Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

A triangle group has a presentation of the form,

$G=\langle a, b; a^{\alpha}, b^{\beta}, c^{\gamma}, abc\rangle, \alpha, \beta, \gamma \geq 2$

(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,

"The conjugacy classes of elements of finite order...are given by the conjugacy classes {$\langle a\rangle$}, {$\langle b\rangle$}, {$\langle c\rangle$}".

This statement is given without proof or reference. I was therefore wondering if someone could provide either a proof or a reference for this?

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...

share|improve this question
    
<shameless self-promotion> If you're interested in a geometrical point of view on conjugacy separability of Fuchsian groups, you might like to look at my paper arxiv.org/abs/0809.2308 . It provides a topological proof that surface groups are conjugacy separable, which also works for the infinite-order elements of Fuchsian groups.</shameless self-promotion> –  HJRW Feb 7 '12 at 10:32
add comment

3 Answers

up vote 6 down vote accepted

According to the survey article H. Zieschang, On Triangle Groups, Russian Mathematical Surveys (October 1976), 31 (5), pg. 226-233, this fact is proved in the Russian paper, H. Zieschang, “Discrete groups of plane motions and plane group images”, Uspekhi Mat. Nauk, 21:3(129) (1966), 195–212. 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).

share|improve this answer
    
Ah, thank you very much –  user6503 Jul 27 '11 at 11:44
add comment

Also you can reference to theorem 2.10 in W.Magnus "Noneuclidean Tesselations and Their Groups" ACADEMIC PRESS New York 1974.

share|improve this answer
add comment

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.)

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

$\chi(O)= 2-(1-1/\alpha)-(1-1/\beta)-(1-1/\gamma)=1/\alpha+1\beta+1/\gamma-1$.

For convenience, we will give the proof in the hyperbolic, ie Fuchsian, case where $\chi(O)<0$; 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'.

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.

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.