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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.

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  • $\begingroup$ 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. $\endgroup$ – Lee Mosher Oct 15 '16 at 17:01
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The Stallings proof is really not difficult. Perhaps you are being put off by the topological language, but it is really purely combinatorial.

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.

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