The Fundamental Theorem of Cyclic Groups states the following:
Let $G$ be a cyclic group generated by $g$ with $|G| = n$ (order of $G$ is $n$). Then:
- For any subgroup $H\le G$, $H$ is cyclic as well
- For any subgroup $H\le G$, $|H|$ (order of $H$) divides $n$
- For each positive divisor $t$ of $n$ there is exactly one subgroup $H\le G$ such that $|H| = t$, and $H = \langle g^{n/t} \rangle$.
I was able to follow the proof given here: http://site.iugaza.edu.ps/mabhouh/files/2011/01/alg1-ch4.pdf
However, I don't quite understand what points (2) and (3) mean. Could you share any intuition you have for these?
To understand this, perhaps it is useful to try to "build" a subgroup yourself. Let $G = \mathbb Z_n$, the integers modulo $n$ --- we can do this since all cyclic groups are the same as $\mathbb Z_n$ (can you see why?). OK, let's try to make a subgroup $H<G$. We first need to take $0$ since all groups need their identity element. Now, let's say we take the element $2$ into the group $H$. We are then forced to also include $4$, and $6$, and so on, in order to make sure that our group $H$ is closed under the group operation. This is the smallest subgroup generated by the element $2$, and so denoted $\langle 2 \rangle$. If $n=10$, can you write down explicitly all its elements?
Now think about the difference if $n=11$. Do you see a difference if $n$ is even or $n$ is odd? What if, instead of picking the element $2$, we picked some other number? What subgroup would it generate then? How does this depend on the choice of $n$? Does the prime factorisation of $n$ somehow matter?
Once you play around with enough examples, and you read the proof again, all the steps should then feel natural and easy to follow. There's no way to convey the intuition for this easily except to ask you to think about these examples yourself, IMO.