Question

give an example of a nontrivial homomorphism Ï• for the given groups,if exists.
Ï•:Z12->Z5

Answer 1

I know by theorem that the image of any subgroup of Z12 must be a subgroup of Z5. So the image of Z12 itself is a subgroup of Z5. But Z5 only has the subgroups of itself or the trivial subgroup, and since we are dealing with nontrivial homomorphisms, then this must mean the image of Z12 is all of Z5.
HOWEVER: where I get confused is that the solution says "But the number of cosets of a subgroup of a finite group must equal the order of the group, and 5 does not divide 12". I don't understand what is meant by that.

Answer 2

Im not sure what that solution means, but I would look at the Kernel of the homomorphism.

Since you have 12 elements mapping to 5, the kernel must be nontrivial. Say there are k elements in the kernel. Since the kernel is a subgroup, the order of the kernel must divide 12, so that means k = 2, 3, 4, or 6. Note I didnt include 1 since that would make the homomorphism injective (cant happen since there arent the same number of elements in the two groups), and I didnt include 12 (that would make the trivial homomorphism).

Now, if the kernel had 2 elements, then every element in Z5 would have 2 things that map to it, giving 10 elements total, but the domain has 12 elements, not 10. Similiarly you can get a contradiction if the kernel had 3, 4, or 6 elements.