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Topic: Showing group is Abelian
Replies: 6   Last Post: Dec 3, 2012 11:49 AM

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magidin@math.berkeley.edu

Posts: 11,141
Registered: 12/4/04
Re: Showing group is Abelian
Posted: Dec 3, 2012 11:49 AM
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On Monday, December 3, 2012 7:17:01 AM UTC-6, Michael Stemper wrote:
> In article <2iqz3wjb4clv$.1vu03te9mzn8p.dlg@40tude.net>, "Brian M. Scott" <b.scott@csuohio.edu> writes:
>

> >On Fri, 30 Nov 2012 14:18:46 +0000 (UTC), Michael Stemper <mstemper@walkabout.empros.com> wrote in <news:k9af86$m35$1@dont-email.me> in alt.math.undergrad:
>
>
>

> >> I'm currently on a problem in Pinter's _A Book of Abstract
>
> >> Algebra_, in which the student is supposed to prove that
>
> >> the (sub)group generated by two elements a and b, such
>
> >> that ab=ba, is Abelian.
>
> >
>
> >> I have an outline of such a proof in my head:
>
>
>

> >> 3. Combine these two facts to show the desired result.
>
> >
>
> >> However, this seems quite messy. I'm also wary that what I
>
> >> do for the third part might end up too hand-wavy.
>
> >
>
> >> Is there a simpler approach that I'm overlooking, or do I
>
> >> need to just dive in and go through all of the details of
>
> >> what I've outlined?
>
> >
>
> >I don't offhand see a simpler approach. (3) isn't really a
>
> >problem: once you've shown that an arbitrary product of m
>
> >p's and n q's is equal to p^m q^n, show that the group is
>
> >isomorphic to Z x Z.
>
>
>
> Sometimes, but I don't think that's always true. Wouldn't the following
>
> conditions be necessary?
>
> 1. a, b have infinite order
>
> 2. a^m = b^n iff m=n=0


True; I think what Brian meant is that you can show that your group is isomorphic to a *quotient* of Z x Z, which in any case suffices to show the group is abelian.

--
Arturo Magidin



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