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

General Linear Groups

Very quickly, before we move on to isomorphisms like I promised, I wanted to throw in a quick plug for the general linear groups. The general linear groups are the matrix groups. I'm just going to introduce this very important group and leave out the details and the proofs because it needs some linear algebra that I'm not going to derive. The only reason I'm going over this group is because it is an excellent group that gives a lot of nice examples. Whenver I use it in the future, though, I'll make sure that if you skip it you won't miss anything since beginners may not know any linear algebra.

Definition: GL(n,F)
Let n be a positive integer and let F be any field. The set GL(n,F) is the set of all n-by-n matrices with elements in F and with a non-zero determinant.

I haven't yet defined a field, but if you don't know what a field is, just replace any field with the real numbers, one of the most common fields. If we just consider the real numbers, R, then GL(n,R) is the set of all of the standard n-by-n matrices that we've always used in linear algebra. As it turns out, this set forms a group under matrix multiplication.

Theorem: General Linear Group
GL(n,F) forms a group under the operation of standard matrix multiplication.

I'm not going to prove this theorem, but as an example, it is pretty clear that GL(n,R) is a group if you're familiar with linear algebra. Every matrix in GL(n,R) is square of dimension n so it has to be closed, all matrix multiplication is associative, the standard identity matrix is the identity element of GL(n,R), and the inverse of any matrix is the standard matrix inverse from linear algebra.

As it turns out, this is a very important and a very cool group. If you don't know enough linear algebra to get this group yet, don't worry about it. I just wanted to throw it out there so that I can use it sparringly in examples later if I want. Tune in next time for an introduction to isomorphisms.

References

Previous Related Post: Properties of Symmetric Groups

Text Reference: Gallain Chapter 2

Wikipedia: General Linear Group

Properties of Symmetric Groups

One thing that I forgot to mention last time is that there is a very simple method of calculating products of permutations, and if I had 5 minutes and a chalk board, I would love to show you. However, I've been trying to think of a way that I could explain it thoroughly and concisely in writing (or in typing, whatever the case may be) but don't think I can do it, so I'm going to omit it instead of dedicating an entire entry to it. You shouldn't need to do any such calcuations while reading my blog (that is, if they're needed I'll just go ahead and give you the answers) so it shouldn't be a problem for now, but if you'd like to know, as always, feel free to ask.

In this post, I'm going to talk about some of the properties of the symmetric groups. I'm not going to prove most of them because the proofs are boring, but my hope is to explain it all well enough so that you can understand these very important groups a little bit better.

Definition: Disjoint Cycles
Let α=(a₁,a₂,...,a_p) and β=(b₁,b₂,...,b_q) be two cycles. Then α and β are called disjoint if their cycles have nothing in common, that is {a_i : 1≤i≤p}∩{b_j : 1≤j≤q}=∅

Theorem: Product of Disjoint Cycles
Every permutation in of a finite set can be expressed as a product of disjoint cycles.

Recall that there are a bunch of different ways to express every permutation. In my last post I noted that (1523) = (134)(25)(12)(34) is one such example. However, what the above theorem says is that there is a "nice" way to write every permutation. What I mean by "nice" is that we like for permutations to be written as the product of cycles that don't have any numbers in common, which is what is meant by disjoint. If π = (1523) then although it is technically correct to write π = (134)(25)(12)(34), some of those cycles have numbers in common, so we prefer π = (1523). The next theorem is another reason its convenient to express a permutation as a product of disjoint cycles.

Theorem: Disjoint Cycles Commute
If α=(a₁,a₂,...,a_p) and β=(b₁,b₂,...,b_q) are two disjoint cycles then α∙β = β∙α.

This theorem is pretty self-explanitory, but is quite nice. It is not an if and only if statement, though, which means that if two cycles commute they are not necessarily disjoint. As an example, (12)∙(21) = e = (21)∙(12) and these cycles are clearly not disjoint. However, though, it is quite rare to find two permutations that commute. In fact, for n>2, Z(S_n) = {e}.

The rest of the post is going to be dedicated to developing an important subgroup of the symmetric groups called the alternating group.

Theorem: Product of 2-Cycles
∀n>1, every permutation in S_n can be written as a product of 2-cycles.

This theorem is best explained through examples. You can verify that (1426) = (16)(12)(14) and that (16243) = (13)(14)(12)(16). If you look at those two examples closely, hopefully you can see a pattern, and that pattern works for every cycle. We can then expand this to permutations of more than one cycles, so that (1347)(265) = (17)(14)(13)(25)(26). In fact, that particular method of turning any permutation into a product of 2-cycles is exactly how that theorem is proven. The proof of the next theorem is actually a little interesting, but it requires induction which I haven't shown you yet, so I'm going to have to leave it out.

Theorem: Always Even or Always Odd
Suppose that α∈S_n and both α = β₁β₂∙∙∙β_p and α = γ₁γ₂∙∙∙γ_q where each β_i and γ_j are 2-cycles. Then either p and q are both odd or p and q are both even.

This theorem says that if a permutation can be written as a product of an even number of 2-cycles then every way of writing that permutation as a product of 2-cycles has an even number of them and similarly, if a permutation can be written as a product of an odd number of 2-cycles then every way of writing that permutation as a product of 2-cycles has an odd number of them. Its hard to give examples of this because I can't show every possible representation of a permutation as 2-cycles (because there are an infinite number of them), however, you can observe that (13)(12)(15) = (1523) = (14)(13)(25)(12)(34). Also, (12)(34) = (12)(34)(12)(12) = (12)(34)(12)(12)(12)(12) and so on. This gives rise to the next definition.

Definition: Even and Odd Permutations
If a permutation can be written as the product of an even number of 2-cycles then that permutation is called even, and if a permutation can be written as the product of an odd number of 2-cycles then that permutation is called odd.

Now that we know the difference between even and odd permutations, we are ready to define the alternating group.

Theorem: A_n is a subgroup of S_n
For n>1, the set of all even permutations of S_n form a subgroup of S_n. This group is called the alternating group and is denoted A_n.

Proof:
First note that e = (12)(12) so e∈A_n and A_n is non-empty. Now, choose α,β∈A_n. Note that since α and β are even, α = α₁α₂∙∙∙α_p and β = β₁β₂∙∙∙β_q where p and q are even integers and each α_i and β_j are 2-cycles. We then get that α∙β = α₁α₂∙∙∙α_p∙β₁β₂∙∙∙β_q so α∙β is the product of p+q 2-cycles and since each of p and q are even then p+q is even, thus α∙β∈A_n. Now, choose γ∈A_n. Then γ = γ₁γ₂∙∙∙γ_m and γ^-1 = γ'₁γ'₂∙∙∙γ'_n where m is even, n is some positive integer, and each γ_i and γ'_j are 2-cycles. I now wish to show that n is even. Naturally we have that e = γ∙γ^-1 = γ₁γ₂∙∙∙γ_m∙γ'₁γ'₂∙∙∙γ'_n so e is the product of m+n 2-cycles. However, e = (12)(12), so e is even and m+n must also be even, but since m is even, n must also be even and thus γ^-1∈A_n. Finally, by the two-step subgroup test, A_n is a subgroup of A_n.

You'll note that S_n is always finite (since Ω_n is always finite), so A_n is always finite and I could have used the finite subgroup test instead of the two-step subgroup test. I wanted to emphasise, though, that the even permutations of the permutation group on any set, Ω (even infinite sets), form a subgroup even if they aren't finite.

That's all the work we'll be doing specifically with the symmetric groups for a while except in a couple of posts I'm going to prove Cayley's Theorem. That's pretty exciting, because whenever a theorem is named after someone, its usually pretty important. Before I do that, though, I'm going to introduce isomorphisms.

References

Previous Related Post: Symmetric Groups

Text Reference: Gallain Chapter 5

Wikipedia: Symmetric Groups

Symmetric Groups

As I mentioned last time, this post is going to be about some of the most important groups in mathematics, the symmetric groups. They are not extremely difficult in that they are more natural than the dihedral groups, but they are rather difficult to explain so bear with me and, as always, ask questions.

I'm going to start using the notation Ω_n to denote the set of integers from 1 to n, or in symbols, Ω_n = {1,2,...,n} and I will use Ω to denote any arbitrary set. I'm going to define a function, σ, that takes Ω₅ to Ω₅ by σ(1)=3, σ(2)=5, σ(3)=4, σ(4)=1, and σ(5)=2. This function, σ, is called a permutation. The reason for this is that it takes each element of Ω₅ to a different element of Ω₅. The best way to think about a permutation, though, is as a way to shuffle, or rearrange the integers. Our σ function reorders Ω₅ in that applying σ to [1,2,3,4,5] gives [σ(1),σ(2),σ(3),σ(4),σ(5)] or [3,5,4,1,2]. A permutation is simply any function that takes shuffles Ω_n around in some order like σ did. As a matter of notation, if the particular permutation is implied, say σ, then we sometimes denote σ(1)=3 by 1→3. Below I'm going to give the formal definition of a permutation, but it is not a very useful definition and you don't need to be too concerned if it doesn't make much sense.

Definition: Permutation
Let Ω be any arbitrary set. A permutation of Ω is a bijection from Ω to Ω.

Like I said, this definition is more formal than the way we traditionally think of permutations. Usually a permutation is not defined as I defined σ simply because its too cumbersome. Instead, one notation is as follows.

The notation above tells where each element of Ω₅ is mapped (the elements in the top line map to the elements directly below them) and gives a good visual representation of the "shuffling." One thing to notice is that σ gets broken down into partitions. Notice that 1→3, 3→4, and 4→1. Similarly, 2→5 and 5→2. This partitions {1,3,4} from {2,5}. Pictorially, we can show this as follows:

Using this as a guideline, we introduce a new notation that we call cycle notation. We take each of these cycles and put them in parenthases in the order of their arrows (the starting point is irrelevant). As an example, σ = (1,3,4)(2,5). We think of the end of one cycle as going back to the beginning of the cycle, so then (1,3,4) = (3,4,1) because they would generate the same picture. However, it should be noted that (1,3,4) ≠ (1,4,3) because 1→3, 3→4, 4→1 is not the same as 1→4, 4→3, 3→1. If n<10 and σ is a permutation of Ω_n, then we generally leave out the commas so that σ = (134)(25).

Hopefully its now clear what a permutation of Ω_n is, and now we need to know how these permutations form a group.

Definition: The Symmetric Group, S_n
Let Ω_n = {1,2,...n} for some positive integer, n, and let S_n be the set of all permutations of Ω_n. Then S_n forms a group under the operation of function composition and is called a Symmetric Group.

I'm not going to formally prove that this is a group, but I am going to talk through the justification. First, though, it is very important to understand the operation of composition. Let σ and τ be permutations of Ω_n. As group elements, we say that σ∙τ is the composition of the two functions, σ∘τ = σ(τ). If you're not familiar with function composition, what this means is that ∀i∈Ω_n, (σ∙τ)(i) = (σ∘τ)(i) = σ(τ(i)). Suppose that σ = (134)(25) and τ = (12)(34) are elements of S₅ and we'd like to calculate σ∙τ. We can first calculate (σ∙τ)(1) = σ(τ(1)) = σ(2) = 5. A similar calculation for the rest of Ω₅ yields that σ∙τ = (134)(25)(12)(34) = (1523)(4) (it may be useful to perform this calculation yourself).

Now that we undestand function composition, I can try to convince you that S_n is a group for each positive integer, n. This is going to require some properties of bijective functions, though, and I'm going to use them without proof or explanation, but if you want an explanation, I'll be happy to provide it. Closure on S_n immediately follows from the fact that composition of bijective functions is also bijective, and associativity come for free because all functions are associative under composition. The identity element is the identity function, ε, that maps each element of Ω_n to itself. Finally, a property of every bijective function is that it is invertible, so every element in S_n has an inverse.

There are a couple more things to tidy up about the symmetric groups. We call a cycle with n elements in it an n-cycle. As an example, (134) is a 3-cycle. Also, any 1-cycle is the identity element. This is easy to see, as (1) only sends 1 to itself. Because of this, we generally omit 1-cycles. Earlier we showed a computation that resulted in σ∙τ = (1523)(4). However, the (4) is usually left off because it does not affect the cycle at all, so we write (1523)(4) = (1523). Thus, as a matter of convention, if a permutation is an element of S_n and its cycles omit some elemets of Ω_n, then it is assumed that the missing elements are mapped to themselves. One last thing that should be noted is that permutations do not have unique cycle structures. In fact, as we saw earlier, if π = (1523) then we can also write it as π = (134)(25)(12)(34). However, as we will see in my next post, every permutation can be expressed uniquely as a product of disjoint cycles (disjoint cycles don't contain any of the same numbers).

Like I said earlier, these symmetric groups are extremely important. A lot of the applications involving symmetric groups are typically deeply embedded within seeminly unrelated things, so it is probably not very informative for me to list them. However, what permutations do is shuffle things around and reorder things (our examples only shuffled numbers, put remember that Ω can be any arbitrary set), so maybe you can at least conceive of the vast usefulness of the group structure of these permutations. In my next post I'm going to discuss some of the elementary properties of symmetric groups.

References

Previous Related Post: Dihedral Groups

Text Reference: Gallain Chapter 5

Wikipedia: Symmetric Groups

The Dihedral Groups

The next two posts are going to be rather long and painful explinations of two very important groups, the dihedral groups and the symmetric groups. I've been trying to tip-toe around these two groups because they're difficult to explain and to understand, however the time has come to dive in. The dihedral groups crop up from time to time and so it is important to understand them, but the nicest part of the dihedral groups is that they're both small and rather complicated so they often make an excellent examples. The symmetric groups, though, are extremely important. I mean, extremely important. There have been entire volumes of books written about them (I'm trying to read one right now, actually) and they crop up all over the place in basically every branch of mathematics. As the title implies, I'm going to start today with the dihedral groups.

Long ago in a previous post I gave the Dihedral group as an example of a group. If you don't remember it, I suggest you go back and read Examples of Groups. Basically, the Dihedral Group of order 8, which I'm going to denote D₄ from now on, is the set of all possible symmetries (rotational and reflective) of a square with the operation of composition - that is if g,h∈D₄ then g∙h is the symmetry that corresponds to performing g and then h. I'm going to use a specific notation for the elements of D₄ and that notation will be D₄ = {e,R₉₀,R₁₈₀,R₂₇₀,F_h,F_r,F_v,F_l}. To be clear what I mean by this, e is the identity, R₉₀ is the clockwise rotation by 90 degrees and R₁₈₀ and R₂₇₀ are defined similarly. The rest of the elements correspond to the symmetries shown below.

There is an easier and more convenient way to view these elements. That is, we start with a square and label the corners and then for each element of D₄ we look at where the symmetry moves the corners. In this manner, each of the elements is shown below.

This is a nice and convenient way to view D₄. However, you don't have to take my word for it. It is a very good exercise to get a piece of paper, label it, and actually do all these symmetries to convince yourself that this is not only correct, but is also a reasonable thing to do. Who knows, I might have even made a typo or two. Now that we've established this nice visualization for the elements, lets explore the operation on D₄. When I introduced the dihedral group, I made the claim that it was closed under the composition operation that I described and in fact, part of convincing you that it was true was convincing you of this closure. I promise that I wasn't lying, but lets take a look at the operation from the standpoint of this new notation. If you take two elements of D₄ and combine them, it is just a matter of performing both symmetries consecutively. As an example, consider R₉₀,F_h∈D₄ and observe below R₉₀∙F_h.

Looking at this diagram and comparing it with the diagrams above of the eight elements of D₄, we see that performing R₉₀ and then F_h is the same as performing just F_r so R₉₀∙F_h = F_r. There are 8 elements in D₄ and so then 64 ways of combining them. Performing each of these 64 operations gives rise to something called a Cayley Table. Its called a Cayley Table because a brilliant British mathematician named Arthur Cayley was the first one to use it, but really its not that special. The elements of D₄ (or of whatever group you'd like) are listed along the top and along the left side and each of the entries in the table corresponds to a product of two elements. If you pick an element along the left side, say a, and an element along the top, say b, then the entry in the table corresponding to that row and column is a∙b (not b∙a). The Cayley Table for D₄ is given below.

That table gives us basically the whole picture of D₄. Indeed, a Cayley Table gives all the information about any particular group and is great for understanding smaller groups, but it is quite obvious that when a group gets large, constructing this table is a very unrealistic task. However, for the sake of learning, there are some interesting things that this table makes visible. You'll notice that every column and every row contains every element of D₄ which is not a coincidence. I'm not going to prove this, but it comes from closure and the existence of inverses. Also, you can see that the table is sort of blocked off into 4 groups. The evenness of the division is coincidental, but the reason for it is that {e,R₉₀,R₁₈₀,R₂₇₀} is a subgroup of D₄ and Cayley Tables tend to illuminate and partition subgroups. Finally, the table is not symmetric in the case of D₄, but the Cayley Table of an arbitrary group is symmetric when the group is abelian and it is not symmetric otherwise.

It is my hope that you've now gathered a pretty good understanding of the dihedral group of order 8, D₄. If not, then feel free to ask me about it - it can certainly be confusing. If you do feel like you understand it, though, then that is certainly an accomplishment and is most likely a sufficient understanding of the dihedral group. However, there is a little bit more that should be said about dihedral groups.

I use the notation, D₄, for the dihedral group of order 8 for a very specific reason, and that's because it is the group of symmetries of the regular 4-sided polygon, which we call a square. (The word 'regular' when applied to a polygon means all of the edges have the same lengths and all the angles are the same.) In the same manner, we can define D_n as the group of symmetries of the regular n-gon. Specifically, D₃ is the group of symmetries of the triangle, D₅ is the group of symmetries of the pentagon, and etc. A general fact about D_n is that it has order 2n. In fact, this brings me to something that is annoying, but needs mentioning. If you leave the comforts of Dihedral Soup, you'll find that the notation of the dihedral groups is not standard. Most people define the dihedral group of order 2n as D_n as I have, but there are people who call it D_2n. I will always use the notation as I've already defined it, but if you do a google search or read a book, I wanted you to know that it might not be consistent.

The last thing that I need to mention is that there is a general form for D_n. If you've never seen the dihedral group before, this is going to be very confusing, but that's okay. I don't plan on using it explicitly in the future, but it is interesting and important, so I'm going to show it to you.

Definition: The dihedral group of order 2n
Choose an integer n≥3. We define the dihedral group of order 2n by D_n = {σⁱ∙τ^j : i,j≥0, σⁿ = e = τ², σ∙τ = τ∙σ^n-1}.

That is a big, giant, bear of a definition, its very confusing, and its very impractical. However, it does work. Consider n=4. If you stare at it long enough, you'll see that according to the above definition, D₄ = {e,σ,σ²,σ³,τ,τ∙σ,τ∙σ²,τ∙σ³}. These elements correspond respectively with the elements that we defined earlier, {e,R₉₀,R₁₈₀,R₂₇₀,F_h,F_r,F_v,F_l}. It is difficult to see that this works, but lets look at it a little bit. We have just two rules, that is that σ⁴ = e = τ² and σ∙τ = τ∙σ³. So then we can calculate σ∙(σ²∙τ) = σ³∙τ. If we translate this into the language of our earlier definition of D₄ we get R₉₀∙F_v = F_l which does correspond to the information in the Cayley Table above. A slightly more difficult calculation is that of (σ∙τ)∙(σ²∙τ). Remember the two rules we have and observe (σ∙τ)∙(σ²∙τ) = (σ∙τ)∙σ∙(σ∙τ) = (τ∙σ³)∙σ∙(τ∙σ³) = τ∙σ⁴∙τ∙σ³ = τ∙e∙τ∙σ³ = τ²∙σ³ = e∙σ³ = σ³. From start to finish, we get (σ∙τ)∙(σ²∙τ) = σ³ which translates into F_r∙F_r = R₂₇₀, which also corresponds to the information in the Cayley Table.

There you have it - a very lenghy explination of the dihedral groups. I do hope it wasn't too bad. I expect there to be some difficulties and some confusion, so as always, feel free to ask questions.

References

Previous Related Post: Recap - Subgroups

Text Reference: Gallain Chapter 1

Wikipedia: Dihedral Group

Planet Math: Dihedral Group

Interactive Demo of Dihedral Groups

Galois

I came across a pretty cool application of group theory the other day and a pretty cool story to go along with it, so I thought I would share it with you. Évariste Galois was a French mathematician in the early 1800's. He is accredited with starting the fields of Galois theory and Galois connections. He is also thought to be the first person to use the word "group" as a mathematical term. Above all that, though, he was a very stupid, and stubborn man.

For almost as long as there has been mathematics, mathematicians have been concerned with the solutions to polynomials. So we studied for a while and came up with a general equation for second degree polynomials which was nice and pretty and everything. Of course, there was no stopping there, so we worked a little harder and came up a similar equation for third degree polynomials and fourth degree polynomials. If you click on the links you'll see that those equations are gigantic and impractical, but nevertheless, there they are. Then mathematicians started working on an equations for fifth degree polynomials but found it to be a lot harder. So they worked and worked and came up with a whole entire branch of mathematics called algebraic geometry to try to find this equation, but they couldn't do it. I guess the mathematicians at that time just sort of assumed that they weren't smart enough yet to figure it out, so they put it off for a while. Then, along came this Galois fella.

Galois was a very excellent mathematician and quite the playboy of his day. He lived during a time of political turmoil in France and he was pretty outspoken about the whole thing, so between that and being a young, hotshot mathematician he was a pretty popular dude. Unfortunately, he was also kind of a jerk and one day he was challenged to a duel at the age of 20. (I don't know by whom and I don't know why, but its not really important.) Well, Galois apparently wasn't that good with a gun and he knew it. Because he was so convinced of his impending death, that night he wrote a letter to a man named Auguste Chevalier detailing all of his work in mathematics. Well, as it happens, staying up all night and writing down a lifetime's worth of mathematics doesn't really help you in a duel, so he died the next day. Chevalier, for whatever reason, never read the letter. He never threw it away either, though, and twenty-four years after receiving it he died and the letter somehow found its way into the hands of another very brilliant mathematician, Joseph Liouville. Liouville did read it and inside was an entirely new approach to the solutions of polynomials which we now call Galois theory. Once Liouville filled in a few of the gaps, Galois' work had produced a brilliant and concise proof showing that there is no general solution to a fifth degree polynomial, which is very cool.

What Galois did was attach a group to a polynomial in a very natural way (that is, natural if you understand the basics of both group theory and algabraic topology) and produced an extremely cool theorem that not only explains why there is no general solution to a fifth degree polynomial, but also explains why there is such a solution for second, third, and fourth degree polynomials.

This is what I love so much about mathematics - not the very stupid people that it seems to attract, but the ingenious breakthroughs that tie seemingly unrelated things together into something beautiful.

Recap - Subgroups

Subgroups are an important part of group theory. I want to review a little bit of the last few posts and try to tie things together a little bit.

A subset is simply a group that's inside of another group. If G is a group and H is a subset of G, then H is a subgroup if H is a group under the operation of G. If G is a group and we're given a subset, H, of G, we learned that there are three test to determine if H is a subgroup of G - the one-step subgroup test, the two-step subgroup test, and the finite subgroup test. One thing that these subgroup tests don't give us is a way to find subgroups inside of a group if we're not given a subset to start with. We did see a couple of ways, though, to find subgroups within any arbitrary group, including the following:

• If G is a group, then Z(G) is a subgroup of G.
• If G is a group, then C(a) is a subgroup of G ∀a∈G.
• If G is a group, then <a> is a subgroup of G ∀a∈G.

We also looked at some particular examples of subgroups. We saw that 2Z is a subgroup of Z. In fact, it is the case that if nZ = {nk : k∈Z} then nZ is a subgroup of Z. It also happens to be the case that nZ = <n>. We also looked at some examples of cyclic subgroups, namely <2> and <3> in Z(6).

As I alluded to before, subgroups are a very important part of group theory - especially the theory of finite groups. In a perfect world, the ultimate goal of group theory would be to be able to describe every single group that there could possibly be. (Note that I said "could possibly be," and not "is." I'm not going to explain it now, but this is a very interesting distinction.) From what I hear, this is an extremely unrealistic goal and will not be completed in my grandchildren's lifetime. However, we do have a small glimpse into this endeavor. There is something called a normal subgroup (which we will learn about eventually - we've got a bit of ground to cover first) and a group that has no normal subgroups is called simple. In arguably the greatest mathematical discovery of the last 25 years, we do know every single imaginable finite simple group. The proof is a series of papers that spans literally 10,000 pages and maybe 10 people in the whole world have read and understood the entire thing, but still, its been done. And if an arbitrary group is a molecule, then simple groups are its atoms - that is to say that every group can be made (in math language we say "is isomorphic to") by combining some of its simple subgroups. So, every imaginable group can be constructed from these finite simple subgroups that we already understand, which is very, very cool.

Before I go, there's probably one last thing I should explain. You might be wondering, if we know all the finite simple groups, and every group can be constructed from finite simple groups, then why is it that we don't know all of the groups? Well, the answer is that even though we know all the atoms (you know, to a reasonable degree), we can't know all the molecules because we don't know all the imaginable ways to construct them. Similarly, in group theory, although we know all the building blocks, we don't know all possible ways of combining them.

References

Previous Related Posts: Center and Centralizers

Center and Centralizers

Something I mentioned early on that I haven't talked about much is commutivity. One of the first things that I defined was the notion of an abelian group - one for which every element of the group commutes with every other element of the group. However, non-abelian groups have certain parts that "act" abelian. The first example of this is called the center of a group.

Definition: Center of a Group
Let G be a group. The center, Z(G), of G is the set of all the elements of G that commute with every element of G. Symbolically, Z(G) = {a∈G : g∙a = a∙g ∀g∈G}.

In an abelian group, every element commutes with every other element. In a non-abelian group, though, only some of the elements commute with every other element. It is these elements that comprise the center of the group. In order to make the notion of a center a little more clear, I would like to introduce a new group, which requires some knowledge of linear algebra. If you're not familiar with matrices then you can skip it for now. This group, called the general linear group, denoted by GL(n,R) where n is a positive integer and R is the real numbers, is the group of all invertible n by n matrices (matrices with non-zero determinant). It is simple to check that GL(n,R) is a group under matrix multiplication with its identity element the n by n identity matrix (which will be denoted by I). Then if x is a real number, it can be checked that (xI)∙M = M∙(xI) for every matrix, M, in GL(n,R). However, in general, it is not true that N∙M = M∙N for every N and M in GL(n,R). It can be shown that Z(GL(n,R)) = {xI : x∈R}.

The center of a group has a lot of interesting properties and uses. One such property is given below.

Theorem: Center is a Subgroup
The center of a group G is a subgroup of G.

Proof:
It is quite clear that Z(G)⊆G and that e∈Z(G) so Z(G) is non-empty. To prove that Z(G) is a subgroup, I will use the two-step subgroup test. First choose a,b∈Z(G). In order to show that a∙b∈Z(G) I must show that a∙b commutes with any arbitrary element of G. As such, choose g∈G. Note that (a∙b)∙g = a∙(b∙g) = a∙(g∙b) = (a∙g)∙b = (g∙a)∙b = g∙(a∙b) so then (a∙b)∙g = g∙(a∙b) and a∙b commutes with g. (That was made possible because both a and b commute with g since they are both in the center of G.) Thus a∙b∈G. Second choose a∈Z(G). In order to show that a^-1∈Z(G) I must show that a^-1 commutes with any arbitrary element of G. As such, choose g∈G. Note that g∙a^-1 = e∙(g∙a^-1) = (a^-1∙a)∙(g∙a^-1) = a^-1∙(a∙g)∙a^-1 = a^-1∙(g∙a)∙a^-1 = (a^-1∙g)∙(a∙a^-1) = (a^-1∙g)∙e = a^-1∙g so then g∙a^-1 = a^-1∙g and a^-1 commutes with g. (This was made possible because a commutes with g since it is in the center of G.) Thus a^-1∈G. Finally, by the two-step subgroup test, Z(G) is a subgroup of G.

I know that's another long and boring proof, but I wrote it out because its a very good example of how to use a subgroup test. As always, feel free to skip the proof if its confusing, but understanding a center and the fact that it is a subgroup is reasonably important. Later, there will be a lot of interesting things that we do with centers and it is crucial that the center is a subgroup (and, in fact, the center turns out to be a normal subgroup, although we haven't gotten there yet).

There is a concept related to the center of a group called a centralizer of an element. The center finds the elements of a group that commute with every single element, whereas the centralizer finds the elements that commute with one single element.

Definition: Centralizer of a in G
Let G be a group and a∈G be fixed. The centralizer of a, C(a) in G is the set of all the elements of G that commute with a. Symbolically, C(a) = {g∈G : a∙g = g∙a}.

The centralizer of a group is not nearly as interesting or as useful as the center, but it is another application of commutivity. I now present the following fact without proof.

Theorem: The Centralizer of an Element is a Subgroup.
Let G be a group. For each a∈G, the centralizer of a in G, C(a), is a subgroup of G.

I'm not going to present the proof here because its very similar to the last proof. There are two other facts that become immediately apparent about the relationship between centers and centralizers. First, ∀a∈G, Z(G)⊆C(a). Second, if a∈G then Z(G) = C(a) if and only if a∈Z(G).

This might not seem all that astounding or interesting, and as of right now it shouldn't. The usefullness of the center of a group (and the centralizer of an element) will come later, but for right now it is sufficient just to understand the definitions.

References

Previous Related Post: Cyclic Subgroups

Text Reference: Gallain Chapter 3

Wikipedia: Center of a Group

Wikipedia: Centralizer of a Group

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

Subgroup Tests

Last time we discovered that if H⊆G, then H is a subgroup only means that H is itself a group under the same operation as G. As an example, I showed that 2Z is a subgroup of Z. But if you remember, the explination was rather frustrating and I implied that there is an easier way to show that something is a subgroup. This method is one of three subgroup tests that I will now show. I'm not going to prove that they work for the sake of time (and boredom), but if someone would like me too, I can.

Theorem: One-Step Subgroup Test
Let G be a group and H be a non-empty subset of G. If ∀a,b∈G, a∙b^-1∈G, then H is a subgroup of G.

Theorem: Two-Step Subgroup Test
Let G be a group and H be a non-empty subset of G. If a∙b∈G whenever a,b∈G and x^-1∈G whenever x∈G, then H is a subgroup of G.

I'll now demonstrate how to use both of these tests to show that 2Z is a subgroup of Z.

Example: One-Step Subgroup Test.
Consider 2Z⊆Z. It is clear that 2Z is non-empty. Choose a,b∈2Z. By the definition of 2Z, a = 2m and b = 2n for some integers m and n. Also, the inverse of b is -b = -2n. Then we have that a+(-b) = 2m-2n = 2(m-n) and since m-n is an integer, 2(m-n)=a+(-b)∈2Z. Since the choice of a and b was arbitrary, 2Z is a subgroup of Z by the one-step subgroup test.

Example: Two-Step Subgroup Test
Consider 2Z⊆Z. It is clear that 2Z is non-empty. First choose a,b∈2Z. By the definition of 2Z, a = 2m and b = 2n for some integers m and n. Thus a+b = 2m+2n = 2(m+n) and since m+n is an integer, 2(m+n)=a+b∈2Z. Second, choose x∈2Z. From the definition of 2Z we have that -x∈2Z. Since the choices of a, b, and x were arbitrary, 2Z is a subgroup of Z by the two-step subgroup test.

You'll notice that both of these proofs looks very similar and prove the same things. However, sometimes it is easier to use one over the other. You can also see where the names come from. In the two-step test one has to show both that the subgroup is closed under its operation and that inverses are contained in the subgroup. In the one-step test both of these steps are essentially done at the same time.

There is one more subgroup test that only works for finite subgroups.

Theorem: Finite Subgroup Test
Let H be a finite, non-empty subset of G. If H is closed under the operation of G, then H is a subgroup of G.

The finite subgroup test is nice when you know that a subgroup is finite, but this isn't always something that one can know. To prove that H⊆G is a subgroup using the finite subgroup test, you show that H is finite and then you show the first step of the two-step test.

References

Previous Related Post: Subgroups

Text Reference: Gallain Chapter 3

Subgroups

As promised, we're now going to learn about subgroups. Subgroups are extremely important in group theory even though the definition is quite simple.

Definition: Subgroup
If G is a group and H is a subset of G, then H is a subgroup of G if H is a group itself under the operation of G.

If H is a subgroup of G, all that means is that H is a group under the operation of G and H lives inside of G - that is every element of H is also an element of G. Like I said before, this is a very simple definition, but the subgroups are often one of the most interesting parts of a group. Lets look at an example of a subgroup.

Example: 2Z is a subgroup of Z
Define Z to be the set of integers. We saw earlier that Z is a group under standard addition. Now lets define 2Z to be the set of all even integers - that is 2Z = {...,-6,-4,-2,0,2,4,6,...}. First, notice that H is a subset of G (we denote this by H⊆G). I now wish to convince you that H is a subgroup of G by showing that H is also a group under standard addition. The properties of identity and associativity are inherited from Z - that is that ∀a,b,c∈2Z, (a+b)+c = a+(b+c) because a,b,c∈Z and similarly, ∀a∈2Z, a+0 = a because a∈Z. Inverses come pretty easily in this case, too, because ∀a∈2Z, -a∈2Z and a+(-a) = 0. Finally, closure comes from the fact that the sum of two odd numbers must be odd, so a,b∈2Z automatically implies that a+b∈2Z. Thus 2Z is a group under standard addition and 2Z is a subgroup of Z.

Notice here that 2Z is itself a group under standard addition. It is a subgroup relative to Z because 2Z⊆Z and because 2Z is a group under the same operation of Z.

As you read the example and my long-winded justification for why 2Z is a subgroup of Z, there are a couple of things that might pop out. First, associativity and identity were automatically inherited and didn't require any justification. Also, the explanation was really awkward and long and cumbersome. There is a much better way to prove when something is a subgroup and that is what I'm going to go over in my next post.

References

Previous Related Post: Early Properties of Groups

Text Reference: Gallain Chapter 3

Wikipedia: Subgroup

Early Properties of Groups

This is going to be the first time we step into some quasi-real mathematics. I'm going to give four properties of groups and their proofs. You might not think the proofs are relevant, but reading proofs and understanding them is generally a useful exercise. I say this because, in general, the beauty of mathematics comes from proofs and from the enormous amount of creative lengths to which the brain can stretch in order to solve a problem. The proofs I'm about to show you are not the result of creative genius. It is necessary, though, to read the simpler proofs in order to understand the harder ones. What can I say? Its a process.

Theorem: Uniqueness of Identity
In a group, G, there is one and only one identity element.

Proof:
It is clear from the definition of a group that there exists at least one identity element of G. I now wish to show that the identity is unique. Suppose that both e and e' are identity elements of G. We know that ∀g∈G, g∙e = g and that ∀g'∈G, g'∙e' = g'. So then the choice of g = e' and g' = e gives us that e'∙e = e' and e∙e' = e. It then follows that both e and e' are equal to e'∙e, so e' = e and the identity is unique.

This proof demonstrates a very prolific proof technique. Almost any time you need to prove that something is unique, you choose two of whatever it is and show that they're equal. The property that identities are unique is neither surprising nor groundbreaking, but showing its truth is necessary and the world of group theory would be much different if it weren't true.

Theorem: Cancellation
If G is a group then ∀a,b,c∈G, b∙a = c∙a implies b = c and a∙b = a∙c implies b = c.

Proof:
Suppose that b∙a = c∙a. Multiplication on the right by a^-1 gives (b∙a)∙a^-1 = (c∙a)∙a^-1 and associativity then gives b∙(a∙a^-1) = c∙(a∙a^-1). From the definition of inverses, b = c as desired. A nearly identical argument gives us that a∙b = a∙c implies b = c (starting with multiplication on the left by a^-1).

The property of cancellation is convenient, but is definitely not obvious. There exist many mathematical universes where it is not true that b∙a = c∙a implies b = c. The first such structure that I will show you (not any time soon, though) is a ring.

Theorem: Uniqueness of Inverses
If G is a group, then ∀a∈G there is a unique a^-1∈G such that a∙a^-1 = e = a^-1∙a where e is the identity of G.

Proof:
Choose an element a∈G. From the definition of a group there exists at least one inverse of a. To show that the inverse is unique, suppose that both b and c are inverses of G. Then a∙b = e and a∙c = e so a∙b = a∙c and canelation gives us that b = c so the inverse of a is unique, as desired.

This is another useful property that may be intuitive, but requires proof. It does give us another common proof technique, though. You'll notice that all I did was start with an assumption and use all the definitions to get to my conclusion. Sometimes you don't need any tricks.

Theorem: Socks-Shoes Property
If G is a group then ∀a,b∈G, (a∙b)^-1 = b^-1∙a^-1.

Proof:
Choose a,b∈G. From the definition of an inverse we note that (a∙b)∙(a∙b)^-1 = e = (a∙b)^-1∙(a∙b). However, we now note that (a∙b)∙(b^-1∙a^-1) = a∙(b∙b^-1)∙a^-1 = a∙a^-1 = e, and (b^-1∙a^-1)∙(a∙b) = b^-1∙(a^-1∙a)∙b = b^-1∙b = e. We then get that (a∙b)∙(b^-1∙a^-1) = e = (b^-1∙a^-1)∙(a∙b) and b^-1∙a^-1 is an inverse of a∙b but since inverses are unique, (a∙b)^-1 = b^-1∙a^-1.

We call this the Socks-Shoes Property because if you put your socks on and then you put your shoes on (like a∙b), then when you want to do the reverse process you have to take your shoes off before you take your socks off (like b^-1∙a^-1). This is a cool property, and is a little bit unexpected.

Next time we'll explore the wonderful world of subgroups.

References

Previous Related Post: Group Attributes

Text Reference: Gallain Chapter 2

Wikipedia: Group