Can any subgroup of a cyclic group $\left$ be presented as $\{x\in \left|x^m=e\}$? Can any subgroup of a cyclic group $\left<a\right>$ be presented as $\{x\in \left<a\right>|x^m=e\}$?
Note that m is a factor of n, where n is the order of $\left<a\right>$.
I found that we can easily construct a subgroup from $\left<a\right>$ using a factor m of n. Then $C_m≡\{x\in \left<a\right>|x^m=e\}$ is a subgroup of $\left<a\right>$ (the proof is simple and I will not write it here).
For example, let $\left<a\right>=\mathbb{Z}_{12}$.
If $m=1$, then $C_m=\{0\}$.
If $m=3$, then $C_m=\{0,4,8\}$.
If $m=4$, then $C_m=\{0,3,6,9\}$.
If $m=12$, then $C_m=\mathbb{Z}_{12}$.
The question is, is the reverse also true? Can every subgroup of $\left<a\right>$ be written as $C_m$? In other words, can we prove that the number of subgroups of $\left<a\right>$ is exactly the number of factors of n?
 A: Let $H$ be a subgroup of $\langle a\rangle$. Then $H$ must be equal to $\langle a^k\rangle$ where $k$ is a factor of $n$.
Note that $C_{n/k}=\{x\in\langle a\rangle\ :x^{n/k}=1\}$.
Let $x\in H$. Then $x=a^{ck}$ for some integer $c$.
Therefore $x^{n/k}=(a^{ck})^{n/k}=(a^{n})^c=1$.
Hence $x\in C_{n/k}$ and we conclude that $H\subseteq C_{n/k}$.
Conversely, let $x\in C_{n/k}$. Since $x\in\langle a \rangle$, it follows that $x=a^c$ for some integer $c$.
Since $(a^c)^{n/k}=x^{n/k}=1$, we have $n$ divides $nc/k$. This implies that $k$ must be divide $c$.
Write $c=rk$ for some integer $r$. We get $x=a^c=(a^k)^r\in H$.
We conclude that $C_{n/k}\subseteq H$.
So we must have $H=C_{n/k}$.
A: On a short note:
Fundamental theorem of cyclic groups
Every subgroup of a cyclic group is cyclic. Moreover, if $|\langle a\rangle|=n,$ then the order of any subgroup of $\langle a\rangle>$ is a divisor of $n$; and, each positive divisor $k $ of $n,$ the group $\langle a\rangle$ has excetly one subgroup of order $k-$ namely, $\langle a^{n/k}\rangle.$
For proof, you can read chapter "Cyclic groups" in Contemporary Abstract Algebra by Joseph Gallian.
