#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.

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