# Abelian group proof

1. Mar 30, 2009

### kathrynag

1. The problem statement, all variables and given/known data
If G is an abelian group, then (ab)^2=a^2b^2 for all a, b in G. Give an example to show that abelian is necessary in the statement of the theorem.

2. Relevant equations

3. The attempt at a solution
Abelian implies commutativity.
a*b=?b*a
a^2b^2=b^2a^2
For example ab=ba must be true for the statement to work.

2. Mar 30, 2009

### waht

take for example a^2 b^2 and expand it to aabb

what can you do if ab is abelian?

3. Mar 30, 2009

### kathrynag

aabb
Implies commutativity so aabb=bbaa
I think there's something with inverse to make this work

4. Mar 30, 2009

### waht

a(ab)b

5. Mar 30, 2009

### kathrynag

Ok so a(ab)b and b(ab)a
so this implies ab=ba

6. Mar 30, 2009

### waht

the definition of abelian is not axb = bxa so can't do that

so instead we can swap the ab

7. Mar 30, 2009

### kathrynag

I don't really understand. You're saying we can swap the ab, but I don't gte what you mean by this.
(ab)ab=(ab)ba

8. Mar 30, 2009

### matt grime

I don't know what waht it making you do. The point was to show that you can find G non-abelian where (ab)^2 is not equal to a^2b^2 for some a and b.

9. Mar 30, 2009

### kathrynag

Yeah and I wanted to prove that abelian implied (ab)^2=a^2b^2
Is there a hint for an a and b to use where the 2 aren't equal?

10. Mar 30, 2009

### Count Iblis

(ab)^2 = a^2 b^2 --->

abab = aabb -------->

cancel the leftmost a and the rightmost b:

ba = ab

11. Mar 30, 2009

### waht

well

a^2b^2 = aabb = a(ab)b

because they are abelian ab = ba

so aabb = a(ba)b = (ab)(ab)

12. Mar 30, 2009

### Count Iblis

But you are supposed to prove that they are Abelian.

13. Mar 30, 2009

### kathrynag

Ok I understnd how to prove it now.
I'm having trouble with the example to show abelain is necessary.

14. Mar 30, 2009

### Count Iblis

It is necessary because:

So, we have that

(ab)^2 = a^2 b^2 implies that ab = ba

This is the same as saying:

Not[ab = ba] implies Not[(ab)^2 = a^2 b^2]

Or, in plain English, if you don't have the Abelian property, you don't have that (ab)^2 = a^2 b^2. So, the Abelian property is a necessary assumption for the identity to hold.

15. Mar 30, 2009

### kathrynag

Hmmm, I think I might see this better if I saw a way where it doesn't work. For example (ab)^2 not equaling a^2b^2. Therefore, not abelian

16. Mar 30, 2009

### Count Iblis

But that's equivalent to Abelian ---> (ab)^2 =a^2b^2

which is the trivial case.

If you know that this is true (so Abelian implies the identity), then should you encounter a case where the identity is not satisfied, you know that the group cannot be Abelian.

17. Mar 30, 2009

### kathrynag

that's what I'm trying to do to find a case now where the identity is not satisfied to show then that the group is not abelian

18. Mar 30, 2009

### matt grime

Um, as count Iblis has pointed out *any* non-abelian group will provide a counter example.

19. Mar 30, 2009

### Count Iblis

You can take some group of matrices. Matrix multiplication does not commute. So, it should be easy to find a particular example where the identity does not hold.

But such a particular example does not show that the fact that the identity does not hold implies that the group is non-abelian. To show that, you need to prove that whenever the identity does not hold, you also have that the group is not abelian.

20. Mar 30, 2009

### kathrynag

Oh, that makes sense.