Showing posts with label cayley table. Show all posts
Showing posts with label cayley table. Show all posts

Saturday, March 6, 2010

Examples of Factor Groups

Today we're going to beat to death an example of a factor group. One of the most common (albeit boring) examples of a factor group is a factor group of a cyclic group. I'm going to use the cyclic group of order 6, Z6. Note that N={0,3} is a subgroup of Z6. As we will see later, N is automatically normal in Z6 because Z6 is abelian. As we learned last time, Z6/N forms a group since N⊳Z6 and it is this group that we're going to inspect. When I defined a factor group, I did it all in terms of a multiplicative group. Z6 is an additive, group, though, so it doesn't really make any sense to use multiplicative notation. Instead of writing cosets as aN, I'm going to be writing them as a+N.
By definition we have that Z6/N = {k+N : k∈Z6}. Normally, it is sufficient just to think of these abstractly, but since Z6 is finite and we are working out an example, we can write all of these out. That is, Z6/N = {0+N,1+N,2+N,3+N,4+N,5+N}. However, not all of these cosets are distinct. For instance, 1+N = {1+0,1+3} = {1,4} and 4+N = {4+0,4+3} = {4,1}, so we see that 1+N = 4+N. However, using one of the first theorems we learned about cosets, we can know when two cosets are equal without actually calculating the cosets. That is, h+N = k+N if h-k∈N. We see that 5-2=3∈N, so then 5+N = 2+N. Similarly, 0+N = 3+N. So then we get that Z6/N = {0+N,1+N,2+N}. Lets look at one example calculation, that is (1+N)+(2+N). The group operation gives us that (1+N)+(2+N) = (1+2)+N = 3+N. This is a perfectly fine calculation, but it would be nicer if we could get our answer as one of the cosets in {0+N,1+N,2+N}. As we already noted, 0+N = 3+N so we can also write (1+N)+(2+N) = 0+N. Through similar calculations we can calculate the Cayley Table of Z6/N.
cyclic factor group cayley table
There are some pretty interesting things about the Cayley Table of the factor group relative to the Cayley Table of the original group. Below is a modified Cayley table for Z6. You'll notice that the elements along the left side and the top are in a strange order and that I've seperated certain blocks of elements with black lines.
altered cyclic group cayley table
What is interesting about the particular groupings in this table is that it shows how a factor group is related to the group itself. What I did was arrange the headings along the top and along the side such that elements in the same coset are adjacent to one another. Then, by blocking off the table as above, what we've basically done is created the Cayley Table for the factor group within the Cayley Table for the original group. If you choose a coset along the side and the top, the corresponding block in the table is the corresponding coset according to the operation in the factor group. Another thing to notice about this particular factor group is that it shows pretty clearly that Z6/N is isomorphic to Z3. Indeed, if you take the Cayley table that we gave for Z6/N and remove "+N" from every entry, you get precisely the cayley table of Z3. For rigor, if φ:Z6/N→Z3 is defined by &phi:(k+N) = k mod 3 ∀k+N∈Z6/N, then you can check that φ is the desired isomorphism. Interestingly enough, if n is an integer, and k is a positive divisor of n, then Zn/<n/k>Zk.
I'm going to give one more example of a factor group. Consider D4 and β = {e,R180}. It is easy to verify that β is a subgroup of D4, but in fact β is a normal subgroup of D4. Thus, D4 forms a factor group. The Cayley Table for this factor group is given below.
dihedral factor group cayley table
It might be useful to work your way through the calculations for that Cayley Table yourself, for practice. When you do that, keep in mind that = R180β, R90β = R270β, Fhβ = Fvβ, and Flβ = Frβ (which are also some calculations you can do yourself). The most interesting thing that can be extracted from this Cayley Table is that D4 is abelian, which is surprising since D4 is non-abelian. It is not the case, however, that all factor groups are abelian. (If you'd like an example of a non-abelian factor group, D6/<σ3>S3 which is non-abelian, but that's a pretty complicated example.)
I hope that by now we've gained some understanding of factor groups, but these normal subgroups might still be a little bit of a mystery. You'll notice that in both of the examples that I gave you, I didn't really do anything to convince you that the subgroups were normal and at this point, if I had wanted to prove it then all I could have done is calculated all of the right and left cosets and shown their equality. This is okay, but is only efficient for very small groups and is completely ineffective for arbitrary groups. Because of this, we've developed a convenient criterion to guarantee normality. I'm not sure that I've introduced the notation aHb yet, but if H⊆G and a,b∈G, then aHb = {ahb : h∈H} (which is very similar to the notation of cosets).
Theorem: Normal Subgroup Test
Let G be a group and N a subgroup of G. N⊳G if and only if ∀x∈G, xNx-1⊆N.
Proof:
Suppose N⊳G. Then ∀x∈G and ∀n∈N, ∃n'∈N such that xn = n'x since xN = Nx. Thus xnx-1=n'∈N and therefore xNx-1⊆N.
Suppose that ∀x∈G, xNx-1⊆N. Choose g∈G. Then letting x=g we get gNg-1⊆N or gN⊆Ng. Conversely, letting x=g-1 we get g-1N(g-1)-1⊆N or g-1Ng⊆N so Ng⊆gN and finally gN = Ng.
This normal subgroup test is the most useful theorem that we have to show that an arbitrary group is a normal subgroup. For example, it just so happens that the center of a group is always a normal subgroup and this is very easy to prove with the normal subgroup test. Also, as it turns out, every subgroup of an abelian group is normal. The reason is easy, because if H is a subgroup of an abelian group, G, then when g∈G and h∈H, xh = hx which gives coset equality.
Theorem: Normal Subgroups of Abelian Groups
Every subgroup of an abelian group is a normal subgroup.
Proof:
Let G be an abelian group and let H be a subgroup of G. Choose g∈G and h∈H. Since G is abelian we get that ghg-1 = hgg-1 = he = h. Thus ghg-1∈H ∀h∈H, which gives that gHg-1⊆H ∀g∈G and H⊳G by the normal subgroup test.
Next time we'll look at Lagrange's Theorem - a very important and useful result in group theory.
References
Previous Related Post: Factor Groups
Text Reference: Gallain Chapter 9
Wikipedia: Quotient Group
Wolfram Mathworld: Quotient Group
Planet Math: Quotient Group

Wednesday, February 24, 2010

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,Fh,Fr,Fv,Fl}. 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.
reflective symmetries of the square
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.
elements of D4
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,Fh∈D4 and observe below R90∙Fh.
example operation on D4
Looking at this diagram and comparing it with the diagrams above of the eight elements of D4, we see that performing R90 and then Fh is the same as performing just Fr so R90∙Fh = Fr. 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.
Cayley Table for D4
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 Dn 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 Dn 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 Dn as I have, but there are people who call it D2n. 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 Dn. 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 Dn = {σi∙τj : i,j≥0, σn = e = τ2, σ∙τ = τ∙σn-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,σ23,τ,τ∙σ,τ∙σ2,τ∙σ3}. These elements correspond respectively with the elements that we defined earlier, {e,R90,R180,R270,Fh,Fr,Fv,Fl}. 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∙Fv = Fl 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 Fr∙Fr = 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