Re: Group theory question

From: Snis Pilbor (snispilbor_at_yahoo.com)
Date: 08/23/04 

Date: 23 Aug 2004 16:05:11 -0700

Well, I guess that would explain why I was not having much luck
proving it! Too bad... but, that is how we learn, yes? Hehe

Thanks for pointing out my blunders! The thing I was working on can
still get through, but will end up being twice as complicated and less
than half as neat without that nice little (false) conjecture...

Sniz Pilbor

mareg@mimosa.csv.warwick.ac.uk () wrote in message news:<cgdb4t$1ns$1@wisteria.csv.warwick.ac.uk>...
> In article <cgd7e8$p8t$1@nntp.itservices.ubc.ca>,
> israel@math.ubc.ca (Robert Israel) writes:
> >In article <dmu0utlsdp.fsf@kuusi.ifi.uio.no>,
> >Jon Haugsand <jonhaug@ifi.uio.no> wrote:
> >>* Snis Pilbor
>
> >>> I have been trying to prove this statement which seems
> >>> intuitively very obvious and which seems like it ought to be very
> >>> straightforward to prove, but which is actually turning out to be
> >>> quite hard...
>
> >>> Conjecture: suppose two binary operations * and . are defined on a
> >>> nonempty set S such that:
> >>> 1. S is a group under * and also under .
> >>> 2. The groups (S under *) and (S under .) are isomorphic
> >>> 3. The identity element of S under * is the identity element of S
> >>> under .
>
> >>> Then: * and . are the same operation
>
> >>What happens if you take any non-abelian group and define x*y=y.x?
> >
> >Or for just about any group, take x*y = f^(-1)(f(x).f(y)) where f is a
> >one-to-one map of S onto itself that is not a group automorphism (under
> >.) but satisfies f(e)=e. There aren't many finite groups with (n-1)!
> >automorphisms (where n is the order of the group), I think.
>
> No - in fact only C1, C2, C3 and C2 x C2 have that property. So those are
> the only groups for which the conjecture is true.
>
> Derek Holt.

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