Equivalence of Definitions of Infinite Cyclic Group

$(1)$ implies $(2)$
Let $G = \gen g$ be an infinite cyclic group by definition 1.

Then by definition:
 * $(1): \quad \forall n \in \Z_{> 0}: n > 0 \implies \nexists a \in G, a \ne e: a^n = e$


 * $\exists a \in G: \exists m, n \in \Z, m \ne n: a^m = a^n$
 * $\exists a \in G: \exists m, n \in \Z, m \ne n: a^m = a^n$

Then by Equal Powers of Finite Order Element:
 * $\exists k \in \Z_{> 0}: k \divides \paren {m - n}$

such that:
 * $a^k = e$

As $m \ne n$, we have that $k \ge 1$.

This contradicts the condition $(1)$ for $G$ to be an infinite cyclic group.

From this contradiction we deduce that:
 * $\forall a \in G: \forall m, n \in \Z: m \ne n \implies a^m \ne a^n$

Thus $G$ is an infinite cyclic group by definition 2.

$(2)$ implies $(1)$
Let $G = \gen g$ be an infinite cyclic group by definition 2.

Then by definition:
 * $\forall a \in G: \forall m, n \in \Z: m \ne n \implies a^m \ne a^n$


 * $\exists a \in G, a \ne e: \exists n \in \Z: a^n = e$
 * $\exists a \in G, a \ne e: \exists n \in \Z: a^n = e$

Then:
 * $a^{2 n} = a^n \circ a^n = e^2 = e = a^n$

That is:
 * $\exists n, 2 n \in \Z: n \ne 2 n: a^n = a^{2 n}$

and so it is not the case that:
 * $\forall a \in G: \forall m, n \in \Z: m \ne n \implies a^m \ne a^n$

From this contradiction we deduce that:
 * $\forall n \in \Z_{> 0}: n > 0 \implies \nexists a \in G, a \ne e: a^n = e$

Thus $G$ is an infinite cyclic group by definition 1.