The kernel of a homomorphism is the set of all elements in the domain that map to the identity element in the co-domain

The kernel of a homomorphism is the set of all elements in the domain that map to the identity element in the co-domain. In other words, if F is a homomorphism from the group G to the group G’, the kernel, denoted Ker F, is the set of all elements x in F such that F(x)=e’ where e’ is the identity of G’.

Application Problem (Created by Ben Mueller)

Define a mapping F such that the following is true:

The domain of F is the group of all integers under multiplication.
The co-domain of F is the multiplicative group Z5. Note: The 5 elements in the group Z5 are {[0],[1],[2],[3],[4]}.
F(x) = [x]

Classify F in as many ways as possible using the terms homomorphism, epimorphism, and isomorphism.
Give the set of elements that are in Ker F. (This set may be an infinite set)

Solution (a) (Provided by Ben Mueller):

For F to be a homomorphism, it must be true that F(xy) = F(x)F(y). Now,

F(x)F(y) = [x][y] = [xy] = F(xy). This verifies that F is a homomorphism.

All of the elements in the group Z5, our co-domain, can be mapped to using our mapping F with an integer in the domain.

F(0) = [0], F(1) = [1], F(2) = [2], F(3) = [3], and F(4) = [4]. Thus, our mapping F is surjective (onto).

Each element in our domain, group of all integers under multiplication, maps to a specific element in Z5. Thus, the mapping F is also injective (1-1).

Since F is a homomorphism that is a bijection (injective and surjective), F can be classified as an isomorphism.

Solution (b)

To find Ker F, we want to find all elements in the domain such that F(x) = e’ where e’ is the identity element in the co-domain. The identity element in our co-domain, Z5 under multiplication, is [1]. Thus, we want to find all integers x such that F(x) = [1].

The integer 1 certainly maps to [1] as F(1) = [1].

Also, the integer 6 maps to 1 as F(6) = [6] and in Z5, [6] = [1].

Also, F(11) = [11] = [1] in Z5.

F(16) = [16] = [1] in Z5.

F(-4) = [-4] = [1] in Z5.

F(-9) = [-9] = [1] in Z5.

From this, it can be observed that all integers that can be expressed in the form 1 +/- 5m where m is an integer are elements in Ker F.

It is actually quite simple to prove this:

F(1 +/- 5m) = [1 +/- 5m] = [1] +/- [5m] = [1] +/- 5[m] = [1] +/- [0] = [1].

Thus, Ker F = {x such that x = 1 +/- 5m where m is an integer} or another way of writing Ker F is {….-14,-9,-4,1,6,11,16,21,26,…}. Note that there is an infinite amount of elements in Ker F.