Proving G is Abelian Given (ab)3 = a3b3 for All a,b in G

[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?

 

[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.

 

[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?

 

[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.

 

[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!