From: mareg@csv.warwick.ac.uk (Dr D F Holt)
Newsgroups: sci.math
Subject: Re: Group with no finite index subgroups
Date: 10 Oct 1996 11:47:27 +0100

In article <53e9ko$q3h@news.rain.org>,
	snowe@rain.org (Nick Halloway) writes:
>
>Benjamin Owens (benclaud@ix.netcom.com) wrote:
>
>: Let G_n be the group with generators s_i 1<=i<=n and relations
>: s_(i+1)*s_i*s_(i+1)^(-1)=(s_i)^2 with s_(n+1)=s_1.
>: prove that any homomorphism of G_n into a finite group is trivial.
>
>provides an example of an infinite group with no proper subgroups
>of finite index, provided you can show that G_n is infinite.  
>
>For n = 2 G_n has order 1.  I suppose for n = 3 this is true also?
>
>What are the usual ways that one would try to find out if factoring
>out a set of relations results in factoring out the whole group?  Or
>how would one generally figure out if the resulting group is finite?

In general the problem of deciding whether a finitely presented group
is finite (or nilpotent, or abelian, or solvable, or just about any other
property you care to mention) has been proved to be theoretically
unsolvable.

However, you can in principal solve the problem in one direction.
If the group is finite, then you can (given enough time), verify
(i.e. prove) this fact, and find the order of the group. The most
practical general algorithm for this is Todd-Coxeter coset enumeration,
and there are excellent computer implementations.
Similarly, there are, in theory,  algorithms that can verify that a
group is abelian, nilpotent, polycyclic, ..., and some of these are even
beginning to be practical for some examples.

It is easy to check mechanically (or even by hand) that the examples
defined above have order 1 for n=2, 3. For n=4, we get the example of
Graham Higman, which is infinite. I do not know how you prove that off-hand,
or even whether it is easy or difficult (I would need to look at the paper).
If anyone knows a quick proof, perhaps they could post it.

Generally, there is no algorithmic procedure to check that a presentation
defines an infinite group, and you have to rely on specific techniques
(such as small cancellation theory, or automatic group theory), which
happen to work for particular examples.

Derek Holt.

==============================================================================

From: geoff@argo.math.ucla.edu.mathnet
Newsgroups: sci.math
Subject: Re: Group with no finite index subgroups
Date: 10 Oct 96 14:42:27

>I do not know how you prove that off-hand,or even whether it is easy
> or difficult (I would need to look at the paper).
Fairly easy using amalgamated free products. See Serre's book Trees
( translated from _SL_2, arbres, amalgames_).

Geoffrey Mess