The Triangle group is the group of reflections of by triangles in a tessellation by congruent triangles. If the angles of the triangles are $\frac \pi l$, $\frac \pi m$, and $\frac \pi n$, then the group is isomorphic too
$$\Delta(l,m,n) = \langle a,b,c \mid a^2=b^2=c^2=(ab)^l = (bc)^n = (ca)^m = 1 \rangle$$
There is of course a group like this corresponding to any tessellation, which will be a subgroup of the proper Isometry group.
My question is, how do we find a group presentation of a tessellation by congruent $k$-gons?
Concretely, if $P$ is a convex polygon in a plane $X$ (a complete simply connected Riemannian surface of constant curvature, i.e. the Euclidean plane, the hyperbolic plane or a round sphere) with the consecutive angles $\frac{\pi}{n_1},..., \frac{\pi}{n_k}$, then the group $G$ of isometries of $X$ generated by isometric reflections in the edges of $P$ has the presentation $$ \langle s_1,...,s_n| s_i^2, (s_i s_{i+1})^{n_i}, i=1,...,k\rangle. $$