C_mncongC_mtimesC_nWhenMNAreRelativelyPrime 2008-04-16 13:49:12 2008-04-17 00:00:02 Proof cyclic group 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 We show that $C_{mn}$, gcd$(m, n)=1$, is isomorphic to $C_m\times C_n$, where $C_r$ denotes the cyclic group of order $r$ for any positive integer $r$. Let $C_m=\langle x\rangle$ and $C_n=\langle y\rangle$. Then the external direct product $C_m\times C_n$ consists of elements $(x^i, y^j)$, where $0\leq i\leq m-1$ and $0\leq j\leq n-1$. Next, we show that the group $C_m\times C_n$ is cyclic. We do so by showing that it is generated by an element, namely $(x, y)$: if $(x, y)$ generates $C_m\times C_n$, then for each $(x^i, y^j)\in C_m\times C_n$, we must have $(x^i, y^j)=(x, y)^k$ for some $k\in\{0, 1, 2, \ldots, mn-1\}$. Such $k$, if exists, would satisfy \begin{eqnarray*} k &\equiv& i\;(mod\;m) \\ k &\equiv& j\;(mod\;n). \end{eqnarray*} Indeed, by the Chinese Remainder Theorem, such $k$ exists and is unique modulo $mn$. (Here is where the relative primality of $m, n$ comes into play.) Thus, $C_m\times C_n$ is generated by $(x, y)$, so it is cyclic. The order of $C_m\times C_n$ is $mn$, so is the order of $C_{mn}$. Since cyclic groups of the same order are isomorphic, we finally have $C_{mn}\cong C_m\times C_n$.