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 Jeffrey, 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 http://mathforum.org/dr.math/
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