Finite subgroups (not finite index, just finite) of the modular group The modular group is commonly described as the group of linear fractional transforms $z \mapsto \displaystyle \frac{az+b}{cz+d}$ with $a,b,c,d$ integers and $ad-bc = 1$. Of course, a great deal is known about the subgroups of the modular group of finite index. But what about just subgroups of finite size?
We know that the modular group is generated by the maps $S: z \mapsto \displaystyle\frac{-1}{z}$ and $T: z \mapsto 1+z$, with $S$ of order 2 and $ST$ of order 3, so these generate groups of size 2 and 3 respectively. But has any work been done on enumerating all possible groups of finite orders?
I would also be interested in knowing about finite subgroups of the extended modular group, which is defined as similar to the modular group but with determinants $ad-bc = \pm 1$.
 A: $\DeclareMathOperator{\PGL}{PGL} \DeclareMathOperator{\GL}{GL} \newcommand{\zz}{\mathbb{Z}} \newcommand{\rr}{\mathbb{R}}$
The answer can also be deduced from elementary algebra, without using that $\operatorname{PSL}(2,\zz)$ is a free product.
A finite subgroup of $\PGL(2,\zz)$ comes from a finite subgroup of $\GL(2,\zz)$. A finite subgroup of $\GL(2,\rr)$ is conjugate to a subgroup of the othogonal group $\mathrm{O}(2,\rr)$, which consists of rotations and reflections. Thus a finite subgroup of $\GL(2,\rr)$ is either cyclic or dihedral. (This is very classical and can be found, for instance, in Michael Artin's Algebra, I think.) But a rotation in dim $2$ is conjugate to an integer matrix if and only if it has order $1$, $2$, $3$, $4$ or $6$. It follows that the finite subgroups of $\GL(2,\zz)$ are either cyclic of orders $1$, $2$, $3$, $4$ or $6$ or dihedral of orders $4$, $6$, $8$, or $12$. This means that the finite subgroups of $\PGL(2,\zz)$ are either cyclic of order $2$ or $3$, or dihedral of order $4$ or $6$ ("dihedral of order $4$" := Klein four group).
Finally, since the normalizer of a rotation group of order $3$, $4$ or $6$ is finite, there are infinitely many conjugates of the subgroups containing one of these, and similarly for the groups containing reflections. Thus $\PGL(2,\zz)$ has also infinitely many subgroups of the above orders, but only finitely many up to conjugacy.
A: ${\rm PSL}(2,\mathbb{Z})$ is well-known to be the free product of a cyclic group of order $2$ and a group of order $3$. By general properties of amalgams, ( which can be found, eg in J-P. Serre's book "Trees") its finite subgroups are precisely those which can be conjugated into either one of the "factors" of the free product, so, in particular, all have order $1,2$ or $3$, and there are only three conjugacy classes of finite subgroups.
A: The classification of finite subgroups of $\text{PGL}_2(\mathbb C)$ is quite classical, and the classification of finite subgroups of $\text{PGL}_3(\mathbb C)$ only slightly less so. For the former, the possibilities are cyclic of arbitary order, dihedral of arbitrary order, the tetrahedral group of order 12, the octohedral group of order 24, and the icosohedral group of order 60. Section 4 of the following paper has a discussion and gives generators for an example of each group.
http://arxiv.org/pdf/1509.06670v2.pdf
Actually, since you've asked about SL$_2(\mathbb C)$, I guess I should also note that the map $\text{SL}_2(\mathbb C)\to\text{PGL}_2(\mathbb C)$ is surjective with kernel equal to $\pm I$, from which you should be able to deduce the answer to your question.
