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

Cayley's Theorem

Today we're going to culminate the last few posts into a theorem that combines the idea of symmetric groups and isomorphisms into something quite useful. I'm going to present a proof because its a good example of a constructive proof. Cayley's theorem claims the existence of something (particularly a permutation group isomorphic to a given group) and most theorems about existence only show a theoretical existence, but constructive proofs show that something exists by actually constructing the thing for which they're claiming existence. The statement and proof of Cayley's Theorem is given below.

Theorem: Cayley's Theorem
Every group is isomorphic to a group of permutations.

Proof:

Before we can prove the isomporphism, we must first build this permutation group. ∀g∈G, define T_g:G→G by T_g(x) = g∙x ∀x∈G. We recall that the definition of a permutation is a bijection from a set to itself. G is a set, T_g is a function from G to G ∀g∈G, and it is easy to verify that T_g is also a bijection ∀g∈G, thus making T_g a permutation on G. Now let G* = {T_g : g∈G}. Then G* is a group under the operation of function composition. I will verify this, but only briefly. Choose T_g,T_h∈G*, let x∈G, and observe that (T_g∘T_h)(x) = T_g(T_h(x)) = T_g(h∙x) = g∙(h∙x) = (g∙h)∙x = T_g∙h(x). Then T_g∘T_h = T_g∙h and T_g∘T_h∈G* which gives closure. Since T_g∘T_h = T_g∙h, it is then clear that T_e is the identity of G* and that (T_g)^-1 = T_g^-1. Finally, all function composition is associative, so G* is a group.

The isomporphism between G and G* is now obvious - define θ:G→G* by θ(g) = T_g ∀g∈G. Suppose that g,h∈G and θ(g) = θ(h). Then T_g = T_h and T_g(e) = T_h(e) so g∙e = h∙e giving g = h. Thus θ is injective. Since G* = {T_g : g∈G}, it is clear that ∀T_g∈G*, θ(g) = T_g so θ is surjective. To see that θ is operation-preserving, observe that ∀a,b∈G, θ(a∙b) = T_a∙b = T_a∘T_b = θ(a)∘θ(b). Finally, θ is an isomporphism and G ≈ G* where G* is a group of permutations, as desired.

Cayley's theorem tells us some neat things. Note that it doesn't say that every group is isomorphic to a symmetric group, but that it is isomorphic to some permutation group. I know I never formally defined a permutation group, but as you might think, its some set of permutations that form a group under function composition, and I did define a permutation. Now suppose G is any arbitrary group, Cayley's Theorem tells us that G ≈ G* where G* is a permutation group and, in particular, G* is a group of permutations on the set G. When we first discussed permutations of symmetric groups, we thought of them as "rearrangements" of the set Ω_n. Similarly, we can think of the permutations in G* as rearrangements of the elements of G. So now we can assign each element of G an integer from Ω_|G| and in this manner, each element of G* corresponds to an element of S_|G|. This correspondence is certainly injective and operation-preserving, but may or may not be surjective. It may require a bit of an imaginative leap of faith, but what this gives us is that G* ≈ H where H is some subgroup of S_|G|. The final result here, is that any group is isomorphic to some subgroup of a symmetric group (that is that G≈G*≈H where H is a subgroup of a symmetric group). This is important because even the most abstract of groups can be represented in a very concrete way. And that is very, very cool.

References

Previous Related Post: Properties of Isomorphisms

Text Reference: Gallain Chapter 6

Wikipedia: Cayley's Theorem

Properties of Isomorphisms

Last time we learned the definition of an isomorphism. Today I intend on explaining the usefulness of this concept of isomorphic groups. The first thing I'm going to do is list a bunch of properties of isomorphisms without proof. You don't need to remember them, but I'd like to use them to help convince you of the vast consequences of this idea of isomorphic groups.

Theorem: Properties of Isomorphisms
Suppose that A and B are groups and φ is an isomorphism between them. Then

• If A and B are finite, then |A| = |B|.
• If e_A∈A is the identity of A and e_B∈B is the identity of B, then φ(e_A) = e_B.
• ∀n∈Z and ∀a∈A, φ(aⁿ) = [φ(a)]ⁿ.
• ∀a,b∈A, a∙b = b∙a if and only if φ(a)∙φ(b) = φ(b)∙φ(a).
• A = <a> if and only if B = <φ(a)>.
• ∀a∈A, |a| = |φ(a)|.
• ∀k∈Z and ∀a∈A, |{x∈A : x^k = a}| = |{x∈B : x^k = φ(a)}|. In other words, the number of solutions to the equation x^k = a is the same as the number of solutions to the equation x^k = φ(a).
• If A and B are finite, then then they have exactly the same number of elements of every order.
• The function φ^-1:B→A is an isomorphism from B to A.
• A is abelian if and only if B is abelian.
• A is cyclic if and only if B is cyclic.
• If H is a subgroup of A, then the set φ(A) = {φ(h) : h∈H} is a subgroup of B.

Normally, when I give a theorem, I then try and explain it, but I'm not really going to do that this time. These 12 facts are more interesting than useful, though they are sometimes useful in proving that groups are not isomorphic. For example, one might conjecture that Z(6) ≈ S₃ because |Z(6)| = 6 = |S₃|, but it would be very easy to calculate the orders of every elements of both groups, and show that these groups fail the 8^th property above.

However, the reason that I listed all of these properties of isomorphisms is to make it evident that isomorphic groups have many properties in common. In fact, isomorphic groups have all of their group-theoretic properties in common. And more than that, the definition of a group-isomorphism was precisely formulated so that isomorphic groups would share every group-theoretic property. What this means is that suppose I have two groups that are isomorphic, but I cannot actually see the individual elements - rather I can only see their interactions. Then I could not tell the difference between them. Because of this, mathematicians tend to think of isomorphic groups as "equal" or as "the same." This is the real power of the idea of isomorphism because it greatly reduces the number of groups that we need to study. For example, it turns out that S₃ ≈ D₃ (this is a very special case - it is not even close to true that S_n ≈ D_n for n≠3). That means that there is no real point in studying both groups in detail because whatever we know about one transfers automatically to the other. As an even broader example, suppose that G is cyclc and |G|=n. Then by a very simple argument we can show that G ≈ Z(n). The cyclic groups, Z(n) are very well understood, so as soon as we know that a group is cyclic, we know everything there is to know about that particular group. As you can see, isomorphisms very much reduce the amount of work that we have to do as group theorists.

Before we leave, there's a couple of other things we should discuss about isomorphisms. Primarily, there are a couple of special sorts of isomorphisms that have their own names.

Definition: Automorphism
An isomorphism from a group G onto itself is called an automorphism on G.

It should be clear that a group is isomorphic to itself. First of all, it makes sense - that is, we defined the notion of isomorphism to be such that isomorphic groups are the same, and clearly a group should be the same as itself. If you'd like to be rigorous, though, if we take a group, G, and define φ:G→G by φ(g)=g ∀g∈G, then it is very simple to show that φ is an isomorphism. This function, φ, shows that for any arbitrary group, G, there is one isomorphism from G to G but there are usually many others. Any of these functions - isomorphisms from a group onto itself - are called automorphisms, and in fact, the set of all possible automorphisms on a group forms a group itself under function composition.

Theorem: Aut(G) forms a group
Let G be any group and define Aut(G) to be the set of all automorphisms on G. Then Aut(G) forms a group under function composition.

Proof:
We denote the composition operation on Aut(G) by ∘.

Closure:

Choose φ,θ∈Aut(G). I wish to show that φ∘θ∈Aut(G) - that is that φ∘θ is an isomorphism from G to G. It is a property of bijections that a composition of bijections is also a bijection. I present this without proof because it is simple to verify. This gives that φ∘θ is a bijection, so we need to show that it is operation preserving. To see this, choose a,b∈G and observe that (φ∘θ)(a∙b) = φ(θ(a∙b)) = φ(θ(a)∙θ(b)) = φ(θ(a))∙φ(θ(b)) = (φ∘θ)(a)∙(φ∘θ)(b) so φ∘θ is operation preserving. Finally, this gives us that φ∘θ∈Aut(G) and Aut(G) is closed under function composition.

Associativity:

Everything in Aut(G) is a function and it is a property of all functions that their composition is associative.

Identity:

Define the function ε:G→G by ε(g)=g ∀g∈G. Now choose φ∈Aut(G) and choose g∈G. Observe that (ε∘φ)(g) = ε(φ(g)) = φ(g) and (φ∘ε)(g) = φ(ε(g)) = φ(g). It then follows that ε∘φ = φ = φ∘ε and ε is the identity element of Aut(G).

Inverses:

Choose φ∈Aut(G). Since φ is an isomorphism from G to G it folows from the first theorem in this post that φ^-1 is also an isomorphism from G to G and so φ^-1∈Aut(G) and by definition we have that φ∘φ^-1 = ε = φ^-1∘φ.

The above 4 properties give us that Aut(G) is a group under the operation of function composition.

I gave the proof because I think its worth reading. Usually I say that you can skip the proof if you'd like, but in this case I think the proof - though not particularly interesting - does give a good idea of the mechanism of the group and how it works. Inside of this group of automorphisms, there are some even more interesting functions called inner automorphisms. I'm not going to go into great detail about these and I'm not going to explain everything. They're sort of strange to understand, the proofs are tedious and boring, and I don't plan on talking about them for a long time, but whenever you're talking about the group of automorphisms, the group of inner automorphisms will probably be in the discussion so its worth defining.

Definition: Inner Automorphism Induced by a
Let G be a group and let a∈G. Then the function φ_a∈Aut(G) defined by φ_a(g) = a∙g∙a^-1 ∀g∈G is called the inner automorphism of G induced by a.

One thing to note is that this is a definition but it requires a bit of proof - that is, it is not obvious that φ_a∈Aut(G). However, it is a pretty easy thing to show.

Theorem: Inn(G) is a Subgroup of Aut(G)
Let Inn(G) be the set of all inner automorphisms on G. That is, if a∈G and φ_a is the inner automorphism induced by a, then Inn(G) = {φ_a : a∈G}. Then Inn(G) is a subgroup of Aut(G).

This theorem can be easily proven by a simple application of a subgroup test.

References

Previous Related Post: Isomorphisms

Text Reference: Gallain Chapter 6

Isomorphisms

Most every branch of mathematics concerns itself with its own special type of mathematical "objects," and there are a lot of them - groups, modules, topological spaces, manifolds, rings, fields, algebras, and vector spaces are just the ones that I can think of off the top of my head - and almost all of them define their own version of the word "isomorphism." In general, an isomorphism is a special type of function between two of the same "objects" that tells us when we can consider them eqivalent. As we will see, the existence of a group-isomorphism between two groups means that we can think of them as the same.

Before learning about isomorphisms, I should talk about bijections. I've used the word bijection before when talking about permutations, but in this post I'm going to need to actually show that things are both injective and surjective, so in order to make sure the meanings of those things are clear, I'm going to define them here.

Definition: Injective
Suppose φ:A→B. φ is called injective if different elements a,b∈A correspond to different elements, φ(a),φ(b)∈B. Symbolically, φ is injective if ∀a,b∈A, φ(a) = φ(b) implies that a = b or equivalently, φ is injective if ∀a,b∈A, a ≠ b implies that φ(a) ≠ φ(b). We call such functions injections.

Definition: Surjective
Suppose φ:A→B. φ is called surjective if the range of φ covers all of B. Symbolically, φ is surjective if ∀b∈B, ∃a∈A such that φ(a) = b. We call such functions surjections.

Definition: Bijective
A function is called bijective if it is both injective and surjective. Such functions are called bijections.

Now that we have the definition of a bijection, we're ready to define a group-isomorphism. Although the technical definition is that of a "group-isomorphism," when the context is known (that is when it is clear that we're talking about groups) I will leave out the word "group" and just refer to them as "isomorphisms."

Definition: Group-Isomorphism
If A and B are groups and φ:A→B is a function from A to B, then φ is an isomorphism if it is a bijection and ∀a,b∈A, φ(a∙b) = φ(a)∙φ(b). If there exists an isomorphism between two groups, A and B, then we say that A and B are isomorphic and write A≈B.

There are two things important in the definition of an isomorphism. First, it must be a bijection, which means that it is both injective and surjective, like always. The second, and more important part, is that it is what we call "operation preserving," which is the condition that ∀a,b∈A, φ(a∙b) = φ(a)∙φ(b). What this says is that if φ is an isomorphism between A and B, then when a,b∈A and φ(a),φ(b)∈B, then it doesn't matter whether you combine them in A or in B, you'll get the same thing on either side of φ. This idea of φ being operation preserving is shown pictorally, below. The dashed arrows represent the group operation and the solid arrows represent the mapping by φ.

In an effort to clarify all of this, here's an example.

Example:
Let A be the group of real numbers under addition and let B be the group of positive real numbers (not including zero) under multiplication. Let φ:A→B be defined by φ(x) = 2^x.

Step 1: φ is injective

Choose x,y∈A such that φ(x) = φ(y). Then we have 2^x = 2^y. Taking log₂ of each side gives log₂(2^x) = log₂(2^y) and by properties of logarithms we get that x = y. Therefore φ is injective.

Step 2: φ is surjective.

Choose y∈B. Let x = log₂(y). We note from the definition of logarithms that x∈A and then φ(x) = 2^log₂(y) = y. Therefor φ is surjective.

Step 3: φ is operation preserving.

Choose x,y∈A. Observe that φ(x+y) = 2^x+y = 2^x∙2^y = φ(x)∙φ(y). Therefore φ is operation preserving.

Steps 1 through 3 verify that φ is indeed an isomorphism and A≈B.

I don't imagine the usefulness of this concept of isomorphism is clear yet, but hopefully you can at least understand the definition. To sum things up a little bit, two groups are isomorphic if each element in one group corresponds to exactly one element in the other group, and combining two elements in the first group is the same as combining their corresponding elements in the second group. As we will see in the next post, there are a lot of very interesting properties that we get out of isomorphisms.

References

Previous Related Post: Recap - Subgroups

Text Reference: Gallain Chapter 6

Wikipedia: Group Isomorphism