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

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magidin@math.berkeley.edu

Posts: 11,145
Registered: 12/4/04
Re: Showing group is Abelian
Posted: Nov 30, 2012 1:57 PM
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On Friday, November 30, 2012 8:18:46 AM UTC-6, 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 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?


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



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