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