Let $G$ be a cyclic group of order $n$. Let $m \in \mathbb{Z}$, $m \leq n$. Find the number of subgroups of $G$ that have order $m$.
My attempt:
Supose that $m\mid n$ that is, there exists $k \in \mathbb{Z}$ such that $n=km$.
$$\langle a^k\rangle=\left\{e, a^k,a^{2k},...,a^{(m-1)k}\right\} \leq G$$
Let us see that this is the only group of order $m$ $$|\langle a^s\rangle|=m \Rightarrow a^{sm}=e \Rightarrow n\mid sm$$
That is, there is an integer $t$ such that $$sm=nt \Rightarrow a^{s}=\left(a^{n/m}\right)^t=(a^k)^t \in \langle a^k\rangle $$
Conclusion:
If $G$ is a cyclic group of order $n$, then for every $m\mid n$ there exists a unique subgroup of $G$ with order $m$.
My question:
What happens if $m$ does not divide $n$?
