It is currently Tue Jun 18, 2019 10:52 pm


All times are UTC [ DST ]




Post new topic Reply to topic  [ 2 posts ] 
Author Message
 Post subject: The group is abelian
PostPosted: Sun Nov 20, 2016 12:19 pm 
Administrator
Administrator
User avatar

Joined: Sat Nov 07, 2015 6:12 pm
Posts: 841
Location: Larisa
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.

_________________
Imagination is much more important than knowledge.
Image


Top
Offline Profile  
Reply with quote  

 Post subject: Re: The group is abelian
PostPosted: Fri May 19, 2017 6:49 am 

Joined: Sat Nov 14, 2015 6:32 am
Posts: 159
Location: Melbourne, Australia
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.

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


Top
Offline Profile  
Reply with quote  

Display posts from previous:  Sort by  
Post new topic Reply to topic  [ 2 posts ] 

All times are UTC [ DST ]


Mathimatikoi Online

Users browsing this forum: No registered users and 2 guests


You cannot post new topics in this forum
You cannot reply to topics in this forum
You cannot edit your posts in this forum
You cannot delete your posts in this forum
You cannot post attachments in this forum

Search for:
Jump to:  
cron
Powered by phpBB® Forum Software © phpBB Group Color scheme created with Colorize It.
Theme created StylerBB.net