
Re: Showing group is Abelian
Posted:
Dec 3, 2012 11:49 AM


On Monday, December 3, 2012 7:17:01 AM UTC6, 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@dontemail.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 handwavy. > > > > > >> 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

