Show that $G$ has finite index subgroup isomorphic to $\Bbb{Z}$ if there exists homomorphism $\Phi:G \rightarrow \Bbb{Z}$ with $\ker(\Phi)$ finite

Question

The task is to show that, for $G$, an infinite group, if there exists homomorphism $\Phi:G \rightarrow \mathbb{Z}$ with $\ker(\Phi)$ finite, then $G$ contains finite index subgroup isomorphic to $\mathbb{Z}$.

I see that taking any $g \in G$ s.t. $\Phi(g) \neq 0$, the subgroup $\langle g\rangle $ must be isomorphic to some subgroup of $\mathbb{Z}$ - which has a form of $n \mathbb{Z}$, so it is isomorphic with $\mathbb{Z}$.

But how can I get a conclusion that $\langle g\rangle $ has finite index in $G$?