Mathematics Asked by user778657 on December 21, 2020

I asked this question here Understanding the hint of a question to show that $G$ is Abelian. but I did not receive answers to all my questions. So I am attaching a trial to the solution of the question which I found online and could not fully understand it.

**First**

Here is the main question: Let $G$ be a finite group such that 3 does not divide $|G|$ and such that the identity $(xy)^3 = x^3 y^3$ holds for all $x,y in G.$ Show that $G$ is abelian.

**And here is the hint I got for the question:**

First show that the map $G rightarrow G$ given by $x mapsto x^3$ is bijective. Then show that $x^2 in Z(G)$ for all $x in G.$

**And here is a trial for the solution:**

**My questions are:**

1-should we show that the map $G rightarrow G$ given by $x mapsto x^3$ is bijective because of the following step in the solution above:

$$(y^2x^2)^3 = (x^2y^2)^3 implies (y^2x^2) = (x^2y^2)$$ because there is no element in $G$ with order $3.$?

2- Also, how the given solution showed that $x^2 in Z(G)$ for all $x in G$? And can this statement be restated as $x^2 in Z(G)$ for all $y in G$?

3- Is the given trial of the solution correct? if not, how can we correct it?

Could anyone help me in answering those questions, please?

Given condition says $x mapsto x^3$ is a group homomorphism, it is injective because its kernel consists of elements of order $3$. It is then bijective by comparing size.

$xy^3x^{-1} = (xyx^{-1})^3 = x^3y^3x^{-3}$, hence $x^2$ commutes with $y^3$, and $y^3$ can be any element in $G$, so $x^2$ is in center for any $x in G$.

Then $x^3y^3 = (xy)^3 = x(xy)^2y$, i.e. $xy=yx$, $G$ is Abelian.

the given trial seems correct btw.

Correct answer by yisishoujo on December 21, 2020

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