$C_{mn}\cong C_m\times C_n$ 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 $C_m=⟨x⟩$ and $C_n=⟨y⟩$. Then the external direct product $C_m\times C_n$ consists of elements $(x^i,y^j)$, where $0\le i\le m-1$ and $0\le j\le n-1$.

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 $k\in \{0,1,2,\mathrm{\dots },mn-1\}$. Such $k$, if exists, would satisfy

$k$$\equiv$$i(modm)$

$k$$\equiv$$j(modn).$

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