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 Post subject: On group theory
PostPosted: Sun Jun 26, 2016 6:48 am 
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Joined: Mon Nov 09, 2015 1:52 pm
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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 theory
PostPosted: Sun Jun 26, 2016 6:49 am 
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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 theory
PostPosted: Sun Jun 26, 2016 6:50 am 
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Joined: Mon Nov 09, 2015 1:36 am
Posts: 454
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 \(2\).

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Grigorios Kostakos


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 Post subject: Re: On group theory
PostPosted: Sun Jun 26, 2016 6:51 am 
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Joined: Mon Nov 09, 2015 1:52 pm
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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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