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

Lagrange's Theorem

A big part of the hard work and the research in group theory is in finding specific types of subgroups inside of finite groups. If you have an arbitrary finite group of order n, then the number of distinct subsets is 2ⁿ. For reference, consider |G|=9. Then the number of subsets of G is 2⁹=512 and each of these is a possible subgroup of G. As you can see, this gets very large very quickly. Lagrange's Theorem, however, gives us a way to drastically cut down the number of possible subgroups of any group. You should read the proof of Lagrange's Theorem, because it makes use of some of the things we've already learned.

Theorem: Lagrange's Theorem
If G is a finite group and H is a subgroup of G, then |H| divides |G|.

Proof:

Let G be a finite group and let H be a subgroup of G. Now, consider the set of all left cosets of H in G, {gH : g∈G}. Note, this set does not necessarily form a factor group because H is not normal, but we still have the cosets. Now choose one representative from each coset such that {g₁H, g₂H,...,g_kH} is the set of distinct left cosets. Then we have that each element of G is a member of one of these cosets, or that

G = g₁H∪g₂H∪⋯∪g_kH.

We know that this union is disjoint - that is if i≠j then g_iH∩g_jH=∅. From this we get that

|G| = |g₁H|+|g₂H|+⋯+|g_kH|.

Also, we know that every coset contains the same number of elements, or |g_iH| = |H|. Finally, this gives us that |G| = k|H| so then k = |G|/|H| and |H| divides |G| as desired.

Lagrange's Theorem gives us that the order of every single subgroup must be a divisor of the order of the original group. What this does is immediately cuts down the number of possible subgroups of any finite group. Earlier I mentioned that if |G|=9 then there are 512 unique subsets of G. Lagrange's Theorem, though, tells us that of these 512 choices, the only subsets that could possibly be subgroups are the ones of order 3, which cuts the choices down to only 84. This might not seem like that big of a deal at this point, but Lagrange's Theorem is definitely the most important and most useful theorem in all of group theory.

I must make a point to emphasise what it is that Lagrange's Theorem does not say. It is easy to think that if k is some divisor of a group, G, then G has some subgroup of order k. This is not at all the case. Lagrange's Theorem only tells us a list of possible group orders and a criteria for when a subset cannot be a subgroup, it is in no way a subgroup test. It would be convenient if the converse of Lagrange's Theorem were true, but it is not. In fact, there are lots of groups that fail the converse. For example, A₄ has no subgroup of order 6.

I've written a lot of group theory proofs and I think about 90% of them use Lagrange's Theorem in one form or another - there are definitely a lot of places where we'd get stuck without it. For now, though, I'm going to give some of the immediate results of Lagrange's Theorem without proving them.

Theorem: |G/N|
Let G be a group and N⊳G. Then |G/N| = |G|/|N|.

Theorem: |a| divides |G|
In a finite group, the order of each element divides the order of the group.

Theorem: Prime Order Groups
Any group of a prime order is cyclic.

Theorem: a^|G| = e
If G is a finite group and a∈G then a^|G| = e.

Theorem: Fermat's Little Theorem
Let a be an integer and p be any prime number. Then a^p mod p = a mod p.

Theorem: Groups of order 2p
Let G be any finite group such that |G| = 2p where p is a prime number greater than 2. Then either G≈Z_2p or G≈D_p.

A few of these things seem to be very far away from Lagrange's Theorem and maybe they are, but I promise that they're all direct consequences. I didn't give the proofs because some of them are quite complicated, but I can write them if someone needs. There are a lot of proofs that start with "Suppose H is a subgroup with the property that..." and ends with "...but then the order of H does not divide the order of G hence contradicting Lagrange's Theorem." It is a very useful, very powerful tool.

References

Previous Related Post: Examples of Factor Groups

Text Reference: Gallain Chapter 7

Wikipedia: Lagrange's Theorem

Wolfram Mathworld: Lagrange's Theorem

Cosets

Up until my last post, I feel like the things I've talked about have flowed pretty linearly. However, math doesn't usually work that way. That often seems counter-intuitive because since first grade everything that we learn builds off of each previous topic, but usually math is all over the place and different topics influence each other in many different ways. This is usually why its so hard to read a book because books, by nature, are linear but math is not. Because of this, I'm going to step back away from isomorphisms and the symmetric groups and talk about cosets, which will eventually get us to normal subgroups and factor groups. There are some properties about cosets that are important to understand that I also need to introduce. These properties may not be obvious, so I'm going to prove them, but you are welcome to just trust me.

Definition: Left and Right Cosets.
Let H be a subgroup of a group, G and let a∈G. We then define a∙H = {a∙h : h∈H} and call it the left coset of H in G containing a and we define H∙a = {h∙a : h∈H} and call it the right coset of H in G containing a. In either case, a is called the coset representative of a∙H or H∙a.

It is noted in the definition that H is a subgroup of G, but even if H is only a subset of G, the notations of a∙H and H∙a are still valid, but they are not called cosets. It will be rare - if ever - that we use a∙H and H∙a when H is not a subgroup in this blog. This definition of a coset is very simple - all you do to calculate a coset, a∙H, is to take every element of H and multiply it on the left by a. What we are mostly concerned with is the interaction between between different cosets. First, though, I'm going to give an example where I calculate all of the left cosets of a subgroup of D₄.

Example: Cosets of the Dihedral Group
Let G = D₄ = {e,R₉₀,R₁₈₀,R₂₇₀,F_h,F_r,F_v,F_l} and let H = {e,F_h}. It is very simple to verify that H is a subgroup. Below I have calculated each of the left cosets of H in G.

e∙H	=	{e,Fh}
R90∙H	=	{R90,Fr}
R180∙H	=	{R180,Fv}
R270∙H	=	{R270,Fl}
Fh∙H	=	{Fh,e}
Fr∙H	=	{Fr,R90}
Fv∙H	=	{Fv,R180}
Fl∙H	=	{Fl,R270}

This is not meant to just demonstrate a routine calculation. You should look at these coset calculations and try to see patterns. There are a lot of things to notice, and those patterns are the properties that I'm going to prove. The first thing that you should notice, that isn't really a theorem, is that most of the cosets are not subgroups - in fact most of them don't even have the identy element. The only cosets that are subgroups of D₄ are the ones that are equal to H.

Theorem:
Let G be a group and H a subgroup of G. Then a∙H = H if and only if a∈H.

Proof:
First, suppose that a∙H = H. Then a=a∙e∈a∙H=H. Next, assume that a∈H. Since H is closed, we get that a∙H⊆H. To show that H⊆a∙H, let h∈H. Note that since a∈H, a^-1∈H, and since h∈H, a^-1∙h∈H. Now we get that h=e∙h=(a∙a^-1)∙h=a∙(a^-1∙h)∈a∙H so H⊆a∙H and a∙H = H.

This property illuminates something very interesting. We noted before that some cosets of H in G from our example are actually equal to H. What this gives us is that this happens precisely when the coset representative is in H which is a very nice condition.

Theorem:
Let G be a group and H a subgroup of G. Then ∀a,b∈G, a∙H = b∙H if and only if a^-1∙b∈H.

Proof:
We observe that a∙H = b∙H if and only if H = (a^-1∙b)∙H and from the previous theorem, H = (a^-1∙b)∙H if and only if a^-1∙b∈H.

This is a bit of an extension of our last property. An immediate consequence is that two cosets are equal when the representative of one lies in the other (that is a∙H = b∙H when a∈b∙H and b∈a∙H). This is pretty cool because it means that if we have a coset of H, then we can choose any element in that coset to be its representative. Which means that if K is some coset of H in G, then ∀a∈K, K = a∙H. This is very, very useful.

Theorem:
Let G be a group and H a subgroup of G. Then ∀a,b∈G, either a∙H = b∙H or a∙H∩b∙H = ∅.

Proof:
Suppose ∃x∈a∙H∩b∙H. Then a^-1∙x∈H so a∙H = x∙H and similarly b^-1∙x∈H so b∙H = x∙H. Finally we have that a∙H = b∙H and so either a∙H = b∙H or a∙H∩b∙H = ∅.

The statement of this property might be a little bit confusing, but what it essentially means is that given two cosets, either they are the same or they have no elements in common. The last property told us how to know if two cosets are the same and this property tells us that if they are not the same, then they are completely distinct. This means that the set of cosets of a particular subgroup partitions G - or basically that if H is a subgroup of G, then every single element of G is in exactly one coset of H.

Theorem:
Let G be a group and H a subgroup of G. Then ∀a,b∈G, |a∙H| = |b∙H|.

Proof:
Define the function, φ:a∙H→b∙H by φ(a∙h)=b∙h ∀a∙h∈a∙H. This is obviously a surjection, and it is an injection because cancellation gives that a∙h=b∙h implies that a=b. Since there exists a bijection between a∙H and b∙H, it then follows that |a∙H| = |b∙H|.

Finally, we have that if H is a subgroup of G, then every coset of H is the exact same size. Now we already saw that the cosets of H partition G, but we now see that these cosets partition G into partitions of the exact same size.

Now we can try to put everything we've learned about cosets together. Suppose G is a group with subgroup H and suppose that K is a coset of H in G. We know that ∀a∈K, K = a∙H so that any element of K can be chosen as its coset representative. We also know that two different cosets are in fact completely disjoint - that is that they have no elements in common - but are the exact same size. You can (and should) go back up to my example of all the cosets of {e,F_h} in D₄ and verify that these four properties hold. One final, very important thing to mention is that even though all of these theorems that I gave were concerned with left cosets, they all have analogous results for right cosets.

We learned a lot about cosets today, but I haven't yet explained how they're useful and it is definitely not obvious. As we'll see, if G is a group and N is a certain type of subgroup (called a normal subgroup), then the sets of all cosets of N in G form a group in itself, called a factor group, and these are extremely important in the world of group theory. Factor groups have a lot of very interesting, very important properties, and there are a lot of proofs in group theory that involve showing a property about a group by inspecting its factor groups, and a lot of these things could not be proved any other way.

References

Previous Related Post: Early Properties of Groups

Text Reference: Gallain Chapter 7

Wikipedia: Cosets

Wolfram Mathworld: Cosets

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