Equivalence of Definitions of Infinite 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 an infinite order element of $G$ by definition 1.

Then by definition there exists no $k \in \Z_{>0}$ such that $x^k = e_G$.

not all $x^r$ are distinct for all $r \in \Z_{>0}$.

Then $x^m = x^n$ for some $m, n \in \Z$ where $m \ne n$.

, let $m > n$.

Then:
 * $x^{m - n} = e^G$

As $m > n$ we have that $m - n \in \Z_{>0}$.

Let $k = m - n$.

and so:
 * $\exists k \in \Z_{>0}$ such that $x^k = e_G$.

But this contradicts the statement that no such $k$ exists.

Hence no such distinct $m$ and $n$ exist.

Thus $x$ is an infinite order element of $G$ by definition 2.

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

Then by definition the powers $x^r$ of $x$ are distinct for all $r \in \Z_{>0}$.

That is, there exist no $x^m = x^n$ for some $m, n \in \Z$ where $m \ne n$.

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

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

But this contradicts the statement that no such $m, n \in \Z_{>0}$ exist.

Thus there can be no $k \in \Z_{>0}$ such that $x^k = e_G$.

Thus $x$ is an infinite order element of $G$ by definition 1.