Given a regular tessellation, i.e. either a platonic solid (a tessellation of the sphere), the tessellation of the euclidean plane by squares or by regular hexagons, or a regular tessellation of the hyperbolic plane.


One can consider its isometry group $G$. It acts on the set of all faces $F$. I want to define a symmetric coloring of the tessellation as a surjective map from $c:F\rightarrow C$ to a finite set of colors $C$, such that for each group element $G$ there is a permutation $p\_g$ of the colors, such that $c(gx)=p\_g\circ c(x)$. ($p:G\rightarrow $Sym$(C)$ is a group homomorphism).


Examples for such colorings are the trivial coloring $c:F\rightarrow \{1\}$ or the coloring of the plane as an infinite chessboard.
The only nontrivial symmetric colorings of the tetrahedron, is the one, that assigns a different color to each face. For the other platonic solids there are also those colorings that assign the same colors only to opposite faces. 


So my question is: Does every regular tessellation of the hyperbolic plane admit a nontrivial symmetric coloring?


I wanted to write a computer program that visualizes those tessellations, but I didn't find a good strategy which colors should be used. So I came up with this question.



Symmetric colorings of regular tessellations
The answer is yes. Moreover, for every two different faces $A$ and $B$ there is a symmetric coloring assigning different colors to $A$ and $B$.


The isometry group $G$ is residually finite, hence here is a normal finite index subgroup $H$ of $G$ that contains no elements (except the identity) sending $A$ to itself or to $B$. Assign a unique color to each orbit of $H$.


The coloring symmetry condition is essentially he following: if $f\in G$ and faces $X$ and $Y$ are of the same color, then so are $f(X)$ and $f(Y)$. Since $X$ and $Y$ are of the same color, there exists $h\in H$ such that $h(X)=Y$. Since $H$ is normal, $h\_1:=f^{-1}hf\in H$. But $h\_1(F(X))=F(Y)$, hence $F(X)$ and $F(Y)$ are of the same color.


