4
$\begingroup$

This question already has an answer here:

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.

$\endgroup$

marked as duplicate by almagest, drhab, tomasz, Morgan Rodgers, user1551 May 12 '16 at 14:17

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

2
$\begingroup$

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

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

$\implies baba = aabb$

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

$\endgroup$
  • 1
    $\begingroup$ Can you please explain more? $\endgroup$ – Qwerty May 12 '16 at 13:22
2
$\begingroup$

Hint:

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

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

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

$\endgroup$
  • 1
    $\begingroup$ can you please explian more? $\endgroup$ – Qwerty May 12 '16 at 13:22
  • $\begingroup$ And please upvote if you liked my question $\endgroup$ – Qwerty May 12 '16 at 13:23
2
$\begingroup$

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.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.