Subgroups of $(\mathbb Z_n,+)$

Question

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.

Answer 1

$\mathbb{Z}_n = \mathbb{Z} / n \mathbb{Z}$, 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 $$\lbrace r\mathbb{Z}/n\mathbb{Z} \ \ | \ \ \ r \mid n \rbrace$$