Equivalence of Definitions of Order of Group Element

Theorem
Let $G$ be a group whose identity is $e$.

Let $x \in G$.

Proof
Let $k$ be the order of $x$ in $G$ according to Definition 1.

Let $l$ be the order of $x$ in $G$ according to Definition 3.

Definition $1$ is equivalent to Definition $3$
Aiming for a contradiction, suppose that $k \ne l$.

Then from Ordering on 1-Based Natural Numbers is Trichotomy either $k \lt l$ or $k \gt l$.

According to Definition 3:
 * $\forall i, j \in \Z: 0 \le i \lt j \lt l \implies x^i \ne x^j$

If $k \lt l$, then letting $i = 0$ and $j = k$ yields a contradiction by Element to Power of Zero is Identity.

If $k \gt l$, then:
 * $\exists i,j \in \Z: 0 \le i \lt j \lt k: x^i = x^j$

But then $x^{j-i} = e$ and $j - i \lt k$, contradicting the fact that $k \in \Z_{\gt 0}$ is the smallest such that $x^k = e$.

Therefore $k = l$.

Definition $1$ is equivalent to Definition $2$
It follows straight away from List of Elements in Finite Cyclic Group that $\left|{\left \langle {a} \right \rangle}\right| = k$:
 * $\left \langle {a} \right \rangle = \left\{{a^0, a^1, a^2, \ldots, a^{k - 1}}\right\}$