Using Bezout's identity to show that $\Bbb Z_{mn} \cong \Bbb Z_m \times \Bbb Z_n$

Let $\gcd(m,n)=1$. Show that $\Bbb Z_{mn} \cong \Bbb Z_m \times \Bbb Z_n$.

Suppose that $\gcd(m,n)=1$. This implies that $mx+ny=1$ for some $x,y \in \Bbb Z$. Or in other words $$mx \equiv 1 \pmod{n} \\ ny\equiv 1 \pmod{m}$$

Now the order of $\Bbb Z_m \times \Bbb Z_n$ is $mn$ so it satisfies to show that $\Bbb Z_m \times \Bbb Z_n$ is cyclic to conclude that it's isomorphic to $\Bbb Z_{mn}$. I'm trying to figure out if $([1]_m,[1]_n)$ has order $mn$ and generates $\Bbb Z_m \times \Bbb Z_n$.

This question has probably been answered here before, but my question is related more in how can I use Bezout's identity to show that $([1]_m,[1]_n)$ has order $mn$?

Solution 1:

1. $mn([1]_m,[1]_n)=([0]_m, [0]_n)$. So the order $k$ of $([1]_m,[1]_n)$ divides $mn$.

2. Since $k([1]_m,[1]_n)=([0]_m, [0]_n)$ we have $k[1]_m=[0]_m$, so $k$ is divisible by $m$. Similarly $k$ is divisible by $n$.

3. Since $mx+ny=1$, we have $mxk+nyk=k$ but the LHS is $mn(x(k/n)+y(k/m))$. So $mn$ divides $k$. Hence $mn=k$.