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 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
http://www.math-atlas.org/index/20-XX.html
dave