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 $a, \beta \in \mathcal{G}$ 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 $\circ \left ( x \right ) \big| \left | \mathcal{G} \right |$ which 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. Thus for all $a,  \beta \in \mathcal{G}$ we have

\begin{align*}
\left ( a \beta \right )^3 &=a\beta a \beta a \beta \\
 &=a^3 \beta^3 \\
 &=a a a  \beta \beta \beta 
\end{align*}

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{align*}
\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{align*}

as well as

\begin{align*}
\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{align*}





The last two relations hold for all  $a, \beta \in \mathcal{G}$. Thus, for all  $a, \beta \in \mathcal{G}$ it holds that:

$$aaaa\beta \beta \beta \beta = a\beta \beta \beta aaa \beta$$

which in turn implies $f\left ( a \beta \right ) = a^3 \beta^3 = \beta^3 a^3 = f \left ( \beta  a \right )$ and since $f$ is $1-1$ we eventually get  $a \beta = \beta a$ proving the claim that $\mathcal{G}$ is abelian.