1. Every subgroup of a cyclic group is itself cyclic.

Let $H\le ⟨a⟩$. If $H$ is nonempty, then $H=⟨a^m⟩$ for the smallest integer $m$ such that $a^m\in H$.

2. The order of any cyclic subgroup divides the order of any cyclic group which contains it.

If $|⟨a⟩|=n$, then $|H|\cdot gcd(n,|H|)=n$.

3. There is exactly one subgroup of the order of each divisor of the order of its container cyclic group.

If $|H|=k$ such that $k|n$, then $H=⟨a^{\frac{n}{k}}⟩$

Taken together, for a cyclic group $⟨a⟩$ of order $n$, every factor $k$ of $n$ generates a unique subgroup $⟨a^{\frac{n}{k}}⟩\le ⟨a⟩$.

Corollary 1

In a finite cyclic group $G$, $|g|$ divides $|G|$

In particular, for $a^k\in ⟨a⟩$,
$|a^k|\cdot gcd(|a|,k)=|a|$

Justification

A reframing of The order of any cyclic subgroup divides the order of any cyclic group which contains it.#Corollary 1,
$|a^m|=\frac{|a|}{gcd(|a|,m)}$

Corollary 2

For $|a|=n$,
$⟨a^i⟩=⟨a^j⟩$ if and only if $gcd(n,i)=gcd(n,j)$

Justification

By Cyclic subgroups are equivalent if one power is expressible as the gcd of the other with the order of the base.,
$⟨a^i⟩=⟨a^{gcd(n,i)}⟩$
$⟨a^j⟩=⟨a^{gcd(n,j)}⟩$
Thus $\{gcd(n,i)=gcd(n,j)\}\Rightarrow \{⟨a^i⟩=⟨a^j⟩\}$

By A group element of finite order generates a finite cyclic subgroup of that order.,
$\{⟨a^{gcd(n,i)}⟩=⟨a^{gcd(n,j)}⟩\}\Rightarrow \{|a^{gcd(n,i)}|=|a^{gcd(n,j)}|\}$
By (2),
$\Rightarrow \{\frac{n}{gcd(n,i)}=\frac{n}{gcd(n,j)}\}\Rightarrow \{gcd(n,i)=gcd(n,j)\}$

Corollary 3

$⟨a^j⟩=⟨a⟩$ if and only if $gcd(j,|a|)=1$

Justification

From Corollary 2,
$|a|=n$ and $gcd(n,n)=1$