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Suppose we have a group $G$ which can be generated by two elements $x$, $y$. Call $H$, $K$, $L$ the subgroups of $G$ generated by $x$, $y$ and $y^{-1}x^{-1}$, respectively.

With these data, we can build up a graph $\mathcal{G}$, by declaring the vertices of $\mathcal{G}$ to be the right-cosets of $H$, $K$ and $L$, and by adding an edge between two right-cosets if they have nonempty intersection. Obviously, $G$ acts faithfully on $\mathcal{G}$.

Geometrically, $\mathcal{G}$ can be seen a planar graph, made by triangles (the vertices of each triangle corresponds to right-cosets of different $H$, $K$, and $L$). Then, acting, e.g., by $x$ means to rotate $\mathcal{G}$ around the vertex $H$. Furthermore, if $\mathcal{G}$ is represented on a constant-curvature surface, then the angle of such a rotation is $2\pi/|H|$. Similarly for $y$ and $y^{-1}x^{-1}$. In other words, $G$ is represented in the group of automorphisms of a regular tiling.

I've heard about this construction years ago during an undergraduate class, but back then I wasn't interested (I even forgot who was the lecturer). Lately, I rediscovered this, and I've spent few days searching the web, but I couldn't find anything resembling what I've explained above. Posting my question here is my last hope! Somehow, it looks such a simple idea, that it is hard to believe that it cannot be found anywhere!

More precisely, my question is the following:

Is there a standard construction to associate a graph $\mathcal{G}$ with a group with two generators $G$, in such a way that

  • $\mathcal{G}$ can be realized as a regular tiling made of triangles on a constant-curvature surface, and

  • $G$ can be seen as a subgroup in the group of automorphisms of $\mathcal{G}$?

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Something's strange here. How do you know that $\mathcal{G}$ is planar? Note that there are 2-generator groups that do not preserve a regular tiling of the plane. – HJRW Apr 16 '12 at 9:58
Where are you getting triangles from? On a side note, this construction is very standard, and is an instance of a Tits geometry. – Steve D Apr 16 '12 at 11:46
@Steve: The triangles are triples of cosets $Ha, Ka, La$ for all $a$. Different $a$ give different triangles provided the intersection of $H, K, L$ is trivial. There is something strange with the question: some phrases seem to mean that $H,K,L$ are finite and some phrases seem to indicate that there are many $H,K,L$. – Mark Sapir Apr 16 '12 at 12:53
Thank you very much, Steve D, you solved my problem! BTW, triangles appear as those complete subgraphs whose three vertices are right-cosets of different types. And, HW, you are right - but I had in my mind a tiling of a constant-curvature surface (plane, sphere, hyperbolic plane). Also, by "regular" I meant not made of regular triangles, but obtained by applying a rigid motion to a prescribed, not necessarily regular, triangle. Sorry for being so imprecise! – Giovanni Moreno Apr 16 '12 at 13:35
@Mark Sapir: thank for your remark about triangles. And, yes, I forgot to mention that there are precisely three subgroups $H$, $K$, and $L$, and also that these subgroups are of finite order. – Giovanni Moreno Apr 16 '12 at 13:39
up vote 4 down vote accepted

I think what you are trying to remember is the triangle group $\langle x,y\mid x^k=y^l=(xy)^m=1\rangle$ (see Wiki). Depending on whether $1/k+1/l+1/m$ is less than, equal to or greater than 1, the group corresponds to a tessellation of a hyperbolic plane, Euclidean plane or a 2-sphere. The tesselation is constructed as you described.

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