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Finite Self-contained proof that finite index subgroup in which $g$ is one element of a basis?

Let $g$ be a nontrivial element of a finitely generated free group $G$. Is there a finite index subgroup $H \subset G$ in which $g$ is one element of a basis?

Andy Putman says this in his answer below that this is an immediate corollary of Marshall Hall's theorem. However, I'm wondering if there is a more "elementary"/"self-contained" way of seeing that such a finite index subgroup exists.

Finite index subgroup in which $g$ is one element of a basis?

Let $g$ be a nontrivial element of a finitely generated free group $G$. Is there a finite index subgroup $H \subset G$ in which $g$ is one element of a basis?

Self-contained proof that finite index subgroup in which $g$ is one element of a basis?

Let $g$ be a nontrivial element of a finitely generated free group $G$. Is there a finite index subgroup $H \subset G$ in which $g$ is one element of a basis?

Andy Putman says this in his answer below that this is an immediate corollary of Marshall Hall's theorem. However, I'm wondering if there is a more "elementary"/"self-contained" way of seeing that such a finite index subgroup exists.

Post Undeleted by Stefan Kohl♦, Andy Putman, Todd Trimble
Post Deleted by user371374
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user371374
user371374
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Finite index subgroup in which $g$ is one element of a basis?

Let $g$ be a nontrivial element of a finitely generated free group $G$. Is there a finite index subgroup $H \subset G$ in which $g$ is one element of a basis?