# Showing a group is abelian

1. Mar 6, 2006

### StatusX

I have a couple questions involving showing a group with certain properties is abelian.

1. For the first, I'm supposed to show that if some group G has the property that (ab)i=aibi for some three consecutive integers i and all a,b in G, then G must be abelian. Using (aba-1)i=abia-1=aibia-i, I've been able to show that a2b=ba2 for all a,b in G, but I can't get any farther.

2. The second is similar. Given that a finite group G has order not divisible by 3, and for every a,b in G, (ab)3=a3b3, show G is abelian. By defining an automorphism on G by sending a to a3, I've been able to show every element has a unique cube root. Using this, I've shown a2b=ba2, as above. But now I'm stuck.

Thanks in advance for any help.

2. Mar 6, 2006

### AKG

Problem 1

Your work doesn't look right. How do you get abia-1 = aibia-i?

Let I = {i-2, i-1, i} be a set of three consecutive integers such that for all a, b in G, and all k in I, (ab)k = akbk

Hints:

1. Starting with (ab)i = aibi, deduce that (ba)i-1 = (ab)i-1

2. Deduce, somehow, that (ba)i-2 = (ab)i-2

3. Mar 6, 2006

### AKG

Note, for problem 2, you don't need to show that x |-> x³ is an automorphism. Simply show that it is injective (it doesn't even need to be a homomorphism). Then, since G is finite, you know the map is surjective, and this is all you need to show that each element has a unique cube root.

4. Mar 6, 2006

### StatusX

Thanks for the help. (ba)i=b(ab)i-1a. It's so simple, and pops out at you if you write the product out, but I was going a completely different way with the problem and would never have thought of it. Both answers follow easily from it.

And by the way, since conjugation by an element is a homomorphism (in fact, an automorphism), then (aba-1)i=abia-1.

5. Mar 6, 2006

### AKG

My issue wasn't with (aba-1)i = abia-1, it was with abia-1 = aibia-i. However, I now see where you got that:

(aba-1)i = aibia-i by the special property given to us
(aba-1)i = abia-1 because it's an easy fact for any group

Using a similar idea for problem 2, you get:

a³b³a-³ = ab³a-1
a²b³ = b³a²

so every square commutes with every cube. But since you can prove that every element is a cube (i.e. it has a cube root), you can say that every square commutes with every element. Not sure where to go from here.

6. Mar 6, 2006

### StatusX

Well, using a similar idea as the one you suggested for the first problem, you can show (ab)2=b2a2. Then you can pull a factor of b on the right side to the other side of a2, and its easy from there.

7. Mar 6, 2006

### AKG

When I first looked at the problem, I did get that (ab)² = b²a². However, I don't see where to go from there. In fact, what do you even mean by:

Then you can pull a factor of b on the right side to the other side of a2, and its easy from there.

8. Mar 6, 2006

### AKG

Oh wait.

(ab)³ = a³b³
ababab = aaabbb
baba = aabb
(ba)² = a²b²

Now this, in conjunction with the fact that the squares commute with everything, gives

(ba)² = b²a²
baba = bbaa
ab = ba []

Is that what you had in mind?

9. Mar 6, 2006

### StatusX

More or less. I actually specifically meant:

(ab)2=b2a2
abab=b(ba2)=b(a2b)=baab
ab=ba