$C_{mn}\cong C_{m}\times C_{n}$ when $m, n$ are relatively prime

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

	$\displaystyle k$	$\displaystyle\equiv$	$\displaystyle i\;(mod\;m)$
	$\displaystyle k$	$\displaystyle\equiv$	$\displaystyle j\;(mod\;n).$

Indeed, by the Chinese Remainder Theorem, such $k$ exists and is unique modulo $m n$ . (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 $m n$ , so is the order of $C_{mn}$ . Since cyclic groups of the same order are isomorphic, we finally have $C_{mn}\cong C_{m}\times C_{n}$ .

Title	$C_{mn}\cong C_{m}\times C_{n}$ when $m, n$ are relatively prime
Canonical name	CmncongCmtimesCnWhenMNAreRelativelyPrime
Date of creation	2013-03-22 17:59:46
Last modified on	2013-03-22 17:59:46
Owner	yesitis (13730)
Last modified by	yesitis (13730)
Numerical id	8
Author	yesitis (13730)
Entry type	Proof
Classification	msc 20A05