Ytukyg

Assuming $(a,b)=1$, I want to construct an isomorphism $phi:Bbb Z_atimesBbb Z_bmapstoBbb Z_{acdot b}$ defined by $phi(1,1)=1$. So I need to prove that for any $(x,y)in Bbb Z_atimesBbb Z_b$ there exists $nin Bbb N$ such that $nequiv x mod a$ and $nequiv ymod b$

For example $(1,2)inBbb Z_2timesBbb Z_3$ is $(5mod2,5mod3)$

I'm not sure how to prove that, I tried using the Bezout equality...

For the other way we need to assume $Bbb Z_atimesBbb Z_b$ is cyclic i.e. $Bbb Z_atimesBbb Z_b =langle(x,y)rangle$ for some $(x,y)$

Since $0inlangle(x,y)rangle$, $exists nin Bbb N~:~nxequiv0mod a~text{and } nyequiv 0mod b$

And here I'm stuck again.

The group $mathbb{Z}_a times mathbb{Z}_b$ is generated by $(1,0)$ and $(0,1)$, so the choices of $phi(1,0)$ and $phi(0,1)$ determine the homomorphism completely. You shall pick those values such that:
$$
begin{align*}
&a cdot phi(1,0) equiv 0 quadbmod ab \
& b cdot phi(0,1) equiv 0 quad bmod ab \
end{align*}
$$
Let's pick:
$$
phi(x,y)=(bx+ay) bmod ab
$$
Bezout's identity ensures that the integers of the form $bx+ay$ are exactly the multiples of $(a,b)=1$.

This implies that $phi$ is a bijection and thus an isomorphism.

First note that the order of an element $(x,y) in mathbb{Z}_a times mathbb{Z}_b$ is $text{lcm}(r_1,r_2)$, where $r_1$ is the order of $x in mathbb{Z}_a$ and $r_2$ is the order of $y in mathbb{Z}_b$. Also note that $text{lcm}(a,b) = ab$ iff $a$ and $b$ are relatively prime.

For the forward direction, we can just prove the contrapositive. So suppose $mathbb{Z}_a times mathbb{Z}_b$ is not cyclic. Then there exists some $r lt ab$ such that $r(1,1) = (0,0)$ otherwise $mathbb{Z}_a times mathbb{Z}_b$ would be cyclic. But this means that $text{lcm}(1,1) le r lt ab$ so that $a$ and $b$ are not relatively prime.

For the other direction, suppose that $mathbb{Z}_a times mathbb{Z}_b$ is cyclic. Now $(1,1) in mathbb{Z}_a times mathbb{Z}_b$ has order $text{lcm}(r_1, r_2)$ where $r_1$ is the order of $1 in mathbb{Z}_a$ and $r_2$ is the order of $1 in mathbb{Z}_b$. Then $text{lcm}(r_1, r_2) = ab$ iff $a$ and $b$ are relatively prime. So if $mathbb{Z}_a times mathbb{Z}_b$ is cyclic, the order of $(1,1)$ is $ab$ since $(1,1)$ has to generate the group and hence $a$ and $b$ must be relatively prime by the previous statement.