# proof of Nielsen-Schreier theorem and Schreier index formula

While there are purely algebraic proofs of the Nielsen-Schreier theorem, a much easier proof is available through geometric group theory.

If $H$ is any subgroup   of $G$, then $H$ also acts freely on $\mathcal{T}$ by restriction. Since groups that act freely on trees are free, $H$ is free.

Moreover, we can obtain the rank of $H$ (the size of the set on which it is free). If $\mathcal{G}$ is a finite graph, then $\pi_{1}(\mathcal{G})$ is free of rank $-\chi(\mathcal{G})-1$, where $\chi(\mathcal{G})$ denotes the Euler characteristic  of $\mathcal{G}$. Since $H\cong\pi_{1}(H\backslash\mathcal{T})$, the rank of $H$ is $\chi(H\backslash\mathcal{T})$. If $H$ is of finite index $n$ in $G$, then $H\backslash\mathcal{T}$ is finite, and $\chi(H\backslash\mathcal{T})=n\chi(G\backslash\mathcal{T})$. Of course $-\chi(G\backslash\mathcal{T})+1$ is the rank of $G$. Substituting, we obtain the Schreier index formula:

 $\mathrm{rank}(H)=n(\mathrm{rank}(G)-1)+1.$
Title proof of Nielsen-Schreier theorem and Schreier index formula ProofOfNielsenSchreierTheoremAndSchreierIndexFormula 2013-03-22 13:56:02 2013-03-22 13:56:02 mathcam (2727) mathcam (2727) 7 mathcam (2727) Proof msc 20E05 msc 20F65 ScheierIndexFormula