TITLE: Is every nonabelian finite simple group a quotient of a triangle group $(a,b,c)$ with $a,b,c$ coprime?
QUESTION [6 upvotes]: 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.

REPLY [2 votes]: 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"