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

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Brian M. Scott

Posts: 1,289
Registered: 12/6/04
Re: Showing group is Abelian
Posted: Nov 30, 2012 2:25 PM
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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:

> 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:

> 1. Show that if xy = yx then (x^-1)y = y(x^-1). This is
> pretty simple.

> 2. Use induction to show that if p and q commute, then any
> product of m p's and n q's is equal to any other,
> regardless of order.

> 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. (Here m and n can be negative, a
product of -3 p's, for instance, being (p^{-1})^3.)

By the way, you might want to take a look at


this would have been a perfect question for the site.


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