Subgroup of Finite Cyclic Group is Determined by Order

Theorem
Let $G = \left \langle {g} \right \rangle$ be a cyclic group whose order is $n$ and whose identity is $e$.

Let $d \mathop \backslash n$, where $d \mathop \backslash n$ in this context means $d$ is a divisor of $n$.

Then there exists exactly one subgroup $G_d = \left \langle {g^{n / d}} \right \rangle$ of $G$ with $d$ elements.

Proof
Let $G$ be generated by $g$, such that $\left|{g}\right| = n$.

From Number of Powers of Cyclic Group Element, $g^{n/d}$ has $d$ distinct powers.

Thus $\left \langle {g^{n / d}} \right \rangle$ has $d$ elements.

Now suppose $H$ is another subgroup of $G$ of order $d$. Then by Subgroup of a Cyclic Group is Cyclic, $H$ is cyclic.

Let $H = \left \langle {y} \right \rangle$ where $y \in G$. Thus $\left|{y}\right| = d$.

Thus $\exists r \in \Z: y = g^r$.

Since $\left|{y}\right| = d$, it follows that $y^d = g^{r d} = e$.

From Equal Powers of Finite Order Element, $n \mathop \backslash r d$.

Thus:
 * $\exists k \in \N: k n = r d = k \left({\dfrac n d}\right) d \implies r = k \left({\dfrac n d}\right)$

So $\dfrac n d \mathop \backslash r$.

Thus $y$ is a power of $g^{n / d}$, so $H$ is a subgroup of $\left \langle {g^{n / d}} \right \rangle$.

Since both $H$ and $\left \langle {g^{n / d}} \right \rangle$ have order $d$, they must be equal.