More About Homomorphisms

Before we talk about the first group isomorphism theorem, I suppose a few examples of homomorphisms are in order. They are often very similar to isomorphisms, as to be expected, but make sepcial notes of their kernels and images.

Example:

Choose n∈N and let φ:Z→Z be given by φ(x) = x mod n ∀x∈Z. Now, suppose x,y∈Z. We can write x = hn + p and y = kn + q where p,q∈Z_n and h,k∈Z. Then

φ(x + y) = x + y mod n = (hn + p) + (kn + q) mod n = p + q mod n

and also

φ(x) + φ(y) = (hn + p mod n) + (kn + q mod n) = p + q mod n

so φ(x+y) = φ(x)+φ(y). Thus φ is operation preserving and is a homomorphism.

The image of φ should be fairly obvious. Indeed, Z_n⊆Z and if x∈Z_n then φ(x) = x giving that Z_n⊆Imφ but by the definition of modular division, nothing in Imφ can be outside of Z_n and so Imφ = Z_n. The kernel of φ, however, is a little more interesting. Suppose that x∈Kerφ. We can write x = hn + p as before but then φ(x) = p = 0 so x = hn giving that the elements in the kernel of φ are the multiples of n, or Kerφ = nZ = {kn : k∈Z}.

Example:

Let G be any group and let N be a normal subgroup of G. Then we can form the factor group, G/N. Let σ:G→G/N be defined by σ(g) = gN ∀g∈G. Choose x,y∈G and note that

σ(ab) = abN = aNbN = σ(a)σ(b)

so σ is operation preserving and thus a homomorphism.

We see that σ is surjective (if K∈G/N simply choose k∈K and σ(k) = K) so that Imσ = G/N. It is also not hard to see the kernel of σ either. Since g→gN, the only elements of G that map to the coset, N, are the elements inside of N, giving that Kerσ = N.

This particular homomorphism crops up quite a bit and as such is given its own special name.

Definition: Canonical Homomorphism
Let G be a group and N a normal subgroup of G. The map σ:G→G/N defined by σ(g) = gN ∀g∈G is called the canonical homomorphism.

Example:

You might recall that the determinant is a homomorphism. Let G = GL(n,R) and let θ:G→R* be defined by θ(A) = det(A) ∀A∈G. (R* is the multiplicative group of real numbers without zero.) You should recall from linear algebra that det(AB) = det(A)det(B) when A and B are square matrices of the same size. Now, choose A,B∈G and observe that

θ(AB) = det(AB) = det(A)det(B) = θ(A)θ(B),

so θ is a homomorphism.

It is not hard to see that θ is surjective, although it may take a little bit of extra linear algebra knowledge. Choose α∈R and let A = α^1/nI_n where I_n is the identity matrix in G. Then φ(A) = det(A) = α. Thus θ is surjective and Imθ = R*. The kernel of θ is a little more complicated. Kerθ consists of all the matrices with determinant equal to one. There is no nice, explicit form for these matrices, but they are used quite a lot and as such this group has its own name.

Definition: Special Linear Group
Let n be an integer and define the set SL(n,R) = {A∈GL(n,R) : det(A) = 1}. Then SL(n,R) forms a subgroup of GL(n,R) and is called the special linear group.

The first group isomorphism theorem gives a relationship between isomorphisms, homomorphisms, kernels, and images. It is an interesting way to find an isomorphism between often unrelated groups.

Theorem: First Group Isomorphism Theorem
Let φ:G→G* be a group homomorphism. Then G/Kerφ ≈ Imφ.

Proof:

Let φ:G→G* be any group homomorphism, let K = Kerφ, and define θ:G/K→Imφ by θ(gK) = φ(g) ∀gK∈G/K. Also call e the identity element of G*

First we must be sure that this is actually a function, that is that the mapping of θ is independent of coset representative. Choose H∈G/K and then choose g,h∈H. Since g∈hK ∃k∈K such that g = hk. Now observe that

θ(gK) = φ(g) = φ(hk) = φ(h)φ(k) = φ(h)e = φ(h) = θ(hK)

so finally θ(gK) = θ(hK) and θ is independent of the choice of coset representative, therefor making it a function.

To show that θ is surjective, choose q∈Imφ. Then clearly ∃p∈G such that φ(p) = q. Finally, θ(pK) = φ(p) = q giving that θ is surjective.

To show that θ is injective, it suffices to show that Kerθ is trivial, or that it contains only the identity of G/K. If we recall, this identity is simply K. Choose gK∈Kerθ. Observe that

e = θ(gK) = φ(g)

which means that g∈K and gK = K. Finally, Kerθ = {K} and θ is injetcive.

Choose aK,bK∈G/K and observe that

θ(aKbK) = φ(ab) = φ(a)φ(b) = &theta(aK)θ(bK)

so that θ is operation preserving.

Finally, we see that θ is an isomorphism and that G/Kerφ ≈ Imφ.

This is a pretty cool theorem and from our examples we arrive at a couple of interesting results. First, Z/nZ ≈ Z_n which can actually be shown without the use of the first isomorphism theorem. The second example gives us that for N⊳G, G/N ≈ G/N which is obvious. The first isomorphism theorem, however, does give a very interesting result from the third example that we saw - that is that GL(n,R)/SL(n,R) ≈ R*, which is pretty cool and rather unexpected.

There is one last thing that I should mention about homomorphisms for the time being, and that is that they can be used as a sort of subgroup test. From the way I introduced and defined SL(n,R) it is very obvious that its a normal subgroup of GL(n,R). However, suppose I'd defined SL(n,R) immediately after defining GL(n,R) and asked you to prove that SL(n,R)⊳GL(n,R) (instead of inventing the special linear group from the kernel of the determinant homomorphism like I did above). It can be done using the normal subgroup test, but its a giant pain with tons of pointless symbol chasing. The easiest way to prove it would be to discover a homomorphism (namely the determinant homomorphism) for which GL(n,R) is the domain and SL(n,R) is the kernel, which automatically gives that SL(n,R)⊳GL(n,R) since all kernels are normal. This is a very common technique for finding normal subgroups.

In the next post, I'm going to be taking a detour from homomorphisms and subgroups and talk about group actions.

References

Previous Related Post: Homomorphisms

Text Reference: Gallain Chapter 10

Wikipedia: First Isomorphism Theorem

The Unapologetic Mathematician: The First Isomorphism Theorem

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