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Group Theory

Date: 02/08/2002 at 16:22:52
From: Jeffrey
Subject: Group Theory

I found this problem in the book _Topics in algebra_ by Herstein. It 
ask you to show that if G is a group s.t., (a*b)^i = a^i*b^i for three 
consecutive integers i for all a,b in G, then G must be abelian.

I'm stumped; any help would be much appreciated.

Date: 02/09/2002 at 15:25:47
From: Doctor Mike
Subject: Re: Group Theory


Where to start?  What to do? 

Let k be the smallest of the three powers in the three given facts.  
My gut tells me (a*b)^(k+1) is a good jumping-off place. Why? Because 
you can interpret it in two useful ways, one that allows you to use 
the fact for k, and one to use the fact for k+1. That's because 
(a*b)^(k+1) equals (a*b)*(a*b)^k. This gives us that a^(k+1)*b^(k+1) 
equals (a*b)*(a^k)*(b^k). Simplify this by multiplying on the left by 
the inverse of a, and multiplying on the right by the inverse of b^k.  
That gives (a^k)*b = b*(a^k). That's not exactly b commuting with a, 
but it IS b commuting with a power of a. A start! 
What I just did was to start with something promising, follow my nose,
and get something that I might be able to use later. A good thing to
do in the middle of a proof is to ask what "givens" I have used that 
I haven't used yet. I've used the facts for k and k+1 to get that b
commutes with a^k. Let's just note that the k+1 and k+2 facts tell me
that b also commutes with a^(k+1). So b commutes with a^k and a^(k+1).
How can we use that? Remember what I did at the beginning? My gut is 
telling me something again. Try looking at b*a^(k+1) = b*a^k*a.
Think about what we did in paragraph 1.  That's all I'm going to say.  

- Doctor Mike, The Math Forum
Associated Topics:
College Modern Algebra

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