# proof of Cauchy’s theorem in abelian case

Suppose $G$ is abelian^{} and the order of $G$ is $h$. Let ${g}_{1}$, ${g}_{2},\mathrm{\dots},{g}_{h}$ be the elements of $G$, and for $i=1,\mathrm{\dots},h$, let ${a}_{i}$ be the order of ${g}_{i}$.

Consider the direct sum^{}

$$H=\underset{i=1}{\overset{h}{\oplus}}\mathbb{Z}/{a}_{i}\mathbb{Z}.$$ |

The order of $H$ is obviously ${a}_{1}{a}_{2}\mathrm{\cdots}{a}_{h}$. We can define a group homomorphism^{} $\theta $ from $H$ to $G$ by

$$({x}_{1},\mathrm{\dots},{x}_{h})\mapsto {g}_{1}^{{x}_{1}}\mathrm{\cdots}{g}_{h}^{{x}_{h}}.$$ |

$\theta $ is certainly surjective^{}. So $|H|=|G|\cdot |\mathrm{ker}(\theta )|$. Since $p$ is a prime factor^{} of $G$, $p$ divides —H—, and therefore must divide one of the ${a}_{i}$’s, say ${a}_{1}$. Then ${g}_{1}^{{a}_{1}/p}$ is an element of order $p$.

Title | proof of Cauchy’s theorem in abelian case |
---|---|

Canonical name | ProofOfCauchysTheoremInAbelianCase |

Date of creation | 2013-03-22 14:30:28 |

Last modified on | 2013-03-22 14:30:28 |

Owner | kshum (5987) |

Last modified by | kshum (5987) |

Numerical id | 7 |

Author | kshum (5987) |

Entry type | Proof |

Classification | msc 20D99 |

Classification | msc 20E07 |