Index of an infinite cyclic subgroup of a free group of infinite rank is infinite

Question

Question: Let $G$ be free of rank at least $2$, and let $H$ be an infinite cyclic subgroup of $G$. Is it true that the index $[G: H]$ of $H$ in $G$ is infinite?

My thought: I know the case when the rank of $G$ is finite, say $n+1$, where $n\geq 1$. If $[G: H]$ were finite, then $H$ would be a free group of rank $[G: H]n+1$, i.e., $H$ would be non-abelian, hence non-cyclic.

I don't know the case when $G$ is of infinite rank.

Answer 1

The following is true:

Claim. Suppose $H$ is a finitely generated subgroup of $G$, and that the index $|G:H|<\infty$. Then $G$ is finitely generated.

If $H$ is a cyclic subgroup of a free group $G$ with $|G:H|<\infty$, then the claim implies that $G$ must be finitely generated. As $H\cong\mathbb{Z}$ is finitely generated, it does not have finite index in any infinite-rank free group $G$, as you require.

To prove the claim, let $\mathbf{x}$ be a generating set for $H$ and let $\mathbf{y}$ be a transversal for $G/H$ (i.e. a set of coset representatives for $G/H$). Then every element of $G$ can be written in the form $yh$ for some $y\in\mathbf{y}$ and some $h\in H$, so $G=\langle \mathbf{x}, \mathbf{y}\rangle$ as required.