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.

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.

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 Z_n/<n/k> ≈ Z_k.

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/β forms a factor group. The Cayley Table for this factor group is given below.

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 eβ = R₁₈₀β, R₉₀β = R₂₇₀β, F_hβ = F_vβ, and F_lβ = F_rβ (which are also some calculations you can do yourself). The most interesting thing that can be extracted from this Cayley Table is that D₄/β 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

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