# Abelian finite group [duplicate]

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.

## marked as duplicate by almagest, drhab, tomasz, Morgan Rodgers, user1551May 12 '16 at 14:17

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

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

$\implies baba = aabb$

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

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

Hint:

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

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

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

• 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

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.