# Subgroup of Finite Index Containing a Given Finitely Generated Subgroup of a Free Group: Problem 12 in 1A in Hatcher.

Problem 1A.12 (Hatcher) Let $F$ be a finitely generated free group and $H$ be a finitely generated subgroup of $F$. Let $x\in F-H$. Show that there is a finite index subgroup $K$ of $F$ such that $H\subseteq K$ and $x\notin K$.

Here is my attempt.

Let $X$ be the wedge sum of appropriately many circles such that the fundamental group of $X$ can be identified with $F$. Since $H$ is a finitely generated subgroup of $F$, we can find a finite sheeted covering $p:\tilde X\to X$ of $X$ such that $p_*(\pi_1(\tilde X, \tilde x_0))=H$, where $\tilde x_0$ is a point in $\tilde X$. We know that $\tilde X$ is also a graph because any covering space of a graph is also a graph.

Let $\gamma$ be a path in $X$ (based at the only vertex in $X$) which corresponds to $x$. Since $x\notin H$, we know that the lift $\tilde \gamma$ of $\gamma$ starting at $\tilde x_0$ in $\tilde X$ is not a loop.

Now Hatcher has given a hint to use the following fact:

Fact. Let $X$ be a wedge sum of $n$ circles, with natural graph structure, and let $\tilde X\to X$ be a covering space with $Y\subseteq \tilde X$ a finite connected subgraph. Show there is a finite graph $Z\supseteq Y$ having the same vertex set as $Y$, such that the projection $Y\to X$ extends to a covering space $Z\to X$.

Back to our problem, we can think of $\tilde \gamma$ as a subgraph of $\tilde X$. By the "fact" above, we can find a connected graph $\tilde Y$ with the same vertex set as those appearing in $\tilde \gamma$ such that the the projection $p:\tilde \gamma\to X$ can be extended to a covering $\tilde Y\to X$.

I am stuck here.

• This proof is from: J. Stallings, Topology of finite graphs. Invent. Math. 71 (1983), no. 3, 551–565. – Moishe Kohan Jun 3 '16 at 0:36

From the generators of $H$, you can construct a (usually incomplete) Schreier graph for $H$ in $F$, in which any reduced word that lies in $H$ will trace out a circuit from the base point. (You can construct this graph very efficiently using the stamndard coset enumeration procedure.)
We know that the word $x$ does not label such a circuit. If it fails to label a path from in the graph from the base point, then add extra vertices to make it do so, and let $V$ be the resulting set of vertices.
Now just complete the graph, by adding extra labelled edges, in any way you like. It is easy to see that you can do that. For example, you could think of the graph as defining incomplete permutations of $K$, one for each free generator of $F$, and so it is just a matter of tending partial bijections to a complete bijection of $K$.
Now the subgroup defined by the labels of all circuits based at $1$ is the required subgroup $K$. Each vertex of the graph corresponds to a coset of $K$ in $F$, so the index $|F:K|$ is equal to $|V|$. You could also think of the graph as defining a homomorphism $F \to {\rm Sym}(V)$.