See cyclic group, order of a group

Let $G$ be a group, and let $a\in G$.
If $|a|=n$, then $⟨a⟩=\{e,a,a^2,\dots ,a^{n-1}\}$, such that $|⟨a⟩|=n$.

Proof

By definition, $⟨a⟩=\{a^k:k\in \mathbb{Z}\}$
$\{e,a,a^2,\dots ,a^{n-1}\}\subseteq \{a^k:k\in \mathbb{Z}\}$]
So $\{e,a,a^2,\dots ,a^{n-1}\}\subseteq ⟨a⟩$

Let $a^k\in ⟨a⟩$
By Division Algorithm 1, $k=qn+r$ with $0\le r<n$
So $a^k=a^{qn+r}=(a^n{)}^q{a}^r=e^q{a}^r=a^r$
Since $0\le r<n$, $a^k\in \{e,a,a^2,\dots ,a^{n-1}\}$
Since $a^k\in ⟨a⟩$ an arbitrary element, $⟨a⟩\subseteq \{e,a,a^2,\dots ,a^{n-1}\}$

Therefore, $⟨a⟩=\{e,a,a^2,\dots ,a^{n-1}\}$

Furthermore, $|⟨a⟩|=n$

QED