MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I need the following information about the quotients of infinite triangle (or von Dyck) groups.

(1) Let $G(l,m,n)$ defined as $S^l$ =$T^m$ = $(ST)^n$ = $E$ is the hyperbolic ($1/l+1/m+1/n<1$) triangle group. What are the additional conditions $S$ and $T$ should satisfy in order to ensure that the given quotients of this group is finite?

(2) Is there exist any reference where such quotient groups have been identified as isomorphic to other famous group?

I am a student of Physics and have very limited knowledge of the group theory, so simple answer will be very helpful.

Thanks in advance.

Ketan Patel

share|cite|improve this question
In particular, your group is finite precisely when $1/l+1/m+1/n>1$. – HJRW May 15 '13 at 11:22
Ketan: Now it is no longer clear what your question is. Are you interested in finite subgroups of these groups? If so, they are all cyclic, whose generators are conjugate to powers of $S$ or $T$ or $ST$. What else do you need to know? (I do not know what "famous group" means.) – Misha May 15 '13 at 12:02
Ketan: are you perhaps interested in finite quotients rather than subgroups. So your question becomes: what relations must I add to obtain a finite group? (It's a question with no very straight-forward answer by the way.) – Nick Gill May 15 '13 at 12:34
Ketan - it appears that Nick is exactly right. You are interested in finite quotients, not finite subgroups. – HJRW May 15 '13 at 13:02
That's an improvement. Now I have a further question. You ask for conditions to ensure that 'the given quotients' are finite, but you haven't given us any quotients! In general, these groups have many finite quotients (because they are residually finite) and also many infinite quotients. Which quotients are you specifically interested in? – HJRW May 15 '13 at 14:35

Your Answer


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

Browse other questions tagged or ask your own question.