Let $G$ be a non-uniform lattice Fuchsian group and let $P$ be a Dirichlet region for $G$. In particular $G$ has parabolic elements, $P$ is not compact and has finite area. We are in the unit disc. Is the following statement true? Is the proof correct? Errors? Counterexamples? Thanks.

STATEMENT: If $G$ is a free group then $P$ is an ideal polygon, that is it has all its vertices on the boundary.

PROOF: let $v$ be a vertex of $P$ and assume that $v$ is an interior point of the disc. Look at the tassellation $g(P)$ with $g \in G$ around the vertex v. Let $P_0, ... , P_n$ be the elements of the tassellation sharing the vertex $v$, ordered counterclockwise, where $P_0 = P$. Such elements are finitely many because the tassellation is locally finite. Moreover $n\geq 2$ because all angles are less than $\pi$. We have elements $g_0 ,..., g_n$ of $G$ with $g_k ( P_k ) = P_{k+1}$ for $k = 0 , ... , n$, where $P_{n+1} = P_0$. This implies that the interior of $g_n...g_1g_0 (P_0)$ overlaps with the interior of $P_0$. The last condition implies that $g_n...g_1g_0 = e$ the identity of $G$, thus we have a non trivial relation (recall $n \geq 2$ ), which contradicts that $G$ is free.

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