If $\gcd(a,b) = 1,$ then why is the set of invertible elements of $\mathbb Z_{ab}$ isomorphic to that of $\mathbb Z_a\times \mathbb Z_b$? If $\gcd(a,b) = 1,$ then why is the set of invertible elements of $\mathbb Z_{ab}$ isomorphic to that of $\mathbb Z_a\times \mathbb Z_b$?
I know the proof that as rings, $\mathbb Z_{ab}$ is congruent to $\mathbb Z_a\times \mathbb Z_b.$ Does this extend to the sets of their invertible (aka relatively prime elements)? If so, why? Is this equivalent to the assertion that $\gcd(a,bc) = 1$ iff $\gcd(a,b) = 1$ and $\gcd(a,c) = 1?$
Note: I selected the answer as the one that does not utilize the Chinese Remainder Theorem, since I was using this to prove the Chinese Remainder Theorem. Thank you for the help!

gcd$(a, bc) = 1 \Rightarrow$ there exists no prime integer $p$ such that $p|a$ and $p|bc \Rightarrow$ there exists no prime integers $r, s$ such that $r|a, r|b$ and $s|a, s|c \Rightarrow$ gcd$(a, b) = 1$ and gcd$(a, c) = 1.$
On the other hand, suppose gcd$(a, b) = 1$ and gcd$(a, c) = 1.$ Let $p$ be a prime integer such that $p|a, p|bc.$ Then $p|b$ or $p|c,$ a contradiction.
