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diff --git a/buch/chapters/50-permutationen/uebungsaufgaben/5001.tex b/buch/chapters/50-permutationen/uebungsaufgaben/5001.tex new file mode 100644 index 0000000..2893adf --- /dev/null +++ b/buch/chapters/50-permutationen/uebungsaufgaben/5001.tex @@ -0,0 +1,121 @@ +Sind die beiden Permutationen +\[ +\sigma_1 += +\begin{pmatrix} +1& 2& 3& 4& 5& 6& 7& 8\\ +8& 6& 5& 7& 2& 3& 4& 1 +\end{pmatrix} +\qquad\text{und}\qquad +\sigma_2 += +\begin{pmatrix} +1& 2& 3& 4& 5& 6& 7& 8\\ +8& 7& 5& 6& 3& 4& 1& 2 +\end{pmatrix} +\] +konjugiert in $S_8$? +Wenn ja, finden Sie eine Permutation $\gamma$ derart, dass +$\gamma\sigma_1\gamma^{-1}=\sigma_2$ + +\begin{loesung} +Die Zyklenzerlegungen von $\sigma_1$ und $\sigma_2$ sind +\begin{center} +\begin{tikzpicture}[>=latex,thick] +\begin{scope}[xshift=-3.3cm] +\node at (-0.25,1.7) {$\sigma_1$}; +\draw (-3.3,-1.3) rectangle (2.8,1.3); +\coordinate (A) at (-2.4,0.5); +\coordinate (B) at (-2.4,-0.5); +\coordinate (C) at (-0.8,0.5); +\coordinate (D) at (-0.8,-0.5); +\coordinate (E) at (0.8,0.5); +\coordinate (F) at (0.8,-0.5); +\coordinate (G) at (1.8,0.5); +\coordinate (H) at (1.8,-0.5); + +\draw[->] (E) to[out=-135,in=135] (F); +\draw[->] (F) to[out=-45,in=-135] (H); +\draw[->] (H) to[out=45,in=-45] (G); +\draw[->] (G) to[out=135,in=45] (E); + +\draw[->] (A) to[out=-180,in=-180] (B); +\draw[->] (B) to[out=0,in=0] (A); + +\draw[->] (C) to[out=-180,in=-180] (D); +\draw[->] (D) to[out=0,in=0] (C); + +\node at (A) [above] {$1$}; +\node at (B) [below] {$8$}; +\node at (C) [above] {$4$}; +\node at (D) [below] {$7$}; +\node at (E) [above left] {$2$}; +\node at (F) [below left] {$6$}; +\node at (H) [below right] {$3$}; +\node at (G) [above right] {$5$}; + +\fill (A) circle[radius=0.05]; +\fill (B) circle[radius=0.05]; +\fill (C) circle[radius=0.05]; +\fill (D) circle[radius=0.05]; +\fill (E) circle[radius=0.05]; +\fill (F) circle[radius=0.05]; +\fill (G) circle[radius=0.05]; +\fill (H) circle[radius=0.05]; +\end{scope} +\begin{scope}[xshift=3.3cm] +\node at (-0.25,1.7) {$\sigma_2$}; +\draw (-3.3,-1.3) rectangle (2.8,1.3); +\coordinate (A) at (-2.4,0.5); +\coordinate (B) at (-2.4,-0.5); +\coordinate (C) at (-0.8,0.5); +\coordinate (D) at (-0.8,-0.5); +\coordinate (E) at (0.8,0.5); +\coordinate (F) at (0.8,-0.5); +\coordinate (G) at (1.8,0.5); +\coordinate (H) at (1.8,-0.5); + +\draw[->] (E) to[out=-135,in=135] (F); +\draw[->] (F) to[out=-45,in=-135] (H); +\draw[->] (H) to[out=45,in=-45] (G); +\draw[->] (G) to[out=135,in=45] (E); + +\draw[->] (A) to[out=-180,in=-180] (B); +\draw[->] (B) to[out=0,in=0] (A); + +\draw[->] (C) to[out=-180,in=-180] (D); +\draw[->] (D) to[out=0,in=0] (C); + +\node at (A) [above] {$3$}; +\node at (B) [below] {$5$}; +\node at (C) [above] {$4$}; +\node at (D) [below] {$6$}; +\node at (E) [above left] {$7$}; +\node at (F) [below left] {$1$}; +\node at (H) [below right] {$8$}; +\node at (G) [above right] {$2$}; + +\fill (A) circle[radius=0.05]; +\fill (B) circle[radius=0.05]; +\fill (C) circle[radius=0.05]; +\fill (D) circle[radius=0.05]; +\fill (E) circle[radius=0.05]; +\fill (F) circle[radius=0.05]; +\fill (G) circle[radius=0.05]; +\fill (H) circle[radius=0.05]; +\end{scope} +\end{tikzpicture} +\end{center} +Da die beiden Permutationen die gleiche Zyklenzerlegung haben, müssen +sie konjugiert sein. +Die Permutation +\[ +\gamma += +\begin{pmatrix} +1&2&3&4&5&6&7&8\\ +6&5&1&4&8&7&2&3 +\end{pmatrix} +\] +bildet die Zyklenzerlegung ab, also ist $\gamma\sigma_1\gamma^{-1}=\sigma_2$. +\end{loesung} |