# Intuition for Fundamental Theorem of Cyclic Groups [duplicate]

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:

1. For any subgroup $$H\le G$$, $$H$$ is cyclic as well
2. For any subgroup $$H\le G$$, $$|H|$$ (order of $$H$$) divides $$n$$
3. 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?

• (2) is Lagrange's theorem. Mar 19 at 10:34
• There's an isomorphism from any cyclic group to $\mathbb{Z}/n\mathbb{Z}$ for some $n$. Have you tried interpreting the results there? It might be more intuitive in that setting. Mar 19 at 10:57
• @user816709 what would be a subgroup in $\mathbb{Z}_n$? Mar 19 at 11:11
• All subgroups of $\mathbb{Z}/n\mathbb{Z}$ are cyclic and are thus of the form $\langle m \rangle$ for some $0\leq m \leq n-1$. Mar 19 at 11:58
• Cyclic group is defined as a group generated by powers of an element (where powers mean repeated group actions). So, for a size $n$ cyclic group, we can write: $$\langle g \rangle = \{ g ,g^2...,g^{n-1}, g^n \}$$ But consider a divisor of the order of group, say $t$, it is clear that $\frac{n}{t} = p$ where $p$ is a positive integer. So, it is clear that the following is also a group: $$\langle g^p \rangle= \{ 1, g^p ,g^{2p}..., g^{n-p},g^{n} \}$$ Mar 19 at 13:12

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 $$H. We first need to take $$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?