# Schreier index formula

Let $F$ be a free group^{} of finite rank, and let $H$ be a subgroup^{} (http://planetmath.org/Subgroup) of finite index in $F$.
By the Nielsen-Schreier theorem, $H$ is free.
The *Schreier index formula* states that

$$\mathrm{rank}(H)=|F:H|\cdot (\mathrm{rank}(F)-1)+1.$$ |

This implies more generally that if $G$ is a group generated by $m$ elements, then any subgroup of index $n$ in $G$ can be generated by at most $nm-n+1$ elements.

Title | Schreier index formula |
---|---|

Canonical name | SchreierIndexFormula |

Date of creation | 2013-03-22 13:56:18 |

Last modified on | 2013-03-22 13:56:18 |

Owner | yark (2760) |

Last modified by | yark (2760) |

Numerical id | 14 |

Author | yark (2760) |

Entry type | Theorem |

Classification | msc 20E05 |

Related topic | ProofOfNielsenSchreierTheoremAndSchreierIndexFormula |