Proof That a Group is AbelianDate: 10/29/2004 at 07:31:12 From: Jackie Subject: prove that group G must be Abelian If G is a finite group whose order is not divisible by 3, and (ab)^3 = a^3b^3 for all a,b in G, prove that G must be Abelian. I think we have to manipulate the given identity, but I don't know how to use the fact that the order of G is not divisible by 3. Date: 10/31/2004 at 07:11:06 From: Doctor Jacques Subject: Re: prove that group G must be abelian Hi Jackie, Assume that G is a finite group whose order is not divisible by 3, and that (ab)^3 = (a^3)(b^3) for all a,b in G. To simplify notations, we will write g' for g^(-1), the inverse of g (an element of G). We will also write the identity element as 1. Consider now the function: f : G -> G f(x) = x^3 The hypothesis on G says that f is a homomorphism. The kernel of f is the set of elements x such that x^3 = 1. This kernel is trivial, because, if x^3 = 1 for some x <> 1, then the cyclic group <x> generated by x is a subgroup of G of order 3, and this is impossible, by Lagrange's theorem, since the order of G is not divisible by 3. As ker f = {1}, f is injective (one-to-one). Because G is finite, this implies that f is a bijection; in particular, every element of G has a unique "cube root", i.e., given x in G, there is exactly one y in G such that x = y^3. By hypothesis, we have: (ab)^3 = (a^3)(b^3) ababab = aaabbb and, by canceling a on the left and b on the right, we find: (ba)^2 = a^2*b^2 [1] Note that, if, instead of [1], we had: (ba)^2 = b^2 * a^2 [2] then we could write: baba = bbaa and, by a similar argument, we could conclude that ab = ba. Consider now the expression: z = (aba')^3 [3] As f is a homomorphism, we have: z = f(aba') = f(a)f(b)f(a') = a^3 * b^3 * (a')^3 On the other hand, if we expand [3] directly, then products (a'a) in the middle cancel out, and we find: z = a*b^3*a' If we compare both expressions of z, we find: a^3 * b^3 * (a')^3 = a * b^3 * a' and, by canceling a on the left and a' on the right, we find: a^2 * b^3 * (a')^2 = b^3 a^2 * b^3 = b^3 * a^2 [4] Now, because f is a bijection, every element x can be written as b^3 for some b. This shows that [4] implies: a^2 * x = x * a^2 [5] for all x in G. In other words, every element of G commutes with any square. If we use that fact in [1] with x = b^2, this gives [2], and, as said before, this allows us to conclude the proof. Does this help? Write back if you'd like to talk about this some more, or if you have any other questions. - Doctor Jacques, The Math Forum http://mathforum.org/dr.math/ |
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