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.

Let $G$ be a group which is free on a set $X$. Any group acts freely on its Cayley graph, and the Cayley graph of $G$ is a $2|X|$-regular tree, which we will call $𝒯$.

If $H$ is any subgroup of $G$, then $H$ also acts freely on $𝒯$ 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 $𝒢$ is a finite graph, then ${\pi }_1(𝒢)$ is free of rank $-\chi (𝒢)-1$, where $\chi (𝒢)$ denotes the Euler characteristic of $𝒢$. Since $H\cong {\pi }_1(H\backslash 𝒯)$, the rank of $H$ is $\chi (H\backslash 𝒯)$. If $H$ is of finite index $n$ in $G$, then $H\backslash 𝒯$ is finite, and $\chi (H\backslash 𝒯)=n\chi (G\backslash 𝒯)$. Of course $-\chi (G\backslash 𝒯)+1$ 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
Canonical name	ProofOfNielsenSchreierTheoremAndSchreierIndexFormula
Date of creation	2013-03-22 13:56:02
Last modified on	2013-03-22 13:56:02
Owner	mathcam (2727)
Last modified by	mathcam (2727)
Numerical id	7
Author	mathcam (2727)
Entry type	Proof
Classification	msc 20E05
Classification	msc 20F65
Related topic	ScheierIndexFormula