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.



Self-contained proof that finite index subgroup in which $g$ is one element of a basis?
Yes, this is a corollary of the following famous theorem of Marshall Hall:


**Theorem**: If $A$ is a finitely generated subgroup of a free group $G$, then there exists a finite-index subgroup $H$ of $G$ containing $A$ such that $A$ is a free factor of $H$, i.e. such that $H = A \ast A'$ for some subgroup $A'$ of $G$.


Just apply this to the cyclic subgroup $A = \langle g \rangle$.


For a proof of Marshall Hall's theorem, see e.g. [this](https://berstein2015.wordpress.com/2015/03/19/free-groups-and-marshall-halls-theorem/) blog post.


