# Is G Abelian?

1. Sep 21, 2008

### e(ho0n3

Problem. Let G be a finite group whose order is not divisible by 3. Suppose that (ab)3 = a3b3 for all a, b in G. Prove that G must be abelian.

Attempt. I know that G has no subgroups of order 3, hence no elements of order 3. Thus, if (ab)3 = a3b3 = e, then ab = e right? I don't know how to proceed. Any tips?

2. Sep 21, 2008

### morphism

It's really hard to give a hint here. OK, let's see, we want to show that ab=ba. Let's look at (ab)^3 and (ba)^3. Try to play with these. Here's a push in the right direction: ababab=(ab)^3=a^3b^3, so that baba=a^2b^2.

3. Sep 21, 2008

### e(ho0n3

I never thought about (ba)3. I had obtained (ba)2 = a2b2, but I didn't do anything with it. Let's see:

From (ba)2 = a2b2 we get that (ba)3 = baa2b2 = ba3b2. And since (ba)3 = b3a3, then ba3b2 = b3a3, so a3b2 = b2a3. Hmm...where can I use this equality?

4. Sep 21, 2008

### morphism

Good progress. Now use the fact that 3 doesn't divide the order of G to show that cb^2=b^2c for all c in G.

5. Sep 21, 2008

### e(ho0n3

To do that, I would show that for any c in G, there is an a in G with c = a3. And to do that, I would show that the mapping f(a) = a3 is 1-1: Suppose f(a) = f(b). Then a3 = b3 and e = b3a-3 = (ba-1)3. Since there are no elements of order 3, then the order of ba-1 must be either 1 or 2. The latter doesn't work so, ba-1 = e and b = a. It follows that f is also onto (since G is finite).

Great. Know what? Hmm... a3b3 = a2ab2b = a2b2ab, the last equality due to the result above. And since a3b3 = (ab)3, we have that a2b2ab = (ab)3, which after cancellation yields ab = ba. Nice.

Thanks a lot!