I've been attempting this every which way and I seem to be going in circles at this point. I feel like the solution is probably very simple, but for whatever reason I am not seeing it. This isn't a homework problem, which is actually why I'm restricting myself from using the Chinese Remainder Theorem even though it may be applicable here. I feel like I either don't have a solid enough grasp of gcd/lcm, isomorphisms, or both, so I'm trying to get to this solution "directly" by using properties of isomorphic functions and gcd.
My attempts so far have not shown much. I've tried showing that there must be a linear combination of a and b which is equivalent to one using the property that an isomorphic function between two rings must send one in the domain to one in the codomain.
This yields me:
$$ f([1]_{ab}) = ([1]_a,[1]_b) $$
implies that, for some $q,x,y,z \in \Bbb Z$
$$ [q]_{ab} = [1]_{ab} \rightarrow q=abx + 1 $$ $$ [q]_{a} = [1]_{a} \rightarrow q=ay + 1 $$ $$ [q]_{b} = [1]_{b} \rightarrow q=bx + 1 $$
But obviously this is a mess and I can really only show that a linear combination of a and b is equivalent to zero.
I've attempted to prove the contrapositive as well by attempting to show that $gcd(a,b) \neq 1$ in some way implies that either surjectivity or injectivity will not hold, but I've similarly not been able to do this without making a mess of things.
So I have two questions. First, is either direction more efficient/require less work? And second, regardless of direction, how do I go about proving this? I don't think I need an explicit step-by-step (though I might not be opposed, I'm at my wits-end with this), but at the very least, what am I not thinking about or what am I thinking about in the wrong way?
Contrapositive is the way to go. Suppose that $gcd(a,b) \neq 1$, then $lcm(a,b) \neq ab$. However, $lcm(a,b) g = 0$ for any $g\in\mathbb Z/a \times \mathbb Z/b$, thus $\mathbb Z/a \times \mathbb Z/b$ is not isomorphic to $\mathbb Z/ab$.