Representation of cell location in hyperbolic plane I want to represent an order-5 square tiling (image from Wikipedia; more text below image):

Obviously for a simple grid I can uniquely refer to a given square by its (x,y) position relative to the origin, in integer coordinates. To get to adjacent cells, one may simply increment or decrement x or y. However, on the hyperbolic plane, things get messy, and it's unclear how to refer to a given cell.
Is there a better visual representation of the order-5 square tiling that would allow intuitive understanding of how to refer to individual cells, and failing that, is there a simple method to give each cell an index (such as (x,y) for the Euclidean plane) that makes it easy to find its neighbors?
 A: A partial answer: For any square there are four “translations” of the hyperbolic plane, each sliding the given square onto one of its four neighbours. These translations generate a group which is finitely generated, namely, by the two translations sending one given square into either one of two adjacent neighbours. (Their inverses will map the given square to the neighbours on the opposite side.) Unfortunately, the given group is non-abelian, but it might be possible (I think) to describe it as the free group on two generators modulo a finite number of relations. I don't know how to determine those relations, though.
This group, call it $G$, acts transitively on the set of squares $S$ in the tiling. So you might expect to be able to label the squares by fixing one ($s_0\in S$), and then labeling $gs_0$ by $g\in G$. But that won't quite work, for the mapping $gs_0\mapsto g$ is not well defined: Translate a square five times, around each of the five edges meeting at one point, and it comes back rotated $90^\circ$. So let $H$ be the stabilizer of $s_0$: It is cyclic of order $4$, and most likely a non-normal subgroup of $G$. Then the quotient $G/H$ labels the squares in the tiling.
Not the answer you hoped for, I am sure, but I don't think it can be done a lot more simply. (And I don't know how to complete the answer.)
