Tell me more ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.
up vote 4 down vote favorite

Suppose a finite group has the property that for every $x, y$, it follows that $(xy)^3 = x^3 y^3$.

How do you prove that it is abelian?

Edit: I recall that the correct exercise needed in addition that the order of the group is not divisible by 3.

share|improve this question
12  
You don't, as the group is not necessarily abelian! The group of upper triangular 3-by-3 matrices with ones along the diagonal and coefficients in the three-element field $\mathbb Z/3\mathbb Z$ has exponent three, so your equation holds, but it is not abelian. – Mariano Suárez-Alvarez♦ Jul 28 '10 at 21:47
(There are lots of examples: the most famous ones are the Burnside groups B(m,3), which you'll find described at en.wikipedia.org/wiki/…; the group in the first comment is B(2,3)) – Mariano Suárez-Alvarez♦ Jul 28 '10 at 21:49
1  
@Mariano, why don't you give that as the answer and then it can be accepted? Otherwise it looks as though nobody has answered the question. – bryn Jul 29 '10 at 6:57
By the way, your statement becomes true if you change 3 by 2. – falagar Jul 29 '10 at 14:54

3 Answers

up vote 5 down vote accepted

On the other hand, if the order of your group is not a multiple of 3 then it must be abelian!

You can read a proof here

share|improve this answer
5  
Urgh! Why do they write proofs in such a complicated way?! – Mariano Suárez-Alvarez♦ Jul 29 '10 at 16:25
Yeah! This is what I had in mind! Thanks! :) – Line Bundle Jul 29 '10 at 18:07
up vote 9 down vote

You don't, as the group is not necessarily abelian! The group of upper triangular 3-by-3 matrices with ones along the diagonal and coefficients in the three-element field $\mathbb {Z}/3\mathbb{Z}$ has exponent three, so your equation holds, but it is not abelian.

There are lots of examples: the most famous ones are the Burnside groups $B(m,3)$: the group I described above is $B(2,3)$.

share|improve this answer
It would be instructive to know what motivated the downvote... – Mariano Suárez-Alvarez♦ Jul 29 '10 at 18:10
It was because I was annoyed that you started a meta thread on me unnecessarily. If you have had objections then you could have left them as comments and I would have read them. – Line Bundle Jul 29 '10 at 18:16
@Line Bundle, you downvoted this answer because I asked, prompted by another question, about policy? – Mariano Suárez-Alvarez♦ Jul 29 '10 at 18:20
Yes, and I later regretted and tried removing the vote; but here one apparently can't do it. If you'll edit your question I can remove it. – Line Bundle Jul 29 '10 at 18:21
Don't worry about it! – Mariano Suárez-Alvarez♦ Jul 29 '10 at 18:24
up vote 2 down vote

I wrote a short proof here .

Steve

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.