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

Question

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 $mn$?

You actually don't need Bezout here. Let $k$ be the order of $([1]_m,[1]_n)$, so $([k]_m,[k]_n) = ([0]_m,[0]_n)$. What can you say about $k$? — Andrea B., Dec 09 '21 at 21:30
What you want to show is that $m\mid x, n\mid x \implies mn\mid x$. This doesn't need Bezout. — WhatsUp, Dec 09 '21 at 21:30
@MarkSaving It sounds like this person is trying to prove CRT. — Andrea B., Dec 09 '21 at 21:31
I'm not trying to prove CRT. I'm trying to get an alternative proof to show that these two groups are isomorphic without using an excplicit homomorphism. — Rico Jello, Dec 09 '21 at 21:34
Yes. Now you want to use the fact that $\gcd(m,n)=1$ to show that $m|k$ and $n|k$ implies $mn|k$. — Andrea B., Dec 09 '21 at 21:38
So this solution at a duplicate uses the Bezout identity... is that what you're looking for? — rschwieb, Dec 09 '21 at 21:52

score 2 · Answer 1 · answered Dec 09 '21 at 21:41

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

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

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