# 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 $\mathcal{T}$.

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 |
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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 |