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 $C_{mn}\cong C_m\times C_n$ .