Fuchsian group with parabolic element I'm interested in this problem: let $\Gamma \subset PSL(2, \mathbb{R})$ a Fuchsian group (i.e. it is a discrete subgroup) which contains the trasnformation $\gamma \colon z \mapsto z+1$ then the quotient $\mathbb{H}/_{\Gamma}$ is not compact.
Here is my attempt: I consider a sequence $\{t_n \}_{n \in \mathbb{N}}$ of increasing real numbers which gives the sequence $\{ z_n\}$ in $\mathbb{H}$ where $z_n:= it_n$. I can consider the segment in $\mathbb{H}$ $s_n$ with endpoints $z_n$ and $\gamma (z_n)$ which in the quotient provide some loops with basepoint $[z_n]=[\gamma(z_n)]$. 
In the hyperbolic plane the segment $s_n$ is easily seen to have length $\frac{1}{|t_n|}$ which converges to zero thus in the quotient the length of the images $[s_n]$ should also coverge to zero and maybe I could prove that the collection of paths $[s_n]$ provides a sequence with no converging subsequence ensuring that the quotient $H/_{\Gamma}$ is not compact.
The problem is that I'm not able to conclude this reasoning because:
1)I don't see how to find this sequence;
2)I don't know explicitly $\Gamma$ hence I cannot be sure if the segments $s_n$ under the quotient have some problematic behavior which undermines my argument.
Thanks for any help or hint.
 A: Jørgensen's inequality says that if two elements $A, B \in SL_2(\mathbb{C})$ generate a non-elementary discrete group, then
$$
|\mathrm{Tr}(A)^2-4| + |\mathrm{Tr}(ABA^{-1}B^{-1})-2| \ge 1
$$
Assuming that $A=\begin{bmatrix} 1& 1 \\ 0 & 1\end{bmatrix}$ and $B=\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ (with $a,b,c,d \in \mathbb{R}$ and $ad-bc=1$) generate a non-elementary discrete group, one has $\mathrm{Tr}(A)=2$ and $\mathrm{Tr}(ABA^{-1}B^{-1})=2+c^2$, so the inequality gives $c^2 \ge 1$.
Then you know that
$$
 \Im \frac{az + b}{cz+d} = \frac{\Im z}{|cz+d|^2} \le \frac{1}{c^2 \Im z} \le \frac{1}{\Im z} < 1
$$
for $\Im z > 1$, so the map associated to the matrix $B$ maps the region above the line of imaginary part $1$ into its complement.
If $A$ and $B$ generate an elementary discrete group in $SL_2(\mathbb{R})$, then $B=\begin{bmatrix} 1 & b \\ 0 & 1 \end{bmatrix}$ with some $b \in \mathbb{R}$. By discreteness there is a smallest positive $b$ that occurs in this group, and taking this together with the first part you can see that the region defined by $0<\Re z < b$ and $\Im z > 1$ is disjoint from all its translates by non-identity group elements, so the quotient is not compact.
A: Here's an approach using hyperbolic geometry. Equip $\mathbb H$ with the hyperbolic metric $$d_\mathbb Hs^2=\frac{|dz|^2}{(\text{Im }z)^2}$$ and set $R=\mathbb H/_{\Gamma}$ which makes $R$ a Riemann surface as well as a hyperbolic surface. Assume for the sake of contradiction that $R$ is compact and $\gamma_0(z)=z+1 \in \Gamma$.
Consider for $a>0$, the line segment $L_a$ joining $ia$ and $1+ia$ in $\mathbb H$. Let $C_a=\pi(L_a)$ be the corresponding closed curve in $R$. We show that the hyperbolic length of $C_a$, namely $l(C_a)\rightarrow 0$ as $a\rightarrow \infty$.  Let $\tilde \gamma_a(t)=t+ia$ and set $\gamma_a(t)=\pi\circ \tilde \gamma_a (t)$. Since $\pi$ is a local isometry by definition of the metrics, $$l(C_a)\leq \int_0^1 |\dot\gamma_a(t)|_{\mathbb H}dt=\int_0^1|d\pi_{\tilde \gamma_a(t)}(\dot{\tilde \gamma_a}(t))|_{\mathbb H}dt=\int_0^1| \dot{\tilde \gamma_a}(t)|_{\mathbb H}dt=\int_0^1\frac{1}{a}dt=\frac{1}{a}\rightarrow 0$$as $a\rightarrow \infty$.
But if $R$ is compact, then we have a sequence $a_n \uparrow\infty $ and $\pi(ia_n)\rightarrow p_0 $ in $R$ . Choose any hyperbolic geodesic ball around $p_0$ say $B$. Then we have $C_{a_n}\subset B$ for all $n$ large. It follows  that any neighbourhood of $p$ is not evenly covered by $\pi$ since $ia_n,ia_n+1\in \pi^{-1}(B)$ map to the same point.
This is a contradiction.
