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

Cyclic Subgroups

If you read a book on group theory they're going to give you all sorts of theorems and properties about cyclic groups and cyclic subgroups. And there are quite a few reasonably useful things that can be shown about cyclic groups with some relatively simple proofs. In all reality, though, none of that is extremely important. What is important is an understanding of what a cyclic group is and how they work. I'll repeat the definition here.

Definition: Cyclic
Let G be a group. Then G is cyclic if there exists an a in G such that for each g in G there is an integer, k, such that a^k = g. In this case, we use the notation G = <a>.

When I first introduced Z(n), I called them the cyclic groups. That is what they're called but I don't want that to be misleading. There do exists plenty of other cyclic groups and I'll give an example of another one.

Example: Another cyclic group
I will use the standard notation of i² = -1. Let G = {1, i, -1, -i} and let the operation on G be standard multiplication. It is easy to verify that G is a group. Now notice that i⁰ = 1, i¹ = i, i² = -1, and i³ = -i. It then follows from the definition of a cyclic group that G is cyclic and G = <i>.

We can generalize this notation of <a>, and this is often where we get the most usefulness out of this concept of cyclic groups. Suppose that G is a group and g∈G. We define <g> = {g^k : k∈Z} = {...,g^-2,g^-1,g⁰,g¹,g²,...}. In regular english, this means that <g> is the set of all elements of the form g^k such that k is any integer. In case it is not clear, if n is positive, then g^-n = (g^-1)ⁿ = (gⁿ)^-1. We now prove that if g∈G then <g> is a subgroup of G.

Theorem: <g> is a subgroup
Let G be a group and g∈G. Then <g> is a subgroup of G and is called the cyclic subgroup generated by g.

Proof:
First we must show that <g> is a subset of G. Choose an element g^k∈<g> where k is an integer. If k = 0 then g^k = e, the identity element of G which is clearly in G. If k ≠ 0 then g^k is either the product of k copies of g (if k is positive) or the product of k copies of g^-1 (if k is negative). But g,g^-1∈G and G is closed under its operation, so g^k∈G and <g>⊆G. I now wish to use the one-step subgroup test to show that <g> is a subgroup of G. Choose g^m,gⁿ∈<g>. Then g^m∙(gⁿ)^-1 = g^m∙g^-n = g^m-n and m-n is an integer so g^m∙(gⁿ)^-1=g^m-n∈<g>. Thus, by the one-step subgroup test, <g> is a subgroup of G.

I know that's kind of a dense proof and might be a little confusing, but the proof isn't that important or groundbreaking. Lets look at an example of one of these cyclic subgroups. Consider Z(6) and 2∈Z(6). I now wish to look at <2>. Every element of <2> is of the form k∙2 for each integer, k and I'd now like to investigate what this set looks like. <2> = {0∙2,1∙2,2∙2,3∙2,4∙2,...} = {0,2,4,0,2,...}. This sequence will continue indefinitely, so <2> = {0,2,4}. Similarly, <3> = {0,3}, and each subsequent cyclic subgroup can be calculated similarly.

Now, if G is a cyclic group, then we have just proven that ∀g∈G, <g> is a subgroup of G. But, by the definition of cyclic groups, we know that there exists some a∈G such that <a> = G. The question then arises as to whether this a is unique. For example, we know that Z(6) = <1> under addition mod 6, but is there any reason there isn't some other k∈Z(6) such that Z(6) = <k>? The answer is no. In fact, Z(6) = <5>. Notice that <5> = {0∙5,1∙5,2∙5,3∙5,4∙5,5∙5,6∙5,7∙5,...} = {0,5,4,3,2,1,0,5,...}. We know that there cannot be anything in <5> that is not in Z(6) (because <5>⊆Z(6)) but we've seen that each element in Z(6) is also in <5>, so <5>=Z(6). This brings us to the following definition.

Definition: Generator of a Cyclic Group
Suppose that G is a cyclic group and a∈G. Then if <a> = G, a is called a generator of G.

In general, most cyclic groups have more than one generator. There is one last thing that I need to say about cyclic subgroups. The examples of cyclic subgroups that I gave were subgroups of groups that were, themselves, cyclic. That was merely a coincidence. Given any arbitrary group, G, and any element a∈G, <a> is a cyclic subgroup of G regardless of the properties of the original group, G.

References

Text Reference: Gallain Chapter 4

Wikipedia: Cyclic Group

Planet Math: Cyclic Group