Finite order elements in $\mathbb{P}(GL_{2}(\mathbb{R}))$ I was playing around with the following problem: Find a function that composed with itself three times gives the identity. After a bit of trial and error I came up with the function $\frac{1}{1-x}$.
Naturally I then tried to find a function that had order four in the group of real functions. Unfortunately to no avail. I then restricted the problem to rational function of degree $1$ in both the numerator and denominator and noticed that those function are isomorphic to the group $\mathbb{P}(GL_{2}(\mathbb{R}))$.
I lastly tried to find matrices in $\mathbb{P}(GL_{2}(\mathbb{R}))$ that had order $4$ but didn't manage to find any.
My questions are the following:
Is there a general formula for a matrix in $\mathbb{P}(GL_{2}(\mathbb{R}))$ with order $n$?
How many distinct cyclic groups of order $n$ are there for each $n$? (distinct as in $\langle-x\rangle \neq \langle\frac{1}{x}\rangle$ but $\langle \frac{1}{1-x} \rangle = \langle\frac{x-1}{x}\rangle$)
 A: The rotation matrix
$$ R(\theta)=\begin{bmatrix}\cos\theta & -\sin\theta \\ \sin\theta & \phantom{-}\cos\theta \end{bmatrix} $$
has order $n$ in $\mathrm{GL}_2\mathbb{R}$ if $\theta=2\pi /n$. If $n$ is odd this remains true for $\pm R(\theta)$ in $\mathrm{PGL}_2\mathbb{R}$, otherwise if $n$ is even then $\pm R(\theta)$ has order $n/2$ in $\mathrm{PGL}_2\mathbb{R}$. (Try to prove these things!)
A: A key fact about elements of $\mathbb P(GL_2(\mathbb R))$ which you probably know is that when you start with a matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix} \in GL_2(\mathbb R)$ and then translate it into function format $\frac{ax+b}{cx+d}$ an amazing thing happens: multiplication of matrices translates into composition of functions. To put this more precisely, the group $\mathbb P(GL_2(\mathbb R))$ is isomorphic to the group of functions of that form (assuming $ad-bc \ne 0$)
So, if you can cook up a matrix $M \in GL_2(\mathbb R)$ which has finite order then its image in the quotient group $\mathbb P(GL_2(\mathbb R))$, has finite order. But you have to be a bit careful because the order of a finite order element might go down when you pass from $GL_2(\mathbb R)$ to its quotient $\mathbb P(GL_2(\mathbb R))$: some power of $M$ might be a diagonal matrix different from the identity matrix.
This is easily controlled, however, by always choosing $M$ to have determinant $+1$, so its only possible diagonalizable powers are $\pm \begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix}$.
To put this another way: Always work in $SL_2(\mathbb R)$ and its quotient $PSL_2(\mathbb R) = SL_2(\mathbb R) / \pm \text{Id}$. What you are taking advantage of here is that there is a natural isomorphism $\mathbb P(GL_2(\mathbb R)) \approx PSL_2(\mathbb R)$.
So now one needs to cook up finite order elements of $SL_2(\mathbb R)$. There's a lot of them, namely the rotation matrices $M_\theta = \begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix}$ where $\theta$ is a rational multiple of $2\pi$.
In general if $\theta = \frac{\pi}{k}$ then $M^k = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}$ which is the smallest power of $M$ that lies in the kernel $\pm\text{Id}$. Taking $k=4$ you get
$$M = \begin{pmatrix} \cos(\pi/4) & \sin(\pi/4) \\ - \sin(\pi / 4) & \cos(\pi/4) \end{pmatrix} = \begin{pmatrix} \sqrt{2}/{2} & \sqrt{2}/{2} \\ - \sqrt{2}/{2} & \sqrt{2}/{2} \end{pmatrix}
$$
and therefore $M$ has order $4$ in $PSL(2,\mathbb R) \approx \mathbb P(GL_2(\mathbb R))$.
When you rewrite this matrix $M$ as a function and clear out the constant factor of $\sqrt{2}/{2}$ in the numerator and denominator you get
$$f(x) = \frac{x+1}{-x+1}
$$
which has order $4$.
And for other even orders $n$ simply take $\theta = \pi/n$, and for odd orders $n$ take $\theta = 2\pi/n$.

To answer your final question, there are uncountably many distinct finite cyclic subgroups of $SL(n,\mathbb R)$: Any one of them is conjugate to uncountably many distinct subgroups.
This follows quickly from the fact that the circle group
$$C = \left\{\begin{pmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{pmatrix} \mid \theta \in \mathbb R\right\}
$$
is a malnormal subgroup of $PSL(2,\mathbb R)$, meaning that $M C M^{-1} \cap C = \{Id\}$ for all $M \in PSL(2,\mathbb R)-C$. In particular $C$ is its own normalizer in $PSL(2,\mathbb R)$, and it has infinite index, hence it has uncountably many distinct conjugates and intersects each of them trivially.

On the other hand, for each order $n$, every finite cyclic subgroup of $PSL(2,\mathbb R)$ is conjugate to a single example, namely the example discussed above with $\theta = \pi/n$ if $n$ is even, and $\theta = 2\pi/n$ if $n$ is odd. Proving this requires new concepts. The proof I like best takes advantage of the fact that $PSL(2,\mathbb R)$ may be identified with the group of orientation preserving isometries of the hyperbolic plane, and each finite cyclic subgroup may be identified with a finite rotation group around a point of the plane.
