Proof

It has been shown in the additive structure of integers that $(\mathbb Z,+)$ is a group.
By the criterion for subgroup, if $a=nk$ and $b=nl$ then $a-b=n(k-l)\in S_n,$ therefore, $S_n$ is a subgroup of $(\mathbb Z,+)$ for ever natural number $n\in\mathbb N.$
It remains to be shown that any subgroup of $(\mathbb Z,+)$ has the required form.
- Let $S$ be any subgroup of $(\mathbb Z,+).$
- If $S=\{0\}$, we have $S=S_0$ and we are done. Therefore, assume $S\neq \{0\}.$
- We have $0\in S$ and, since $(S,+)$ is closed under addition, we have $-a\in S$ for all $a\in S.$
- Let $n\in S$ be the smallest¹ positive integer.
- Since $n\in S$, we have $nk\in S$ for all $k\in\mathbb Z,$ therefore $S_n\subset S.$
- On the other hand, any $a\in S$ can be written as $a=nk+r$ with $k\in\mathbb Z$ and $0\le r < k$, by the division with quotient and remainder.
- Since $a-r=nk,$ and $nk\in S_n,$ we have $a-r\in S_n.$
- But $r=0,$ since $r > 0$ would contradict the choice of $n$ as the smallest positive integer of $S.$
- Therefore, $a\in S_n,$ which shows $S\subset S_n.$
- Altogether, we haver shown $S=S_n$ by the equality of sets.

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