The group is abelian

Let $\mathcal{G}$ be a finite group such that $\left ( \left | \mathcal{G} \right | , 3 \right ) =1$ . If for the elements $a, \beta \in \mathcal{G}$ it holds that

$\left ( a \beta \right )^3 = a^3 \beta^3$

then prove that $\mathcal{G}$ is abelian.

Solution

Let $x \in \mathcal{G}$ such that $x^3=e$ . If $x \neq e$ then the order of $x$ would be $3$ . This would immediately imply that the order of $x$ would divide the order of the group $\mathcal{G}$ . This is an obscurity due to the data of the exercise. Thus $x=e$ . As $\left ( a \beta \right )^3 = a^3 \beta^3$ we conclude that the mapping $f(x)=x^3$ is an $1-1$ group homomorphism.

Therefore forall $a, \beta \in \mathcal{G}$ we have $\beta a \beta a = a a \beta \beta$ or equivelantly $\left ( \beta a \right ) ^2 = a^2 \beta^2$ . Taking advantage of the last relation we get that:

$\begin{aligned} \left ( a \beta \right )^4 &=\left ( \left ( a \beta \right )^2 \right )^2 \\ &= \left ( \beta^2 a^2 \right )^2\\ &= \left ( a^2 \right )^2 \left ( \beta^2 \right )^2 \\ &= a^4 \beta^4\\ &= a a a a \beta \beta \beta \beta \end{aligned}$

as well as

$\begin{aligned} \left ( a \beta \right )^4 &=a \beta a \beta a \beta a \beta \\ &=a \left ( \beta a \right )^3 \beta\\ &= a \beta^3 a^3 \beta \\ &= a \beta \beta \beta a a a \beta \end{aligned}$