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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.

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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)

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what about a proof by looking at covering spaces? –  solea Nov 19 '12 at 0:38
    
Can be that, can be with directly by Tietze transformations or even by actions of (free) groups on trees. Nevertheless, this all is old history for me and I don't think I'll get into this here. –  DonAntonio Nov 19 '12 at 2:32
    
I'm basically interested to know how to show this by use of covering spaces.?? –  solea Nov 19 '12 at 5:38
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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.

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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 $M=kn-k+1$.

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