# $C_{mn}\cong C_{m}\times C_{n}$ 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 $mn$. (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 $C_{mn}\cong C_{m}\times C_{n}$.

Title $C_{mn}\cong C_{m}\times C_{n}$ when $m,n$ are relatively prime CmncongCmtimesCnWhenMNAreRelativelyPrime 2013-03-22 17:59:46 2013-03-22 17:59:46 yesitis (13730) yesitis (13730) 8 yesitis (13730) Proof msc 20A05