Abelian finite group

Question

This is a (maybe be simple) problem from Group Theory, but being a beginner, I am unable to take even a first step forward.

Let $G$ be a finite group whose order is not divisible by $3$.Suppose that $(ab)^3=a^3b^3\ \ $ $\forall\ \ a,b\in G$. I am to prove that $G$ must be an abelian group.

Please help.

score 2 · Answer 1 · answered May 12 '16 at 13:06

2

Hint:

$(ab)^3=a^3b^3$

$\Rightarrow (ab)(ab)(ab)=aaabbb$

$\Rightarrow a(ba)(ba)b=a(aa)(bb)b$

answered May 12 '16 at 13:06

kalpeshmpopat

3,362

1

can you please explian more? – Qwerty May 12 '16 at 13:22
And please upvote if you liked my question – Qwerty May 12 '16 at 13:23

score 2 · Accepted Answer · answered May 12 '16 at 13:10

2

$ababab = (ab)^3 = a^3b^3 = aaabbb$

$\implies a^{-1}abababb^{-1} = a^{-1}aaabbbb^{-1}$

$\implies baba = aabb$

Use now the fact that the order of the group is not divisible by 3.

answered May 12 '16 at 13:10

Luísa Borsato

2,356
12
17

1

Can you please explain more? – Qwerty May 12 '16 at 13:22

score 2 · Answer 3 · answered May 12 '16 at 13:55

The given condition implies that $ \phi : x \to x^3 $ is an endomorphism of $ G $. On the other hand, $ \phi(x) = x^3 = e $ implies that $ x = e $ as the group cannot have an element of order 3, so $ \phi $ has trivial kernel, and is therefore an automorphism.

$ \mu(x) = g x g^{-1} $ is also an automorphism for any $ g $, which means that

$$ g x^3 g^{-1} = \mu(x^3) = \mu(x)^3 = \phi(\mu(x)) = \phi(g x g^{-1}) = g^3 x^3 g^{-3} $$

or $ x^3 g^2 = g^2 x^3 $, so that any square commutes with any cube, and therefore any element of the group $ G $ as $ \phi $ is surjective. Now, we have

$$ (xy)(xy)^2 = (xy)^3 = x^3 y^3 = x x^2 y y^2 = (xy)(x^2 y^2) $$

and cancellation yields that $ yx = xy $, as desired.

Abelian finite group

3 Answers3