Sorry about asking so many questions, but I am a bit further on in my classification, and I am up to the group $G := \langle a, b \ | \ a^2, b^3, (ab)^7, [a,b]^{10}, ([a,b]^4b)^7 \rangle$. It has no small quotients (up to 500000), and I suspect that it is simple. Is there a way to check for simplicity in infinite groups?

UPDATE (edit by YC)

The original question about the group $G$ has been answered below. Now I am up to the two groups $H := \langle a, b \ | \ a^2, b^3, (ab)^7, [a,b]^{10}, ([a,b]^4b)^8 \rangle$ and $I := \langle a, b \ | \ a^2, b^3, (ab)^7, [a,b]^{10}, ([a,b]^4b)^9 \rangle$. I need to know if they are trivial, finite (the order would be good), or infinite.

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    $\begingroup$ Is the group infinite? $\endgroup$ – Mark Sapir Aug 22 '13 at 4:54
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    $\begingroup$ I would not trust a calculator, especially if you do not really understand what it says. $\endgroup$ – Mark Sapir Aug 22 '13 at 5:01
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    $\begingroup$ Answer "0" may mean "infinite", or it may mean "stop asking these questions, go read a group theory book, and here is $\sin(\pi)$." $\endgroup$ – Mark Sapir Aug 22 '13 at 5:34
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    $\begingroup$ You should not have deleted the definition of the group $G$, because that has made the two answers to your original question incomprehensible. $\endgroup$ – Derek Holt Aug 29 '13 at 19:38
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    $\begingroup$ Thomas, why exactly are you crunching through all these two-generator groups? As HJRW has pointed out here mathoverflow.net/questions/139991/… you seem to be on an open-ended mission and it is not clear what the goal is $\endgroup$ – Yemon Choi Aug 29 '13 at 20:09

In fact your group is trivial. Here are two different computations with Magma, the first using coset enumeration over the subgroup $\langle ab \rangle$, and the second using the Knuth-Bendix completion algorithm.

 > G<a,b>:=Group<a,b|a^2,b^3,(a*b)^7,(a,b)^10, ((a,b)^4*b)^7 >;
 > H:=sub<G|a*b>;
 > time Index(G,H:CosetLimit:=100000000,Hard:=true); 
 Time: 27.730

 > time  R := RWSGroup(G:MaxRelations:=100000, TidyInt:=1000);
 Time: 90.350
 > Order(R);
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    $\begingroup$ Being trivial is a good reason for not having many quotients. $\endgroup$ – Jérémy Blanc Aug 24 '13 at 16:55
  • $\begingroup$ Wait, so why did magma return 0 as the order when I asked for it? $\endgroup$ – Thomas Aug 26 '13 at 1:25
  • $\begingroup$ Will this program work on the online calculator? Also, what about the other groups I mentioned? $\endgroup$ – Thomas Aug 26 '13 at 1:46
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    $\begingroup$ The answer 0 just means that the program did not complete. That can cause some confusion, but no group can have order 0, so it cannot be interpreted as a valid answer. I think it needs too much memory for the Magma calculator. I've made no progress with your other two groups, so they could be trivial too or they could be infinite. They don't appear to have any reasonably sized finite simple quotients, so I have no ideas about how they might be proved infinite. $\endgroup$ – Derek Holt Aug 26 '13 at 8:16
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    $\begingroup$ The paper: G. Havas and D.F. Holt. On Coxeter's families of group presentations. Journal of Algebra 324 (2010), 1076--1082, which you can find at homepages.warwick.ac.uk/~mareg/download/papers/coxpres/… might be useful as a starting point, and for further references. $\endgroup$ – Derek Holt Aug 29 '13 at 13:52

I think Mark Sapir's cautionary joke that magma is just returning a value of $sin(\pi)$ is actually surprisingly accurate. The "Order" function on a finitely presented group, denoted by magma as GrpFP, Magma returns positive integer if the group is computed to be finite, "Infinity" if the group is known to be infinite (e.g. a map to $\mathbb{Z}$ exists), and "0" when its certificates of infinite order cannot be established, coset enumeration exhausts memory, and magma can't determine the order of the group.


EDIT: The OP uses $[a,b]=a\cdot b^{-1} \cdot a^{-1} \cdot b$, my explanation below is $[a,b]=a\cdot b \cdot a^{-1} \cdot b^{-1}$. The computation of Order with the OP's definition returns a 0, as noted in the comments below.

Having said all that, the magma computation I ran gave a different answer for $G$. Magma is saying $G$ is trivial. I used a machine at the University of Texas, which might have more available memory for coset enumeration than the online magma calculator.

$ > G<a,b>:=Group<a,b|a^\wedge 2,b^\wedge 3,(a*b)^\wedge 7,(a*b*a^\wedge -1*b^\wedge -1)^\wedge 10,((a*b*a^\wedge -1*b^\wedge -1)^\wedge 4*b)^\wedge 7\ >;\\ > Order(G);\\ 1\\ $

However, the orders of $H$ and $I$ don't seem to be computable with the available memory, so they might be infinite and they might not be.

  • $\begingroup$ [a,b]=ab^-1ab, not abab^-1, I'm not sure, but it may make a difference. $\endgroup$ – Thomas Aug 22 '13 at 11:44
  • $\begingroup$ I checked on magma, and the presentation you used does result in the trivial group, so they are different. $\endgroup$ – Thomas Aug 22 '13 at 12:41
  • $\begingroup$ I also checked what you said earlier, and you're right, magma should show infinity, but it shows zero because it couldn't figure it out. Is there a way to check whether it is infinite? Also, are there any other groups theory calculators that you recommend? $\endgroup$ – Thomas Aug 22 '13 at 14:25
  • $\begingroup$ I think you may be able to check that the group normally generated by $[a,b]^{10}, ([a,b]^4 b)^n$ for $n\geq 7$ is infinite index in $\langle a, b| a^2,b^3,c^7\rangle$, which is an index 2 subgroup of a hyperbolic triangle group. $\endgroup$ – Neil Hoffman Aug 22 '13 at 14:51
  • $\begingroup$ That it is what? Infinite? Simple? Also, how will I be able to check? Some group theory program? By hand? $\endgroup$ – Thomas Aug 22 '13 at 14:54

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