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I have the group $\mathbb{Z}/18\mathbb{Z}$ with the operation of sum of classes of rest. I have to find, for each element, the order and the cyclic subgroup generated.

My attempt:

  1. $\mathbb{Z}/18\mathbb{Z}=\{0,1,.......,17\}$, order of $\mathbb{Z}/18\mathbb{Z}$ is $18$.
  2. $(\mathbb{Z}/18\mathbb{Z},+)$ is a cyclic group generated by [1]. order(k[1])=order([k])=${{18}\over{MCD(18,k)}}$.

So I have order([1])=18,order([2])=9...... how can I find elements of <[3]>?

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First, there is a bijection between that subgroups of $\mathbb Z$ that contain $18\mathbb Z$ and the subgroups of $\mathbb Z/18\mathbb Z$. Therefore if you find the subgroup of $\mathbb Z$ that contain $18\mathbb Z$, the rest goes on...

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