It seemingly is a fact that a discrete subgroup of $\mathrm{PSL}_2(\mathbb{R})$ acting on hyperbolic space cannot be compact if it contains a parabolic element.
I was wondering if the following proof works:
Let $\Gamma$ be our group and $T \in \Gamma$ be parabolic. Then we can conjugate $\Gamma$ in $\mathrm{PSL}_2(\mathbb{R})$ such that $T':=gTg^{-1} \in g\Gamma g^{-1}=: \Gamma'$ is of the form $z\overset{T'}{\mapsto} z+a$ for some $a\in\mathbb{R}$.
By discreteness we can choose such $T$ such that $a$ is minimal amongst parabolic elements of $\Gamma$.
Therefore, we can choose a fundamental domain $D$ of $\Gamma'$ such that the lines $i\mathbb{R}_+$ and $a + i\mathbb{R}_+$ are part of the boundary in the sense that there is some $y$ such that for $P_y := \{ c + id : c \in [0,a], d>y\}$, we have $P_y \cap \overline{D} = P_y$, where $\overline{D}$ is the closure of $D$.
For the sequence $x_n := i(y+n) \in D $, $x_n \rightarrow \infty$ and thus $D$ is not compact. It follows that $\Gamma$ is not compact as well.
I feel like the iffy step is choosing such a Fundamental domain.
Does it need additional justification?
Is additional justification also needed if we start the proof under the assumption of $\Gamma$ being finitely generated or of finite volume?
Similar question, different approach: Fuchsian group with parabolic element
First of all, a discrete subgroup $\Gamma$ of a locally compact topological group $G$ is called cocompact (not compact!) if $G/\Gamma$ is compact. In the setting of subgroups of $G= PSL(2, {\mathbb R})$, a subgroup $\Gamma< G$ is cocompact if and only if $\Gamma$ acts properly discontinuously on ${\mathbb H}^2$ with compact quotient space ${\mathbb H}^2/\Gamma$.
Lemma. If a discrete subgroup $\Gamma< PSL(2, {\mathbb R})$ is cocompact, then $\Gamma$ contains no parabolic elements.
Proof. Suppose that $\Gamma$ contains a parabolic element $\gamma$. Then there exists a sequence $x_n\in {\mathbb H}^2$ such that $$ \lim_{n\to\infty} d(x_n, \gamma x_n)=0. $$ Since $\Gamma$ is cocompact, there exists a compact subset $C\subset {\mathbb H}^2$ which intersects every $\Gamma$-orbit in at least one point. Thus, there exists a sequence $g_n\in \Gamma$ such that $g_n(x_n)=y_n\in C$. Since $C$ is compact, after pasing to a subsequence, we can assume that the sequence $(y_n)$ converges to some point $y\in C$. Note that for $\alpha_n:= g_n \circ \gamma\circ g_n^{-1}\in \Gamma$ we have $$ \lim_{n\to\infty} d(y_n, \alpha_n(y_n))= d(x_n, \gamma(x_n))= 0. $$ In particular, for every neighborhood $U$ of $y$ there exists $N$ such that for all $n\ge N$, $\alpha_n(y)\in U$. But, since $\Gamma$ acts on the hyperbolic plane properly discontinuously, there exists a neighborhood $U$ of $y$ such that for every $g\in \Gamma$ either $gU\cap U=\emptyset$ or $g(y)=y$ (i.e. $g$ is elliptic). But the elements $\alpha_n$ are conjugate to $\gamma$, hence, are not elliptic. This is a contradiction. qed