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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$.1vu03te9mzn8p.dlg@40tude.net>, "Brian M. Scott" <b.scott@csuohio.edu> writes:
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> >On Fri, 30 Nov 2012 14:18:46 +0000 (UTC), Michael Stemper <mstemper@walkabout.empros.com> wrote in <news:k9af86$m35$1@dont-email.me> in alt.math.undergrad:
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> >> I'm currently on a problem in Pinter's _A Book of Abstract
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> >> Algebra_, in which the student is supposed to prove that
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> >> the (sub)group generated by two elements a and b, such
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> >> that ab=ba, is Abelian.
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> >
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> >> I have an outline of such a proof in my head:
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> >> 3. Combine these two facts to show the desired result.
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> >
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> >> However, this seems quite messy. I'm also wary that what I
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> >> do for the third part might end up too hand-wavy.
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> >
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> >> Is there a simpler approach that I'm overlooking, or do I
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> >> need to just dive in and go through all of the details of
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> >> what I've outlined?
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> >
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> >I don't offhand see a simpler approach. (3) isn't really a
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> >problem: once you've shown that an arbitrary product of m
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> >p's and n q's is equal to p^m q^n, show that the group is
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> >isomorphic to Z x Z.
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> Sometimes, but I don't think that's always true. Wouldn't the following
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> conditions be necessary?
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> 1. a, b have infinite order
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> 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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