Date: 09/14/2005 at 21:42:55 From: George Subject: Abelian groups If a and b are any elements of a group G and (ab)^3 = a^3*b^3, is G necessarily Abelian? I know that (ab)^(-1) = a^(-1)*b^(-1). We use this repeatedly, and I think it might be involved in answering this question.
Date: 09/15/2005 at 09:47:47 From: Doctor Vogler Subject: Re: Abelian groups Hi George, Thanks for writing to Dr. Math. No, it is not necessarily Abelian. For example, there is a non-Abelian group of order 27 all of whose elements have order 3 (except, of course, the identity). It has a name, in fact, the "Heisenberg group over Z/3Z" and can be described as the 3-by-3 matrices of the form [ 1 a b ] [ 0 1 c ] [ 0 0 1 ] where a, b, and c are elements of the finite field of size 3, namely Z/3Z. It can also be written with the presentation <x, a, b: x^3 = a^3 = b^3 = 1, ab = ba, xa = abx, xb = bx> and is, in fact, generated by a and x. If you have any questions about this or need more help, please write back and show me what you have been able to do, and I will try to offer further suggestions. - Doctor Vogler, The Math Forum http://mathforum.org/dr.math/
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