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Aufgabe 2
Es sei \( G=\left\{\sigma_{1}, \sigma_{2}, \sigma_{3}, \sigma_{4}, \sigma_{5}, \sigma_{6}\right\} \) die Gruppe der Permutationen auf der Menge \( M=\{1,2,3\} \)
\begin{tabular}{|c|c|c|c|}
\hline\( m \) & 1 & 2 & 3 \\
\hline\( \sigma_{1}(m) \) & 1 & 2 & 3 \\
\hline\( \sigma_{2}(m) \) & 1 & 3 & 2 \\
\hline\( \sigma_{3}(m) \) & 2 & 1 & 3 \\
\hline\( \sigma_{4}(m) \) & 3 & 2 & 1 \\
\hline\( \sigma_{5}(m) \) & 2 & 3 & 1 \\
\hline\( \sigma_{6}(m) \) & 3 & 1 & 2 \\
\hline
\end{tabular}
Das neutrale Element der Gruppe ist \( \sigma_{1} \), da \( \sigma_{1}(m)=m \), für alle \( m \in M \) gilt.
a) \( \left(\sigma_{2} \circ \sigma_{3}\right) \circ \sigma_{4}^{-1} \)
\( \begin{aligned} \sigma_{2} \circ \sigma_{3}=\sigma_{2} \circ \sigma_{3}(m) & =\sigma_{2}(2)=3 \\ & =\sigma_{2}(1)=1 \\ & =\sigma_{2}(3)=2 \\ & =\sigma_{6} \\ \sigma_{6} \circ \sigma_{4}^{-1}=\sigma_{6} \circ \sigma_{4}^{-1}(m) & =\sigma_{6}(1)=3 \\ & =\sigma_{6}(2)=1 \\ & =\sigma_{6}(3)=2 \\ & =\sigma_{6} \\ \left(\sigma_{2} \circ \sigma_{3}\right) \circ \sigma_{4}^{-1} & =\sigma_{6} \end{aligned} \)
b) \( \left(\sigma_{4}^{-1} \circ \sigma_{2}\right) \circ \sigma_{3} \)
\( \begin{aligned} \sigma_{4}^{-1} \circ \sigma_{2}=\sigma_{4}^{-1} \circ \sigma_{2}(m) & =\sigma_{4}^{-1}(1)=1 \\ & =\sigma_{4}^{-1}(3)=3 \\ & =\sigma_{4}^{-1}(2)=2 \\ & =\sigma_{2} \\ \sigma_{2} \circ \sigma_{3}=\sigma_{2} \circ \sigma_{3}(m) & =\sigma_{2}(2)=3 \\ & =\sigma_{2}(1)=1 \\ & =\sigma_{2}(3)=2 \\ & =\sigma_{6} \\ & \\ \left(\sigma_{4}^{-1} \circ \sigma_{2}\right) \circ \sigma_{3} & =\sigma_{6} \end{aligned} \)
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c) \( \left(\sigma_{2}^{-1} \circ \sigma_{5}\right) \circ \sigma_{2} \)
\( \begin{aligned} \sigma_{2}^{-1} \circ \sigma_{5}=\sigma_{2}^{-1} \circ \sigma_{5}(m) & =\sigma_{2}^{-1}(2)=3 \\ & =\sigma_{2}^{-1}(3)=2 \\ & =\sigma_{2}^{-1}(1)=1 \\ & =\sigma_{4} \\ & \\ \sigma_{4} \circ \sigma_{2}=\sigma_{4} \circ \sigma_{2}(m) & =\sigma_{4}(1)=3 \\ & =\sigma_{4}(3)=1 \\ & =\sigma_{4}(2)=2 \\ & =\sigma_{6} \\ \left(\sigma_{2}^{-1} \circ \sigma_{5}\right) \circ \sigma_{2} & =\sigma_{6} \end{aligned} \)
d) \( \left(\sigma_{2} \circ \sigma_{3}\right) \circ \sigma_{5}^{-1} \)
\( \begin{aligned} \sigma_{2} \circ \sigma_{3}=\sigma_{2} \circ \sigma_{3}(m) & =\sigma_{2}(2)=3 \\ & =\sigma_{2}(1)=1 \\ & =\sigma_{2}(3)=2 \\ & =\sigma_{6} \\ \sigma_{6} \circ \sigma_{5}^{-1}=\sigma_{6} \circ \sigma_{5}^{-1} & =\sigma_{6}(2)=1 \\ & =\sigma_{6}(3)=2 \\ & =\sigma_{6}(1)=3 \\ & =\sigma_{1} \\ \left(\sigma_{2} \circ \sigma_{3}\right) \circ \sigma_{5}^{-1} & =\sigma_{1} \end{aligned} \)
Die Permutationsgruppe ist nicht abelsch, da z.B. \( \sigma_{2} \circ \sigma_{3} \neq \sigma_{3} \circ \sigma_{2} \).
\( \begin{aligned} \sigma_{2} \circ \sigma_{3}=\sigma_{2} \circ \sigma_{3}(m) & =\sigma_{2}(2)=3 \\ & =\sigma_{2}(1)=1 \\ & =\sigma_{2}(3)=2 \\ & =\sigma_{6} \\ \sigma_{3} \circ \sigma_{2}=\sigma_{3} \circ \sigma_{2}(m) & =\sigma_{3}(1)=2 \\ & =\sigma_{3}(3)=3 \\ & =\sigma_{3}(2)=1 \\ & =\sigma_{5} \end{aligned} \)
