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 $n$.