FiniteSubgroup 2009-06-01 11:06:43 2009-06-01 19:40:11 Theorem subgroup % 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} \textbf{Theorem.}\, A non-empty finite subset $K$ of a group $G$ is a subgroup of $G$ if and only if \begin{align} xy \in K \quad\mbox{for all}\quad x,\,y \in K. \end{align} \emph{Proof.}\, The condition (1) is apparently true if $K$ is a subgroup.\, Conversely, suppose that a nonempty finite subset $K$ of the group $G$ satisfies (1).\, Let $a$ and $b$ be arbitrary elements of $K$.\, By (1), all (\PMlinkescapetext{positive}) powers of $b$ belong to $K$.\, Because of the finiteness of $K$, there exist positive integers $r,\,s$ such that $$b^r \;=\; b^s, \quad r \;>\; s\!+\!1.$$ By (1), $$K \;\ni\; b^{r-s-1} \;=\; b^{r-s}b^{-1} \;=\; eb^{-1} \;=\; b^{-1}.$$ Thus also\, $ab^{-1} \in K$,\, whence, by the theorem of the \PMlinkid{parent entry}{1045}, $K$ is a subgroup of $G$.\\ \textbf{Example.}\, The multiplicative group $G$ of all nonzero complex numbers has the finite multiplicative subset \,$\{1,\,-1,\,i,\,-i\}$,\, which has to be a subgroup of $G$.