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Non-Abelian Groups

Date: 02/11/2003 at 19:11:43
From: Jeremy
Subject: Abelian groups

Given a group in which every a satisfies a^3 = 1, is that group 
abelian?

I am fairly certain that I cannot prove that all groups with a^3 = 1 
are abelian, but I cannot disprove it or come up with a 
counterexample.


Date: 02/12/2003 at 08:35:10
From: Doctor Jacques
Subject: Re: Abelian groups

Hi Jeremiah,

There is a non-abelian group in which every non-unit element has 
order 3.

It is the subgroup of S_9 generated by the following permutations:

  a = (123)(654)
  b = (147)(258)(369)

If you plot the numbers 1 through 9 in a square:

  123
  456
  789

- a permutes the first row cyclically to the right and the second to 
the left.
- b permutes all columns cyclically down.

This group is not abelian (ab <> ba), and has order 27.

To prove that the group is of order 27 and every element satisfies 
x^3 = 1, you may try to make a multiplication table (this involves 
some tedious work, but is guaranteed to get the result).

It is also possible to prove this by geometric considerations using 
the square representation as outlined above.

Here are a few hints for achieving this (*** this is NOT a rigorous 
proof ****) - you may want to prove each statement separately.

All permutations preserve rows; therefore, if the group is called G, 
there is a homomorphism f : G -> S_3. As the row permutations are 
generated by the element b (an even permutation), the image of f is 
A_3 = C_3 (the cyclic group of order 3).

The kernel K of f is the subgroup of permutations that leave each 
element in its row.

K is generated by a and c = b^(-1)ab = (456)(798).

Any element of G can be written uniquely as a^i*b^j*c^k, where 
0 <= i, j, k <= 2.

G has 27 elements.

Any element of g satistfies x^3 = 1.

This is a particular case of a general problem, known as "Burnside's 
problem". For more information, please read:

   World of Mathematics - Eric Weisstein
   http://mathworld.wolfram.com/BurnsideProblem.html 

Please write back if you want to discuss this further.

- Doctor Jacques, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
College Modern Algebra

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