In a textbook, I found two different solved problems, However after analyzing both the problems and their solutions, I am confused. Following are the problems and their solution. Q1. Find the number of homomorphisms from S3 (permutation group on 3 symbols) in Z3 (group of residue class module 3).
Solution (as per book): Only one homomorphism (Trivial homomorphism)
Explanation (as per book): If there exists a non-trivial homomorphism of S3 in Z3 then |kerf|=2. And since kerf is also a normal subgroup of S3, That implies there exists a normal subgroup of order 2 in S3, which is not possible. Hence only trivial homomorphism exists.
Q2. Find the number of homomorphisms from Z8 (group of residue class module 8) into Z6 (group of residue class module 6).
Solution: Two homomorphisms (Trivial one and f(x)=3x)
Explanation: Try to map 1 in Z8 with the elements in Z6. Let 1 of Z8 is mapped to "x" of Z6. Now, O(x)= 1 or 2. For O(x)= 1, its trivial homomorphism, for O(x)= 2, its f(x)=3x.
Now my confusion is, if we try to solve the second problem with the same logic as in 1st problem, We will get only one homomorphism, i.e., trivial one.( by that logic if there exists a non-trivial homomorphism, then |kerf|= 8/6, which is impossible, since order of a group must be a natural number.) So, which explanation is right and which is wrong?