Permutationen hinzugefügt

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diff --git a/buch/chapters/50-permutationen/uebungsaufgaben/5001.tex b/buch/chapters/50-permutationen/uebungsaufgaben/5001.tex
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+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}