From: rusin@vesuvius.math.niu.edu (Dave Rusin)
Subject: Re: Combinatorial group question
Date: 19 Aug 1999 15:08:18 GMT
Newsgroups: sci.math
Keywords: How many groups satisfy some relations on their generators?

In article <01bee9f1$00cd5780$439508d1@TheBlueWizard>,
The Blue Wizard <TheBlueWizard@pressroom.com> wrote:
>Consider a/some finite group(s) generated by distinct elements with a set
>of relations on these
>elements.
>
>In other words, we obtain the group(s) via this generating form < g1, g2,
>g3, .... | rel1, rel2, rel3, ... >,
>where g1, g2, g3, ... are the elements and rel1, rel2, rel3, ... are
>relations involving g1, g2, g3, ...
>
>Now a question:  Is it possible to construct a generating form such that it
>produces more than one,
>but finitely many nonisomorphic finite groups?  Any particular example of
>such cases?

"produces"?

Let me see if I understand. You have a group  G  which contains some
elements  { g_i }, and you know
        (1) These elements generate  G
	(2) They satisfy all the relations in  { rel_j }.

So for example you might have a cyclic group  G  of order 3, with one
generator  g  which therefore satisfies the relation  g^6 = e  (the "6" is 
not a typo!)

Conversely, if  G'  is any other group generated by a single element  g'
which happens to satisfy  (g')^6 = 1, then  G'  must be isomorphic to
a cyclic group of order  3  (as above), or 6, or 2, or 1.


If I've understood your situation correctly, what you're observing is
that given a set  { rel_j }  of words in an alphabet of letters  { g_i }
(and their inverses), one may construct the abstract group defined by
these generators and relations by taking the free group  F  on  { g_i }
modulo the smallest normal subgroup  N  of  F  containing all of { rel_j }.
(N  may be all of  F  -- there's no algorithmic way to tell for sure,
but that's another story).

Then what's true is that any group  G  is a homomorphic image of  F/N
iff  G  can be generated by a set of elements  { g'_i } which
satisfy all the relations  { rel_j }.

In the example I gave first (one generator, one relation)  F/N  is
cyclic of order  6, so the groups  G  which have a generator whose
6th power is the identity element  have to be quotients of  C_6,
which then means something isomorphic to  C_6, C_3, C_2, or C_1.

So if you want "more than one but only finitely many" isomorphism
classes, you require precisely that  F/N  have finite order greater than 1.

This is "combinatorial group theory"; see 
     index/20-XX.html

dave