Exercise 1.2.8 (Order of the cyclic subgroup of $D_{2n}$ generated by $r$)

Question

Find the order of the cyclic subgroup of $D_{2n}$ generated by $r$ (cf Exercise 27 of Section 1).

Richard Ganaye · Accepted Answer

\begin{proof} 
Since the order of $r$ is $n$, 
$$\langle r \rangle = \{e, r, r^2, r^3, \ldots, r^{n-1} \}.$$
The elements $e, r, r^2, r^3, \ldots, r^{n-1}$ are distinct, otherwise $r^i = r^j$, where $0 \leq i < j <n$ implies that $r^{j-i} = e$, where $0 < j-i < n$, in contradiction with $|r| = n$. Therefore
$$|\langle r \rangle | = n.$$
\end{proof}

Exercise 1.2.8 (Order of the cyclic subgroup of $D_{2n}$ generated by $r$)

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Exercise 1.2.8 (Order of the cyclic subgroup of $D_{2n}$ generated by $r$)

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