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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}

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.

 

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