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Ameyah 2018-10-12 17:43:52 +02:00
commit af18a2093d
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@ -43,7 +43,7 @@ Sei $G$ eine Gruppe, $g\in G$.
\item $m=\infty\Rightarrow\ord(g)=\infty$: klar
\item $m<\infty\Rightarrow\langle g\rangle=\{g^k\mid 0\le k<m\}$: Für $k\in\whole$ schreibe $k=qm+r$ mit $q,r\in\whole$ und $0\le r<m$
\begin{align}
g^k=g^{qm+r}=(\underbrace{g^m}_{=1})^q\cdot g^r=g^r\in{^1,g,...,g^{m-1}}\notag
g^k=g^{qm+r}=(\underbrace{g^m}_{=1})^q\cdot g^r=g^r\in\{1,g,...,g^{m-1}\}\notag
\end{align}
\end{itemize}
\end{proof}

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