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Denote by $B(m,n)$ the free Burnside group with $m$ generators and order $n$, i.e., $$B(m,n):=\langle x_1,\ldots,x_m\mid w(x_1,\ldots,x_m)^n=1\ \forall w\rangle,$$ where $w(x_1,\ldots,x_m)$ is any word.

On the other hand, the von Dyck group admits the presentation $$D(a,b,c):=\langle x,y\mid x^a=y^b=(xy)^c=1\rangle.$$ So, $B(2,n)$ is "covered" by $D(n,n,n)$, in the sense that the latter projects over the former, since it has the same generators but less relations: $x$, $y$, and $xy$ are just particular words $w(x,y)$.

Question 1: Has the group epimorphism $D(n,n,n)\to B(2,n)$ been observed before? And, if yes, to which purpose?

Question 2: Any idea how to generalize it for $m>2$, i.e., how to speak of "higher-dimensional von Dyck groups"?

Thanks in advance!

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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

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