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$$ \langle a,b,c|a^2=b^2=c^2=(ab)^p=(bc)^q=(ca)^r=1\rangle =\Delta(p,q,r) $$ This is a presentation of a triangle group $\Delta(p,q,r)$, a special kind of Coxeter group.

EDIT In fact, these are called extended triangle groups, by G. Jones and D. Singerman in Maps, hypermaps and triangle groups...

What about the following presentation: $$ \langle a,b,c|a^2=b^2=c^2=(ab)^p=(bc)^q=(ca)^r=(abc)^s=1\rangle $$ Do these groups have a name and where are they treated?

The presentation in question are motivated by this and that...

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  • $\begingroup$ Have you come across generalised triangle groups? What I call a triangle group is the group of transformations of the tiled plane (no reflections); Wikipedia calls these von Dyck groups. Such groups have presentation $\langle x, y, x; x^p, y^q, z^r, xyz\rangle$, and here $ab=x$, $bc=y$ and $ca=z$. A generalised triangle group is a group with presentation $\langle x, y, x; x^p, y^q, z^r, W(x, y, z)\rangle$. Fine and Rosenberger wrote a whole book motivated by these groups and their generalisations (one-relator products). Jim Howie has also written a lot about them. $\endgroup$ – user1729 Jul 8 '16 at 10:36
  • $\begingroup$ @user1729 you mean $<x,y,{\bf z};\dots>$? Do you have some explicite references or links to them? $\endgroup$ – draks ... Jul 8 '16 at 13:17
  • $\begingroup$ Yes, $x, y, z$. When I google the first link is to a paper of Jim Howie (macs.hw.ac.uk/~jim/preprint27.pdf). The references look extensive. I am pretty sure he gave a talk on these at a conference in 2012 I was at. The book of Fine and Rosenberger is "Algebraic generalizations of discrete groups: a path to combinatorial group theory through one-relator products". $\endgroup$ – user1729 Jul 8 '16 at 14:02
  • $\begingroup$ @user1729 why not posting this as an answer... $\endgroup$ – draks ... Jul 12 '16 at 5:11
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I haven't come across a name for this family in full generality, but the special case in which $p=2$ was defined and studied by Coxeter in his paper

H. S. M. Coxeter, The abstract groups $G^{ m, n, p}$, Trans. Amer. Math. Soc. 45 (1939), 73-150.

where (in your notation) the group is called $G^{q,r,s}$.

Also, when $s$ is even, your group has a subgroup of index $2$ with presentation $\langle x,y \mid x^p=y^q=(xy)^r=[x,y]^{s/2} \rangle$.

These groups were studied in the same paper by Coxeter, and denoted $(p,q,r;s/2)$.

Both of these families have been extensively studied since then, in particular concerning their finiteness. They are generally infinite for sufficiently large values of the parameters, and there is just a handful of remaining cases for which their finiteness is still unknown.

A few years ago Havas and I showed, using a big computer calculation, that $(2,3,13;4)$ is finite of order $358\,848\,921\,600$. So your group with $(p,q,r,s) = (2,3,13,8)$ has twice that order.

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    $\begingroup$ +1 great the paper is available online and for free $\endgroup$ – draks ... Feb 4 '16 at 14:57
  • $\begingroup$ hmmm: when I replace $A,B,C$ in the definition of $G^{m,n,p}$ by $A=x/y, \; B=y/z,\; C=z/x$ and use that $x^2=y^2=z^2=1$ (so they are self-inverse) I get: $\langle (xy)^m=(yz)^n=(zx)^p=x^2=y^2=z^2=(\frac xy \frac yz \frac zx)^2=1 \rangle$, where the triple product is trivial. Now it looks just like the triangle group. What have I done (wrong)? $\endgroup$ – draks ... Feb 4 '16 at 19:24
  • $\begingroup$ Maybe you are interested: The presentation in question are motivated by this and that... $\endgroup$ – draks ... Feb 4 '16 at 19:50
  • $\begingroup$ Help you with what? $\endgroup$ – Derek Holt Feb 9 '16 at 9:51

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