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

This is a cross-post of my question here on MathOverflow.

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?

Apparently this is a special case of Marshall Hall's theorem, and there is a well-known proof of this more general result due to Stallings. However, I'm wondering if there is a more "elementary"/"self-contained" way of seeing that such a finite index subgroup exists.

• Can you give any more details on what you mean by "elementary" / "self-contained"? Are there any particular elements of proof which you wish not to allow? I ask because, as said in Putman's answer on MO and in @DerekHolt's answer here (with both of which I agree), the proofs are already pretty easy. – Lee Mosher Oct 15 '16 at 17:01

In your situation, it can be summarized as follows. Suppose that $G$ is free on the set $X$, and that $g$ has length $n$ as a reduced word in $X^{\pm 1}$. It is easy to construct a map $X \to S_n$ that extends to a homomorphism $\theta:G \to S_n$ with $\theta(g) = (1,2,\ldots,n)$. Then the inverse image image of the stabilizer of $1$ in $\theta(G)$ has index $n$ in $G$ and has $g$ as a Schreier generator.