Find all subgroups of a cyclic group with 12 elements and show why the list is complete.

My intuition here is that the list of subgroups will have order of divisors of 12. So there's 1, 2, 3, 4, 6, and 12, thus there will be 6 subgroups. I am unclear how to explicitly find these subgroups though.

  • $\begingroup$ Have you encountered the theorem that all subgroups of a cyclic group are cyclic? $\endgroup$ – Aaron Montgomery Oct 26 '17 at 0:00
  • $\begingroup$ Two hints: 1.All subgroups of cyclic groups are cyclic and 2. The order of a subgroup divides the order of the group (Lagrange's theorem) $\endgroup$ – Theo C. Oct 26 '17 at 0:00
  • $\begingroup$ We have done absolutely nothing with subgroups yet. All we have is the definition of a subgroup. $\endgroup$ – RZB Oct 26 '17 at 0:08
  • $\begingroup$ Think of the cyclic group as a clock with $n$ "hour" positions... here, $12$. Lagrange's Theorem states the order of the order of the subgroup divides the order of the group (as D. Beec states). So for $n = 2$ you have group elements at the "top" and "bottom" of the clock (the 12 o'clock and 6 o'clock positions). For $n = 3$ you have 12 o'clock, 4 o'clock and 8 o'clock. You do the rest... $\endgroup$ – David G. Stork Oct 26 '17 at 0:27

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