# proof of Fermat’s little theorem using Lagrange’s theorem

###### Theorem.

If $a\mathrm{,}p\mathrm{\in}\mathrm{Z}$ with $p$ a prime and $p\mathrm{\nmid}a$, then ${a}^{p\mathrm{-}\mathrm{1}}\mathrm{\equiv}\mathrm{1}\phantom{\rule{veryverythickmathspace}{0ex}}\mathrm{(}\mathrm{mod}p\mathrm{)}$.

###### Proof.

We will make use of Lagrange’s Theorem: Let $G$ be a finite group^{} and let $H$ be a subgroup^{} of $G$. Then the order of $H$ divides the order of $G$.

Let $G={(\mathbb{Z}/p\mathbb{Z})}^{\times}$ and let $H$ be the multiplicative subgroup of $G$ generated by $a$ (so $H=\{1,a,{a}^{2},\mathrm{\dots}\}$). Notice that the order of $H$, $h=|H|$ is also the order of $a$, i.e. the smallest natural number^{} $n>1$ such that ${a}^{n}$ is the identity^{} in $G$, i.e. ${a}^{h}\equiv 1modp$.

By Lagrange’s theorem $h\mid |G|=p-1$, so $p-1=h\cdot m$ for some $m$. Thus:

$${a}^{p-1}={({a}^{h})}^{m}\equiv {1}^{m}\equiv 1modp$$ |

as claimed. ∎

Title | proof of Fermat’s little theorem using Lagrange’s theorem |
---|---|

Canonical name | ProofOfFermatsLittleTheoremUsingLagrangesTheorem |

Date of creation | 2013-03-22 14:23:53 |

Last modified on | 2013-03-22 14:23:53 |

Owner | alozano (2414) |

Last modified by | alozano (2414) |

Numerical id | 4 |

Author | alozano (2414) |

Entry type | Proof |

Classification | msc 11-00 |

Related topic | LagrangesTheorem |