Finding all the subgroups of a cyclic group 
Theorem (1): If $G$ is a finite cyclic group of order $n$ and $m \in \mathbb{N}$, then $G$ has a subgroup of order $m$ if and only if $m | n$. Moreover for each divisor $m$ of $n$, there is exactly one subgroup of order $m$ in $G$.

The above theorem states that for any finite cyclic group $G$ of order $n$, the subgroups of $G$ (and also their orders) are in a one-to-one correspondence with the set of divisors of $n$. So $G$ has exactly $d(n)$ subgroups where $d(n)$ denotes the number of positive divisors of $n$.
Now the following example was given in my class notes and I paraphrase them below.

Example: Consider $G = (\mathbb{Z}_{12}, +)$. We list all subgroups of $G$. Note that $G$ is cyclic because $G = \langle \bar{1} \rangle$ and $|G| = 12$. Note also that $12$ has $6$ positive divisors, those being $\{1, 2, 3, 4, 6, 12\}$, hence $G$ has exactly six subgroups those being $$\langle \bar{1} \rangle; \ \
\langle (\bar{1})^2 \rangle; \ \
\langle (\bar{1})^3 \rangle;\ \
\langle (\bar{1})^4 \rangle;\ \
\langle (\bar{1})^6 \rangle;\ \
\langle (\bar{1})^{12} \rangle\ \
$$

Now my question is that how did the authors of these class notes know that the subgroups would be of the form $\langle (\bar{1})^{d} \rangle$  where $d$ is a positive divisor of $12$? It leads me to make the following conjecture

Conjecture: Given a finite cyclic group $G = \langle x \rangle$ of order $n$, all subgroups of $G$ are of the form $\langle x^d \rangle$ where $d$ is a positive divisor of $n$.

 A: The conjecture above is true. To prove it we need the following result:

Lemma: Let $G$ be a group and $x \in G$. If $o(x) = n$ and $\operatorname{gcd}(m, n) = d$, then $o(x^m) = \frac{n}{d}$

Here now is a proof of the conjecture.
Proof: Let $G = \langle x \rangle$ be a finite cyclic group of order $n$, then we have $o(x) = n$. 
Choose a subgroup $H \leq G$, by theorem $(1)$ mentioned in the question above, $|H| = m$ where $m$ is some divisor of $n$. Since $m | n$ (and both $m$ and $n$ are positive integers), there exists a $d \in \mathbb{N}$ such that $md = n \iff \frac{n}{d}=m$. Note also that $d$ is a divisor of $n$.
By the above lemma and the fact that $\operatorname{gcd}(d, n) = d$ (since $d$ is a divisor of $n$) it follows that $o(x^d) = \frac{n}{d} = m$. Hence the subgroup $\langle x^d \rangle$ has order $m$. But since by theorem $(1)$ there is only one subgroup of order $m$ in $G$ we must have $H = \langle x^d \rangle$. Thus any subgroup of $G$ is of the form $\langle x^d \rangle$ where $d$ is a positive divisor of $n$. $\ \square$

The above conjecture and its subsequent proof allows us to find all the subgroups of a cyclic group once we know the generator of the cyclic group and the order of the cyclic group.
