If $\operatorname{ord}(g) = n$ then $g^{k} = e$ iff $k$ is an integer multiple of $n$, the powers $g^{0}, \dots, g^{n-1}$ are distinct, and $\langle g \rangle$ has exactly $n$ elements; if $g$ has infinite order then $g^{j} = g^{k}$ only for $j = k$

$e^{x} = e$ for every $x \in \mathbb{Z}$. For natural exponents the set of $t$ with $e^{t} = e$ contains $0$, since $e^{0} = e$, and is closed under $\sigma$, since $e^{\sigma(t)} = e^{t} e = e\,e = e$; induction gives it for all naturals. For $x < 0$ write $-x = \iota(k)$; then $e^{x} = (e^{k})^{-1} = e^{-1} = e$.

Assume $\operatorname{ord}(g) = n$ with $n \ge 1$. Then $n \in S_g$, so $g^{n} = e$, and no natural $s$ with $1 \le s < n$ satisfies $g^{s} = e$, since $n$ is the least element of $S_g$. Also $\iota(n) > 0$, because $n \ge 1$ and $\iota$ preserves the order.

Infinite order. Assume $S_g = \varnothing$ and suppose $g^{j} = g^{k}$ with $j, k \in \mathbb{Z}$ and $j \ne k$. Put $x := j - k$, so $x \ne 0$ and $g^{x} = g^{j} g^{-k} = g^{j}(g^{k})^{-1} = e$. Also $g^{-x} = (g^{x})^{-1} = e^{-1} = e$. By totality one of $x$ and $-x$ is positive; call it $y$, so $g^{y} = e$ and $y = \iota(s)$ with $s \in \mathbb{N}$ and $s \ne 0$, hence $s \ge 1$. Then $s \in S_g$, contradicting $S_g = \varnothing$.

The "if" half of claim 1: if $k = q\iota(n)$ for some $q \in \mathbb{Z}$, then $g^{k} = g^{\iota(n) q} = (g^{\iota(n)})^{q} = (g^{n})^{q} = e^{q} = e$.

The "only if" half. Suppose $g^{k} = e$. Divide: $k = q\iota(n) + r$ with $0 \le r < \iota(n)$, legitimate since $\iota(n) > 0$. Then $e = g^{k} = g^{q\iota(n)} g^{r} = (g^{n})^{q} g^{r} = e^{q} g^{r} = g^{r}$. Since $0 \le r$, we have $r = \iota(s)$ for a unique $s \in \mathbb{N}$, and $\iota(s) < \iota(n)$ forces $s < n$, because otherwise $n \le s$ and $\iota$ would give $\iota(n) \le \iota(s)$, contradicting antisymmetry. So $g^{s} = e$ with $s < n$.

Claim 2. Let $i, j \in \mathbb{N}$ with $i < n$, $j < n$ and $g^{i} = g^{j}$. By trichotomy we may assume $i \le j$, interchanging the names if necessary, and then $i + t = j$ for some $t \in \mathbb{N}$. Now $g^{i} g^{t} = g^{i+t} = g^{j} = g^{i} = g^{i} e$, so $g^{t} = e$ by cancellation. Moreover $t \le j$, since $t + i = j$, and $j < n$, so $t < n$. If $t \ge 1$ then $t \in S_g$ with $t < n$, impossible; so $t = 0$ and $i = j$.

Therefore in the infinite-order case $g^{j} = g^{k}$ forces $j = k$. Moreover $\langle g \rangle$ is then not finite: a bijection $\psi : \langle g \rangle \to m$ with $m \in \mathbb{N}$ would make $k \mapsto \psi(g^{\iota(k)})$ a map $\sigma(m) \to m$, and that map is injective, since $\iota$ is injective, distinct integer exponents give distinct powers by step 1.3, and $\psi$ is injective; but claim 1 of the pigeonhole principle forbids an injection $\sigma(m) \to m$.

In step 2.2 the case $s \ge 1$ is impossible, since it would put $s$ in $S_g$ below its least element; hence $s = 0$, so $r = 0$ and $k = q\iota(n)$. With step 2.1 this is claim 1.

Every integer power of $g$ is one of $g^{0}, \dots, g^{n-1}$: given $x \in \mathbb{Z}$, divide $x = q\iota(n) + r$ with $0 \le r < \iota(n)$, write $r = \iota(s)$ with $s \in \mathbb{N}$ and $s < n$ as in step 2.2, and compute $g^{x} = (g^{n})^{q} g^{r} = e^{q} g^{s} = g^{s}$.

Claim 3. By [L2] and step 3.2, $\langle g \rangle = \{\, g^{s} : s \in \mathbb{N},\ s < n \,\}$. The map $\varphi : n \to \langle g \rangle$ with $\varphi(s) = g^{s}$ is well defined, the elements of the natural number $n$ being exactly the naturals $s < n$; it is surjective by the displayed description and injective by step 2.3. So $\varphi$ is a bijection, $\langle g \rangle \approx n$, and $\langle g \rangle$ is finite with $|\langle g \rangle| = n = \operatorname{ord}(g)$, the value $|\langle g \rangle|$ being the unique natural equinumerous with $\langle g \rangle$.

Claims 1, 2 and 3 are steps 3.1, 2.3 and 4.1, and the infinite-order statement is steps 1.3 and 2.4.