The strategy is twofold: first verify that each $n\mathbb{Z}$ is indeed a subgroup, then use the well-ordering principle (via minimality) to show no other subgroups exist. The key idea is that if a subgroup contains any nonzero element, the Division Algorithm forces it to be generated by the smallest positive element it contains. **Step 1: Each $n\mathbb{Z}$ is a subgroup.** For any $n \geq 0$, the [set](/page/Set) $n\mathbb{Z} = \{nk : k \in \mathbb{Z}\}$ is a subgroup of $(\mathbb{Z}, +)$: it contains $0 = n \cdot 0$; if $na, nb \in n\mathbb{Z}$ then $na + nb = n(a+b) \in n\mathbb{Z}$; and if $na \in n\mathbb{Z}$ then $-(na) = n(-a) \in n\mathbb{Z}$. **Step 2: Every subgroup has this form.** [claim:Subgroups Determined By Smallest Positive Element] Let $H \leq \mathbb{Z}$. If $H \neq \{0\}$, then $H = n\mathbb{Z}$ where $n$ is the smallest positive element of $H$. [/claim] [proof] Since $H \neq \{0\}$, there exists a nonzero $h \in H$. If $h < 0$, then $-h \in H$ with $-h > 0$, so $H$ contains at least one positive element. By the well-ordering principle, let $n$ be the smallest positive element of $H$. Since $n \in H$ and $H$ is closed under addition and inverses, $n\mathbb{Z} \subseteq H$. For the reverse inclusion, let $m \in H$. By the Division Algorithm, write $m = qn + r$ with $0 \leq r < n$ and $q \in \mathbb{Z}$. Then: \begin{align*} r = m - qn = m + (-qn). \end{align*} Since $m \in H$ and $-qn \in n\mathbb{Z} \subseteq H$, closure gives $r \in H$. But $0 \leq r < n$ and $n$ is the smallest positive element of $H$, so $r = 0$. Therefore $m = qn \in n\mathbb{Z}$, giving $H \subseteq n\mathbb{Z}$. [/proof] **Step 3: Conclusion.** If $H = \{0\}$, then $H = 0\mathbb{Z}$. Otherwise, Step 2 gives $H = n\mathbb{Z}$ for the minimal positive $n \in H$. So every subgroup of $\mathbb{Z}$ is of the form $n\mathbb{Z}$ for some $n \geq 0$.