After proving the theorem, the author proceeds to say:
As a consequence of the theorem:
1- We see that all homomorphic images of G can be determined using G.
I can't see how this follows. That assumes that we know the factor groups which contain the cosets of a normal group which is a kernel of a homomorphism which we are supposed to find in the first place!
2- We know that the number of homomorphic images of a cyclic group G of order n is the number of divisors of n, since there is exactly one subgroup of G (and therefore on factor group of G) for each divisor of n. (Be careful: The number of homomorphisms of a cyclic group of order n need not be the same as the number of divisors of n, since different homomorphisms can have the same image).
But again, that assumes that different homomorphisms can't have the same kernel with different images and thus a factor group can't be isomorphic to different homomorphic images! But that is not correct!
What I'm missing?
Best Answer
By the theorem, any two homomorphisms with the same kernel have isomorphic images.
For the second part, cyclic groups have unique normal subgroups of any order dividing the order of the group.
It's safe to say that the author's comments are to be taken to mean "up to isomorphism". Thus if two homomorphisms differ by an isomorphism, their images are isomorphic. So we regard their images as the same. They are still different homomorphisms though, if the isomorphism isn't the identity. Notice the author said "homomorphic images" are determined. Not that the homomorphisms are determined.
In the second case, he even cautions to distinguish the two (homomorphism and homomorphic image).