Let's say the group is generated by three involutions $a$, $b$, and $c$ such that $ord(ab)= 2$, $ord(bc)=3$, and $ord(ac)=m$. Under which conditions is it isomorphic to the rank 3 Coxeter group $(2, 3, m)$?
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1$\begingroup$ Isn't this the definition of the rank 3 Coxeter group via generators and relations? $\endgroup$– LSpiceCommented Aug 12, 2015 at 1:54
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2$\begingroup$ This shows that the group generated by a b and c is a quotient of the rank 3 Coxeter group, but not necessarily that it is the group itself. Edit: I mean, say you have a group and you pick three elements out of it and you only know that they are involutions and the orders of their pairs. Is that enough to ensure that they do not satisfy any conditions that makes the group they generate a proper quotient of the Coxeter group and not the group itself? $\endgroup$– user76098Commented Aug 12, 2015 at 1:57
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3$\begingroup$ Once $m>5$ there are quotients of the $(2,3,m)$ group in which the generators have the same order. For example, such finite quotients of the $(2,3,7)$ group are precisely the Hurwitz groups. $\endgroup$– Noam D. ElkiesCommented Aug 12, 2015 at 2:07
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1$\begingroup$ I don't think there's a full classification known or expected, but they include the group ${\rm PSL}_2({\bf F}_q)$ for all prime powers $q\equiv \pm 1 \bmod 7$ as well as ${\rm PSL}_2({\bf F}_7)$, and also various more exotic examples. See for example Conder's 1990 survey in Bull. AMS: projecteuclid.org/euclid.bams/1183555884 $\endgroup$– Noam D. ElkiesCommented Aug 12, 2015 at 2:36
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5$\begingroup$ Except for very small $m$ ($m\le 6$), this Coxeter group is SQ-universal and hence it is hopeless to classify its quotients (in particular there are uncountably many). Also it is a intrinsic condensation group and in particular it is not possible to detect the Coxeter group just by checking that finitely many elements are nontrivial. $\endgroup$– YCorCommented Aug 12, 2015 at 7:56
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