# Can the finiteness of a Burnside group with two generators be checked algorithmically by using Fuchsian von Dyck groups?

BIG EDIT of the previous question "Coverings of the free Burnside groups", never answered.

In the paper http://link.springer.com/article/10.1007/BF00046586 (last section) there is an interesting statement concerning the finiteness of the 2-generator Burnside groups $B(2,n)$ (let's assume $n\geq 4$). More precisely, it is claimed that, once $B(2,n)$ is identified with the quotient of the Fuchsian von Dyck group $D(n,n,n)$ (see, e.g. von dyck groups and solvability for its definition) by its $n^\textrm{th}$ powers subgroup $K_n:=D(n,n,n)^n$, then $B(2,n)$ acts on the Riemann surface $\frac{\mathbb{H}}{K_n}$ and, hence, if the latter is compact, the group is finite.

QUESTION 1: what is known about the surface $\frac{\mathbb{H}}{K_4}$? Should be compact, since $B(2,4)$ has order 1024.

(The "planar case" $\frac{\mathbb{R}^2}{K_3}$ is a nice toy model: one obtains a tiled hexagon whose edges correspond to the 27 elements of $B(2,3)$; see, e.g., my paper.)

The above-cited paper is 30-year old, never quoted, and the author himself wouldn't give me explanations. I've found closely related questions on MO (see, e.g., this and this one), but I cannot convince myself about the effectiveness of the proposed algorithm for checking the finiteness of $B(2,n)$, based on the computation of the fundamental domain of $K_n$. In particular, the paper proposes to construct the standard polygon of $K_n$ and check whether its sides close a compact domain.

QUESTION 1 (equivalent formulation): is the fundamental domain of $K_4$ compact?

The proposed algorithm is the following: we fix an origin $o\in\mathbb{H}$, we start to enumerate the elements of $K_n$, and, for any $g\in K_n$ generated by the enumeration, we construct the geodesic line $\ell_g$ perpendicular to the geodesic segment $[o,g\cdot o]$ through its middle point; if the $\ell_g$'s produced so far form a closed polygon, then the algorithm stops.

QUESTION 2 (main): is this algorithm effective? has been used, e.g., to check the still unknown finiteness of $B(2,5)$?

Any link or reference to current algorithmic methods for the two-generator Burnside problem are quite welcome!

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The most natural generalization is: Take a finite graph on $m$ vertices; declare each vertex $v$ to be a generator; take relators $v^n=1$ for every vertex and $(vw)^n=1$ for every pair of vertices corresponding to an edge. Then von Dyck groups $D(n,n,n)$ correspond to the case of a graph consisting of a single edge. All this is quite trivial, the question is what are you going to do with such groups. –  Misha Jan 17 '13 at 6:38
Misha, thanks for the information: does the group you've mentioned have a name? And yet Question 1 is still unanswered... Since von Dyck groups can be seen as groups of isometries of constant-curvature surfaces, the epimorphism I'm interested in allows to regard free Burnside groups as groups of transformations of suitable orbifolds: in this perspective, the generalization of the von Dyck group I'm looking for should give a group of isometries of some geometrical object. Does this graph-construction of your admit such an interpretation? –  G_infinity Jan 17 '13 at 17:33

The answer to your Question 1 is certainly 'yes', for the reason you already explained: $\mathbb{H}^2/K_4$ is a Riemann surface, and its quotient by the finite group $B(2,4)$ is compact.

I don't know about Question 2, and I haven't looked at the paper that you link to, but it seems worth pointing out that the algorithm you are interested in is an instance of a much more general phenomenon. Indeed, finite groups are recursively recognizable among all recursively presented groups. More precisely:

Proposition 1: There is an algorithm that takes as input a recursively presented group $\Gamma$ and terminates if and only if $\Gamma$ is finite.

(Note that $B(m,n)$ is clearly recursively presentable.)

Let me quickly sketch the proof. There is a short exact sequence

$1\to R\to F\to \Gamma\to 1$

where $F$ is the free group on the generators of $\Gamma$. The assertion that $\Gamma$ is recursively presented means that the subgroup $R\subseteq F$ is recursively enumerable, and $\Gamma$ is finite if and only if $R$ is of finite index in $F$.

So Proposition 1 is a special case of the more general:

Proposition 2: There is an algorithm that takes as input a recursively enumerable subgroup $R$ of a free group $F$ and terminates if and only if $R$ is of finite index in $F$.

Although you can think about this in terms of Riemann surfaces, I prefer to think about graphs. Putting the algorithm into a combinatorial setting will make it easier to code, and any geometric group theorist will tell you that the geometries of Riemann surfaces and graphs are very similar, anyway.

Let's think of $F$ as the fundamental group of a finite graph $X$, and let $Y\to X$ be the covering map corresponding to $R$. Then Proposition 2 boils down to the following facts:

1. You can recursively build $Y$, using the famous 'folding algorithm' from Stallings' paper 'Topology of finite graphs', exhibiting a sequence of immersions $Y_n\to X$ that factor as $Y_n\to Y\to X$.

2. You can recognize if at any time the immersion $Y_n\to X$ is an actual covering map.

This is all quite straightforward and standard. I certainly don't see any particular obstruction to attempting the computation in the case of $B(2,5)$

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Perhaps I should add something to explain the connection with the cited paper. Since $B(2,n)$ is a quotient of the von Dyck group $D(n,n,n)$, which is the fundamental group of a hyperbolic (for $n>3$) 2-orbifold $O$, you can replace $F$ by $D(n,n,n)$ and $X$ by $O$, and play the same game. –  HJRW Sep 7 at 10:15
Your almost cleared out all my doubts - yet I need time to fully digest your answer. Now, let us focus on your last sentence: "I certainly don't see any particular obstruction to attempting the computation in the case of $B(2,5)$". Believe me, I've been trying for years to find out what is the state-of-the-art concerning the finiteness of $B(2,5)$, but in vain. Can you point out somebody to whom I may ask? Why there are a lot of computers employed to discover the 'last' digit of $\pi$, or the 'biggest' prime, but none cares about finiteness of $B(2,5)$? Is there at least the algorithm written? –  G_infinity Sep 8 at 5:13
@G_infinity, various experts on Burnside groups are (or have been) active on MO. Mark Sapir is one. The difference with the search for digits of $\pi$ or prime numbers is that these are finite problems: there are upper bounds on the time it will take to find the next one, and the task is to reduce the run time to something reasonable. On the other hand, if $B(2,5)$ is infinite then the algorithm I described will run forever. –  HJRW Sep 8 at 22:31