SubgroupsOfFiniteCyclicGroup 2009-06-02 10:39:14 2009-06-02 17:02:05 Theorem cyclic group % 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 \theoremstyle{definition} \newtheorem*{thmplain}{Theorem} Let $n$ be the order of a finite cyclic group $G$.\, For every positive \PMlinkname{divisor}{Divisibility} $m$ of $n$, there exists one and only one subgroup of order $m$ of $G$.\, The group $G$ has no other subgroups.\\ \emph{Proof.}\, If $g$ is a generator of $G$ and\, $n = mk$,\, then $g^k$ generates the subgroup $\langle g^k\rangle$, the order of which is equal to the order of $g^k$, i.e. equal to $m$.\, Any subgroup $H$ of $G$ is cyclic (see \PMlinkid{this entry}{4097}).\, If\, $|H| = m$,\, then $H$ must have a generator of order $m$; thus apparently\, $H = \langle g^{\pm k}\rangle = \langle g^k\rangle$.