[proofplan]
We prove the result by measuring the first time a positive power of $g$ returns $x$ to itself. Since $g$ has order $n$, the set of positive return times is nonempty, so it has a least element $m$. We show that the orbit consists exactly of the $m$ points $x,g \cdot x,\dots,g^{m-1}\cdot x$, and then use the [division algorithm](/theorems/725) applied to $n$ by $m$ to prove that $m \mid n$.
[/proofplan]
[step:Define the first positive return time of $x$ under powers of $g$]
Let
\begin{align*}
R = \{k \in \mathbb{N} : g^k \cdot x = x\}
\end{align*}
be the set of positive return times of $x$ under the action of powers of $g$. Since $\operatorname{ord}(g)=n$, we have $g^n=e_G$, where $e_G$ is the identity element of $G$. Hence
\begin{align*}
g^n \cdot x = e_G \cdot x = x,
\end{align*}
so $n \in R$. Thus $R$ is nonempty. By the [well-ordering principle](/theorems/721) for $\mathbb{N}$, let
\begin{align*}
m = \min R.
\end{align*}
Then $m \in \mathbb{N}$ and $g^m \cdot x=x$.
[/step]
[step:Show that the orbit has exactly $m$ elements]
We claim that
\begin{align*}
\langle g \rangle \cdot x = \{g^r \cdot x : r \in \{0,1,\dots,m-1\}\}.
\end{align*}
First let $h \cdot x \in \langle g \rangle \cdot x$. Since $h \in \langle g \rangle$, there exists $k \in \mathbb{Z}$ such that $h=g^k$. By the division algorithm, choose $q \in \mathbb{Z}$ and $r \in \{0,1,\dots,m-1\}$ such that
\begin{align*}
k = qm+r.
\end{align*}
Because $g^m \cdot x=x$, the element $(g^m)^q$ also fixes $x$: if $q \ge 0$ this follows by repeated application, and if $q<0$ it follows because the inverse of an element fixing $x$ also fixes $x$. Therefore
\begin{align*}
h \cdot x = g^k \cdot x = g^{qm+r}\cdot x = g^r \cdot ((g^m)^q \cdot x)=g^r \cdot x.
\end{align*}
Thus every point of $\langle g \rangle \cdot x$ appears among $x,g \cdot x,\dots,g^{m-1}\cdot x$. The reverse inclusion follows because each $g^r$ lies in $\langle g \rangle$.
It remains to show that these $m$ displayed points are distinct. Suppose $a,b \in \{0,1,\dots,m-1\}$ and $g^a \cdot x=g^b \cdot x$. If $a<b$, then applying $g^{-a}$ to both sides gives
\begin{align*}
x = g^{b-a}\cdot x.
\end{align*}
But $b-a \in \mathbb{N}$ and $b-a<m$, contradicting the minimality of $m$. The case $b<a$ is identical after exchanging $a$ and $b$. Hence $a=b$.
Therefore
\begin{align*}
|\langle g \rangle \cdot x| = m.
\end{align*}
[guided]
The number $m$ is the first positive power of $g$ that sends $x$ back to itself. We want to show that this first return time is exactly the size of the orbit.
First, every element of the cyclic subgroup $\langle g \rangle$ is some integer power of $g$. Thus an arbitrary point of the orbit has the form $g^k \cdot x$ for some $k \in \mathbb{Z}$. Divide $k$ by $m$: there are $q \in \mathbb{Z}$ and $r \in \{0,1,\dots,m-1\}$ such that
\begin{align*}
k = qm+r.
\end{align*}
Since $g^m \cdot x=x$, applying $g^m$ repeatedly still fixes $x$, and applying its inverse repeatedly also fixes $x$. Hence $(g^m)^q \cdot x=x$ for every $q \in \mathbb{Z}$. Therefore
\begin{align*}
g^k \cdot x = g^{qm+r}\cdot x = g^r \cdot ((g^m)^q \cdot x)=g^r \cdot x.
\end{align*}
So every orbit point is one of
\begin{align*}
x,\ g\cdot x,\ g^2\cdot x,\ \dots,\ g^{m-1}\cdot x.
\end{align*}
Now we check that no two of these points are the same. Suppose $0 \le a<b\le m-1$ and $g^a\cdot x=g^b\cdot x$. Applying $g^{-a}$ to both sides gives
\begin{align*}
x=g^{b-a}\cdot x.
\end{align*}
Here $b-a$ is a positive integer strictly smaller than $m$. This contradicts the definition of $m$ as the least positive return time. Hence the listed $m$ points are distinct.
Thus the orbit under $\langle g \rangle$ has exactly $m$ elements:
\begin{align*}
|\langle g \rangle \cdot x|=m.
\end{align*}
[/guided]
[/step]
[step:Prove that the orbit size divides $n$]
Since $m \in R$, we have $g^m \cdot x=x$. Apply the division algorithm to $n$ by $m$: choose $q \in \mathbb{Z}$ and $r \in \{0,1,\dots,m-1\}$ such that
\begin{align*}
n=qm+r.
\end{align*}
Because $g^n=e_G$, we have
\begin{align*}
x=g^n\cdot x=g^{qm+r}\cdot x=g^r\cdot ((g^m)^q\cdot x)=g^r\cdot x.
\end{align*}
If $r>0$, then $r \in R$ and $r<m$, contradicting the minimality of $m$. Hence $r=0$, so $m \mid n$.
From the previous step, $|\langle g \rangle \cdot x|=m$. Therefore
\begin{align*}
|\langle g \rangle \cdot x| \mid n.
\end{align*}
In particular, $\langle g \rangle \cdot x$ is finite. This proves the theorem.
[/step]
