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- H22)\\
- $(\mathds{Z}_{20}^{*}, \cdot) = (\{1,3,7,9,11,13,17,19\},\cdot)$\\
- $|\mathds{Z}_{20}^{*}| = 8$\\
- \\
- \begin{tabular}{|l|l|l|l|l|l|l|l|l|}
- \hline
- $\cdot$ & 1 & 3 & 7 & 9 & 11 & 13 & 17 & 19 \\ \hline
- 1 & 1 & 3 & 7 & 9 & 11 & 13 & 17 & 19 \\ \hline
- 3 & 3 & 9 & 1 & 7 & 13 & 19 & 11 & 17 \\ \hline
- 7 & 7 & 1 & 9 & 3 & 17 & 11 & 19 & 13 \\ \hline
- 9 & 9 & 7 & 3 & 1 & 19 & 17 & 13 & 11 \\ \hline
- 11 & 11 & 13 & 17 & 19 & 1 & 3 & 7 & 9 \\ \hline
- 13 & 13 & 19 & 11 & 17 & 3 & 9 & 1 & 7 \\ \hline
- 17 & 17 & 11 & 19 & 13 & 7 & 1 & 9 & 3 \\ \hline
- 19 & 19 & 17 & 13 & 11 & 9 & 7 & 3 & 1 \\ \hline
- \end{tabular}\\
- \\
- \\
- H23)\\
- a)\\
- \underline{Behauptung:}\\
- Sind e und f neutrale Elemente, so folgt $e=f$.\\
- \underline{Beweis:}\\
- Zu zeigen: $e=f$\\
- $e=(g'*g)$ und $f=(g'*g)$ \hspace{1cm} $|(GA3)$\\
- also $e=(g'*g)=f$\\
- sodass $e=f$\\
- $\square$ \\
- \\
- b)\\
- \underline{Behauptung:}\\
- Sind h und k invers zu g, so ist $h=k$.\\
- \underline{Beweis:}\\
- Seien $g*h=e$ und $g*k=e$.\\
- Zu zeigen: $h=k$\\
- $h=h*e \hspace{1cm}| e=g*k$\\
- = $h*g*k \hspace{1cm} |g*h=e$ (GA3)\\
- = $e*k$ \hspace{1cm} $|(GA1)$\\
- = $k$ \hspace{1cm} $|(GA2)$\\
- also $h=k$\\
- $\square$ \\
- \\
- B6)\\
- $\mathds{Z}_{20}^{*} \leq \mathds{Z}_{20}^{*}$\\
- $(\{1\},\cdot) \leq \mathds{Z}_{20}^{*}$\\
- $(\{1,9\}\cdot) \leq \mathds{Z}_{20}^{*}$\\
- $(\{1,11\}\cdot) \leq \mathds{Z}_{20}^{*}$\\
- $(\{1,19\}\cdot) \leq \mathds{Z}_{20}^{*}$\\
- $(\{1,2,7,9\}\cdot) \leq \mathds{Z}_{20}^{*}$\\
- $(\{1,9,13,17\}\cdot) \leq \mathds{Z}_{20}^{*}$\\
- $(\{1,9,11,19\}\cdot) \leq \mathds{Z}_{20}^{*}$\\
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