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.
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.