ProofOfCauchysTheoremInAbelianCase 2004-07-29 15:22:13 2004-07-29 17:44:02 Proof Cauchy's theorem 0 % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here Suppose $G$ is abelian and the order of $G$ is $h$. Let $g_1$, $g_2,\ldots, g_h$ be the elements of $G$, and for $i=1,\ldots, h$, let $a_i$ be the order of $g_i$. Consider the direct sum \[ H = \bigoplus_{i=1}^h \mathbb{Z}/a_i\mathbb{Z}. \] The order of $H$ is obviously $a_1a_2\cdots a_h$. We can define a group homomorphism $\theta$ from $H$ to $G$ by \[ (x_1,\ldots,x_h) \mapsto g_1^{x_1}\cdots g_h^{x_h}. \] $\theta$ is certainly surjective. So $|H| = |G|\cdot|\ker(\theta)|$. 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 $p$.