Every subgroup of $(\mathbb{Z}, +)$ is $\langle n \rangle = n\mathbb{Z}$ for exactly one natural number $n$

$(\mathbb{Z},+,0)$ is an abelian group: addition is an associative and commutative binary operation, $0$ is a two-sided identity, and every $x$ has the two-sided inverse $-x$.

In this group the power $g^{x}$ is the ring product $gx$. For $x = \iota(k)$ with $k \in \mathbb{N}$, the function $k \mapsto g\,\iota(k)$ satisfies $g\,\iota(0) = g \cdot 0 = 0$ and $g\,\iota(\sigma(k)) = g(\iota(k) + 1) = g\,\iota(k) + g$, which are exactly the two defining equations of $k \mapsto g^{k}$ written additively; by the uniqueness in [L7] the two functions agree. For $x < 0$, writing $-x = \iota(k)$, we get $g^{x} = -(g^{k}) = -(g\,\iota(k)) = g(-\iota(k)) = gx$.

Discreteness: if $0 < x$ in $\mathbb{Z}$ then $1 \le x$. Indeed $x \ge 0$, so $x = \iota(k)$ with $k \ne 0$, hence $0 < k$ and $1 = \sigma(0) \le k$ in $\mathbb{N}$; applying $\iota$ gives $1 \le x$.

Existence, main case. Suppose $H \ne \{0\}$ and pick $h \in H$ with $h \ne 0$. Then $-h \in H$ as well, and by totality one of $h$, $-h$ is positive; so the set $P$ of positive elements of $H$ is nonempty.

Hence $\langle g \rangle = \{\, g^{x} : x \in \mathbb{Z} \,\} = \{\, gk : k \in \mathbb{Z} \,\} = g\mathbb{Z}$ for every $g \in \mathbb{Z}$; in particular $g\mathbb{Z}$ is a subgroup.

Every element of $P$ is nonnegative, hence of the form $\iota(k)$; so $T := \{\, k \in \mathbb{N} : \iota(k) \in P \,\}$ is a nonempty subset of $\mathbb{N}$. Let $k_0$ be its least element and put $n := \iota(k_0) \in P$.

For $g > 0$, $g$ is the least positive element of $g\mathbb{Z}$. Indeed $g = g \cdot 1 \in g\mathbb{Z}$ and $g > 0$; and if $gk > 0$ then $k > 0$, since $k = 0$ gives $gk = 0$ and $k < 0$ gives $0 < -k$, hence $0 < g(-k) = -(gk)$ and so $gk < 0$. Then $1 \le k$ by step 1.3, so $0 \le k - 1$ and $0 \le g(k-1) = gk - g$, that is $g \le gk$.

Existence, trivial case. If $H = \{0\}$ then $H = 0\mathbb{Z} = \langle 0 \rangle = \langle \iota(0) \rangle$, since $0 \cdot k = 0$ for every $k$.

$n$ is the least element of $P$: for $p \in P$ write $p = \iota(k)$ with $k \in T$; then $k_0 \le k$, and applying $\iota$, which preserves the order, gives $n \le p$.

$\langle n \rangle \subseteq H$, since $n \in H$ and $\langle n \rangle$ is the smallest subgroup containing $n$.

Uniqueness. Suppose $\iota(m)\mathbb{Z} = \iota(n)\mathbb{Z} = H$ with $m, n \in \mathbb{N}$. If $\iota(n) = 0$ then $H = \{0\}$, so $\iota(m) \in H$ forces $\iota(m) = 0$ and hence $m = 0 = n$ by injectivity. Otherwise $\iota(n) > 0$, so $H \ne \{0\}$ and likewise $\iota(m) > 0$; by step 2.3 both $\iota(m)$ and $\iota(n)$ are the least positive element of $H$, hence equal, and $m = n$ by injectivity.

$H \subseteq n\mathbb{Z}$. Let $h \in H$. Since $n > 0$, [L8] gives $h = qn + r$ with $0 \le r < n$. Now $qn \in n\mathbb{Z} = \langle n \rangle \subseteq H$ by step 2.1 and step 3.3, so $r = h - qn \in H$ because $H$ is closed under inverses and under addition. If $r$ were positive it would lie in $P$ and satisfy $r < n$, contradicting step 3.2 together with antisymmetry; so $r = 0$ and $h = qn \in n\mathbb{Z}$.

Combining, $H = n\mathbb{Z} = \langle n \rangle$ with $n = \iota(k_0)$, which with step 3.1 proves existence for every subgroup.

Every subgroup of $(\mathbb{Z},+)$ is therefore $\langle \iota(n) \rangle = \iota(n)\mathbb{Z}$ for exactly one $n \in \mathbb{N}$, and in particular is cyclic.