Since the group $G := \langle a, b \ | \ a^2, b^3, (ab)^7, [a,b]^{10} \rangle$ seems to have resisted attacks of some powerful programs, I will turn to a group that seems to be a bit easier to analyse. The group

$H := \langle a, b, c \ | \ a^2, b^2, c^2, (ab)^2, (ac)^3, (bc)^7, (abc)^{19} \rangle$

only has one of the simple groups that are a quotient of G, J1. Therefore it seems it will be easier to analyse. Adding the relation (abcbc)^i give the trivial group for all i less than 25 except for 15, where it gives J1. What is the group

$I := \langle a, b, c \ | \ a^2, b^2, c^2, (ab)^2, (ac)^3, (bc)^7, (abc)^{19}, (abcbc)^{25} \rangle$?

I have checked on magma using knuth bendix, and it takes too long, so magma stops the calculation.

  • "Adding the relation" ?? Please edit. Also, I added a tag. – Jim Humphreys Sep 17 '13 at 19:43
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    $J_1$ is not the only simple quotient of $G$. Another one is ${\rm PSL}(2,113)$. – Derek Holt Sep 17 '13 at 20:24
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    Your group $I$ is trivial. I did a coset enumeration over the subgroup $\langle abcbc \rangle$ and got index $1$. – Derek Holt Sep 17 '13 at 21:10
  • @Derek: Indeed. -- Checking this is even quite quick -- in GAP this takes less than half a minute on my laptop. – Stefan Kohl Sep 17 '13 at 21:28
  • Taking the relation $(abcbc)^{26}$ instead of $(abcbc)^{25}$, the group is still trivial, though the verification is somewhat harder -- making the GAP job grow to something like 700MB before returning the answer that there is only one coset. – Stefan Kohl Sep 17 '13 at 21:51

As you may know, in the family of groups

$H^{m,n,p} := \langle a, b, c \ | \ a^2, b^2, c^2, (ab)^2, (ac)^m, (bc)^n, (abc)^p \rangle$

considered by Coxeter, there is only one remaining case for which finiteness has not been decided, $H^{3,7,19}$, which is your group $H$. It is possible that $H \cong J_1 \times {\rm PSL}(2,113)$, since only two finite simple quotients are known. The kernel of the homomorphism onto $J_1$ is perfect. I haven't quite proved it, but I think the kernel onto ${\rm PSL}(2,113)$ is also perfect.

A homomorphism onto ${\rm PSL}(2,113)$ is induced by the map onto ${\rm SL}(2,113)$

$$a \mapsto \left(\begin{array}{rr}58& 57\\ 50& 55\end{array}\right), b \mapsto \left(\begin{array}{rr}55& 69\\ 2& 58\end{array}\right), c \mapsto \left(\begin{array}{rr}100& 67\\ 43& 13\end{array}\right).$$

The image of $abcbc$ under this map has order $57$.

With the ACE coset enumerator, I have been able to show that adding the extra relation $(abcbc)^n$ gives the trivial group for $n=25,26,27,28,29$, but I've got stuck on $n=30$.

  • Oh, cool, I stumbled upon an open problem! Adding the relation (abcbc)^30 at least gives J1, so it will be harder. – Thomas Sep 18 '13 at 8:09
  • @Derek, that is a fantastic answer. Where did Coxeter consider this family of groups? What's so special about the ones that are left? (Maybe I should ask this as a separate question on MO...) – Nick Gill Sep 18 '13 at 8:33
  • @Nick Gill: My paper with George Havas, G. Havas and D.F. Holt. On Coxeter's families of group presentations. Journal of Algebra 324 (2010), 1076--1082 (which you can find on my webpage) describes the current state of knowledge of several families of groups considered by Coxeter, and includes references. I am not aware of any further advances since then. – Derek Holt Sep 18 '13 at 9:23
  • @Derek, I looked at the paper. Am I being dumb or do you actually prove that the group $H^{3,8,13}$ is finite? (You say in the answer above that the status of this group is unresolved, but your paper seems to deal with it. I could be wrong, though, because you write the presentations differently in the paper than in this answer and I'm not certain things are equivalent.) – Nick Gill Sep 18 '13 at 10:07
  • @Nick: No, I'm being dumb, not you! I only looked at the introduction to the paper and not at its contents! Yes, we proved that $H^{3,8,13}$ is finite - it's effectively the same proof as for the $(2,3,13;4)$ group. I will edit the post. – Derek Holt Sep 18 '13 at 11:54

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