Let \(F\) be a finitely generated free group, \(H \subset F\) a finitely generated subgroup, and \(x \in F - H\). Show there is a subgroup \(K\) of finite index in \(F\) such that \(K \supset H\) and \(x \notin K\). [Apply Exercise 10.]
Solution
Realize \[ F=\pi_1(R,r_0) \] where \(R\) is a finite rose, and let \[ p:\widetilde R_H\to R \] be the connected covering corresponding to \(H\). Choose a base vertex \(\tilde r_0\) above \(r_0\).
1. Since \(H\) is finitely generated, there is a finite connected subgraph\[ Y_0\subseteq\widetilde R_H \]containing \(\tilde r_0\) and closed lifts of loops representing a finite generating set of \(H\).
Choose generators \[ h_1,\dots,h_k \] of \(H\) and reduced edge loops in \(R\) representing them. Their lifts beginning at \(\tilde r_0\) are closed because each \(h_i\in H\). The union of these finitely many finite lifted edge loops is a finite connected subgraph \(Y_0\).
2. Let \(\gamma_x\) be a reduced loop in \(R\) representing \(x\) and let \(\widetilde\gamma_x\) be its lift beginning at \(\tilde r_0\).Its endpoint \(\tilde r_1\) is distinct from \(\tilde r_0\).
A based loop in \(R\) lifts closed at \(\tilde r_0\) exactly when its class lies in the subgroup corresponding to the pointed cover, namely \(H\). Since \(x\notin H\), the lift of \(\gamma_x\) is not closed.
3. Let\[ Y=Y_0\cup\operatorname{im}(\widetilde\gamma_x). \]Then \(Y\) is a finite connected subgraph containing both distinct vertices \(\tilde r_0,\tilde r_1\).
Both pieces are finite and contain \(\tilde r_0\), so their union is finite and connected. The endpoint \(\tilde r_1\) belongs to the lifted path and is distinct from \(\tilde r_0\) by 2.
4. By Exercise 10, \(Y\) is contained in a finite covering graph\[ q:Z\to R \]with exactly the same vertex set as \(Y\).Let\[ K=q_*\pi_1(Z,\tilde r_0)\le F. \]Then \(K\) has finite index in \(F\).
Exercise 10 completes the finite labeled subgraph \(Y\) to a finite-sheeted covering without adding vertices. Since \(Y\) is connected and already contains every vertex of \(Z\), the graph \(Z\) itself is connected. Hence the subgroup corresponding to the pointed cover \(Z\to R\) has index equal to the finite cardinality of the fiber, so \([F:K]<\infty\).
5. One has\[ H\subseteq K. \]
Each chosen generator loop \(h_i\) has a closed lift in \(Y_0\subseteq Y\subseteq Z\) based at \(\tilde r_0\). Hence \(h_i\in K\) for every \(i\), so the subgroup they generate, namely \(H\), is contained in \(K\).
6. One has\[ x\notin K. \]
The lift of \(\gamma_x\) in \(Z\) beginning at \(\tilde r_0\) is forced by uniqueness of path lifting to be the same path \(\widetilde\gamma_x\) already contained in \(Y\). Its endpoint is \(\tilde r_1\ne\tilde r_0\). Since Exercise 10 added no vertices and did not change existing edges, this lift remains nonclosed. Therefore \(x\notin q_*\pi_1(Z,\tilde r_0)=K\).
7. Thus\[ \boxed{H\subseteq K\le F,\qquad [F:K]<\infty,\qquad x\notin K.} \]
Combine <1>4–<1>6.