# group theoretic proof of Wilson’s theorem

Here we present a group theoretic proof of it.

Clearly, it is enough to show that
$(p-2)!\equiv 1\phantom{\rule{veryverythickmathspace}{0ex}}(modp)$ since $p-1\equiv -1\phantom{\rule{veryverythickmathspace}{0ex}}(modp)$.
By Sylow theorems^{}, we have that $p$-Sylow subgroups of ${S}_{p}$, the
symmetric group^{} on $p$ elements, have order $p$, and the number ${n}_{p}$ of
Sylow subgroups is congruent to 1 modulo $p$. Let $P$ be a Sylow subgroup
of ${S}_{p}$. Note that $P$ is generated by a $p$-cycle. There are $(p-1)!$ cycles
of length $p$ in ${S}_{p}$. Each $p$-Sylow subgroup contains $p-1$ cycles
of length $p$, hence there are $\frac{(p-1)!}{p-1}=(p-2)!$ different
$p$-Sylow subgrups in ${S}_{p}$, i.e. ${n}_{P}=(p-2)!$. From Sylow’s Second
Theorem, it follows that $(p-2)!\equiv 1\phantom{\rule{veryverythickmathspace}{0ex}}(modp)$,so $(p-1)!\equiv -1\phantom{\rule{veryverythickmathspace}{0ex}}(modp)$.

Title | group theoretic proof of Wilson’s theorem |
---|---|

Canonical name | GroupTheoreticProofOfWilsonsTheorem |

Date of creation | 2013-03-22 13:35:27 |

Last modified on | 2013-03-22 13:35:27 |

Owner | ottocolori (1519) |

Last modified by | ottocolori (1519) |

Numerical id | 10 |

Author | ottocolori (1519) |

Entry type | Proof |

Classification | msc 11-00 |