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?
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Shaun
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What is your calculation for $(s r)^{2} = s r s r$? – Andreas Caranti Dec 13 '21 at 13:12
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I thought it would be $r^{2}$ because $s^{2}$ = $1$ but this must be wrong as my calculations are not correct. – Dec 13 '21 at 13:19
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You should have learned what $s r s = s^{-1} r s$ is. – Andreas Caranti Dec 13 '21 at 19:54
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@AndreasCaranti yes I understand the mistake I made now. $srsr = r^{-1}r = 1$ – Dec 30 '21 at 12:46