Yes - your reasoning is correct as far as it goes.

Any homomorphism will produce a kernel, and that kernel will be a normal subgroup. The kernel can't be {e}, because otherwise, since those two groups have the same (finite) order, a kernel of {e} would correspond to a homomorphism that's actually an isomorphism - and because one group is cyclic and the other isn't, they can't be isomorphic. You're absolutely right. (Aside: Another way to quickly see that they're not isomorphic is to observe that one is abelian while the other isn't.)

You've looked at the case of the kernel being A3, and the case of the kernel being {e}, which you eliminated. There's another normal subgroup of S3 that you haven't considered: S3 itself. Which homomorphisms, if any, would have that as a kernel? (Hint: don't overthink this - it's easy.)

In trying to find the homomorphisms of S3 to Z6, you've broken it down according to:

1) Kernel = {e}: No homomorphisms.

2) Kernel = S3: ??? homomorphisms.

3) Kernel = A3: ??? homomorphisms.

That's a good start, but you'll still need to then examine which homomorphisms, if any, exist in those cases (well, only case 3 requires any work). Just because A3 is a normal subgroup of S3 does *not* guarantee that there's a homomorphism from S3 to Z6 having A3 as a kernel. Also, at the other extreme, there might be several different homomorphisms from S3 to Z6 having A3 as their kernel.

Ex: Is there a homomorphism from S3 to Z3 having kernel A3? No. If there were, then the image of that

homomorphism would be isomorphic to S3/A3, which is the group of order 2. But, by Lagrange's theorem, Z3 doesn't have a subgroup of order 2.

So your reasoning so far is excellent, but you've still got more work in front of you for the problem of finding the homomorphisms between those two groups.