I've been trying to work out the cyclic subgroups of $D_{8}$ out by hand and thought I could do it but I've run into something I don't understand.
$D_{8}$ = $\{1,r,r^{2},r^{3},s,sr,sr^{2},sr^{3}\}$
I found $\{1\}, \{1,r,r^{2},r^{3}\}, \{1,r^{2}\}$
I thought it would be the same process for $s$ but I'm confused...
First of all there was $\{1,s\}$ which makes sense but then I tried to find the cyclic subgroup generated by $<sr>$ and got $\{1, sr, r^{2}, sr^{3}\}$ which is wrong as it should be $\{1,sr\}$ and the same applies for $<sr^{3}>$ being $\{1,sr^{3}\}$.
For some reason $\{1,sr^{2}\}$ was still correct for $<sr^{2}>$ though. Can anyone explain?
2026-03-29 11:43:37.1774784617
Working out cyclic subgroups of $D_{8}$
21 Views Asked by user984940 https://math.techqa.club/user/user984940/detail AtRelated Questions in GROUP-THEORY
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