Fix several typos
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@ -403,10 +403,10 @@ $e = 0, g^n = n \cdot g, \inv g = -g$
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$a^r = e, b^s = e \implies a^{rs} = e, b^{rs} = e \overset{\text{abelsch}}{\implies} (ab)^rs = e
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\implies \ord_G(ab) \mid rs$ \\
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Angenommen: $\ord_G(ab) < rs$. Dann $\exists p \in \Primes: (ab)^{\frac{rs}p} = e$. \\
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Sei oBdA $p\mid r (\implies p \nmid r, \text{ weil }(r, s) = 1)$
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Sei oBdA $p\mid r (\implies p \nmid s, \text{ weil }(r, s) = 1)$
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\[
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(ab)^{\frac{rs}p} = a^{\frac{rs}p} \cdot b^{\frac{rs}p} = (a^s)^{\frac rp} \cdot
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{\underbrace{(b^s)}_{=e}}^{\frac{rs}p} = (a^s)^{\frac{rs}p} = e
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{\underbrace{(b^s)}_{=e}}^{\frac{r}p} = (a^s)^{\frac{r}p} = e
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\]
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Die Ordnung von $a$ und daher von $a^s$ ist aber $r$ und daher sicher $>
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\frac rp$. \qed
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