I was working on a question on the no. of homomorphisms from S3 to Z/6Z.

My approach was as follows. if f is a homomorphism from G-> G', ker(f) = {g: f(g) = e'} should be a normal subgroup of G.

I recalled from a theorem that the only non-trivial normal subgroup of S3 is A3, the alternating group on 3 letters.

So I guessed that there could only be one homomorphism. I have two questions.

Can we count the homomorphism which has {e} has its kernel?

Would it not imply that it is also an isomorphism contradicting the fact that S3 is isomorphic to Z/6Z?

I doubt the contradiction because S3 cannot be isomorphic to Z/6Z because Z/6Z is in turn isomorphic to Z6, which is a cyclic group. But I guess S3 is not cyclic.