
Re: Showing group is Abelian
Posted:
Nov 30, 2012 1:57 PM


On Friday, November 30, 2012 8:18:46 AM UTC6, Michael Stemper wrote: > 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 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?
That's pretty much it. Probably the best way to handle your final step is to do a more general statement to be proven by induction. Namely:
Let a_1,...,a_n be elements in a group that commute pairwise (that is, a_ia_j = a_ja_i for all i,j). Then for every permutation s of {1,...,n},
a_1*...*a_n = a_{s(1)}*...*a_{s(n)}.
Then use the characterization of the subgroup generated by a and b as the set of all finite works in a,b,a^{1}, and b^{1}, together with your first observation, to show what you want.
Alternatively: prove that the set of all elements of the form a^nb^m with n and m integers forms a subgroup, and that multiplication is given by
(a^nb^m)(a^rb^s) = a^{n+r}b^{m+s}
Since the set includes a and b, it includes the subgroup they generate; and it is clearly contained in the subgroup they generate. Then observe the product is commutative.
 Arturo Magidin

