Groups containing Fuchsian groups Dear all,
my question is: Suppose we have a Fuchsian group G and another group H, containing G with finite index. Must H be a Fuchsian group? Can you give a proof? Or do you know any counter examples?
Best
Ali K.
[Edit: Ali K's answer below indicates that he is interested in the case in which $H$ is also a subgroup of $\operatorname{PSL}_2(\mathbb{R})$, so I have answered the question with that additional hypothesis.  However the question as stated, though it may not be the OP's intent, is probably more interesting, so I do not wish to officially change it. -- PLC]
 A: Well, if your definition of a Fuchsian group is a finitely generated, discrete subgroup of $PSL_2(\mathbb{R})$, then the answer is 'no'.  $H$ could be the direct product of $G$ with an arbitrary finite group $B$, and this isn't in general a Fuchsian group.  (If $G$ acts discretely on the hyperbolic plane then the stabiliser of a point would have to contain $B$, but point stabilisers in $PSL_2(\mathbb{R})$ are cyclic or dihedral.)
More interestingly, it is true that Fuchsian groups can be characterised by their actions on their boundaries, as proved by Tukia, Casson--Jungreis and Gabai.  In particular, I believe it follows that any group quasi-isometric to a Fuchsian group is virtually a Fuchsian group.
A: Ali K has further asked how to show that if $G \subset H \subset \operatorname{PSL}_2(\mathbb{R})$ with $G$ Fuchsian and $[H:G] < \infty$, then $H$ is Fuchsian.
Recall that a subgroup of $\operatorname{PSL}_2(\mathbb{R})$ is Fuchsian iff it is discrete in the subgroup topology.  Thus the following suffices:
Lemma: Let $H$ be a Hausdorff topological group with a finite index discrete subgroup $G$.  Then $H$ is itself discrete.
Proof: $G$ is a locally compact sugroup of the Hausdorff topological group $H$, so $G$ is therefore closed.  Since $G$ has finite index in $H$, the complement of $G$ is a finite union of closed cosets of $G$, so $G$ is also open.  The trivial subgroup $\{e\}$ is open in $G$ and $G$ is open in $H$, so $\{e\}$ is open in $H$, i.e., $H$ is discrete.
Remark: The Hausdorff condition is necessary here (and obviously satisfied, since $\operatorname{PSL}_2(\mathbb{R})$ is Hausdorff): otherwise let $H$ be a nontrivial finite group endowed with the trivial topology (i.e., the only open subsets are $\varnothing$, $H$) and let $G = \{e\}$.
A: Let us suppose that H is indeed a subgroup of PSL2. How do you then prove the claim?
