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

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 Michael Stemper Posts: 671 Registered: 6/26/08
Re: Showing group is Abelian
Posted: Dec 3, 2012 8:17 AM
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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

--
Michael F. Stemper
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Date Subject Author
11/30/12 Michael Stemper
11/30/12 magidin@math.berkeley.edu
11/30/12 Brian M. Scott
12/3/12 Michael Stemper
11/30/12 Brian M. Scott
12/3/12 Michael Stemper
12/3/12 magidin@math.berkeley.edu

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