I have a really quick question. I am interested in $G=SL\_2(\mathbb{Z}[1/p\_1,...,1/p\_n])$, where $p\_1$,..., $p\_n$ are prime numbers. Since $G$ is a subgroup of $SL\_2(\mathbb{R})$, it acts in the hyperbolic plane $\mathbb{H}$. My question is: are all the isotropy groups of this action finite?


Thank you



The action of an S-arithmetic group on the hyperbolic plane
This answer addresses the question posed in the comments. 


It follows from Bass-Serre theory that every finite subgroup of $G$ is conjugate into $SL\_2(\mathbb{Z})$ (see Section II.1.4 of Serre's book [Trees](http://link.springer.com/book/10.1007/978-3-642-61856-7)). So maximal finite subgroups will be order $4$ or $6$, generated by an element of trace $0$ or $1$, and thus property (M) holds. More specifically, to every $p\_i$ is associated an action on a tree. A finite subgroup of $SL\_2(\mathbb{Q})$ has to fix a vertex of this tree, and by Bass-Serre theory is conjugate into $SL\_2(\mathbb{Z}\_{(p\_i)})$. Applying this to each $i$, one sees that a finite group is conjugate into $SL\_2(\mathbb{Z})$, for which the finite subgroups are well-known to be cyclic. 


The second property (NM) is false in general. Consider the matrix $\left[\begin{array}{cc}a & b \\-b & a\end{array}\right]$, $a^2+b^2=1$, then it normalizes the maximal finite subgroup generated by $\left[\begin{array}{cc}0 & 1 \\-1 & 0\end{array}\right]$. Such matrices are plentiful coming from Pythagorean triples. 


On the other hand, for certain denominators $p\_i$, I think property (NM) will hold. If $-1$ and $-3$ are not quadratic residues $(\mod p\_i)$ for all $i$, then I think that the stabilizers of maximal finite subgroups will be trivial. Equivalently, $p\_i\equiv -1 (\mod 12)$ for all $i$. 


