Question

Find all possible homomorphisms between Z and Z5.

Answer 1

Answer: A homomorphism φ : Z → Z5 is determined by φ(1) since φ(n) = n · φ(1) for every n ∈ Z. Also, for any a ∈ Z5, we can get a homomorphism Z → Z5 taking 1 to a by sending n to the reduction mod 5 of an. So, there are Five homomorphisms φ : Z → Z5, one for each value in Z5.

If φ(1) = 0, we get the zero map. Its kernel is all of Z and its image is {0}.

If φ(1) = 1, our map is just reduction mod 5, which is clearly surjective; that is, its image is all of Z5. We see that an element is sent to zero if and only if it’s a multiple of 5, so ker(φ) = 5Z.

If φ(1) = 2, our map takes n to the reduction of 2n mod 5. The image is generated by 2, which is <2> = {0, 2, 4}. The kernel is the set of elements of n such that 2n is a multiple of 5. This is the set of even integers, 2Z.

If φ(1) = 3, our map takes n to the reduction of 3n mod 5. The image is generated by φ(1) = 3 which is <3> = {0, 3}. The kernel is the set of elements of n such that 3n is a multiple of 5.

If φ(1) = 4, our map takes n to the reduction of 4n mod 5. The image is generated by φ(1) = 4 and so is all of Z5 (so it’s surjective). Since 4n is a multiple of 5 if and only if n is a multiple of 5, the kernel is 5Z.