The problem is to define all subgroups of $(\mathbb Z_n,+), n \in \mathbb N$. My guess is if n is prime number, then there is only trivial subgroups. If n is not prime, then I can factorize it, and every prime divisor will generate it's own subgroup in $(\mathbb Z_n,+)$. That means that $(\mathbb Z_n,+) \cong (\mathbb Z_h,+) \times (\mathbb Z_k,+) \times \dots$ , $h,k \in \mathbb N$ are the prime factors of $n$. The problem is that I don't know how to prove it. It's pretty easy to show that, for example, in $(\mathbb Z_6,+)$ $ [2]_6$ and $[3]_6$ generate their own subgroups and $[1]_6$ generate entire $(\mathbb Z_6,+)$, but I don't know how to show that $[5]_6$ do the same, except by showing it all: $[5]^2_6 = [4]_6$ and so on.

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    $\begingroup$ You're right, if $n$ is prime, then the only subgroups are $\mathbb{Z}_n$ and $\{0\}$. However, it's not always true that $\mathbb{Z}_n$ is the direct product like you've written. It's true if the factors are single prime factors. But $\mathbb{Z}_4$, for example, is not $\mathbb{Z}_2 \times \mathbb{Z}_2$ (which is the Klein four-group). $\endgroup$ – M. Vinay Jun 7 '14 at 13:01
  • $\begingroup$ An element generates the same group as its inverse, so in the example you gave, $[1]_6$ and $[5]_6$, being inverses, certainly generate the same subgroup. $\endgroup$ – M. Vinay Jun 7 '14 at 13:08
  • $\begingroup$ I saw that, but I thought it was somehow connected with the fact that $4 = 2^2$ and have only one factor. Can you show me example of $\mathbb Z_n$, where $n$ can be factorized into two or more factors and still $(\mathbb Z_n,+) \cong (\mathbb Z_h,+) \times (\mathbb Z_k,+) \times \dots$ isn't true? $\endgroup$ – Dark Archon Jun 7 '14 at 13:10
  • $\begingroup$ It happens only when at least one factor is repeated. Are you asking an example of this with some factor(s) repeated but at least two distinct factors? $\endgroup$ – M. Vinay Jun 7 '14 at 13:11
  • $\begingroup$ Yes, I would be thankful for that. $\endgroup$ – Dark Archon Jun 7 '14 at 13:15

$\mathbb{Z}_n = \mathbb{Z} / n \mathbb{Z}$, so by the correspondence theorem its subgroups are in bijection with the subgroups of $\mathbb{Z} $, $r\mathbb{Z} $ such that $n\mathbb{Z} \subseteq r\mathbb{Z} \subseteq \mathbb{Z}$ .

But $n\mathbb{Z} \subseteq r\mathbb{Z} \Longleftrightarrow r \mid n$, and so the subgroups of $\mathbb{Z}_n $ are $$\lbrace r\mathbb{Z}/n\mathbb{Z} \ \ | \ \ \ r \mid n \rbrace$$

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    $\begingroup$ Could you explain the meaning of expression $\mathbb Z_n = \mathbb Z/n\mathbb Z$? $\endgroup$ – Dark Archon Jun 7 '14 at 13:18
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    $\begingroup$ It is the quotient group (en.wikipedia.org/wiki/Quotient_group). Briefly (and specifically in this case), consider the collection of subsets of $\mathbb{Z}$ defined as $\mathbb{Z}/n\mathbb{Z} = \{n\mathbb{Z} + k \mid k \in \mathbb{Z}\}$ (these subsets are called the cosets of the subgroup $n\mathbb{Z}$). This collection itself forms a group under the addition defined by $(n\mathbb{Z} + k) + (n\mathbb{Z} + l) = n\mathbb{Z} + (k + l)$. Check this out carefully and you will see that $\mathbb{Z}/n\mathbb{Z}$ under this addition behaves exactly like $\mathbb{Z}_n$ under $+_n$. $\endgroup$ – M. Vinay Jun 7 '14 at 13:40

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