Torsion in triangle groups - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T19:38:14Z http://mathoverflow.net/feeds/question/70990 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/70990/torsion-in-triangle-groups Torsion in triangle groups AL 2011-07-22T14:33:59Z 2012-02-07T15:06:29Z <p>A triangle group has a presentation of the form,</p> <p>$G=\langle a, b; a^{\alpha}, b^{\beta}, c^{\gamma}, abc\rangle, \alpha, \beta, \gamma \geq 2$</p> <p>(I believe that these are also called von Dyke groups, or "ordinary" triangle groups, with triangle groups being something slightly different, but names are beside the point). I have been reading the Fine and Rosenberg paper which proves that these groups are conjugacy separable ("Conjugacy separability of Fuchsian groups and related questions"; the proof, and the statement below, can also be found in their book, "Algebraic generalizations of discrete groups"), and in it the authors state,</p> <p>"The conjugacy classes of elements of finite order...are given by the conjugacy classes {$\langle a\rangle$}, {$\langle b\rangle$}, {$\langle c\rangle$}".</p> <p>This statement is given without proof or reference. I was therefore wondering if someone could provide either a proof or a reference for this?</p> <p>I understand where the comment comes from - it is the obvious generalisation of the one-relator groups case (here we are dealing with one-relator products, which generalise one-relator groups). However, I cannot seem to find a proof of the statement in the literature, although I am sure it must be there. Unless, of course, I am simply missing something and the result is obvious...</p> http://mathoverflow.net/questions/70990/torsion-in-triangle-groups/70997#70997 Answer by James for Torsion in triangle groups James 2011-07-22T15:32:33Z 2011-07-22T15:32:33Z <p>According to the survey article <a href="http://iopscience.iop.org/0036-0279/31/5/R18" rel="nofollow">H. Zieschang, On Triangle Groups, Russian Mathematical Surveys (October 1976), 31 (5), pg. 226-233</a>, this fact is proved in the Russian paper, <a href="http://mi.mathnet.ru/eng/umn/v21/i3/p195" rel="nofollow">H. Zieschang, “Discrete groups of plane motions and plane group images”, Uspekhi Mat. Nauk, 21:3(129) (1966), 195–212</a>. I've been unable to locate an English translation of the latter. However, Fine and Rosenberger discuss it in Chapter 4 of their book (see Theorem 4.3.2).</p> http://mathoverflow.net/questions/70990/torsion-in-triangle-groups/87729#87729 Answer by Dan for Torsion in triangle groups Dan 2012-02-06T22:38:37Z 2012-02-06T22:38:37Z <p>Also you can reference to theorem 2.10 in W.Magnus "Noneuclidean Tesselations and Their Groups" ACADEMIC PRESS New York 1974.</p> http://mathoverflow.net/questions/70990/torsion-in-triangle-groups/87785#87785 Answer by HW for Torsion in triangle groups HW 2012-02-07T10:29:13Z 2012-02-07T15:06:29Z <p>In fact it's easy, and hopefully enlightening, to give a direct proof. Here's one, using the theory of orbifolds (which happens to be how I think about it). (NB I suspect it's not how Fine and Rosenberger think about it.)</p> <p>Your triangle group is the fundamental group of an orbifold $O$ with underlying space a 2-sphere and cone points of order $\alpha,\beta,\gamma$. This orbifold has Euler characteristic</p> <p>$\chi(O)= 2-(1-1/\alpha)-(1-1/\beta)-(1-1/\gamma)=1/\alpha+1\beta+1/\gamma-1$.</p> <p>For convenience, we will give the proof in the hyperbolic, ie Fuchsian, case where $\chi(O)&lt;0$; a similar proof can be given in the Euclidean ($\chi(O)=0$) case. It seems clear that the statement is false in the spherical (ie $\chi(O)>0$) case, since then $G$ is finite. For more information about orbifolds, see Peter Scott's survey article `The geometries of 3-manifolds'.</p> <p>The orbifold $O$ is the quotient of the hyperbolic plane $\mathbb{H}^2$ by your triangle group $G$. That is to say, $G$ acts properly discontinuously and cocompactly, but not freely, on $\mathbb{H}^2$. The cone points on $O$ are precisely the images of the points in $\mathbb{H}^2$ with non-trivial stabilizers. Each cone point on $O$ has a preimage in $\mathbb{H}^2$ whose stabilizer is generated by one of the generators $a,b$ or $c$. Call these preimages $x_a,x_b,x_c$ respectively. </p> <p>Now suppose that $g\in G$ has finite order. By the classification of isometries of $\mathbb{H}^2$, it follows that $g$ fixes a point $y$ in $\mathbb{H}^2$. Thus $y$ has non-trivial stabilizer, and so for some $h\in G$, $y=hx_a$ or $hx_b$ or $hx_c$; wlog, let's say $y=hx_a$. Therefore $h^{-1}gh$ stabilizers $x_a$, and so $g$ is conjugate into $\mathrm{Stab}_G(x_a)=\langle a\rangle$, as required. </p>