# Showing that a group $G$ such that 3 does not divide $|G|$ is Abelian.

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?

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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