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Here by triangle group $(a,b,c)$ I mean the group with presentation $$\langle x,y \;|\; x^a = y^b = (xy)^c = 1\rangle$$

In other words, for every finite simple nonabelian group $G$, do there exist pairwise coprime integers $a,b,c$ such that $G$ is generated by $x,y$ with $|x| = a$, $|y| = b$, and $|xy| = c$?

I'd also be interested in any result where we relax the "pairwise-coprime" condition to the condition that $(|x|,|y|)\cdot (|x|,|xy|)\cdot (|y|,|xy|)$ is small.

Will Chen
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  • The fact that any finite simple group is 2-generated (a corollary of the classification) immediately implies we get a quotient of a triangle group. – S. Carnahan Jul 28 '15 at 01:12
  • I believe most finite simple groups are generated by elements of order 2 and 3, which addresses the 2nd part of your question. http://mathoverflow.net/a/59300/1345 – Ian Agol Jul 28 '15 at 03:56
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    I would be very surprised if this were not true, but I am not aware that it a known result. I believe that for a suitably chosen definition of "most" (Lie-type groups of sufficiently large rank), it is true for most simple groups with $(a,b,c)=(2,3,7)$. See https://en.wikipedia.org/wiki/Hurwitz%27s_automorphisms_theorem#Examples_of_Hurwitz.27s_groups_and_surfaces You could try asking Bob Guralnick or Martin Liebeck. – Derek Holt Jul 28 '15 at 08:18
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    You should look at Claude Marion's lovely paper On triangle generation of finite groups of Lie type which studies triangle groups $(a,b,c)$ where $a,b$ and $c$ are prime. You may well be able to extract what you need from that paper. – Nick Gill Jul 28 '15 at 09:33
  • @NickGill's reference, clickably: Marion - On triangle generation of finite groups of Lie type (published version). – LSpice Feb 24 '20 at 12:39

1 Answers1

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If my memory is correct, it is a consequence of Thompson's N-group paper that a finite group $G$ is solvable if and only it not possible to find three elements $(x,y,z)$ of pairwise coprime orders, not all $1$, with $xyz = 1_{G}.$ It follows from this that every minimal non-Abelian finite simple group $G$ is a quotient of the type of "triangle group" asked for. For there must be such a triple of elements $(x,y,y^{-1}x^{-1})$ of pairwise coprime orders. Hence $\langle x,y \rangle$ is not a solvable group, so must be all of $G$, and $G$ is a quotient of a group of the required form.

As has been remarked in comments, there is an extensive literature of generating pairs for finite simple groups, and it appears extremely likely that the answer to this question is "yes"

Geoff Robinson
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    This result (about minimal FSG's) is a special case of Theorem B of A NEW SOLVABILITY CRITERION FOR FINITE GROUPS by Dolfi, Guralnick, Herzog and Praeger. They also state a conjecture in the final section which pertains directly to the OP's original question. It's slightly too long for me to write down in a comment, but you can find it here - http://arxiv.org/pdf/1105.0475.pdf – Nick Gill Jul 29 '15 at 09:03
  • @NickGill : Yes, probably, thanks. I think the version I stated was already a corollary of Thompson's N-group paper, which dates to late 60s or early 70s, so Thompson has priority on that particular result, or it may be P.X. Gallagher. Of course, people have done much better since using CFSG, including the paper you mention. – Geoff Robinson Jul 29 '15 at 09:07