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The group is abelian

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Tolaso J Kos
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The group is abelian

#1

Post by Tolaso J Kos » Sun Nov 20, 2016 12:19 pm

Here is an exercise that caught my attention the other day.

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.
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Riemann
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Re: The group is abelian

#2

Post by Riemann » Fri May 19, 2017 6:49 am

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{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.
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$
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