Gamma_1$ of infinite index. It is a general fact about general Fuchsian groups that the limit set of a nontrivial normal subgroup $\Gamma_2\triangleleft \Gamma_1$ equals the limit set of $\Gamma_1$. Since our group $\Gamma_1$ is of the first kind, so is the normal subgroup $\Gamma_2$. On the other hand, area is multiplicative under isometric coverings between Riemannian surfaces: If $S_2\to S_1$ is an isometric covering of degree $d$ of Riemannian surfaces, then $$ Area(S_2)=d Area(S_1). $$ (The same holds for manifolds in all dimensions, but the area would mean volume.) In our case, the degree of the covering map $$ {\mathbb H}^2/\Gamma_2\to {\mathbb H}^2/\Gamma_1 $$ equals the index $|\Gamma_1: \Gamma_2|=\infty$. Hence, $$ Area({\mathbb H}^2/\Gamma_2)=\infty. $$
An element of a Fuchsian group is either elliptic, parabolic, or hyperbolic.
If $g$ is elliptic, then the group generated by $g$ is finite.
If $g$ is parabolic then there is only one limit point for $g^n z$, independent of $z$. For example if $g(z) = z+1$, then this limit point is $\infty$ regardless of what $z$ we start at. So then the group generated by $g$ is not Fuchsian of the first kind.
If $g$ is hyperbolic, then there are only 2 limit points for $g^n z$, independent of $z$. For example, if $g(z) = 2z$ then these limit points are 0 and $\infty$. So again the group generated by $g$ is not Fuchsian of the first kind.
Best Answer
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.
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, $$\lim_{i \to \infty} \delta_i \cdot z = \lim_{i \to \infty} \gamma_i \cdot z = x $$