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