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 Post subject: On group theoryPosted: Sun Jun 26, 2016 6:48 am
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Joined: Mon Nov 09, 2015 1:52 pm
Posts: 426
1. Let $\displaystyle{n\in\mathbb{N}\,\,,n\geq 3}$ . Prove that there exists a finite group $\displaystyle{\left(G,\cdot\right)}$ which contains two elements $\displaystyle{x\,,y}$ of order $\displaystyle{2}$ but their product $\displaystyle{x\cdot y}$ is of order $\displaystyle{n}$ .

2. Give an example of an infinite group $\displaystyle{\left(G,\cdot\right)}$ having the property :
Each element of $\displaystyle{G}$ is of finite order.

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 Post subject: Re: On group theoryPosted: Sun Jun 26, 2016 6:49 am
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Joined: Mon Nov 09, 2015 1:52 pm
Posts: 426
Answer for the first one :

Consider the finite group $\displaystyle{\left(GL_{2}(\mathbb{Z_{n}}),\cdot\right)}$ .

Let $\displaystyle{A=\begin{pmatrix} -1 &1 \\ 0& 1 \end{pmatrix}\,\,,B=\begin{pmatrix} -1 &0\\ 0&1 \end{pmatrix}\in GL_{2}(\mathbb{Z}_{n})}$ .

We have that :

$\displaystyle{A^2=\begin{pmatrix} -1 &1 \\ 0&1 \end{pmatrix}\cdot \begin{pmatrix} -1&1 \\ 0&1 \end{pmatrix}=\begin{pmatrix} 1&0 \\ 0&1 \end{pmatrix}=I_{2}}$

and

$\displaystyle{B^2=\begin{pmatrix} -1 &0 \\ 0&1 \end{pmatrix}\cdot \begin{pmatrix} -1&0 \\ 0&1 \end{pmatrix}=\begin{pmatrix} 1&0 \\ 0&1 \end{pmatrix}=I_{2}}$

so : $\displaystyle{o(A)=o(B)=2}$

However,

$\displaystyle{A\cdot B=\begin{pmatrix} 1&1 \\ 0&1 \end{pmatrix}}$

and by induction $\displaystyle{(A\cdot B)^{m}=\begin{pmatrix} 1&m \\ 0&1 \end{pmatrix}\,\,,\forall\,m\in\mathbb{N}}$

so : $\displaystyle{o(A\cdot B)=n}$ .

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 Post subject: Re: On group theoryPosted: Sun Jun 26, 2016 6:50 am
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Joined: Mon Nov 09, 2015 1:36 am
Posts: 457
Location: Ioannina, Greece
Answer for the second: $\mathbb{Z}_2\times\mathbb{Z}_2\times\mathbb{Z}_2\times\cdots$
Every non-zero element has inverse itself and, therefore, has order $2$.

_________________
Grigorios Kostakos

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 Post subject: Re: On group theoryPosted: Sun Jun 26, 2016 6:51 am
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Joined: Mon Nov 09, 2015 1:52 pm
Posts: 426
Thank you Grigorios. More analytically :

The set $\displaystyle{\prod_{n=1}^{\infty}\mathbb{Z}_{2}}$ contains all the sequences

$\displaystyle{x:\mathbb{N}\longrightarrow \mathbb{Z}_{2}}$ .

If $\displaystyle{x\,,y\in\prod_{n=1}^{\infty}\mathbb{Z}_{2}}$, then we define

$\displaystyle{x+y:\mathbb{N}\longrightarrow \mathbb{Z}_{2}}$ by $\displaystyle{(x+y)_{n}=x_{n}+y_{n}}$ .

Then, the pair $\displaystyle{\left(\prod_{n=1}^{\infty}\mathbb{Z}_{2},+\right)}$ is an abelian group with the sequence

$\displaystyle{\mathbb{O}:\mathbb{N}\longrightarrow \mathbb{Z}_{2}\,\,,\mathbb{O}(n)=\left[0\right]_{2}}$

be the zero element. If $\displaystyle{n\in\mathbb{N}}$ then we define :

$\displaystyle{x_{n}:\mathbb{N}\longrightarrow \mathbb{Z}_{2}\,\,,x_{n}(i)=\left[1\right]_{n}\,,i=n\,\,\,,x_{n}(i)=\left[0\right]_{n}\,\,,i\neq n}$

and thus :

$\displaystyle{\left\{x_{n}\in\prod_{k=1}^{\infty}\mathbb{Z}_{2}: n\in\mathbb{N}\right\}\subseteq \prod_{k=1}^{\infty}\mathbb{Z}_{2}}$

which means that $\displaystyle{\left|\mathbb{Z}_{2}\times \mathbb{Z}_{2}\times...\right|=\infty}$ .

Finally, if $\displaystyle{x\in\prod_{n=1}^{\infty}\mathbb{Z}_{2}-\left\{\mathbb{O}\right\}}$ , then

$\displaystyle{\forall\,n\in\mathbb{N}: 2\,x_{n}=x_{n}+x_{n}=\left[0\right]_{2}=\mathbb{O}(n)\implies 2\,x=\mathbb{O}}$ .

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