[proofplan] We use the definition of the cyclic subgroup generated by $g$ as the set of all integer powers of $g$. If $g$ has finite order $n$, the [division algorithm](/theorems/725) reduces every power $g^k$ to one of $e,g,\dots,g^{n-1}$, and the equality criterion for powers shows these $n$ elements are distinct. If $g$ has infinite order, the same equality criterion shows that the map $k \mapsto g^k$ from $\mathbb{Z}$ into $\langle g \rangle$ is injective and surjective, so the generated subgroup is infinite. [/proofplan]

[step:Reduce the generated subgroup to integer powers] Let $e \in G$ denote the identity element of $G$. By definition of the subgroup generated by one element, \begin{align*} \langle g \rangle=\{g^k : k \in \mathbb{Z}\}. \end{align*} Thus computing $|\langle g \rangle|$ is the same as counting the distinct elements among the integer powers of $g$. [/step]

[step:Count the distinct powers when $g$ has finite order]Assume $\operatorname{ord}(g)=n<\infty$. Then $n \in \mathbb{N}$ and $g^n=e$. Let \begin{align*} R=\{0,1,\dots,n-1\}\subset \mathbb{Z}. \end{align*} For every $k \in \mathbb{Z}$, the division algorithm gives integers $q \in \mathbb{Z}$ and $r \in R$ such that \begin{align*} k=qn+r. \end{align*} Using the laws of integer powers in a group and $g^n=e$, we obtain \begin{align*} g^k=g^{qn+r}=(g^n)^q g^r=e^q g^r=g^r. \end{align*} Therefore \begin{align*} \langle g \rangle=\{g^r : r \in R\}. \end{align*} It remains to show that these $n$ elements are distinct. Let $a,b \in R$ and suppose $g^a=g^b$. Since $\operatorname{ord}(g)=n<\infty$, [citetheorem:8238] gives \begin{align*} n \mid (a-b). \end{align*} Because $a,b \in \{0,1,\dots,n-1\}$, we have \begin{align*} -(n-1)\le a-b\le n-1. \end{align*} The only multiple of $n$ in this interval is $0$, so $a-b=0$, hence $a=b$. Thus $\{g^r:r \in R\}$ has exactly $n$ elements, and consequently \begin{align*} |\langle g \rangle|=n=\operatorname{ord}(g). \end{align*}[/step]

[guided]Assume $\operatorname{ord}(g)=n<\infty$. This means $n$ is the least positive integer such that $g^n=e$. The central counting idea is that every exponent can be reduced modulo $n$, because multiplying the exponent by $n$ contributes a factor of $g^n=e$. Define the set of allowed remainders by \begin{align*} R=\{0,1,\dots,n-1\}\subset \mathbb{Z}. \end{align*} For an arbitrary exponent $k \in \mathbb{Z}$, the division algorithm gives integers $q \in \mathbb{Z}$ and $r \in R$ satisfying \begin{align*} k=qn+r. \end{align*} Using the group laws for integer powers, together with $g^n=e$, we get \begin{align*} g^k=g^{qn+r}=(g^n)^q g^r=e^q g^r=g^r. \end{align*} So every element of $\langle g \rangle$ is one of the finitely many elements \begin{align*} e,g,g^2,\dots,g^{n-1}. \end{align*} Equivalently, \begin{align*} \langle g \rangle=\{g^r : r \in R\}. \end{align*} Now we must check that this list has no repetitions. Let $a,b \in R$ and suppose $g^a=g^b$. Since $g$ has finite order $n$, the finite-order part of [citetheorem:8238] applies and gives \begin{align*} g^a=g^b \iff n \mid (a-b). \end{align*} Hence $n \mid (a-b)$. But because $a$ and $b$ both lie between $0$ and $n-1$, their difference satisfies \begin{align*} -(n-1)\le a-b\le n-1. \end{align*} The only integer in this interval divisible by $n$ is $0$. Therefore $a-b=0$, so $a=b$. Thus the elements $g^0,g^1,\dots,g^{n-1}$ are exactly the elements of $\langle g \rangle$, and they are pairwise distinct. There are $n$ of them, so \begin{align*} |\langle g \rangle|=n=\operatorname{ord}(g). \end{align*}[/guided]

[step:Show the generated subgroup is infinite when $g$ has infinite order] Assume $\operatorname{ord}(g)=\infty$. Define the map \begin{align*} \psi:\mathbb{Z}\to \langle g \rangle,\quad k\mapsto g^k. \end{align*} The map $\psi$ is surjective by the definition of $\langle g \rangle$. If $a,b \in \mathbb{Z}$ and $\psi(a)=\psi(b)$, then $g^a=g^b$. Since $\operatorname{ord}(g)=\infty$, [citetheorem:8238] gives $a=b$. Hence $\psi$ is injective. Therefore $\psi$ is a bijection from the infinite set $\mathbb{Z}$ onto $\langle g \rangle$, so $\langle g \rangle$ is infinite. Thus \begin{align*} |\langle g \rangle|=\infty=\operatorname{ord}(g). \end{align*} [/step]

[step:Combine the finite and infinite cases] The finite-order case proves $|\langle g \rangle|=\operatorname{ord}(g)$ whenever $\operatorname{ord}(g)<\infty$. The infinite-order case proves that both sides are infinite whenever $\operatorname{ord}(g)=\infty$. These two cases exhaust all possibilities for $g$, so \begin{align*} |\langle g \rangle|=\operatorname{ord}(g) \end{align*} with the stated infinite interpretation. [/step]