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Proof That a Group is Abelian

Date: 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/ 
Associated Topics:
College Modern Algebra

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