Răspuns:
i)
[tex]\sigma^2=\begin{pmatrix}1 & 2 & 3\\3 & 1 & 2\end{pmatrix}\begin{pmatrix}1 & 2 & 3\\3 & 1 & 2\end{pmatrix}=\begin{pmatrix}1 & 2 & 3\\2 & 3 & 1\end{pmatrix}[/tex]
[tex]\sigma^3=\sigma^2\sigma=\begin{pmatrix}1 & 2 & 3\\2 & 3 & 1\end{pmatrix}\begin{pmatrix}1 & 2 & 3\\3 & 1 & 2\end{pmatrix}=\begin{pmatrix}1 & 2 & 3\\1 & 2 & 3\end{pmatrix}=e[/tex]
Atunci
[tex]\sigma^n=\begin{cases}e, & n=3k\\\sigma, & n=3k+1\\\sigma^2, & n=3k+2\end{cases}[/tex]
deci mulțimea puterilor lui [tex]\sigma[/tex] este
[tex]\{e,\sigma,\sigma^2\}[/tex]
ii)
[tex]x\sigma=\sigma x\Leftrightarrow x(\sigma(k))=\sigma(x(k)), \ \forall k\in\{1,2,3\}[/tex]
Dacă [tex]x(1)=1\Rightarrow\sigma(x(1))=\sigma(1)=3[/tex]
[tex]x(\sigma(1))=x(3)\Rightarrow x(3)=3[/tex]
[tex]\sigma(x(3))=\sigma(3)=2, \ x(\sigma(3))=x(2)\Rightarrow x(2)=2\Rightarrow x=e[/tex]
Dacă [tex]x(1)=2\Rightarrow\sigma(x(1))=\sigma(2)=1, \ x(\sigma(1))=x(3)\Rightarrow x(3)=1[/tex]
[tex]\sigma(x(3))=\sigma(1)=3, \ x(\sigma(3))=x(2)\Rightarrow x(2)=3[/tex]
Deci [tex]x=\begin{pmatrix}1 & 2 & 3\\2 & 3 & 1\end{pmatrix}=\sigma^2[/tex]
Dacă [tex]x(1)=3\Rightarrow \sigma(x(1))=\sigma(3)=1, \ x(\sigma(1))=x(3)\Rightarrow x(3)=1[/tex]
[tex]\sigma(x(3))=\sigma(1)=3, \ x(\sigma(3))=x(2)\Rightarrow x(2)=3[/tex], ceea ce nu se poate.
Deci [tex]x\in\{e, \sigma^2\}[/tex]
iii)
Dacă [tex]x(1)=1\Rightarrow x(x(1))=\sigma(1)\Rightarrow x(1)=3[/tex], contradicție.
Dacă
[tex]x(1)=2\Rightarrow x(x(1))=\sigma(1)\Rightarrow x(2)=3\Rightarrow x(x(2))=\sigma(2)\Rightarrow x(3)=1\Rightarrow\\\Rightarrow x=\begin{pmatrix}1 & 2 & 3\\2 & 3 & 1\end{pmatrix}[/tex]
Dacă [tex]x(1)=3\Rightarrow x(x(1))=\sigma(1)\Rightarrow x(3)=3[/tex], contradicție.
Deci
[tex]x=\begin{pmatrix}1 & 2 & 3\\2 & 3 & 1\end{pmatrix}[/tex]
Explicație pas cu pas:
