# proof of Cauchy’s theorem

Let $G$ be a finite group^{}, and suppose $p$ is a prime divisor^{} of $|G|$.
Consider the set $X$ of all $p$-tuples $({x}_{1},\mathrm{\dots},{x}_{p})$
for which ${x}_{1}\mathrm{\cdots}{x}_{p}=1$.
Note that $|X|={|G|}^{p-1}$ is a multiple^{} of $p$.
There is a natural group action^{} of
the cyclic group^{} $\mathbb{Z}/p\mathbb{Z}$ on $X$
under which $m\in \mathbb{Z}/p\mathbb{Z}$ sends
the tuple $({x}_{1},\mathrm{\dots},{x}_{p})$
to $({x}_{m+1},\mathrm{\dots},{x}_{p},{x}_{1},\mathrm{\dots},{x}_{m})$.
By the Orbit-Stabilizer Theorem, each orbit contains exactly $1$ or $p$ tuples.
Since $(1,\mathrm{\dots},1)$ has an orbit of cardinality $1$,
and the orbits partition^{} $X$,
the cardinality of which is divisible by $p$,
there must exist at least one other tuple $({x}_{1},\mathrm{\dots},{x}_{p})$
which is left fixed by every element of $\mathbb{Z}/p\mathbb{Z}$.
For this tuple we have ${x}_{1}=\mathrm{\dots}={x}_{p}$,
and so ${x}_{1}^{p}={x}_{1}\mathrm{\cdots}{x}_{p}=1$,
and ${x}_{1}$ is therefore an element of order $p$.

## References

- 1 James H. McKay. Another Proof of Cauchy’s Group Theorem, American Math. Monthly, 66 (1959), p119.

Title | proof of Cauchy’s theorem |
---|---|

Canonical name | ProofOfCauchysTheorem |

Date of creation | 2013-03-22 12:23:30 |

Last modified on | 2013-03-22 12:23:30 |

Owner | yark (2760) |

Last modified by | yark (2760) |

Numerical id | 8 |

Author | yark (2760) |

Entry type | Proof |

Classification | msc 20E07 |

Classification | msc 20D99 |