How to prove $f:G \rightarrow \mathbb{Z}$, $f(a^k)=k$ is bijective?

Question

The following isomorphism

$f:G \rightarrow \mathbb{Z}$, $f(a^k)=k$

is involved in the proof of the theorem

Let $G$ be cyclic, $|G|= \infty$, then $G \cong \mathbb{Z}$.

However,

my source does not show, why $f$ is bijective.

How is this seen?

Answer 1

Suppose that $|<a>|$ (order of group generated by $a$) is infinite. Then, define $f$ as in your post, then $f(a)=1$, $f(e)=0$ (where $e$ is the identity in $<a>$) and $f(a^{-1})=-1$. Consider the group $(\mathbb{Z},+)$, and compare its behavior with that of $<a>$, what can you conclude?