We know that given a cyclic group $G$ with order $n$, the order of the subgroup with generator $g^m$ is found by gcd$(m,n) = d \Rightarrow \frac{n}{d} = $ order.
My question is can we just work in reverse from here to find the generators, given that we just know the order? If so, is there a more elegant way to doing this?
For example, consider this problem: Find all the orders, and their respective generators, of the subgroups of a cyclic group $G$, with order 16.
So, the subgroup orders of $G$ are just the divisors: $2,4,8$.
Let $B,C,D$ be subgroups of $G$ with orders $|B| = 2, |C| = 4, |D| = 8$.
Then the generators $g^m$ are $$ \begin{align} B & : g^8\\ C & : g^4,g^{12}\\ D & : g^2, g^6, g^{10}, g^{14} \end{align} $$
There is in fact a neat trick. In your example, suppose we want to quickly find the generators of a subgroup of order 8. We consider all the numbers less than 8 that are relatively prime to 8: 1, 3, 5, and 7. Now, multiplying these numbers by $\frac{16}{8} = 2$ gives you 2, 6, 10, 14. See any similarities? We can do the same for a subgroup of any order in general.