My last question on the quotients of the group $$H := \langle a, b, c \ | \ a^2, b^2, c^2, (ab)^2, (ac)^3, (bc)^7, (abc)^{19} \rangle$$ couldn't be completely answered, because the finiteness of the group is unknown, I will now look at the next group: $$G := \langle a, b, c \ | \ a^2, b^2, c^2, (ab)^2, (ac)^3, (bc)^7, (abc)^{21} \rangle.$$ I couldn't find whether this group was finite or not, but I know that it has the quotient PSL(2,43). What other quotients are there? Which of the groups $$I := \langle a, b, c \ | \ a^2, b^2, c^2, (ab)^2, (ac)^3, (bc)^7, (abc)^{21}, (abcbc)^i \rangle$$ are trivial (for i up to 16 the groups are trivial), and which are not?
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$\begingroup$ In Derek's (fantastic) answer to your previous question, he asserted that we know which members of the Coxeter family are finite, except for $H^{3,7,19}$. So I would have thought your group $G$ would have therefore been covered, surely? You should refer to his paper: dropbox.com/s/zybsgnicajh46ps/… $\endgroup$– Nick GillCommented Oct 11, 2013 at 8:50
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$\begingroup$ It has been covered in the sense that we know it is infinite, but we don't know much more than that. $\endgroup$– ThomasCommented Oct 14, 2013 at 8:08
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1 Answer
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The group $G$ is proved infinite in the paper
M. Edjvet and A. Juhàsz, The groups $G^{m,n,p}$, J. Algebra 319 (2008), 248-266.
The proof is geometric using pictures (which are similar to van Kampen diagrams), and doesn't provide any information about the quotients of $G$.
As far as I know, ${\rm PSL}(2,43)$ is the only known finite quotient. I checked this for all simple groups up to order $10^9$, and also that it is the only ${\rm PSL}(2,q)$ that is a quotient. The kernel of the homomorphism onto ${\rm PSL}(2,43)$ is perfect, so there are no further quotients to be easily found there.
For your other question, I got one further than you did, and proved using a big coset enumeration over the subgroup $\langle abc \rangle$ that the quotient is trivial for $i=17$. I am trying $i=18$, but I am not sure where this is leading, because you will inevitably get stuck at some point.
answered Oct 11, 2013 at 11:36
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$\begingroup$ Oh, interesting. It seems that beyond a certain point with these groups, all of them have some unanswered questions, and their structure is not fully known. Are there any infinite quotients? They might be easier to work with. $\endgroup$– ThomasCommented Oct 11, 2013 at 16:36
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1$\begingroup$ AFAIK, the only known infinite quotient is the group itself! $\endgroup$ Commented Oct 11, 2013 at 16:43
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$\begingroup$ Oh. This seems to be a problem with these groups. We know they're infinite, but we have almost no idea about their structure. $\endgroup$– ThomasCommented Oct 13, 2013 at 2:39