In article <firstname.lastname@example.org>, "Brian M. Scott" <email@example.com> writes: >On Fri, 30 Nov 2012 14:18:46 +0000 (UTC), Michael Stemper <firstname.lastname@example.org> wrote in <news:email@example.com> 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 #include <Standard_Disclaimer> No animals were harmed in the composition of this message.