Order of elements in Cyclic Groups

Short, sweet, to the point. More of the number theory aspects.

Let $G$ be a cyclic group of order $n$ and $a$ generate $G$, ie $\mid ⟨a⟩\mid =n$. Then for $b=a^k\in ⟨a⟩$, order of b is $\frac{n}{gcd(n,k)}$

Practice:

In ${\mathbb{Z}}_{12}$ through this we know the order of 1 is $\frac{12}{gcd(1,12)}=12$.
order of 2 is 6
order of 5 is also 12 so $⟨5⟩=⟨1⟩$
This adds more intuition to what we proved in All Ideals in integers mod n

This explains why all subgroups of ${\mathbb{Z}}_n$ are devisors of n, because:
For any generated subgroup $⟨a⟩$ we know $a=1^a$ (additively) so

order of $a=\frac{n}{gcd(n,a)}$; or the number of elements in $⟨a⟩$.

So if we are in ${\mathbb{Z}}_{12}$ then we see that $a=\frac{12}{gcd(12,a)}$ and from here, we know all possible subgroups of ${\mathbb{Z}}_{12}$ are all cyclic. However, 2,4,6 have the same gcd so $⟨6⟩=⟨4⟩=⟨2⟩$ as they all have 6 elements. This breaks everything down to $a$ just being the devisors of 12 as the other devisor will be $gcd(12,a)$ for all elements.