Finding the rank of subgroups of free groups? How do you find the rank of a subgroup(of finite index) of a free group?
I was thinking of looking at the fundamental group of a graph.
 A: Assuming the free group $\,F\,$ has rank $\,n\,$ and the index of the subgroup $\,K\leq F\,$  is $\,k\,$, the group $\,K\,$ is free on $\,k(n-1)+1\,$ generators.
You can find this formula in several books, for example in the classic Magnus-Karrass-Solitar (theorem 2.10), or in Lyndon-Schupp (Proposition 3.9)
A: You can use the Euler characteristic. For a free group of rank n, this is easily seen to $n-1$. For a subgroup of index $k$ we get Euler characteristic $k(n-1)$. Hence the rank $M$ satisfies  $M-1=k(n-1)$ or $M=kn-k+1$.
A: Take a wedge of $n$ spheres. This has fundamental group the free group on $n$ generators $F_n$. Let $G$ be a subgroup of $F_n$ of index $k$. Associated to $G$ is a $k$-fold connected covering space $P$ of the wedge of spheres. This covering space is again a 1-dimensional CW complex (or a graph, if you prefer) with $k$ vertices and $nk$ edges. A maximal tree in this graph has $k-1$ edges. Contracting this maximal tree, we obtain a wedge of $nk-(k-1)$ spheres. Hence the fundamental group of $P$ is free on $nk-k+1$ generators. But by definition of the cover associated to $G$, this fundamental group is isomorphic to $G$ and we are home.  
