Homomorphisms

Group homomorphisms are very closely related to group isomorphisms. In fact, it turns out that every isomorphism is a homomorphism and as such, homomorphisms can be viewed as a generalization of isomorphisms. These very important functions are fundamental in the world of algebra and are one of the most important tools that we have. As we'll see in this and in upcoming posts, the application of homomorphisms to various groups gives a lot of useful results.

Definition: Group Homomorphism
A function, φ, from a group G to a group G* is a group homomorphism if it preserves the group operation - that is if ∀a,b∈G, φ(ab) = φ(a)φ(b).

A homomorphism is a function that has the operation preserving property that we described for isomorphisms without regard to bijectiveness. Of course, homomorphisms can be bijective, explaining why each isomorphism is a homomorphism. This means that it is easier for a function to be a homomorphism than an isomorphism, but it also means that homomorphisms have more interesting properties. Isomorphisms are used to identify groups as isomorphic, but not for much else, whereas homomorphisms provide us with a lot of usefullness, mostly as a result of the following sets.

Definition: Kernel
The kernel of a homomorphism, φ:G→G*, is the set {g∈G : φ(g)=e} and it is denoted by Ker(φ) or Kerφ.

Definition: Image
The image of a homomorphism, φ:G→G*, is the set {φ(x) : x∈G} and is denoted by Im(φ) or Imφ.

The first thing to notice here is that if φ:G→G* is a homomorphism, then Kerφ⊆G and Imφ⊆G*. The kernel is all of the things in G that are mapped to the identity, and the image is all of the things in G* that can be attained by a mapping of φ. It is important to note that these sets are trivial in the case that φ is actually an isomorphism - that is that in this case, Kerφ={e} and Imφ=G*. This does give another convenient way to check if a function is an isomorphism. Instead of proving that φ is bijective, it suffices to show that Kerφ={e} and Imφ=G* (and, of course, that φ is operation preserving). However, we get much more than that. It turns out that both the image and the kernel are subgroups.

Theorem: Kernels are Subgroups
Let φ:G→G* be a group homomorphism. Then Kerφ is a subgroup of G.

Proof:

Let φ:G→G* be a group homomorphism and choose a,b∈Kerφ. We then have that φ(ab^-1) = φ(a)φ(b)^-1 = e(e^-1) = e and thus ab^-1∈Kerφ. Therefore, by the one-step subgroup test, Kerφ is a subgroup of G.

Theorem: Images are Subgroups
Let φ:G→G* be a group homomorphism. Then Imφ is a subgroup of G*.

Proof:

Let φ:G→G* be a group homomorphism and choose x,y∈Imφ. Then ∃a,b∈G such that φ(a)=x and φ(b)=y. Now, φ(ab^-1) = φ(a)φ(b)^-1 = xy^-1 and thus xy^-1∈Imφ. Therefore, by the one-step subgroup test, Imφ is a subgroup of G*.

In those proofs I used a fact that I have not yet proved - specifically that φ(b^-1 = φ(b)^-1. This is true, though, and is an easy thing to prove. The fact that these two sets are subgroups is important and very convenient, but we get even more than that.

Theorem: Kernels are Normal
Let φ:G→G* be a group homomorphism. Then Kerφ⊳G.

Proof:

Let φ:G→G* be a group homomorphism and choose g∈G and n∈Kerφ. Then φ(gng^-1) = φ(g)φ(n)φ(g)^-1 = φ(g)eφ(g)^-1 = φ(g)φ(g)^-1 = e thus giving that gng^-1∈Kerφ. Since the choice of n∈Kerφ was arbitrary, gKerφg^-1⊆Kerφ for g∈G. Thus, by the normal subgroup test, Kerφ⊳G

It turns out that images are not normal subgroups. This is not all that surprising when you really think about what an image is in relation to a codomain. However, the fact that kernels are normal subgroups is a surprisingly wonderful fact and we get a lot of mileage out of it. In the next post, we'll see a very useful result called the first group isomorphism theorem that uses the normality of kernels in a very natural way.

References

Previous Related Post: Properties of Isomorphisms

Text Reference: Gallain Chapter 10

Wikipedia: Group Homomorphism

Wolfram Mathworld: Group Homomorphism

Planet Math: Group Homomorphism