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

  • $\begingroup$ The answer is no, in general. Why do you want to know? $\endgroup$ – Venkataramana Feb 23 '16 at 0:03
  • $\begingroup$ Actually I would like to verify that those groups satisfy the following two properties: (M) every finite subgroup is contained in a maximal finite subgroup and (NM) If M is a finite maximal subgroup then N(M) its normalizer it is equal to M. $\endgroup$ – user88026 Feb 23 '16 at 1:02

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). 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 $i$.


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