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Being given $Z^*_{11} = \left\langle 2 \right\rangle$, find all its subgroups.

I know from a theorem that $\left\langle n/k\right\rangle$ is a unique subgroup for all $k|n$, but 2 doesn't fit, yet still generates the entire group. What is a method that I can use to find all subgroups of a certain group?

Gary
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Alan Raven
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  • Do you know that this group is $U(11)\cong C_{10}$, and that a cyclic group of order $n$ has a unique subgroup fo each divisor $d\mid n$? See here. – Dietrich Burde May 29 '20 at 08:30
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    Minor comment: when you write something like "$\langle n/k\rangle$ is a unique subgroup for all $k \mid n$", it seems like you are confusing every cyclic group to be $\Bbb Z/n\Bbb Z$. That is true only up to isomorphism. Here you should view $2$ as the generator of $Z_{11}^*$ of order $10$ and get the subgroups to be $\langle 2^{n/k}\rangle$ for $k \mid n$ with $n = 10$. – Aryaman Maithani May 29 '20 at 09:23
  • Thank you! I have managed to find the solution from this and the duplicate question's answer. – Alan Raven May 29 '20 at 09:28

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