I have the following group presentation:

$G=\left\langle a,b,c\ |\ a^2,b^{11},(ab)^{4},(ab^2)^6,ab^2abab^{-1}abab^{-2}ab^2ab^{-1},c^2,(ac)^3,(bc)^2\right\rangle$

Is $G$ finite?

GAP's Size(G) runs out of memory pretty quickly. No surprise there. I also tried using the ideas in the answer here to look for homomorphisms onto certain small simple groups, with no luck. And I used LowIndexSubgroups() to look for subgroups up to index $30$ with no results.

It's also worth noting that:

$a$ and $b$ generate a subgroup isomorphic to $M_{11}$ (The Mathieu Group on $11$ points)

$a$ and $c$ generate a subgroup isomorphic to $D_6$ (The Dihedral group of size $6$)

$b$ and $c$ generate a subgroup isomorphic to $D_{22}$ (The Dihedral group of size $22$)

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    $\begingroup$ Just out of curiosity, where did this presentation come from? $\endgroup$ – lulu Feb 14 '18 at 2:00
  • $\begingroup$ @lulu I've been exploring some finite groups presented with two generators, then tacking on $c$ and the relations involving $c$. Sometimes you do get a finite group. $\endgroup$ – Josh B. Feb 14 '18 at 2:03
  • $\begingroup$ Let us continue this discussion in chat. $\endgroup$ – verret Feb 14 '18 at 3:56
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    $\begingroup$ Yes, if you leave out one of the letters, the other two will present a group as listed in the question. But if you combine such presentations, it is not guaranteed -- see the example here -- that such groups survive as subgroup. A more stupid example is to take a free cyclic group generated by $a$ and then add relations $b^2$, $abab$ and $ba^2b$. Suddenly the infinite subgroup $<a>$ is only cyclic of order $2$. $\endgroup$ – ahulpke Feb 14 '18 at 4:01
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    $\begingroup$ I checked that it has no proper subgroups up to index $40$, and no simple images up order up to $500$ million. I don't have any further ideas of things to try. The general problem of deciding whether a group defined by a finite presentation is trivial, finite or infinite, etc is known to be undecidable, so it is unsurprising that presentations arise from time to time which are difficult to resolve. $\endgroup$ – Derek Holt Feb 14 '18 at 9:44

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