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Data publicarii: 24 Octombrie, 2014

EXERCITIUL 6

Suport teoretic:

Permutari,substitutii,compunerea permutarilor,transpozitii,permutarea identica,

compunerea functiilor.

Enunt:

Sa se scrie sub forma de produs de transpozitii urmatoarea substitutie:

\sigma=\begin{pmatrix}1&2&3&4&5\\3&5&2&1&4\end{pmatrix}.\sigma=\begin{pmatrix}1&2&3&4&5\\3&5&2&1&4\end{pmatrix}.

Raspuns:

\sigma=(34)\circ(23)\circ(14).\sigma=(34)\circ(23)\circ(14).  

Rezolvare:

Fie transpozitia

{\tau}_1=(14)\;si\;{\sigma}\circ{\tau_1}={\tau}_1=(14)\;si\;{\sigma}\circ{\tau_1}= {\begin{pmatrix}1&2&3&4&5\\3&5&2&1&4\end{pmatrix}}\circ{(14)}{\begin{pmatrix}1&2&3&4&5\\3&5&2&1&4\end{pmatrix}}\circ{(14)} =\begin{pmatrix}1&2&3&4&5\\1&5&2&3&4\end{pmatrix}.=\begin{pmatrix}1&2&3&4&5\\1&5&2&3&4\end{pmatrix}.

Fie transpozitia

{\tau}_2=(23)\;si\;{\sigma}\circ{\tau_1}\circ{\tau_2}={\tau}_2=(23)\;si\;{\sigma}\circ{\tau_1}\circ{\tau_2}= {\begin{pmatrix}1&2&3&4&5\\1&5&2&3&4\end{pmatrix}}\circ{(23)}{\begin{pmatrix}1&2&3&4&5\\1&5&2&3&4\end{pmatrix}}\circ{(23)} =\begin{pmatrix}1&2&3&4&5\\1&2&4&3&5\end{pmatrix}=\begin{pmatrix}1&2&3&4&5\\1&2&4&3&5\end{pmatrix} =(34)=\tau_3.=(34)=\tau_3.

Deci am obtinut:

{\sigma}\circ{\tau_1}\circ{\tau_2}=\tau_3.{\sigma}\circ{\tau_1}\circ{\tau_2}=\tau_3.

Compunand, la dreapta, cu  \tau_2\tau_2 ,  apoi cu  \tau_1\tau_1 , se obtine,succesiv:

{\sigma}\circ{\tau_1}\circ{\tau_2}\circ{\tau_2}\circ{\tau_1}={\tau_3}\circ{\tau_2}\circ{\tau_1}{\sigma}\circ{\tau_1}\circ{\tau_2}\circ{\tau_2}\circ{\tau_1}={\tau_3}\circ{\tau_2}\circ{\tau_1} \Leftrightarrow\Leftrightarrow

\Leftrightarrow\Leftrightarrow {\sigma}\circ{\tau_1}\circ{e}\circ{\tau_1}={\tau_3}\circ{\tau_2}\circ{\tau_1}{\sigma}\circ{\tau_1}\circ{e}\circ{\tau_1}={\tau_3}\circ{\tau_2}\circ{\tau_1} \Leftrightarrow\Leftrightarrow

\Leftrightarrow\Leftrightarrow {\sigma}\circ{\tau_1}\circ{\tau_1}={\tau_3}\circ{\tau_2}\circ{\tau_1}{\sigma}\circ{\tau_1}\circ{\tau_1}={\tau_3}\circ{\tau_2}\circ{\tau_1} \Leftrightarrow\Leftrightarrow  

\Leftrightarrow\Leftrightarrow {\sigma}\circ{e}={\sigma}\circ{e}= {\tau_3}\circ{\tau_2}\circ{\tau_1}{\tau_3}\circ{\tau_2}\circ{\tau_1} \Leftrightarrow\Leftrightarrow

\Leftrightarrow\Leftrightarrow \sigma={\tau_3}\circ{\tau_2}{\circ}{\tau_1}.\sigma={\tau_3}\circ{\tau_2}{\circ}{\tau_1}. \Leftrightarrow\Leftrightarrow \sigma=(34)\circ(23)\circ(14).\sigma=(34)\circ(23)\circ(14).  

Observatie:

In calculele de mai sus, s-a tinut cont de faptul ca permutarile sunt functii (bijective),

compunerea acestora fiind asociativa si necomutativa. In plus, compunerea unei

transpozitii cu ea insasi este egala cu "e" (permutarea identica).


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