This is taken from Keith Conrad's expository paper on subgroups of cyclic groups.
Let $G$ be a cyclic group, with a generator $g$. For a subgroup $H \subset G$, we will show $H=\langle g^n\rangle$ for some $n\geq0$, so $H$ is cyclic. (...) The idea is that every element in a subgroup of the form $\langle g^n\rangle$ looks like $g^{nk}$, so $n$ is the smallest positive exponent among the elements of the subgroup.
I'm not sure how to think about this... If we take the cyclic group with ten elements $(e,g^1,g^{-1},\ldots, g^5)$ and build the subgroup $\langle g^4\rangle=\{g^4,g^{-2},g^2,g^{-4},e\}$ so $g^{4*3}=g^2$ and $2<4$. Am I making some mistake?
Note that it says it will show $H=\langle g^n\rangle$ for some $n > 0$. That is, it is only necessary to find a single $n$, not every $n$ that will work. The claim is that if $H \ne \{e\}$, it is sufficient to choose $$n = \min \{k \in \Bbb N^+\mid g^k \in H\}$$
Since this set cannot be empty (I'll leave it to you to figure out why), it always has a least element. For any $g^k \in H$, we can write $k = qn + r$ for some $q$ and $0 \le r < n$. But $g^r = (g^n)^{-q}g^k \in H$, and since $n$ is the smallest positive exponent of $g$ in $H$, it must be that $r = 0$. I.e., $g^k = (g^n)^q \in \langle g^n \rangle$.
It doesn't matter if there are other values $n_1$ such that $\langle g^{n_1} \rangle = H$. (In your example, $n = 2$ and $n_1 = 4$.) All that matters is that the $n$ chosen can be proven to work.