Find direct isometries In hyperbolic space. Let H be the upper half of the complex plane model for hyperbolic geometry. let $p,q$ be points in the plane. show how to find a direct Isometry F of H such that $F(p)=q$. Now suppose $p\prime,q\prime$ are points unit hyperbolic distance from $p,q$ respectively. Show how to find a direct isometry G such that $G(p)=q,G(p\prime)=q\prime$
I know the direct isometries of H are mobius transforms corresponding to the elements of SL(2,$\mathbb{R}$). This would suggest to find F I simply need to find a matrix in SL(2,$\mathbb{R}$) which transforms p to q. This is relatively easy to do for a specific example. I am confused as to how the question changes for finding G. It seems as though you must now simply find a matrix that satisfies both conditions. Im sure Im missing something but I dont know what. Any advice would be much appreciated.
 A: You might be thinking that the image of a point in the complex plane (viewed as a vector in $\mathbb{R}^2$) under a Mobius transformation is the same as its image under the corresponding linear map.  But this is not true!  
So, while it is easy to write down a matrix which takes one pair of vectors, $(p,q)$, to another pair of vectors, $(p',q')$, this is not what you are seeking.
A standard approach to this type of problem is to use families of nice transformations to move one point (or a pair of points) to the other point (or pair of points).  In the upper half plane model, the isometries $f(z) = z + c$, $c$ real, and the isometries $g(z) = kz$, $k$ real and positive, are the two families which are easiest to use.  There are also rotations.  You won't need rotations to solve the first problem (and this is why it is easier than the second problem).  Hint for the first problem: Find a mobius transformation mapping $p$ to $i$.  Then use two such transformations to map $p$ to $q$.
For the second problem, you need to use rotations or inversions in circles.  These are harder to write down explicitly in the upper half plane model.  But you see them easily in the disk model.  Using the transformation $M(z) = (z-i)/(z+i)$ you can go from the upper half space model to the disk model.  You should compute the inverse of $M(z)$ so you can go back and forth between the two models.  The simplest rotations in the disk model have the form $h(z) = az$, where $a$ is a unit complex number.
Depending on how explicit you want the formula to be, this can quite challenging.  But if you are just trying to understand why this is always possible to do, then you have all the ingredients now.
