# 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.

• Note that Andy's post answered the question in its original formulation.
– YCor
Commented Oct 15, 2016 at 3:03
• Marshall Hall's theorem is a basic piece of geometric group theory, and its proof is not difficult. I recommend learning it. Indeed, it would be useful to you if you went through the proof specializing everything to the case you are after. You'll find it very enlightening and elementary (and also completely algorithmic). The blog post I pointed you to is a good place to start. Commented Oct 15, 2016 at 3:05
• The fact you are asking for basically is Marshall Hall's theorem: the special case of a cyclic subgroup is no easier than the general case. As Andy says, you should learn the proof. I blogged about it here: ldtopology.wordpress.com/2008/12/01/… .
– HJRW
Commented Oct 15, 2016 at 16:10
• Perhaps the problem is that the language used in the Stallings proof is very topological, although the proof is essentially combinatorial and can be worded more algebraically. For example, what is known by many as Stallings Folding would be referred to as the coincidence routine in coset enumeration by people who work in computational group theory. Commented Oct 15, 2016 at 19:05
• You don't need folding to prove Marshall Hall's theorem, anyway. This is a common misconception.
– HJRW
Commented Oct 22, 2016 at 5:19

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 $A = \langle g \rangle$.