Show that there are only finitely many subgroups of $F$ in which $H$ can be of finite index. Result- Let $H$ be a finitely generated subgroup of the free group $F$. Show that there are only finitely many subgroups of $F$ in which $H$ can be of finite index.
I encountered a similar result in Group theory Book by Bogopolski (pg 120), 
B- "The number of subgroups of a finite index n in a finitely generated group is finite."
Does the Bogopolski result (B) also implies Result, the way Bogopolski proved B does not help in proving Result. So what should be the approach to prove Result.
 A: By the Schreier index formula, for subgroups of finite index in free groups, there is a bound $n$, say, on the index $|G:H|$ for subgroups $G$ of $F$ that contain $H$ with finite index. Also, there can be only finitely many subgroups between $H$ and $G$. So it is enough to show that the set of such groups $G$ that are maximal in $F$ subject to containing $H$ with finite index is finite.
If $|G:H| \le n$, then the core $K := \cap_{g \in G} g^{-1}Hg$ of $H$ in $G$ is normal in $G$ with $|G:K| \le n!$. So, for fixed $H$, there are only finitely many possible cores $K$. Hence, if there infinitely many subgroups $G$ of $F$  maximal with respect to containing $H$ with finite index, then there must be two such groups, $G_1$ and $G_2$ with the same core $K$. The result is easy when $H$ is trivial, so assume not. So $K$ is also nontrivial.
It is a standard result that, for a finitely generated nontrivial subgroup $K$ of a free group $F$, $|N_F(K):K|$ is finite. (Equivalently, a nontrivial normal subgroup of infinite index in a free group cannot be finitely generated.) So , since $K \unlhd \langle G_1,G_2 \rangle$, $|\langle G_1,G_2 \rangle:K|$ is finite, which contradicts the maximality of $G_1$ and $G_2$.
