Explicație pas cu pas:
Cautam cel mai mic k nenul astfel incat sa avem: [tex] \sigma_2^k=e [/tex].
Calculam: [tex] \sigma_2^2=\sigma_2*\sigma_2 [/tex].
[tex]\sigma_2^2=\left(\begin{array}{ccccc}1&2&3&4&5\\2&1&4&5&3\end{array}\right) *\left(\begin{array}{ccccc}1&2&3&4&5\\2&1&4&5&3\end{array}\right) =\left(\begin{array}{ccccc}1&2&3&4&5\\1&2&5&3&4\end{array}\right)[/tex]
Calculam: [tex] \sigma_2^3=\sigma_2*\sigma_2^2 [/tex].
[tex]\sigma_2^3=\left(\begin{array}{ccccc}1&2&3&4&5\\2&1&4&5&3\end{array}\right) *\left(\begin{array}{ccccc}1&2&3&4&5\\1&2&5&3&4\end{array}\right) =\left(\begin{array}{ccccc}1&2&3&4&5\\2&1&3&4&5\end{array}\right)[/tex]
Calculam: [tex] \sigma_2^4=\sigma_2*\sigma_2^3 [/tex].
[tex]\sigma_2^4=\left(\begin{array}{ccccc}1&2&3&4&5\\2&1&4&5&3\end{array}\right) *\left(\begin{array}{ccccc}1&2&3&4&5\\2&1&3&4&5\end{array}\right) =\left(\begin{array}{ccccc}1&2&3&4&5\\1&2&4&5&3\end{array}\right)[/tex]
Calculam: [tex] \sigma_2^5=\sigma_2*\sigma_2^4 [/tex].
[tex]\sigma_2^5=\left(\begin{array}{ccccc}1&2&3&4&5\\2&1&4&5&3\end{array}\right) *\left(\begin{array}{ccccc}1&2&3&4&5\\1&2&4&5&3\end{array}\right) =\left(\begin{array}{ccccc}1&2&3&4&5\\2&1&5&3&4\end{array}\right)[/tex]
Calculam: [tex] \sigma_2^6=\sigma_2*\sigma_2^5 [/tex].
[tex]\sigma_2^6=\left(\begin{array}{ccccc}1&2&3&4&5\\2&1&4&5&3\end{array}\right) *\left(\begin{array}{ccccc}1&2&3&4&5\\2&1&5&3&4\end{array}\right) =\left(\begin{array}{ccccc}1&2&3&4&5\\1&2&3&4&5\end{array}\right) =\e[/tex]
Cum cel mai mic k care satisface relatia data este 6 si observand ca puterile lui [tex] \sigma [/tex] se repeta din 6 in 6, inseamna ca [tex] k\in M_6 [/tex] si, deci, 6 | k.
