On group theory
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On group theory
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.
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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Re: On group theory
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 : \(\displaystyle{o(A\cdot B)=n}\) .
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 : \(\displaystyle{o(A\cdot B)=n}\) .
- Grigorios Kostakos
- Founder
- Posts: 461
- Joined: Mon Nov 09, 2015 1:36 am
- Location: Ioannina, Greece
Re: On group theory
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 \(2\).
Every non-zero element has inverse itself and, therefore, has order \(2\).
Grigorios Kostakos
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- Community Team
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- Joined: Mon Nov 09, 2015 1:52 pm
Re: On group theory
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
\(\displaystyle{\forall\,n\in\mathbb{N}: 2\,x_{n}=x_{n}+x_{n}=\left[0\right]_{2}=\mathbb{O}(n)\implies 2\,x=\mathbb{O}}\) .
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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