Let \(x\) be a nontrivial element of a finitely generated free group \(F\). Show there is a finite-index subgroup \(H \subset F\) in which \(x\) is one element of a basis. [Exercises 4 and 10 may be helpful.]
Solution
Let \[ F=\pi_1(R,r_0) \] for a finite rose \(R\), and let \(x\in F\) be nontrivial. Write \[ x=uyu^{-1} \] with \(y\) cyclically reduced and nontrivial. It is enough to find a finite-index subgroup \(H'\le F\) in which \(y\) is primitive, since then \[ H=uH'u^{-1} \] has the same finite index and has \(x=uyu^{-1}\) as a basis element.
Consider the connected covering \[ p:R_y\to R \] corresponding to the cyclic subgroup \[ \langle y\rangle\le F. \] Choose the base vertex \(\widetilde r_0\) corresponding to this subgroup.
1. The loop representing \(y\) lifts to a closed reduced edge path in \(R_y\) whose image is an embedded circle \(C\subseteq R_y\) through \(\widetilde r_0\).
Let \(T\to R\) be the universal covering tree. Because \(y\) is cyclically reduced, the geodesic from the identity vertex to \(y\) lies on the invariant axis \(A\) of the deck transformation \(y\). The cover \(R_y\) is the quotient \[ T/\langle y\rangle. \] The quotient \[ C=A/\langle y\rangle \] is a circle embedded in \(R_y\): if two points of the axis have the same image in the quotient, they differ by a power of \(y\), so the only identifications on \(A\) are precisely those defining the circle quotient. Since the identity vertex lies on \(A\), its image \(\widetilde r_0\) lies on \(C\), and traversing \(C\) once represents \(y\).
2. Choose a finite connected subgraph\[ Y\subseteq R_y \]containing \(C\).
The embedded circle \(C\) is a finite edge circuit because it is the quotient of one finite translation segment of the axis. Thus we may take \(Y=C\), or enlarge it by finitely many edges if desired.
3. By Exercise 10, \(Y\) is contained in a finite-sheeted covering\[ q:Z\to R \]with no new vertices.Let\[ H'=q_*\pi_1(Z,\widetilde r_0). \]Then \(H'\) has finite index in \(F\) and \(y\in H'\).
Exercise 10 gives a finite covering extending the labeled graph \(Y\). Since the circle \(C\subseteq Y\subseteq Z\) is a closed lift of the loop \(y\) at \(\widetilde r_0\), the class \(y\) belongs to the subgroup corresponding to the cover. Finite-sheetedness gives \([F:H']<\infty\).
4. The element \(y\) is primitive in \(H'\).
The finite covering graph \(Z\) contains the subgraph \[ C\cong S^1 \] through the basepoint, and the loop \(C\) represents \(y\). By Exercise 4, the fundamental group \(\pi_1(Z,\widetilde r_0)=H'\) has a free basis containing the element represented by \(C\). Thus \(y\) is one member of a basis of \(H'\).
5. Therefore\[ \boxed{H=uH'u^{-1}\le F} \]has finite index and has \(x=uyu^{-1}\) as a basis element.
Conjugation by \(u\) is an isomorphism \(H'\to H\) carrying any free basis of \(H'\) containing \(y\) to a free basis of \(H\) containing \(uyu^{-1}=x\).