Equivalence of Definitions of Finite Order Element

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

Let $x \in G$ be an element of $G$.

$(1)$ implies $(2)$
Let $x$ be a finite order element of $G$ by definition 1.

Then by definition there exists $k \in \Z_{>0}$ such that $x^k = e_G$.

Consider some $m, n \in \Z_{>0}$ such that $m = n + k$.

Thus $x$ is a finite order element of $G$ by definition 2.

$(2)$ implies $(1)$
Let $x$ be a finite order element of $G$ by definition 2.

That is, there exists $m, n \in \Z_{>0}$ such that $x^m = x^n$ but $m \ne n$.

, suppose that $m > n$.

Let $k = m - n$.

From $x^m = x^n$ it follows from Powers of Group Elements that:
 * $x^k = x^{m - n} = x^m x^{-n} = x^n x^{-n} = e_G$

Thus there exists $k \in \Z_{>0}$ such that $x^k = e_G$.

Thus $x$ is a finite order element of $G$ by definition 1.