SUBGROUPS OF PRODUCT GROUPS. A Thesis. Presented to the. Faculty of. San Diego State University. In Partial Fulfillment

Transcription

1 SUBGROUPS OF PRODUCT GROUPS A Thesis Presented to the Faculty of San Diego State University In Partial Fulfillment of the Requirements for the Degree Master of Arts in Mathematics by Robin Michael Devitt-Ryder Fall 2014

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3 iii Copyright c 2014 by Robin Michael Devitt-Ryder

4 iv DEDICATION Dedicated to my love Marianna. She supports me and stands by my side. For this I am forever indebted to her.

5 v ABSTRACT OF THE THESIS Subgroups of Product Groups by Robin Michael Devitt-Ryder Master of Arts in Mathematics San Diego State University, 2014 Our goal is to count the number of subgroups of the direct product of finite cyclic groups, Z m Z n. When m and n are relatively prime, Z m Z n is isomorphic to Z mn. Counting the number of subgroups of Z mn is a simple exercise in combinatorics. However, if m and n are not relatively prime, then we use Goursat s (lesser-known) Theorem to find the number of subgroups of Z m Z n. We may also extend the usage of this theorem to cases beyond the direct product of finite cyclic groups.

6 vi TABLE OF CONTENTS PAGE ABSTRACT... LIST OF FIGURES... ACKNOWLEDGMENTS... v vii viii CHAPTER 1 INTRODUCTION The Problem This Paper BACKGROUND Cyclic Groups Product Groups MAIN RESULT Subgroups of Z m Z n Counting Subgroups Goursat s Theorem Generalized cases: beyond Z m Z n CONCLUSION Conclusion Further Studies BIBLIOGRAPHY... 40

7 vii LIST OF FIGURES PAGE Figure 3.1. Subgroup Lattice of Z 7 Z Figure 3.2. Subgroup Lattice of Z p Z p Figure 3.3. Subgroup Lattice of Z 5 Z Figure 3.4. Subgroup Lattice of Z 9 Z Figure 3.5. Subgroup Lattice of Z 4 Z Figure 3.6. Subgroup Lattice of Z 4 Z Figure 3.7. Subgroup Lattice of A 4 and D

8 viii ACKNOWLEDGMENTS I would like to thank Dr. Interlando for his assitance in helping me complete my thesis in a timely manner and for accommodating my many specific needs. I would also like to thank Dr. Hui for his (near constant) academic assistance and for being on my thesis committe. Finally, I would like to thank Dr. Root for being on my thesis committee even though we had never even met before. That was nice of you. All three of you.

9 1 CHAPTER 1 INTRODUCTION In this chapter, we introduce our motivation for the task at hand: counting subgroups. Some expository information is provided as well as an introduction for the contents of this paper. 1.1 THE PROBLEM How many subgroups are there in the direct product Z m Z n? There doesn t seem to be much of a general answer among basic algebra texts. Without resorting to an advanced graduate text, most students will not have much enlightenment on this problem. The only insight that most undergraduate students are given is the fact that Z m Z n is isomorphic to Z mn but only when gcd(m, n) = 1. This is unfortunately not the case when gcd(m, n) 1. Additionally, subgroups of Z m Z n are not always of the form H K where H is a subgroup of Z m and K is a subgroup of Z n. Example To illustrate this, we look at the subgroups of Z 2 Z 4. This group has 8 subgroups. However, Z 2 only has 2 subgroups while Z 4 has 3 subgroups. Indeed, if we list the subgroups of Z 2 Z 4 we see that not they are not all trivial: (1) {(0, 0)}, (2) {(0, 0), (1, 0)}, (3) {(0, 0), (0, 2)}, (4) {(0, 0), (1, 2)}, (5) {(0, 0), (1, 0), (1, 2), (0, 2)}, (6) {(0, 0), (0, 1), (0, 2), (0, 3)}, (7) {(0, 0), (1, 1), (0, 2), (1, 3)}, (8) {(0, 0), (0, 1), (0, 2), (0, 3), (1, 0), (1, 1), (1, 2), (1, 3)}.

10 Note that subgroups (4) and (7) are not of the form H K where H is a subgroup of Z 2 and K is a subgroup of Z 4. 2 To tackle this problem, we use a little-known theorem by Édouardo Goursat. His theorem resolves the problem quite nicely in a slightly roundabout way. Instead of examining the direct product Z m Z n itself, we work with the quotients and subquotients of Z m and Z n individually. This is a straight forward task and quite simple to do with the integer groups Z k. While this does not immediately produce a single, closed formula for calculating the number of subgroups of Z m Z n, it does give us a definitive process for completing this task (we will produce some formulas that speed up this counting process). Although finding the subgroups of a given group is one of the most basic problems in abstract algebra, classic textbooks such as Hungerford [9] and Dummit and Foote [4] do not even mention Goursats Theorem. Lang [10, pg. 75] briefly describes it as an exercise. This is despite the fact that Goursat s Theorem is quite approachable for beginning algebraists. Typically, Goursat s Theorem (often referred to as a lemma) is used in a more general setting related to the isomorphism theorems and applied in some Galois Theory. Goursat s Theorem is rarely used in such concrete terms (such as counting subgroups of Z m Z n ). This paper will attempt to fill this role. Beginning algebraists can learn how to work with cyclic groups and quotient groups by utilizing Goursat s Theorem while more advanced students can better understand the structures of product groups. 1.2 THIS PAPER The author s contribution to this issue is to state an explicit method for calculating the number of subgroups of Z m Z n. Even though subgroups are studied and classified extensively in algebra classes, teachers rarely discuss classifying subgroups of Z m Z n. This is despite that fact that many students broach the issue (the author included). Much of this paper is based off of information that can be found in [12]. However, this paper goes further in depth. More examples are provided with the work so that readers may

11 3 follow along. Additionally, all proper expository information is provided. Because of the calculatory nature of the solution, this paper should be accessible to any student who has completed an introductory course on abstract algebra. We will first begin with the fundamentals of cyclic groups and calculating their subgroups. From there, we will move on to the main theorem of the paper and explain how to calculate our results. Afterwards, we will prove Goursat s Theorem and attempt to generalize our results. For a more in-depth study into the classification of subgroups, the author suggests [1]. That paper, however, is much more dense than [12] and [6], both of which contain the foundations for this paper.

12 4 CHAPTER 2 BACKGROUND In an introductory course on abstract algebra, one learns to calculate all the subgroups of Z m Z n when m and n are coprime. This group is isomorphic to Z mn, so this task is fairly straight-forward. Simply list all the divisors of mn. There is one subgroup for each divisor. So the total number of subgroups of Z mn is equal to the number of divisors of mn. This is the same number of subgroups for Z m Z n. Naturally, the next question a student tends to ask at this point is how many subgroups there are of Z m Z n when m and n are not coprime. From here, the naive approach of listing the subgroups manually easily leads to errors (and more importantly, is far more time consuming). Luckily, there is another way. This method involves using Goursat s Theorem. Before discussing this theorem, we need to cover the fundamentals of cyclic groups. This will allow us to prove several results about cyclic groups (including the results mentioned in the previous paragraph). 2.1 CYCLIC GROUPS Our focus in this paper will be the study of cyclic groups. We begin by defining cyclic groups. From there, we will move on to proving several useful properties about cyclic groups. All of the following definitions, theorems and lemmas are from [6]. Definition Let a be an element of a group G. Then a = {a n n Z} is called the subgroup of G generated by a. Note that a n is simply the notation used to indicate the operation of the group (it may not necessarily be multiplication). Equivalently, some literature uses the notation na instead (it may not necessarily be addition). The operation should be clear from the context.

13 Definition A group, G, is cyclic if there exists an element a in G such that G = a. We say that the group G is generated by a. 5 Example Z 6 = 1 because 1 1 = 1, 1 2 = = 2, 1 3 = = 3, 1 4 = = 4, 1 5 = = 5 and 1 6 = = 0. Additionally, Z 6 = 5 because 5 1 = 5, 5 2 = = 4, 5 3 = = 3, 5 4 = = 2, 5 5 = = 1 and 5 6 = = 0. Note, however, that Z 6 does not equal 2 because 2 = {0, 2, 4}. The same reasoning demonstrates that Z 6 does not equal 3 nor 4. We say that Z 6 is generated by 1 and by 5. The previous example showed that a cyclic group can have more than one generator. Example Z 5 = 1 = 2 = 3 = 4. Notice that in this last example every positive integer less than 5 is a generator of Z 5. This brings us to our first theorem regarding cyclic groups. Theorem If p is prime, then any of 1, 2,..., p 1 is a generator of Z p. Example Z has two generators: -1 and 1. For given any positive integer n Z, we have that n = } 1 + {{ + 1 } = 1 n. n times To generate any negative integer, we simply use negative exponents. So if k is a positive integer, k = 1 k. The same reasoning applies when we use -1 as the generator. The following is an example of a non-cyclic group.

14 6 Example The group GL(2, Z) = a c b d a, b, c, d Z and ad bc = 1 is a group under matrix multiplication. We will show that this group is not cyclic. Let B and C be matrices in this group. Since this group is not abelian, we have that BC CB. Assume, by way of contradiction, that this group is cyclic. Then there exists a matrix A such that B = A m and C = A n for some integers m and n. But this means that BC = A m A n = A m+n = A n+m = A n A m = CB, which is a contradiction. Thus, this group is not cyclic. Now we proceed with some general theorems regarding cyclic groups. This will give us the ground work necessary to understand the simplicity of calculating subgroups of the integer groups Z n. Once this framework has been established, we will move on to the more challenging task of calculating the subgroups of the direct product Z m Z n. Theorem Let G be a finite group. Let a be in G. If a has order n in G, then a = {e, a, a 2,..., a n 1 } and a j = a k if and only if n divides j k. We will only be dealing with finite groups in this paper. Therefore, it is useful to have this theorem for the fact that a single group can have multiple generators (as in the previous examples). Corollary For any a in G, a = a. The next corollary is a special case of Theorem Corollary Let a be an element of order n in G. If a k = e, then n divides k. This is useful to know because it is essentially saying that multiplication in the group a is calculated by addition modulo n. So if we have i + j (mod n) k, then a i a j = a k. So given any group G and any element a in G, multiplication in a is the same as addition in Z n whenever a = n. Additionally, if G is not a finite group and a has infinite order, then multiplication in a works the same as addition in Z because a i a j = a i+j. This means that

15 7 we can safely say that all cyclic groups are isomorphic to some integer group Z n or Z. Also, if a group is not cyclic then it is not isomorphic to some Z n nor Z. This also indicates that there is only one cyclic group of each order n, namely, Z n. Finally, since Z n and Z are commutative, every cyclic group is commutative and every non-commutative group is not cyclic. These are powerful and, more importantly, useful properties to have under our belt. The following theorem not only tells us exactly how may subgroups a finite cyclic has but also how to find these subgroups. Theorem (The Fundamental Theorem of Cyclic Groups). This has three parts. (i) Every subgroup of a cyclic group is cyclic. (ii) If a = n, then the order of any subgroup of a is a divisor of n. (iii) For each positive divisor k of n, the group a has exactly one subgroup of order k - namely, a n/k. Corollary For each positive divisor k of n, the set n/k is the unique subgroup of Z n of order k. Moreover, these are the only subgroups of Z n. This is extremely convenient. Here is an example to illustrate how we can use this. Example The subgroups of Z 24. By Theorem and Corollary , we have that the subgroups of Z 24 are 1 = {0, 1, 2,..., 23} order 24, 2 = {0, 2, 4,..., 22} order 12, 3 = {0, 3, 6,..., 21} order 8, 4 = {0, 4, 8, 12, 16, 20} order 6, 6 = {0, 6, 12, 18} order 4, 8 = {0, 8, 16} order 3, 12 = {0, 12} order 2, 24 = {0} order 1. Furthermore, these are the only subgroups of Z 24.

16 8 The next theorem will invoke Euler s totient function, φ(d). This is used to calculate how many integers between 1 and d are relatively prime to d. To use Euler s totient function, simply rewrite d into its prime decomposition, d = p k 1 1 p kn n. Then we have that φ(d) = φ(p k 1 1 p kn n ) = φ(p k 1 1 ) φ(p kn n ) = (p k 1 1 p k ) (p kn n p kn 1 n ). Theorem If d is a positive divisor of n, then the number of elements of order d in a cyclic group of order n is φ(d). As a quick illustration of this, again consider the group Z 24. Here, 6 is a positive divisor of 24. So the number of elements in Z 24 that have order 6 is equal to φ(6) = φ(2 3) = φ(2)φ(3) = ( )( ) = (2 1)(3 1) = 2. So, by Theorem , there are only 2 elements of order 6 in Z 24. Those elements are 4 and 20. Corollary In a finite group, the number of elements of order d is divisible by φ(d). Theorem and Corollary are particularly useful when we are trying to calculate the number of subgroups of Z n for very large values of n. Since we have the properties of φ(d) that φ(p k ) = p k p k 1 and φ(mn) = φ(m)φ(n) (when m and n are relatively prime), we have a quick process for calculating φ(d). For further information regarding basic group theory and cyclic groups, see [6]. For a more advanced graduate level text, see [4].

17 9 2.2 PRODUCT GROUPS The previous section gave us the ground work necessary to find the subgroups of finite, cyclic groups. The aim of this section is to find the subgroups of the direct product of finite, cyclic groups. It is here that the main body of the work will follow. Definition Let G 1 and G 2 be groups. Then the external direct product of G 1 and G 2 is the group G 1 G 2 = {(g 1, g 2 ) g 1 G 1 and g 2 G 2 }. This may also be notated G 1 G 2. Additionally, this will be referred to simply as the direct product. (To avoid confusion, there is a similar object known as the internal direct product. We do not need to concern ourselves with that in this paper.) The group operation of G 1 G 2 is done component-wise. This gives us that (a, b) (c, d) = (a 1 c, b 2 d), where 1 is the operation in G 1 and 2 is the operation in G 2. Probably the most common example of this is the Cartesian plane, R R. We may look at a subgroup of this: Z Z. This is the Cartesian plane but only considering the integers and the operation of addition. Thus, we have that for ( 1, 2) and (2, 3) in Z Z, ( 1, 2) + (2, 3) = ( 1 + 2, 2 + 3) = (1, 5). Going back to the integer groups that we reviewed in the last section, we have the direct product Z m Z n = {(a, b) a is in Z m and b is in Z n }. So, for example, we have that Z 3 Z 4 is the set of coordinates {(0, 0), (1, 0), (2, 0), (0, 1), (1, 1), (2, 1), (0, 2), (1, 2), (2, 2), (0, 3), (1, 3), (2, 3)}.

18 10 Since the operation (in this case, addition) is done component-wise, we have that (0, 2) + (1, 3) = (0 + 1, 2 + 3) = (1, 1). Now, we already know what the subgroups of Z n look like. The subgroups of Z m Z n will follow suit in a similar fashion. To illustrate this, consider the product Z 4 Z 8. We have that 2 generates a subgroup of Z 4 and that 2 generates a different subgroup of Z 8. For clarity, we will denote the subgroups as 2 4 and 2 8. Thus, we have that is the set of coordinates {(0, 0), (0, 2), (0, 4), (0, 6), (2, 0), (2, 2), (2, 4), (2, 6)}, which is itself a subgroup of Z 4 Z 8. So, for instance, we have that (2, 4) + (2, 6) = (2 + 2, 4 + 6) = (0, 2). Our goal is to calculate the number of subgroups of Z m Z n. Under the right conditions, this is quite easy. This gives us our first theorem of the section. Theorem Let m and n be relatively prime integers. Then Z m Z n is isomorphic to Z mn. This is convenient because Corollary gave us a quick and efficient method for finding all the subgroups of Z mn. We simply need to find all the divisors of mn. And since Z mn is isomorphic to Z m Z n, we quickly know how many subgroups there are. Example The subgroups of Z 2 Z 3. We know from Theorem that this group is isomorphic to Z 6. Corollary gives us that the subgroups of Z 6 will be 0, 1, 2, and 3. Furthermore, these 4 subgroups are the only subgroups of Z 6. Since the groups are isomorphic, that means their subgroups will also be isomorphic. The 4 subgroups of Z 2 Z 3

19 are 1 = 1 1 order 6, 2 = 0 1 order 3, 3 = 1 0 order 2, 0 = 0 0 order Of course, not all subgroups of Z m Z n need to necessarily be the direct product of subgroups of Z m and Z n. The following example can be found in [12]. Example The subgroups of Z 9 Z 9. In the following section we will use specific methods to calculate that there are 23 subgroups of Z 9 Z 9 (for now we will just accept this as true). Two such subgroups are the groups {(0, 0), (3, 3), (6, 6)} and {(0, 0), (6, 3), (3, 6)}. Notice the if we combine them into a single subgroup, we would have 3 3. However, individually they each form a different subgroup of Z 9 Z 9 that is not the direct product of subgroups of Z 9. The difference with Example is that m and n were not relatively prime. Clearly, this makes a world of difference. Our primary objective with this paper is to calculate the number of subgroups of Z m Z n (for the remainder of the paper we will not list out each subgroup explicitly). We know from Theorem and Corollary that finding the number of subgroups of Z n is quite straight-forward. We simply calculate the number of divisors of n. Does this method work for arbitrary Z m Z n? The answer is no. To illustrate this, we know that the number of subgroups for Z 9 is equal to 3. One could assume, then, that various combinations of these 3 subgroups would form all the subgroups of Z 9 Z 9. The maximum number of possible combinations then would be 3 3 = 9 subgroups. However, as was already stated above, we know that the number of subgroups of Z 9 Z 9 is actually 23. So we must use some other method for finding the correct number of subgroups of Z m Z n. For this task we will use Goursat s Theorem.

20 12 CHAPTER 3 MAIN RESULT Our primary focus for this paper will be in this chapter. Here we introduce Goursat s Theorem and explain how it applies to counting subgroups. We then construct an algorithm that enables us to use Goursat s Theorem in a practical manner. We then prove Goursat s Theorem so that we may use some of the results for generalization into further topics of study. 3.1 SUBGROUPS OF Z m Z n Édouard Goursat was a French mathematician who lived between the mid 1800s through the early 1900s. He authored about a dozen books, most of which were about analysis in some form or another. He is best known for the Cauchy-Goursat Theorem, which is usually just called the Cauchy Integral Theorem. That theorem states that a complex function integrated over a closed, counter-clockwise oriented curve is equal to 0. Here, we will use a lesser known theorem of his that relates to group theory. This theorem was first proved in 1889 in [7]. It is as follows. Theorem (Goursat s Theorem). Let G 1 and G 2 be groups. Then there exists a bijection between the set of all subgroups of G 1 G 2 and the set of all triples (A/B, C/D, ϕ) where A/B is a subquotient of G 1, C/D is a subquotient of G 2 and ϕ : A/B C/D is an isomorphism. Other variations of this theorem may be found in [3], [8] and [1]. However, the author believes that the version presented here is the most accessible to the widest audience. Furthermore, it gives a direct method for its usage. Since such a bijection exists between the set of subgroups of Z m Z n and the set of triples (A/B, C/D, ϕ), we can easily construct an algorithm to calculate all the subgroups of Z m Z n when m and n are not coprime.

21 13 First, we need to find all the divisors of m. This allows us to find all the subquotients, A/B, of Z m. Then find the divisors of n. This allows us to find all the subquotients, C/D, of Z n. Then we see which maps are isomorphisms from selected ϕ : A/B C/D. This is easy in the case of integer groups: we simply need the domains and the ranges of each mapping to have the same order. Essentially, these are just automorphisms. From here, we count how many automorphisms exist for each particular mapping. Lastly, we add up the total number of isomorphisms. This gives us the total number of subgroups of Z m Z n. The following examples will illustrate this process. Our first example will be the direct product of two prime numbered groups with the same cardinality. Example The subgroups of Z 7 Z 7. In order to implement Goursat s Theorem, we need to find all the subquotients of Z 7. Before we can do this, we need to find all the subgroups of Z 7. Since 7 is prime, this is easy. There are only two subgroups of Z 7 : the group itself and the trivial group {0}. Since Z 7 is abelian, every subgroup of Z 7 is normal. So constructing subquotients is simple. This quickly gives us only three possible subquotients of Z 7. Those are Z 7 /Z 7, Z 7 /{0} and {0}/{0}. We need to find all the triples of the form (A/B, C/D, ϕ) as stated above. That means we will have isomorphisms of the form ϕ : A/B C/D where A/B is a subquotient of Z 7 (the first group of the direct product) and C/D is a subquotient of Z 7 (the latter group of the direct product). And since they are isomorphisms, A/B and C/D must have the same number of elements. Since Z 7 /Z 7 and {0}/{0} both only have one element, we can have up to five

22 14 isomorphisms. Those are (1) ϕ : Z 7 / Z7 Z 7 / Z7 (2) ϕ : Z 7 / Z7 {0} / {0} (3) ϕ : {0} / {0} {0} / {0} (4) ϕ : {0} / {0} Z 7 / Z7 (5) ϕ : Z 7 / {0} Z 7 / {0}, where (5) is an isomorphism from a group with 7 elements to a group with 7 elements. We can simplify these. Since Z 7 /Z 7 and {0}/{0} are both isomorphic to Z 1, we may replace isomorphisms (1), (2), (3) and (4) each with the automorphism ϕ 1 : Z 1 Z 1. And since Z 7 /{0} is isomorphic to Z 7, we may replace isomorphism (5) with the automorphism ϕ 2 : Z 7 Z 7. Note here that ϕ 2 is actually 6 distinct automorphisms. Finally, we tally up the total number of morphisms. We have automorphisms (1), (2), (3) and (4) plus the 6 distinct automorphisms that make up (5). That is, we have (6 1) = 10 isomorphisms from subquotients of Z 7 to subquotients of Z 7. By Goursat s Theorem, there is a bijection between the set of all subgroups of Z 7 Z 7 and the set of these 10 isomorphisms. Therefore, there must be 10 subgroups of Z 7 Z 7. The following is a subgroup lattice of Z 7 Z 7. Here, 6 of the Z 7 subgroups in the middle row are isomorphic to the group generated by (1, 1 k ) where k = 1,..., 6. The two remaining groups will be the groups generated by (1, 0) and (0, 1). (Here, the choice of 1 was arbitrary. We could have chosen any generator of Z 7. That is, we could have chosen any integer 1 through 6.) See Figure 3.1 for the subgroup lattice of Z 7 Z 7.

23 15 Z 7 Z 7 Z 7 Z 7 Z 1 Z 1 Figure 3.1. Subgroup Lattice of Z 7 Z 7 The previous example illustrates a general method for finding the number of subgroups of Z p Z p, where p is a prime number. The subquotients for Z p will always be Z p /Z p, Z p /{0} and {0}/{0}. This will always give us the same 5 isomorphisms from A/B to C/D. Those are (1) ϕ : Z p / Zp Zp / Zp (2) ϕ : Z p / Zp {0} / {0} (3) ϕ : {0} / {0} {0} / {0} (4) ϕ : {0} / {0} Zp / Zp (5) ϕ : Z p / {0} Zp / {0}. Notice that since Z p /Z p and {0}/{0} are both isomorphic to Z 1, we may replace isomorphisms (1) through (4) with ϕ 1 : Z 1 Z 1. And since Z p /{0} is isomorphic to Z p, we may replace isomorphism (5) with ϕ p : Z p Z p. Note that ϕ p is actually p 1 distinct

24 16 automorphisms. So there are 4 isomorphisms for ϕ 1 and there are p 1 isomorphisms for ϕ p. Adding this up means that are 4 + (p 1) = p + 3 isomorphisms from subquotients A/B to subquotients C/D. By Goursat s Theorem, that means that there are p + 3 subgroups of the direct product of Z p Z p. As it turns out, it is fairly straight forward to calculate the actual subgroups themselves of Z p Z p directly. There will be p + 3 subgroups total with one subgroup being {0} (of order 1) and one subgroup being Z p Z p (the group itself). That means there are p + 1 distinct subgroups of order p. These can be easily calculated. Each subgroup will be cyclic and of the form (a, a k ) where we have a fixed a {1,..., p 1} and k = 1,..., p 1. The final two subgroups are (0, a) and (a, 0). The general subgroup lattice looks very similar to the previous example s subgroup lattice (see Figure 3.2). Z p Z p p + 1 subgroups Z 1 Z 1 Figure 3.2. Subgroup Lattice of Z p Z p Example The subgroups of Z 5 Z 25. This is similar to the last example except now we are raising one of the prime numbered groups to a power greater than 1. The subgroups of Z 5 are Z 5 and {0}. This means that the subquotients for Z 5 are Z 5 /Z 5, Z 5 /{0} and {0}/{0}.

25 17 The subgroups of Z 25 are Z 25, 5 and {0}. This means that the subquotients for Z 25 are Z 25 /Z 25, 5 / 5, Z 25 / 5, We may simplify these by noting that 5 /{0}, Z 25 /{0} and {0}/{0}. Z 25 /Z 25 = Z5 /Z 5 = 5 / 5 = {0}/{0} = Z1 and This gives us 2 basic isomorphisms. They are Z 25 / 5 = 5 /{0} = Z 5 /{0} = Z 5. ϕ 1 : Z 5 Z 5 and ϕ 2 : Z 1 Z 1. Now we need to simply count how many isomorphisms there are from the domain to the range. The domain has 1 subquotient isomorphic to Z 5 while the range has 2 subquotients isomorphic to Z 5. This gives a total of 1 2 = 2 possible mappings. Each of these mappings is technically an automorphism of Z 5. There are 4 possible mappings from Z 5 to itself, so that gives us a total of 2 4 = 8 isomorphisms of order 5. The domain has 2 subquotients isomorphic to Z 1 while the range has 3 subquotients isomorphic to Z 1. Thus, there are 2 3 = 6 possible isomorphisms of order 1. Adding the two together gives us = 14 total isomorphisms from subquotients of Z 5 to subquotients of Z 25. By Goursat s Theorem, there is a bijection from every subgroup of Z 5 Z 25 to each of these isomorphisms. Therefore, there are 14 subgroups of Z 5 Z 25. See Figure 3.3 for the subgroup lattice of Z 5 Z 25. The following example is from [12]. Example The subgroups of Z 9 Z 9. Now we have both prime-powered groups raised to a power greater than 1.

26 18 Z 5 Z 25 Z 25 Z 25 Z 5 Z 5 Z 1 Z 1 Figure 3.3. Subgroup Lattice of Z 5 Z 25 The subgroups of Z 9 are Z 9, 3 and {0}. This means that there 6 subquotients of Z 9. They are Z 9 /Z 9 = Z1, 3 / 3 = Z1, Z 9 / 3 = Z3, 3 / 0 = Z3, Z 9 / 0 = Z9, 0 / 0 = Z1. Since our group is Z 9 Z 9, the isomorphisms will map the subquotients of Z 9 (the former group) to the subquotients of Z 9 (the latter group). This gives us 3 basic isomorphisms: ϕ 1 : Z 9 Z 9, ϕ 2 : Z 3 Z 3, ϕ 3 : Z 1 Z 1. Now we need to simply count how many isomorphisms there are from the domain to the range. The domain and range both have only 1 subquotient isomorphic to Z 9. Also, Z 9 has 6 automorphisms. Therefore, ϕ 1 has a total of 1 6 = 6 isomorphisms. The domain and range both have 2 subquotients isomorphic to Z 3. So that means that there are 2 2 = 4 possible mappings from the domain to the range. But since Z 3 has 2 possible automorphisms, that

27 19 gives ϕ 2 a total of 4 2 = 8 possible mappings. Finally, the domain and range both have 3 subquotients isomorphic to Z 1, so there are 3 3 = 9 possible mappings of ϕ 3 from the domain to the range (Z 1 obviously has only 1 automorphism). Adding these numbers together, there are a total of = 23 isomorphisms from subquotients of Z 9 to subquotients of Z 9. By Goursat s Theorem, there is a bijection from every subgroup of Z 9 Z 9 to each of these isomorphisms. Therefore, there are 23 subgroups of Z 9 Z 9. See Figure 3.4 for the subgroup lattice of Z 9 Z 9. Z 9 Z 9 Z 9 Z 3 Z 3 Z 9 Z 9 Z 9 Z 3 Z 3 Z 1 Z 1 Figure 3.4. Subgroup Lattice of Z 9 Z 9

28 20 Example The subgroups of Z 4 Z 6. The subgroups of Z 4 are {0}, 2 and Z 4. This means that the subquotients of Z 4 are Z 4 /Z 4 = Z1, 2 / 2 = Z1, Z 4 / 2 = Z2, 2 /{0} = Z 2, Z 4 /{0} = Z 4, {0}/{0} = Z 1. The subgroups of Z 6 are {0}, 3, 2, Z 6. This means that the subquotients of Z 6 are Z 6 /Z 6 = Z1, 2 / 2 = Z1, Z 6 / 2 = Z2, 2 /{0} = Z 3, Z 6 / 3 = Z3, 3 / 3 = Z1, Z 6 /{0} = Z 6, 3 /{0} = Z 2 and {0}/{0} = Z 1. This gives us 2 basic isomorphisms: ϕ 1 : Z 2 Z 2, ϕ 2 : Z 1 Z 1, each of which has only 1 possible automorphism. Thus, we need only count the possible combinations of mappings from the subquotients of Z 4 to the subquotients of Z 6. For ϕ 1, there are 2 2 = 4 possible combinations of mappings from Z 2 (the domain) to Z 2 (the range). For ϕ 2, there are 3 4 = 12 possible combinations of mappings from Z 1 (the domain) to Z 1 (the range). This gives us a grand total of = 16 subgroups of Z 4 Z 6. See Figure 3.5 for the subgroup lattice of Z 4 Z 6. Example The subgroups of Z 4 Z 8. The subgroups of Z 4 are {0}, 2 and Z 4. This means that the subquotients of Z 4 are Z 4 /Z 4 = Z1, 2 / 2 = Z1, Z 4 / 2 = Z2, 2 /{0} = Z 2, Z 4 /{0} = Z 4, {0}/{0} = Z 1.

29 21 Z 4 Z 6 Z 4 Z 3 Z 2 Z 6 Z 2 Z 2 Z 1 Z 1 Figure 3.5. Subgroup Lattice of Z 4 Z 6 The subgroups of Z 8 are {0}, 2, 4, Z 8. This means that the subquotients of Z 8 are Z 8 /Z 8 = Z1, 2 / 4 = Z2, Z 8 / 2 = Z2, 2 /{0} = Z 4, Z 8 / 4 = Z4, 4 / 4 = Z1, Z 8 /{0} = Z 8, 4 /{0} = Z 2 2 / 2 = Z1, {0}/{0} = Z 1.

30 22 This gives us 3 basic isomorphisms: ϕ 1 : Z 4 Z 4, ϕ 2 : Z 2 Z 2, ϕ 3 : Z 1 Z 1. Now we need to count how many isomorphisms there are from the domain to the range. The domain has only 1 subquotient isomorphic to Z 4 while the range has 2 subquotients isomorphic to Z 4. Therefore, there is a total of 1 2 = 2 possible combinations of mapping from Z 4 (the domain) to Z 4 (the range). Moreover, Z 4 has 2 automorphisms. So that means that there is a total of 2 2 = 4 possible isomorphisms for ϕ 1. The domain has 2 subquotients isomorphic to Z 2 while the range has 3 subquotients isomorphic to Z 2. So that means that there are 2 3 = 6 possible mappings from the domain to the range. But since Z 2 has 1 possible automorphism, that gives ϕ 2 a total of 6 1 = 6 possible isomorphisms. Finally, the domain has 3 subquotients isomorphic to Z 1 while the range has 4 subquotients isomorphic to Z 1. Thus, there are 3 4 = 12 possible mappings of ϕ 3 from the domain to the range (Z 1 obviously has only 1 automorphism). Adding the total numbers of isomorphisms together we get a total of = 22 isomorphisms from subquotients of Z 4 to subquotients of Z 8. By Goursat s Theorem, there is a bijection from every subgroup of Z 4 Z 8 to each of these isomorphisms. Therefore, there are 22 subgroups of Z 4 Z 8. See Figure 3.6 for the subgroup lattice of Z 4 Z 8. Now that we have seen this process completed several times, we summarize it. The following is the algorithm that we have for calculating the number of subgroups of Z m Z n : Step 1: Step 2: Find all the divisors j 1 < j 2 < < j s of m where j 1 = 1 and j s = m. Also, find all the divisors k 1 < k 2 < < k t of n where k 1 = 1 and k t = n. Calculate all the possible subquotients of Z m. To do this, we take every combination pair of divisors, j a and j b, and calculate the quotient group j a / j b if and only if j a divides j b and j a j b. Repeat this process for all the possible subquotients of Z n. Each of these subquotients is isomorphic to some Z i where i is a positive integer.

31 23 Z 4 Z 8 Z 4 Z 4 Z 2 Z 8 Z 4 Z 2 Z 8 Z 4 Z 4 Z 2 Z 2 Z 1 Z 1 Figure 3.6. Subgroup Lattice of Z 4 Z 8 Step 3: Step 4: We have isomorphisms ϕ i : Z i Z i such that the domain is a subquotient of Z m and the range is a subquotient of Z n. Count how many isomorphisms exists for each applicable i by multiplying the number of subquotients in the domain times the number of subquotients in the range. Multiply this result by the number of automorphisms that exist for Z i. Add up the results for each i from Step 3. This is the total number of subgroups of Z m Z n. 3.2 COUNTING SUBGROUPS The algorithm that we used in the previous section can be generalized to work for any cross product of groups. In general, it can be difficult to come up with a formula to calculate the number of subgroups of product groups. That is, it can be difficult unless we restrict ourselves to certain conditions. Here, we can create formulas for a few different cases of

32 24 Z m Z n. Using this knowledge, we can come up with a formula to quickly count the total number of subgroups for arbitrary Z m Z n. We already know that when m and n are relatively prime that Z m Z n = Zmn. This means that the number of subgroups is equivalent to the number of divisors of mn. Therefore, we need a function that gives the number of divisors of an integer. For n = p k 1 1 p kr r, we have d(n) = r (k i + 1). i=1 For example, let n = 24 = So here, a 1 = 3 and a 2 = 1. This gives us that d(24) = 2 (a i + 1) i=1 = (3 + 1)(1 + 1) = 4 2 = 8. Now we may use Goursat s Theorem explicitly when m and n are relatively prime. Let A Z m where A is a divisor of m and let C Z n where C is a divisor of n. Then each subgroup of Z m Z n corresponds to the triple (A/A, C/C, ϕ). Our choice of quotient groups is restricted because A and C are relatively prime. Thus, each ϕ will be the identity isomorphism. This means that every subgroup of Z m Z n has the form A C. Now we may use the divisor function to count the number of divisors of A and C. This gives us the following result from [12]. Theorem Let m and n be relatively prime with prime factorizations m = p a 1 1 p ar r and n = q b 1 1 q bs s. Then the number of subgroups of Z m Z n is d(m) d(n) = r s (a i + 1) (b j + 1). i=1 j=1 Corollary If p and q are distinct primes, then the number of subgroups of Z p a Z q b is (a + 1)(b + 1).

33 25 Now we look at how many subgroups of Z p r Z p s where r s. First, we need to count how many isomorphisms there are of order p k between subquotients of Z p r and subquotients of Z p s for each k such that 0 k r. This will give us the total number of subgroups of Z p k. For instance, Z p r has r + 1 subquotients of order 1. These subquotients are Z p r/z p r = p r 1 / p r 1 = = 0 / 0 = Z 1. Z p r has r subquotients of order p. These subquotients are Z p r/ p = p / p 2 = = p r 1 / 0 = Z p. Z p r has r 1 subquotients of order p 2, etc. Generally, Z p k has r k + 1 subquotients of order p k. To illustrate this, we count the subquotients of Z 16. Here, 16 = p r where p = 2 and r = 4. Then there are r + 1 = 5 subquotients of order 1: Z 1 = Z16 /Z 16 = 2 / 2 = 4 / 4 = 8 / 8 = 0 / 0 = p 0 / p 0 = p 1 / p 1 = p 2 / p 2 = p 3 / p 3 = p 4 / p 4. There are r = 4 subquotients of order p: Z p = Z16 / 2 = 2 / 4 = 4 / 8 = 8 / 0 = p 0 / p 1 = p 1 / p 2 = p 2 / p 3 = p 3 / p 4. There are r 1 = 3 subquotients of order p 2 : Z p 2 = Z16 / 4 = 2 / 8 = 4 / 0 = p 0 / p 2 = p 1 / p 3 = p 2 / p 4. There are r 2 = 2 subquotients of order p 3 : Z p 3 = Z16 / 8 = 2 / 0 = p 0 / p 3 = p 1 / p 4. Finally, there are r 3 = 1 subquotients of order p 4 : Z 16 = Z16 / 0 = p 0 / p 4.

34 26 We want the total number of isomorphisms from subquotients of Z p r to subquotients of Z p s. We can choose a subquotient of Z p r in r k + 1 different ways and we can choose a subquotient of Z p s in s k + 1 different ways. Additionally, the number of automorphisms of each is simply φ(p k ) when k > 0 (and only 1 automorphism if k = 0). This means that the total number of isomorphisms for 0 k r is (r k + 1)(s k + 1)φ(p k ) = (r k + 1)(s k + 1)(p k p k 1 ). If k = 0 then the total is simply (r + 1)(s + 1). So this means that the total number of subgroups of Z p r Z p s is (r + 1)(s + 1) + r (r k + 1)(s k + 1)(p k p k 1 ). k=1 Using the methods found in [11], we may clean this up algebraically to equal (r + s + 1) r p k 2 k=0 The series on the left is the finite geometric series r k=0 r kp k. k=1 p k = pr+1 1 p 1. The series on the right can be cleaned up by noting that (p 1) Dividing both sides by p 1 yields r k=1 r ( ) p kp k = rp r+1 r 1 p. p 1 k=1 kp k = rpr+1 p 1 pr+1 p (p 1) 2. Substituting these both into our original formula gives us that the number of subgroups of Z p r Z p s is ( ) p r+1 1 (r + s + 1) p 1 More simplification gives us the following theorem. ( ) rp r+1 2 p 1 pr+1 p. (p 1) 2

35 Theorem Let p be prime. Let r and s be nonnegative integers such that r s. Then the number of subgroups of Z p r Z p s is p r+1 [(s r + 1)(p 1) + 2] [(s + r + 3)(p 1) + 2] (p 1) 2. We can use Theorem to come up with formulas for some general cases. The number of subgroups of Z 2 r Z 2 s is 2 r+1 (s r + 3) (s + r + 5). The number of subgroups of Z 2 r Z 2 r is 2 r+1 (3) 2r 5. The number of subgroups of Z 3 r Z 3 r is 3 r+1 r 2. The number of subgroups of Z p r Z p r is pr+1 (p + 1) 2r(p 1) 3p + 1 (p 1) 2. The number of subgroups of Z p Z p is p + 3. The number of subgroups of Z 1 Z p s = Zp s is s Example Examples and are directly computed using Theorem We have that the number of subgroups of Z 7 Z 7 is = 10 while the number of subgroups of Z 9 Z 9 is = 12. Example Example is directly computed using Theorem where r = 1 and s = 2. Then we have that the number of subgroups of Z 5 Z 25 is [( )(5 1) + 2] [( )(5 1) + 2] (5 1) 2 = 14. Example Example is directly computed using the special case of Theorem where our group orders are 2 r and 2 s. Here, r = 2 and s = 3. Then we have that the number of subgroups of Z 4 Z 8 is ( ) ( ) = 22. Our next result will use the following well-known theorem.

36 28 Theorem (The Fundamental Theorem of Finite Abelian Groups). Every finite Abelian group is isomorphic to the direct product of cyclic subgroups of prime-powered order. In other words, any finite abelian group can be written, up to isomorphism, as Z p1 r 1 Z p2 r 2 Z ps rs where the p i primes are not necessarily distinct and the r i are positive integers. This allows us to rewrite our product Z m Z n. For if m = p a 1 1 p ar r and n = p b 1 1 p br r with a i, b j = 0, 1,..., then we have that Z m Z n = ( Zp1 a 1 Z p1 b 1 ) ( Zpr ar Z pr br ). For our purposes, this is analogous to factoring by primes. The following theorem may be found in [13]. In it, L(G) denotes the subgroup lattice of group G while L(G) denotes the order of the subgroup lattice (thus, telling us how many subgroups there are in G). Therefore, L(G 1 G 2 ) denotes the number of subgroups of G 1 G 2. Theorem (Suzuki s Theorem). Let G 1 and G 2 be finite groups. Then L(G 1 G 2 ) = L(G 1 ) L(G 2 ) and L(G 1 G 2 ) = L(G 1 ) L(G 2 ) if and only if G 1 and G 2 are relatively prime. This means that we may apply Suzuki s Theorem to Theorem to obtain the culmination of our efforts. The following is attributed to Petrillo [12]: Theorem Let m and n be positive integers. Let p 1,..., p k denote the list of distinct primes dividing the product mn so that m = p r 1 1 p r k k and n = p s 1 1 p s k k. Then L(Z m Z n ) = k i=1 L(Z p r i i Z p s i i ). This allows us to break up Z m Z n into various groups. From there, we use the formulas of Theorem to quickly count the subgroups of Z m Z n.

37 29 Example We redo Example 3.1.5, the number of subgroup of Z 4 Z 6. Note that 4 = 2 2 and 6 = 2 3. Then L(Z 4 Z 6 ) = L(Z 2 2 Z 2 ) L(Z 1 Z 3 ) = 8 2 = 16. Example The subgroups of Z 18 Z 30 (this example is from [12]). First note that 18 = and 30 = Then by Theorem we have that L(Z 18 Z 30 ) = L(Z 2 Z 2 ) L(Z 3 2 Z 3 ) L(Z 1 Z 5 ) = = 100. Example The subgroups of Z 2940 Z 630. Note that 2940 = and 630 = Then L(Z 2940 Z 630 ) = L(Z 2 2 Z 2 ) L(Z 3 Z 3 2) L(Z 5 Z 5 ) L(Z 7 2 Z 7 ) = = So while we do not have a simple closed formula that automatically counts the number of subgroups of arbitrary Z m Z n, we may use Theorem 3.2.7, Theorem and Suzuki s Theorem. This will be the fastest method that we may construct. 3.3 GOURSAT S THEOREM It is here that we shall provide a proof of Goursat s Theorem, given in steps through Lemma 3.3.1, Lemma and Theorem This way, the less mathematically mature students can follow along and, more or less, learn to prove it themselves. Additionally, this

38 30 could make a great exercise for an algebra class. The primary reference for this process is [11] but the presentation is similar to [2]. Also see [1] for a similar presentation where it is referred to as Goursat s Lemma. Let G and H be groups and let S be the set of all subgroups of G H. Let T be the set of all triples (A/B, C/D, ϕ) where A/B is a subquotient of G, C/D is a subquotient of H and ϕ is an isomorphism from A/B to C/D. Then the following subgroup exists. Lemma Let (A/B, C/D, ϕ) be a triple in T and define U ϕ = {(g, h) A C ϕ(gb) = hd}. Then U ϕ is a subgroup of G H. Proof. We have that (e G, e H ) A C where e G A is the identity element for G and e H C is the identity element for H. By definition of U ϕ, we have that ϕ(e G B) = ϕ(b) = D = e H D, so (e G, e H ) U ϕ is the identity element. Let (a, c) U ϕ. Note that a 1 A and c 1 C. This means that (a 1, c 1 ) = (a, c) 1 A C. Let (a 1, c 1 ), (a 2, c 2 ) A C. Then we have (a 1, c 1 ) (a 2, c 2 ) = ϕ(a 1 B)ϕ(a 2 B) = c 1 Dc 2 D = c 1 c 2 D (by coset properties) = ϕ(a 1 a 2 B) = (a 1 a 2, c 1 c 2 ), so A C = U ϕ is closed under the operation. Therefore, it is indeed a subgroup of G H. Example Lemma gives us a method to explicitly find the subgroups of G 1 G 2. Looking back at Example 1.1.1, we saw that there are two subgroups of Z 2 Z 4 that are not

39 31 of the form H K where H is a subgroup of Z 2 and K is a subgroup of Z 4. Those subgroups are {(0, 0), (1, 2)} and {(0, 0), (1, 1), (0, 2), (1, 3)}. We can apply Lemma to Z 2 Z 4 to show how we obtain the second group. We will look at the isomorphism ϕ : Z 2 / 0 Z 4 / 2. By the lemma, we want to find the group U ϕ = {(g, h) Z 2 Z 4 ϕ(g + 0 ) = h + 2 }. The most that we know about ϕ is that it is an isomorphism that maps cosets to cosets. Therefore, there are only four distinct possible mappings of ϕ : (i) = which simplifies to {0} {0, 2}, (ii) = which simplifies to {0} {1, 3}, (iii) = which simplifies to {1} {0, 2}, (iv) = which simplifies to {1} {1, 3}. Clearly ϕ can not map {0} to {0, 2} and to {1, 3}. So this means that there must be two distinct possibilities for how ϕ maps. Those are {0} {0, 2} {0} {1, 3} ϕ 1 = or ϕ 2 = {1} {1, 3} {1} {0, 2}. The elements in Z 2 Z 4 that satisfy ϕ 1 are S 1 = {(0, 0), (1, 1), (0, 2), (1, 3)}. The elements in Z 2 Z 4 that satisfy ϕ 2 are S 2 = {(0, 1), (0, 3), (1, 0), (1, 2)}. Note, however, that S 2 is not actually a subgroup of Z 2 Z 4 (it s not even a group). Therefore, ϕ 1 must have been the mapping from Z 2 / 0 to Z 4 / 2. This means that U ϕ1 = {(0, 0), (1, 1), (0, 2), (1, 3)}. A similar process can be used to show how to construct the subgroup {(0, 0), (1, 2)}.

40 32 Lemma For a subgroup U in S, let A U = {g G (g, h) U for some h H}, B U = {g G (g, 1) U}, C U = {h H (g, h) U for some g G}, D U = {h H (1, h) U}, and define the mapping ϕ U : A U /B U C U /D U by Then ϕ U (gb U ) = hd U whenever (g, h) U. (i) A U is a subgroup of G and C U is a subgroup of H, (ii) B U is a normal subgroup of A U and D U is a normal subgroup of C U and (iii) ϕ U is a group isomorphism. Proof. (i) First we show that A U is a subgroup of G. Note that e G is the identity for G. Moreover, (e G, h) U for some h H, so e G A U, and A U contains the identity element. Since g, g 1 G, we have that if (g, h) U for some h H, then (g 1, h) U by properties of U being a subgroup of G H. So g, g 1 A U, and A U contains inverses. Let g 1, g 2 A U. Then (g 1, h 1 ) U for some h 1 H and (g 2, h 2 ) U for some h 2 H. Note that h 1 h 2 H. Then (g 1 g 2, h 1 h 2 ) U. So g 1 g 2 A U. So A U is closed and A U is a subgroup of G. Now we show that C U is a subgroup of H. Note that e H is the identity for H. Moreover, (g, e H ) U for some g G, so e H C U, and C U contains the identity element. Since h, h 1 H, we have that if (g, h) U for some g G, then (g, h 1 ) U by properties of U being a subgroup of G H. So h, h 1 C U, and C U contains inverses. Let h 1, h 2 C U. Then (g 1, h 1 ) U for some g 1 G and (g 2, h 2 ) U for some g 2 G. Note that g 1 g 2 G. Then (g 1 g 2, h 1 h 2 ) U. So h 1 h 2 C U. So C U is closed and C U is a subgroup of H. (ii) First we show that B U is normal in A U. Note that B U is a subgroup of A U where h = 1 (the identity element of H). To show it s normal, we need to show that ab U = B U a for every a A U. Note that ab U = {ag 1 A U (ag 1, 1) U} and B U a = {g 2 a A U (g 2 a, 1) U}. Since A U is a group, ag 1 = g 2 a for some g 1, g 2 A U. Thus, ab U = B U a, and B U is normal in A U. Now we show that D U is normal in C U. Note that D U is a subgroup of C U where g = 1 (the identity element of G). To show it s normal, we need to show that cd U = D U c for

41 33 every c C U. Note that cd U = {ch 1 C U (1, ch 1 ) U} and D U c = {h 2 c C U (1, h 2 c) U}. Since C U is a group, ch 1 = h 2 c for some h 1, h 2 C U. Thus, cd U = D U c, and D U is normal in C U. (iii) Now we show that ϕ U is an isomorphism. First assume that ϕ U (g 1 B U ) = ϕ U (g 2 B U ). Note that ϕ 1 U ϕ U(g 1 B U ) = ϕ 1 U ϕ U(g 2 B U ) if and only if g 1 B U = g 2 B U, which we have already shown to be true because B U is normal in A U. Therefore, ϕ U is 1-to-1. To show onto, note that ϕ 1 U (hd U) = ϕ 1 U (ϕ U(gB U )) = gb U. Finally, note that ϕ U (g 1 B U )ϕ U (g 2 B U ) = h 1 D U h 2 D U = h 1 h 2 D U = ϕ(g 1 g 2 B U ), so ϕ U preserves the operation. Thus, ϕ U is an isomorphism. The following theorem is equivalent to Goursat s Theorem as stated at the beginning of this paper. Here, S is the set of all subgroups of G 1 G 2. Also, T is the set of all the isomorphisms of subquotients of G 1 and G 2. Theorem Define the mapping α : S T and β : T S by α(u) = (A U /B U, C U /D U, ϕ U ) and Then α is a bijection with inverse β. β(a/b, C/D, ϕ) = U ϕ. Proof. First we show that α is 1-to-1. Let α(u) = α(v ). That means that (A U /B U, C U /D U, ϕ U ) = (A V /B V, C V /D V, ϕ V ). This implies that each component of the triples are equivalent. That is, A U /B U = A V /B V, C U /D U = C V /D V and ϕ U = ϕ V. Because α is well-defined, the corresponding subgroups U and V must have been the same subgroup. So α is 1-to-1. To show that α is onto, note that α 1 ((A U /B U, C U /D U, ϕ U )) = α 1 (α(u)) = U. Since α is both 1-to-1 and onto, it is a bijection. Thus, α 1 = β is its inverse.

42 3.4 GENERALIZED CASES: BEYOND Z m Z n. 34 We can have the direct product of groups G 1 G 2 where G 1 and G 2 are not cyclic groups. The next example will use the following definitions. Definition The group of even permutations of n symbols is called the alternating group of degree n. It is denoted by A n. The property of a permutation being even or odd is apparent when the group permutations are written in cycle notation. For further information on permutations, see [6]. Definition Let n 3. The group of symmetries of a regular n-polygon is called the dihedral group of order 2n. It is denoted D n. So, for example, D 3 represents all the symmetries of a triangle. Again, see [6] for more information. Example The subgroups of A 4 D 4. This example does not involve integer groups as we have shown throughout the rest of this paper. Instead, we have A 4, which is the alternating group on 4 elements and D 4, which is the symmetries of a square. These groups do not have easily defined subgroups the way the integer groups do. This makes calculating the subquotients a little more tricky. The group A 4 has 12 elements. The divisors of 12 are 1, 2, 3, 4, 6 and 12. However, a subgroup of each of these orders need not necessarily exist. Indeed, there is no subgroup of order 6. Thus, our subquotients will have order 1, 2, 3, 4 or 12. The group D 4 has 8 elements. The divisors of 8 are 1,2,4 and 8. Thus, our subquotients will have order 1, 2, 4 or 8. Since we are looking for isomorphisms between subquotients, we need only concern ourselves with subquotients of orders 1, 2 and 4. Note that A 4 has 10 subquotients of order 1 (one for each subgroup). Additionally, D 4 also has 10 subquotients of order 1. Therefore, there are = 100 possible combinations of mappings from subquotients of order 1 from A 4 to subquotients of order 1 from D 4 (there is only 1 isomorphism between subgroups of order 1).

43 35 Note that A 4 has 6 subquotients of order 2 and D 4 has 15 subquotients of order 2. This gives us a total of 6 15 = 90 possible combinations of mappings from subquotients of order 2 from A 4 to subquotients of order 2 from D 4 (there is only 1 isomorphism between subgroups of order 2). A 4 has only one subquotient of order 4. It is isomorphic to Z 2 Z 2, also known as the Klein 4-group. D 4 has a total of 4 subquotients of order 4. However, one of these subquotients is cyclic, so it can t be isomorphic to the Klein 4-group (which is not cyclic). Thus, D 4 has 3 subquotients isomorphic to Z 2 Z 2. This gives us a total of 1 3 = 3 possible combinations of mappings from subquotients of A 4 of order 2 to subquotients of D 4 of order 2. And since there are 6 automorphisms of the Klein 4-group, there are 3 6 = 18 possible mappings of order 4. This gives us a total of = 208 isomorphisms from subquotients of A 4 to subquotients of D 4. By Goursat s Theorem, there is a bijection from every subgroup of A 4 D 4 to each of these isomorphisms. Therefore, there are 208 subgroups of A 4 D 4. A subgroup lattice of A 4 D 4 will not be provided. However, Figure 3.7 contains the subgroup lattices of A 4 and D 4 individually. Now we count the number of subgroups of the direct product of A B C. This is clearly done in an inductive manner. Example The subgroups of Z 2 Z 4 Z 8. Our initial assumption is that we can proceed inductively. First we count the number of isomorphisms from Z 2 to Z 4, then we multiply this by the number of isomorphisms of Z 8. Proceeding in this fashion gives us that there are 30 isomorphisms total. Thus, Goursat s Theorem should give us that there are 30 subgroups of Z 2 Z 4 Z 8. However, Goursat s Theorem only applies to the direct product of two groups. This being the case, we would need to count the number of isomorphisms from Z 2 Z 4 to Z 8 or count the number of isomorphisms from Z 2 to Z 4 Z 8. Unfortunately, calculating all the