Is it Possible to Show a Group is Abelian with Certain Properties?

[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.

 

[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 = a^kb^k

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

 

[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.

 

[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.

 

[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.

 

[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.

 

[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.

 

[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?

 

[StatusX]

More or less. I actually specifically meant:

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