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I am trying to learn a bit of abstract algebra and Galois theory on my own. I have been watching some videos and doing some readings, in particular I was watching this video: https://youtu.be/pdYe4BKcm74

When at minute 23:10 he says that one of the properties is that the modulo between two consecutive groups of the chain needs to be cyclic. Going back to the examples shown in the video, at some point the chain of relations that he creates is

$$S_4 \supseteq D_4 \supseteq K \supseteq Z_2 \supseteq E$$

which means, for example, that $D_4/K_4$ must be cyclic. I can't convince myself of this. I tried to do the Cayley table but I couldn't get it to be cyclic. Am I misunderstanding something or I just couldn't figure out the right way of doing it?

Thanks for the help!

EDIT: I think my question is different because I am not asking for general conditions but rather for a specific "bruteforce" computing to prove that that's actually cyclic. I mean, how does that quotient group looks like?

Bernard
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dd95
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  • I edited the question to explain why I feel like this is different from the one you linked – dd95 Jun 08 '19 at 14:23
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    Please note that $D_{4}$ is not mormal in $S_{4}$, and in fact in the video the chain starts with $D_{4}$, Anyway, $D_{4}/K_{4}$ has order $2$, so it is cyclic, as a group of prime order. – Andreas Caranti Jun 08 '19 at 14:43
  • If in the $D_4$ there are 8 elements and in $K_4$ there are 4, I understand that 8/4 = 2, but what are those 2 elements? I guess they are 2 cosets? – dd95 Jun 08 '19 at 14:50
  • The cosets of $K_4$ in $D_4$ is precisely $K_4$ and its complement. – user10354138 Jun 08 '19 at 14:51
  • https://math.stackexchange.com/questions/84632/subgroup-of-index-2-is-normal –  Jun 08 '19 at 14:53
  • That actually confuses me, because $K_4$ is not cyclic – dd95 Jun 08 '19 at 14:53
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    $K_4$ need not be cyclic for $D_4/K_4$ to be cyclic. $D_4/K_4$ is the group consisting of all cosets of $K_4$ in $D_4$, and since the order of the group $D_4/K_4$ is 2, then $D_4/K_4$ is isomorphic to $C_2$ which is cyclic. (up to isomorphism there is only one group of order 2, and that group is $C_2$). – Locally unskillful Jun 10 '19 at 14:17

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