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, and z = (010111)t have been received. Then Hx =   , Hy =   1 1  1  1 1 , Hz = 0  1 0  0  . Hence, x has an error in the third bit and z has an error in the fourth bit. The transmitted codewords for x and z must have been (110110) and (010011), respectively. The syndrome of y does not occur in any of the columns of the matrix H, so multiple errors must have occurred to produce y. Coset Decoding We can use group theory to obtain another way of decoding messages. A linear code C is a subgroup of Zn 2. Coset or standard decoding uses the cosets of C in Zn 2 to implement maximum-likelihood decoding. Suppose that C is an (n, m)-linear code. A coset of C in Zn 2 is written in the form x + C, where x ∈ Zn 2. By Lagrange’s Theorem (Theorem 6.5), there are 2n−m distinct cosets of C in Zn 2. Example 18. Let C be the (5, 3)-linear code given by the parity-check matrix.4 EFFICIENT DECODING 137 C Cosets (00000) (01101) (10011) (11110) (10000) + C (10000) (11101) (00011) (01110) (01000) + C (01000) (00101) (11011) (10110) (00100) + C (00100) (01001) (10111) (11010) (00010) + C (00010) (01111) (10001) (11100) (00001) + C (00001) (01100) (10010) (11111) (10100) + C (00111) (01010) (10100) (11001) (00110) + C (00110) (01011) (10101) (11000) Table 8.5. Cosets of C The code consists of the codewords (00000) (01101) (10011) (11110). There are 25−2 = 23 cosets of C in Z5 are
listed in Table 8.5. 2, each with order 22 = 4. These cosets Our task is to ﬁnd out how knowing the cosets might help us to decode a message. Suppose that x was the original codeword sent and that r is the n-tuple received. If e is the transmission error, then r = e+x or, equivalently, x = e + r. However, this is exactly the statement that r is an element in the coset e + C. In maximum-likelihood decoding we expect the error e to be as small as possible; that is, e will have the least weight. An n-tuple of least weight in a coset is called a coset leader. Once we have determined a coset leader for each coset, the decoding process becomes a task of calculating r + e to obtain x. Example 19. In Table 8.5, notice that we have chosen a representative of the least possible weight for each coset. These representatives are coset leaders. Now suppose that r = (01111) is the received word. To decode r, we ﬁnd that it is in the coset (00010) + C; hence, the originally transmitted codeword must have been (01101) = (01111) + (00010). A potential problem with this method of decoding is that we might have to examine every coset for the received codeword. The following proposition gives a method of implementing coset decoding. It states that we can associate a syndrome with each coset; hence, we can make a table that designates a coset leader corresponding to each syndrome. Such a list is called a decoding table. 138 CHAPTER 8 ALGEBRAIC CODING THEORY Syndrome Coset Leader (000) (001) (010) (011) (100) (101) (110) (111) (00000) (00001) (00010) (10000) (00100) (01000) (00110) (10100) Table 8.6. Syndromes for each coset Proposition 8.16 Let C be an (n, k)-linear code given by the matrix H and suppose that x and y are in Zn 2. Then x and y are in the same coset of C if and only if Hx = Hy. That is, two n-tuples are in the same coset if and only if
their syndromes are the same. Proof. Two n-tuples x and y are in the same coset of C exactly when x − y ∈ C; however, this is equivalent to H(x − y) = 0 or Hx = Hy. Example 20. Table 8.6 is a decoding table for the code C given in Example 18. If x = (01111) is received, then its syndrome can be computed to be Hx =    0 1 . 1 Examining the decoding table, we determine that the coset leader is (00010). It is now easy to decode the received codeword. Given an (n, k)-block code, the question arises of whether or not coset decoding is a manageable scheme. A decoding table requires a list of cosets and syndromes, one for each of the 2n−k cosets of C. Suppose that we have a (32, 24)-block code. We have a huge number of codewords, 224, yet there are only 232−24 = 28 = 256 cosets. Exercises 1. Why is the following encoding scheme not acceptable? Information: Codeword: 0 000 1 001 2 010 3 011 4 101 5 110 6 111 7 000 8 001 EXERCISES 139 2. Without doing any addition, explain why the following set of 4-tuples in Z4 2 cannot be a group code. (0110) (1001) (1010) (1100) 3. Compute the Hamming distances between the following pairs of n-tuples. (a) (011010), (011100) (c) (00110), (01111) (b) (11110101), (01010100) (d) (1001), (0111) 4. Compute the weights of the following n-tuples. (a) (011010) (b) (11110101) (c) (01111) (d) (1011) 5. Suppose that a linear code C has a minimum weight of 7. What are the error-detection and error-correction capabilities of C? 6. In each of the following codes, what is the minimum distance for the code? What is the best situation we might hope for in connection with error detection and error correction? (a) (011010) (011100) (11
0111) (110000) (b) (011100) (011011) (111011) (100011) (000000) (010101) (110100) (110011) (c) (000000) (011100) (110101) (110001) (d) (0110110) (0111100) (1110000) (1111111) (1001001) (1000011) (0001111) (0000000) 7. Compute the null space of each of the following matrices. What type of (n, k)-block codes are the null spaces? Can you ﬁnd a matrix (not necessarily a standard generator matrix) that generates each code? Are your generator matrices unique? (a) (bc) (d 140 CHAPTER 8 ALGEBRAIC CODING THEORY 8. Construct a (5, 2)-block code. Discuss both the error-detection and error- correction capabilities of your code. 9. Let C be the code obtained from the null space of the matrix 01111 10101 01110 00011 Decode the message if possible. 10. Suppose that a 1000-bit binary message is transmitted. Assume that the probability of a single error is p and that the errors occurring in diﬀerent bits are independent of one another. If p = 0.01, what is the probability of more than one error occurring? What is the probability of exactly two errors occurring? Repeat this problem for p = 0.0001. 11. Which matrices are canonical parity-check matrices? For those matrices that are canonical parity-check matrices, what are the corresponding standard generator matrices? What are the error-detection and error-correction capabilities of the code generated by each of these matrices? (a) (bc) (d 12. List all possible syndromes for the codes generated by each of the matrices in the previous exercise. 13. Let Compute the syndrome caused by each of the following transmission errors. (a) An error in the ﬁrst bit EXERCISES 141 (b) An error in the third bit (c) An error in the last bit (d) Errors in the third and fourth bits 14. Let C be the group code in Z3 2 deﬁned by the codewords (000) and (111). Compute the cosets of H in Z3 2. Why
was there no need to specify right or left cosets? Give the single transmission error, if any, to which each coset corresponds. 15. For each of the following matrices, ﬁnd the cosets of the corresponding code C. Give a decoding table for each code if possible. (a) (bc) (d 16. Let x, y, and z be binary n-tuples. Prove each of the following statements. (a) w(x) = d(x, 0) (b) d(x, y) = d(x + z, y + z) (c) d(x, y) = w(x − y) 17. A metric on a set X is a map d : X × X → R satisfying the following conditions. (a) d(x, y) ≥ 0 for all x, y ∈ X; (b) d(x, y) = 0 exactly when x = y; (c) d(x, y) = d(y, x); (d) d(x, y) ≤ d(x, z) + d(z, y). In other words, a metric is simply a generalization of the notion of distance. Prove that Hamming distance is a metric on Zn 2. Decoding a message actually reduces to deciding which is the closest codeword in terms of distance. 18. Let C be a linear code. Show that either the ith coordinates in the codewords of C are all zeros or exactly half of them are zeros. 19. Let C be a linear code. Show that either every codeword has even weight or exactly half of the codewords have even weight. 142 CHAPTER 8 ALGEBRAIC CODING THEORY 20. Show that the codewords of even weight in a linear code C are also a linear code. 21. If we are to use an error-correcting linear code to transmit the 128 ASCII characters, what size matrix must be used? What size matrix must be used to transmit the extended ASCII character set of 256 characters? What if we require only error detection in both cases? 22. Find the canonical parity-check matrix that gives the even parity check bit code with three information positions. What is the matrix for seven information positions? What are the corresponding standard generator matrices? 23. How many check positions are needed for a single error-correcting code with 20 information
positions? With 32 information positions? 24. Let ei be the binary n-tuple with a 1 in the ith coordinate and 0’s elsewhere and suppose that H ∈ Mm×n(Z2). Show that Hei is the ith column of the matrix H. 25. Let C be an (n, k)-linear code. Deﬁne the dual or orthogonal code of C to be C ⊥ = {x ∈ Zn 2 : x · y = 0 for all y ∈ C}. (a) Find the dual code of the linear code C where C is given by the matrix   b) Show that C ⊥ is an (n, n − k)-linear code. (c) Find the standard generator and parity-check matrices of C and C ⊥. What happens in general? Prove your conjecture. 26. Let H be an m × n matrix over Z2, where the ith column is the number i written in binary with m bits. The null space of such a matrix is called a Hamming code. (a) Show that the matrix generates a Hamming code. What are the error-correcting properties of a Hamming code? (b) The column corresponding to the syndrome also marks the bit that was in error; that is, the ith column of the matrix is i written as a binary number, and the syndrome immediately tells us which bit is in error. If the received word is (101011), compute the syndrome. In which bit did the error occur in this case, and what codeword was originally transmitted? EXERCISES 143 (c) Give a binary matrix H for the Hamming code with six information positions and four check positions. What are the check positions and what are the information positions? Encode the messages (101101) and (001001). Decode the received words (0010000101) and (0000101100). What are the possible syndromes for this code? (d) What is the number of check bits and the number of information bits in an (m, n)-block Hamming code? Give both an upper and a lower bound on the number of information bits in terms of the number of check bits. Hamming codes having the maximum possible number of information bits with k check bits are called perfect. Every possible syndrome except 0 occurs as a column. If the number of information bits is less
than the maximum, then the code is called shortened. In this case, give an example showing that some syndromes can represent multiple errors. Programming Exercises Write a program to implement a (16, 12)-linear code. Your program should be able to encode and decode messages using coset decoding. Once your program is written, write a program to simulate a binary symmetric channel with transmission noise. Compare the results of your simulation with the theoretically predicted error probability. References and Suggested Readings [1] Blake, I. F. “Codes and Designs,” Mathematics Magazine 52 (1979), 81–95. [2] Hill, R. A First Course in Coding Theory. Oxford University Press, Oxford, 1990. [3] Levinson, N. “Coding Theory: A Counterexample to G. H. Hardy’s Conception of Applied Mathematics,” American Mathematical Monthly 77 (1970), 249–58. [4] Lidl, R. and Pilz, G. Applied Abstract Algebra. 2nd ed. Springer, New York, 1998. [5] MacWilliams, F. J. and Sloane, N. J. A. The Theory of Error-Correcting Codes. North-Holland Mathematical Library, 16, Elsevier, Amsterdam, 1983. [6] Roman, S. Coding and Information Theory. Springer-Verlag, New York, 1992. [7] Shannon, C. E. “A Mathematical Theory of Communication,” Bell System Technical Journal 27 (1948), 379–423, 623–56. [8] Thompson, T. M. From Error-Correcting Codes through Sphere Packing to Simple Groups. Carus Monograph Series, No. 21. Mathematical Association of America, Washington, DC, 1983. [9] van Lint, J. H. Introduction to Coding Theory. Springer, New York, 1999. 9 Isomorphisms Many groups may appear to be diﬀerent at ﬁrst glance, but can be shown to be the same by a simple renaming of the group elements. For example, Z4 and the subgroup of the circle group T generated by i can be shown to be the same by demonstrating a one-to-one correspondence between the elements of the two groups and between the group operations. In such a case we say that the groups are isomorphic. 9.
1 Deﬁnition and Examples Two groups (G, ·) and (H, ◦) are isomorphic if there exists a one-to-one and onto map φ : G → H such that the group operation is preserved; that is, φ(a · b) = φ(a) ◦ φ(b) for all a and b in G. If G is isomorphic to H, we write G ∼= H. The map φ is called an isomorphism. ∼= i, deﬁne a map φ : Z4 → i by φ(n) = in. Example 1. To show that Z4 We must show that φ is bijective and preserves the group operation. The map φ is one-to-one and onto because φ(0) = 1 φ(1) = i φ(2) = −1 φ(3) = −i. Since φ(m + n) = im+n = imin = φ(m)φ(n), 144 9.1 DEFINITION AND EXAMPLES the group operation is preserved. 145 Example 2. We can deﬁne an isomorphism φ from the additive group of real numbers (R, +) to the multiplicative group of positive real numbers (R+, ·) with the exponential map; that is, φ(x + y) = ex+y = exey = φ(x)φ(y). Of course, we must still show that φ is one-to-one and onto, but this can be determined using calculus. Example 3. The integers are isomorphic to the subgroup of Q∗ consisting of elements of the form 2n. Deﬁne a map φ : Z → Q∗ by φ(n) = 2n. Then φ(m + n) = 2m+n = 2m2n = φ(m)φ(n). By deﬁnition the map φ is onto the subset {2n : n ∈ Z} of Q∗. To show that the map is injective, assume that m = n. If we can show that φ(m) = φ(n), then we are done. Suppose that m > n and assume that φ(m) = �
�(n). Then 2m = 2n or 2m−n = 1, which is impossible since m − n > 0. Example 4. The groups Z8 and Z12 cannot be isomorphic since they have diﬀerent orders; however, it is true that U (8) ∼= U (12). We know that U (8) = {1, 3, 5, 7} U (12) = {1, 5, 7, 11}. An isomorphism φ : U (8) → U (12) is then given by → 11. The map φ is not the only possible isomorphism between these two groups. We could deﬁne another isomorphism ψ by ψ(1) = 1, ψ(3) = 11, ψ(5) = 5, ψ(7) = 7. In fact, both of these groups are isomorphic to Z2 × Z2 (see Example 14 in Chapter 3). Example 5. Even though S3 and Z6 possess the same number of elements, we would suspect that they are not isomorphic, because Z6 is abelian and 146 CHAPTER 9 ISOMORPHISMS S3 is nonabelian. To demonstrate that this is indeed the case, suppose that φ : Z6 → S3 is an isomorphism. Let a, b ∈ S3 be two elements such that ab = ba. Since φ is an isomorphism, there exist elements m and n in Z6 such that φ(m) = a and φ(n) = b. However, ab = φ(m)φ(n) = φ(m + n) = φ(n + m) = φ(n)φ(m) = ba, which contradicts the fact that a and b do not commute. Theorem 9.1 Let φ : G → H be an isomorphism of two groups. Then the following statements are true. 1. φ−1 : H → G is an isomorphism. 2. \|G\| = \|H\|. 3. If G is abelian, then H is abelian. 4. If G is cyclic, then H is cyclic. 5. If G has a subgroup of order n, then H has a subgroup of order n. Proof. Assertions (1) and (
2) follow from the fact that φ is a bijection. We will prove (3) here and leave the remainder of the theorem to be proved in the exercises. (3) Suppose that h1 and h2 are elements of H. Since φ is onto, there exist elements g1, g2 ∈ G such that φ(g1) = h1 and φ(g2) = h2. Therefore, h1h2 = φ(g1)φ(g2) = φ(g1g2) = φ(g2g1) = φ(g2)φ(g1) = h2h1. We are now in a position to characterize all cyclic groups. Theorem 9.2 All cyclic groups of inﬁnite order are isomorphic to Z. Proof. Let G be a cyclic group with inﬁnite order and suppose that a is a generator of G. Deﬁne a map φ : Z → G by φ : n → an. Then φ(m + n) = am+n = aman = φ(m)φ(n). 9.1 DEFINITION AND EXAMPLES 147 To show that φ is injective, suppose that m and n are two elements in Z, where m = n. We can assume that m > n. We must show that am = an. Let us suppose the contrary; that is, am = an. In this case am−n = e, where m − n > 0, which contradicts the fact that a has inﬁnite order. Our map is onto since any element in G can be written as an for some integer n and φ(n) = an. Theorem 9.3 If G is a cyclic group of order n, then G is isomorphic to Zn. Proof. Let G be a cyclic group of order n generated by a and deﬁne a map φ : Zn → G by φ : k → ak, where 0 ≤ k < n. The proof that φ is an isomorphism is one of the end-of-chapter exercises. Corollary 9.4 If G is a group of order p, where p is a prime number, then G is isomorphic to Zp. Proof. The proof is a direct result of Cor
ollary 6.7. The main goal in group theory is to classify all groups; however, it makes sense to consider two groups to be the same if they are isomorphic. We state this result in the following theorem, whose proof is left as an exercise. Theorem 9.5 The isomorphism of groups determines an equivalence relation on the class of all groups. Hence, we can modify our goal of classifying all groups to classifying all groups up to isomorphism; that is, we will consider two groups to be the same if they are isomorphic. Cayley’s Theorem Cayley proved that if G is a group, it is isomorphic to a group of permutations on some set; hence, every group is a permutation group. Cayley’s Theorem is what we call a representation theorem. The aim of representation theory is to ﬁnd an isomorphism of some group G that we wish to study into a group that we know a great deal about, such as a group of permutations or matrices. Example 6. Consider the group Z3. The Cayley table for Z3 is as follows 148 CHAPTER 9 ISOMORPHISMS The addition table of Z3 suggests that it is the same as the permutation group G = {(0), (012), (021)}. The isomorphism here is 0) = (012) = (021). Theorem 9.6 (Cayley) Every group is isomorphic to a group of permutations. Proof. Let G be a group. We must ﬁnd a group of permutations G that is isomorphic to G. For any g ∈ G, deﬁne a function λg : G → G by λg(a) = ga. We claim that λg is a permutation of G. To show that λg is one-to-one, suppose that λg(a) = λg(b). Then ga = λg(a) = λg(b) = gb. Hence, a = b. To show that λg is onto, we must prove that for each a ∈ G, there is a b such that λg(b) = a. Let b = g−1a. Now we are ready to deﬁne our group G. Let G = {λg : g ∈ G}.
We must show that G is a group under composition of functions and ﬁnd an isomorphism between G and G. We have closure under composition of functions since (λg ◦ λh)(a) = λg(ha) = gha = λgh(a). λe(a) = ea = a Also, and (λg−1 ◦ λg)(a) = λg−1(ga) = g−1ga = a = λe(a). We can deﬁne an isomorphism from G to G by φ : g → λg. The group operation is preserved since φ(gh) = λgh = λgλh = φ(g)φ(h). 9.2 DIRECT PRODUCTS 149 It is also one-to-one, because if φ(g)(a) = φ(h)(a), then ga = λga = λha = ha. Hence, g = h. That φ is onto follows from the fact that φ(g) = λg for any λg ∈ G. The isomorphism g → λg is known as the left regular representation of G. Historical Note Arthur Cayley was born in England in 1821, though he spent much of the ﬁrst part of his life in Russia, where his father was a merchant. Cayley was educated at Cambridge, where he took the ﬁrst Smith’s Prize in mathematics. A lawyer for much of his adult life, he wrote several papers in his early twenties before entering the legal profession at the age of 25. While practicing law he continued his mathematical research, writing more than 300 papers during this period of his life. These included some of his best work. In 1863 he left law to become a professor at Cambridge. Cayley wrote more than 900 papers in ﬁelds such as group theory, geometry, and linear algebra. His legal knowledge was very valuable to Cambridge; he participated in the writing of many of the university’s statutes. Cayley was also one of the people responsible for the admission of women to Cambridge. 9.2 Direct Products Given two groups G and H, it is possible to construct a new group from the Cartesian product of G and H, G × H. Conversely, given a large group, it is sometimes possible to
decompose the group; that is, a group is sometimes isomorphic to the direct product of two smaller groups. Rather than studying a large group G, it is often easier to study the component groups of G. External Direct Products If (G, ·) and (H, ◦) are groups, then we can make the Cartesian product of G and H into a new group. As a set, our group is just the ordered pairs (g, h) ∈ G × H where g ∈ G and h ∈ H. We can deﬁne a binary operation on G × H by (g1, h1)(g2, h2) = (g1 · g2, h1 ◦ h2); that is, we just multiply elements in the ﬁrst coordinate as we do in G and elements in the second coordinate as we do in H. We have speciﬁed the 150 CHAPTER 9 ISOMORPHISMS particular operations · and ◦ in each group here for the sake of clarity; we usually just write (g1, h1)(g2, h2) = (g1g2, h1h2). Proposition 9.7 Let G and H be groups. The set G × H is a group under the operation (g1, h1)(g2, h2) = (g1g2, h1h2) where g1, g2 ∈ G and h1, h2 ∈ H. Proof. Clearly the binary operation deﬁned above is closed. If eG and eH are the identities of the groups G and H respectively, then (eG, eH ) is the identity of G × H. The inverse of (g, h) ∈ G × H is (g−1, h−1). The fact that the operation is associative follows directly from the associativity of G and H. Example 7. Let R be the group of real numbers under addition. The Cartesian product of R with itself, R × R = R2, is also a group, in which the group operation is just addition in each coordinate; that is, (a, b) + (c, d) = (a + c, b + d). The identity is (0, 0) and the inverse of (a, b) is (−a, −b). Example 8. Consider Z2 × Z2 =
{(0, 0), (0, 1), (1, 0), (1, 1)}. Although Z2 × Z2 and Z4 both contain four elements, it is easy to see that they are not isomorphic since for every element (a, b) in Z2 × Z2, (a, b) + (a, b) = (0, 0), but Z4 is cyclic. The group G × H is called the external direct product of G and H. Notice that there is nothing special about the fact that we have used only two groups to build a new group. The direct product n i=1 Gi = G1 × G2 × · · · × Gn of the groups G1, G2,..., Gn is deﬁned in exactly the same manner. If G = G1 = G2 = · · · = Gn, we often write Gn instead of G1 × G2 × · · · × Gn. Example 9. The group Zn 2, considered as a set, is just the set of all binary n-tuples. The group operation is the “exclusive or” of two binary n-tuples. For example, (01011101) + (01001011) = (00010110). This group is important in coding theory, in cryptography, and in many areas of computer science. 9.2 DIRECT PRODUCTS 151 Theorem 9.8 Let (g, h) ∈ G × H. If g and h have ﬁnite orders r and s respectively, then the order of (g, h) in G × H is the least common multiple of r and s. Proof. Suppose that m is the least common multiple of r and s and let n = \|(g, h)\|. Then (g, h)m = (gm, hm) = (eG, eH ) (gn, hn) = (g, h)n = (eG, eH ). Hence, n must divide m, and n ≤ m. However, by the second equation, both r and s must divide n; therefore, n is a common multiple of r and s. Since m is the least common multiple of r and s, m ≤ n. Consequently, m must be equal to n. Corollary 9.9 Let (g1,..., gn) ∈ Gi. If gi has �
��nite order ri in Gi, then the order of (g1,..., gn) in Gi is the least common multiple of r1,..., rn. Example 10. Let (8, 56) ∈ Z12 × Z60. Since gcd(8, 12) = 4, the order of 8 is 12/4 = 3 in Z12. Similarly, the order of 56 in Z60 is 15. The least common multiple of 3 and 15 is 15; hence, (8, 56) has order 15 in Z12 × Z60. Example 11. The group Z2 × Z3 consists of the pairs (0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (1, 2). ∼= Z6. In this case, unlike that of Z2 × Z2 and Z4, it is true that Z2 × Z3 We need only show that Z2 × Z3 is cyclic. It is easy to see that (1, 1) is a generator for Z2 × Z3. The next theorem tells us exactly when the direct product of two cyclic groups is cyclic. Theorem 9.10 The group Zm × Zn is isomorphic to Zmn if and only if gcd(m, n) = 1. ∼= Zmn, then gcd(m, n) = 1. To Proof. Assume ﬁrst that if Zm × Zn show this, we will prove the contrapositive; that is, we will show that if gcd(m, n) = d > 1, then Zm × Zn cannot be cyclic. Notice that mn/d is divisible by both m and n; hence, for any element (a, b) ∈ Zm × Zn, (a, b) + (a, b) + · · · + (a, b) mn/d times = (0, 0). 152 CHAPTER 9 ISOMORPHISMS Therefore, no (a, b) can generate all of Zm × Zn. The converse follows directly from Theorem 9.8 since lcm(m, n) = mn if and only if gcd(m, n) = 1. Corollary 9.11 Let n1,..., nk be positive integers. Then k i=
1 Zni ∼= Zn1···nk if and only if gcd(ni, nj) = 1 for i = j. Corollary 9.12 If m = pe1 1 · · · pek k, where the pis are distinct primes, then Zm ∼= Z pe1 1 × · · · × Z p. ek k Proof. Since the greatest common divisor of pei proof follows from Corollary 9.11. i and pej j is 1 for i = j, the In Chapter 13, we will prove that all ﬁnite abelian groups are isomorphic to direct products of the form Z pe1 1 × · · · × Z p ek k where p1,..., pk are (not necessarily distinct) primes. Internal Direct Products The external direct product of two groups builds a large group out of two smaller groups. We would like to be able to reverse this process and conveniently break down a group into its direct product components; that is, we would like to be able to say when a group is isomorphic to the direct product of two of its subgroups. Let G be a group with subgroups H and K satisfying the following conditions. • G = HK = {hk : h ∈ H, k ∈ K}; • H ∩ K = {e}; • hk = kh for all k ∈ K and h ∈ H. 9.2 DIRECT PRODUCTS Then G is the internal direct product of H and K. Example 12. The group U (8) is the internal direct product of H = {1, 3} and K = {1, 5}. 153 Example 13. The dihedral group D6 is an internal direct product of its two subgroups H = {id, r3} and K = {id, r2, r4, s, r2s, r4s}. It can easily be shown that K ∼= S3; consequently, D6 ∼= Z2 × S3. Example 14. Not every group can be written as the internal direct product of two of its proper subgroups. If the group S3 were an internal direct product of its proper subgroups H and K, then one of the subgroups, say H, would have to have order 3. In this case H is the subgroup {(1), (123), (132)}. The subgroup K must have order
2, but no matter which subgroup we choose for K, the condition that hk = kh will never be satisﬁed for h ∈ H and k ∈ K. Theorem 9.13 Let G be the internal direct product of subgroups H and K. Then G is isomorphic to H × K. Proof. Since G is an internal direct product, we can write any element g ∈ G as g = hk for some h ∈ H and some k ∈ K. Deﬁne a map φ : G → H × K by φ(g) = (h, k). The ﬁrst problem that we must face is to show that φ is a well-deﬁned map; that is, we must show that h and k are uniquely determined by g. Suppose that g = hk = hk. Then h−1h = k(k)−1 is in both H and K, so it must be the identity. Therefore, h = h and k = k, which proves that φ is, indeed, well-deﬁned. To show that φ preserves the group operation, let g1 = h1k1 and g2 = h2k2 and observe that φ(g1g2) = φ(h1k1h2k2) = φ(h1h2k1k2) = (h1h2, k1k2) = (h1, k1)(h2, k2) = φ(g1)φ(g2). 154 CHAPTER 9 ISOMORPHISMS We will leave the proof that φ is one-to-one and onto as an exercise. Example 15. The group Z6 is an internal direct product isomorphic to {0, 2, 4} × {0, 3}. We can extend the deﬁnition of an internal direct product of G to a collection of subgroups H1, H2,..., Hn of G, by requiring that • G = H1H2 · · · Hn = {h1h2 · · · hn : hi ∈ Hi}; • Hi ∩ ∪j=iHj = {e}; • hihj = hjhi for all hi ∈ Hi and hj ∈ Hj. We will leave the
proof of the following theorem as an exercise. Theorem 9.14 Let G be the internal direct product of subgroups Hi, where i = 1, 2,..., n. Then G is isomorphic to i Hi. Exercises 1. Prove that Z ∼= nZ for n = 0. 2. Prove that C∗ is isomorphic to the subgroup of GL2(R) consisting of matrices a of the form b −b a 3. Prove or disprove: U (8) ∼= Z4. 4. Prove that U (8) is isomorphic to the group of matrices 1 −1 1 0 0 1, 0 0 −1 −1 0, 0 1, 0 0 −1. 5. Show that U (5) is isomorphic to U (10), but U (12) is not. 6. Show that the nth roots of unity are isomorphic to Zn. 7. Show that any cyclic group of order n is isomorphic to Zn. 8. Prove that Q is not isomorphic to Z. 9. Let G = R \ {−1} and deﬁne a binary operation on G by a ∗ b = a + b + ab. Prove that G is a group under this operation. Show that (G, ∗) is isomorphic to the multiplicative group of nonzero real numbers. EXERCISES 155 10. Show that the matrices   form a group. Find an isomorphism of G with a more familiar group of order 6. 11. Find ﬁve non-isomorphic groups of order 8. 12. Prove S4 is not isomorphic to D12. 13. Let ω = cis(2π/n) be a primitive nth root of unity. Prove that the matrices A = ω 0 0 ω−1 and B = 0 1 1 0 generate a multiplicative group isomorphic to Dn. 14. Show that the set of all matrices of the form ±1 n 1 0 B =, where n ∈ Zn, is a group isomorphic to Dn. 15. List all of the elements of Z4 × Z2. 16. Find the order of each of the following elements. (a) (3, 4) in Z4 × Z6 (b) (6, 15, 4) in Z30
× Z45 × Z24 (c) (5, 10, 15) in Z25 × Z25 × Z25 (d) (8, 8, 8) in Z10 × Z24 × Z80 17. Prove that D4 cannot be the internal direct product of two of its proper subgroups. 18. Prove that the subgroup of Q∗ consisting of elements of the form 2m3n for m, n ∈ Z is an internal direct product isomorphic to Z × Z. 19. Prove that S3 × Z2 is isomorphic to D6. Can you make a conjecture about D2n? Prove your conjecture. [Hint: Draw the picture.] 20. Prove or disprove: Every abelian group of order divisible by 3 contains a subgroup of order 3. 156 CHAPTER 9 ISOMORPHISMS 21. Prove or disprove: Every nonabelian group of order divisible by 6 contains a subgroup of order 6. 22. Let G be a group of order 20. If G has subgroups H and K of orders 4 and 5 respectively such that hk = kh for all h ∈ H and k ∈ K, prove that G is the internal direct product of H and K. 23. Prove or disprove the following assertion. Let G, H, and K be groups. If G × K ∼= H × K, then G ∼= H. 24. Prove or disprove: There is a noncyclic abelian group of order 51. 25. Prove or disprove: There is a noncyclic abelian group of order 52. 26. Let φ : G1 → G2 be a group isomorphism. Show that φ(x) = e if and only if x = e. 27. Let G ∼= H. Show that if G is cyclic, then so is H. 28. Prove that any group G of order p, p prime, must be isomorphic to Zp. 29. Show that Sn is isomorphic to a subgroup of An+2. 30. Prove that Dn is isomorphic to a subgroup of Sn. 31. Let φ : G1 → G2 and ψ : G2 → G3 be isomorphisms. Show that φ−1 and ψ ◦ φ are both isomorphisms. Using these results, show that the
isomorphism of groups determines an equivalence relation on the class of all groups. 32. Prove U (5) ∼= Z4. Can you generalize this result to show that U (p) ∼= Zp−1? 33. Write out the permutations associated with each element of S3 in the proof of Cayley’s Theorem. 34. An automorphism of a group G is an isomorphism with itself. Prove that complex conjugation is an automorphism of the additive group of complex numbers; that is, show that the map φ(a + bi) = a − bi is an isomorphism from C to C. 35. Prove that a + ib → a − ib is an automorphism of C∗. 36. Prove that A → B−1AB is an automorphism of SL2(R) for all B in GL2(R). 37. We will denote the set of all automorphisms of G by Aut(G). Prove that Aut(G) is a subgroup of SG, the group of permutations of G. 38. Find Aut(Z6). 39. Find Aut(Z). 40. Find two nonisomorphic groups G and H such that Aut(G) ∼= Aut(H). 41. Let G be a group and g ∈ G. Deﬁne a map ig : G → G by ig(x) = gxg−1. Prove that ig deﬁnes an automorphism of G. Such an automorphism is called an inner automorphism. The set of all inner automorphisms is denoted by Inn(G). EXERCISES 157 42. Prove that Inn(G) is a subgroup of Aut(G). 43. What are the inner automorphisms of the quaternion group Q8? Is Inn(G) = Aut(G) in this case? 44. Let G be a group and g ∈ G. Deﬁne maps λg : G → G and ρg : G → G by λg(x) = gx and ρg(x) = xg−1. Show that ig = ρg ◦ λg is an automorphism of G. The isomorphism g → ρg is called the right regular representation of G. 45. Let G be
the internal direct product of subgroups H and K. Show that the map φ : G → H × K deﬁned by φ(g) = (h, k) for g = hk, where h ∈ H and k ∈ K, is one-to-one and onto. 46. Let G and H be isomorphic groups. If G has a subgroup of order n, prove that H must also have a subgroup of order n. 47. If G ∼= G and H ∼= H, show that G × H ∼= G × H. 48. Prove that G × H is isomorphic to H × G. 49. Let n1,..., nk be positive integers. Show that k i=1 Zni ∼= Zn1···nk if and only if gcd(ni, nj) = 1 for i = j. 50. Prove that A × B is abelian if and only if A and B are abelian. 51. If G is the internal direct product of H1, H2,..., Hn, prove that G is isomor- phic to i Hi. 52. Let H1 and H2 be subgroups of G1 and G2, respectively. Prove that H1 × H2 is a subgroup of G1 × G2. 53. Let m, n ∈ Z. Prove that m, n = d if and only if d = gcd(m, n). 54. Let m, n ∈ Z. Prove that m ∩ n = l if and only if l = lcm(m, n). 55. Groups of order 2p. In this series of exercises we will classify all groups of order 2p, where p is an odd prime. (a) Assume G is a group of order 2p, where p is an odd prime. If a ∈ G, show that A must have order 1, 2, p, or 2p. (b) Suppose that G an element of order 2p. Prove that G isomorphic to Z2p. Hence, G is cyclic. (c) Suppose that G does not contain an element of order 2p. Show that G must contain an element of order p. Hint: Assume that G does not contain an element of order p. 158 CHAPTER 9 ISOMORPHIS
MS (d) Suppose that G does not contain an element of order 2p. Show that G must contain an element of order 2. (e) Let P be a subgroup of G with order p and y ∈ G have order 2. Show that yP = P y. (f) Suppose that G does not contain an element of order 2p and P = z is a subgroup of order p generated by z. If y is an element of order 2, then yz = zky for some 2 ≤ k < p. (g) Suppose that G does not contain an element of order 2p. Prove that G is not abelian. (h) Suppose that G does not contain an element of order 2p and P = z is a subgroup of order p generated by z and y is an element of order 2. Show that we can list the elements of G as {ziyj \| 0 ≤ i < p, 0 ≤ j < 2}. (i) Suppose that G does not contain an element of order 2p and P = z is a subgroup of order p generated by z and y is an element of order 2. Prove that the product (ziyj)(zrys) can be expressed as a uniquely as zmyn for some non negative integers m, n. Thus, conclude that there is only one possibility for a non-abelian group of order 2p, it must therefore be the one we have seen already, the dihedral group. Sage Sage can quickly determine if two permutation groups are isomorphic, even though this should, in theory, be a very diﬃcult computation. 10 Normal Subgroups and Factor Groups If H is a subgroup of a group G, then right cosets are not always the same as left cosets; that is, it is not always the case that gH = Hg for all g ∈ G. The subgroups for which this property holds play a critical role in group theory: they allow for the construction of a new class of groups, called factor or quotient groups. Factor groups may be studied by using homomorphisms, a generalization of isomorphisms. 10.1 Factor Groups and Normal Subgroups Normal Subgroups A subgroup H of a group G is normal in G if gH = Hg for all g ∈ G. That is, a normal subgroup of a group G is one in which the right and left cosets are precisely the
same. Example 1. Let G be an abelian group. Every subgroup H of G is a normal subgroup. Since gh = hg for all g ∈ G and h ∈ H, it will always be the case that gH = Hg. Example 2. Let H be the subgroup of S3 consisting of elements (1) and (12). Since (123)H = {(123), (13)} and H(123) = {(123), (23)}, H cannot be a normal subgroup of S3. However, the subgroup N, consisting of the permutations (1), (123), and (132), is normal since the cosets of N are N = {(1), (123), (132)} (12)N = N (12) = {(12), (13), (23)}. 159 160 CHAPTER 10 NORMAL SUBGROUPS AND FACTOR GROUPS The following theorem is fundamental to our understanding of normal subgroups. Theorem 10.1 Let G be a group and N be a subgroup of G. Then the following statements are equivalent. 1. The subgroup N is normal in G. 2. For all g ∈ G, gN g−1 ⊂ N. 3. For all g ∈ G, gN g−1 = N. Proof. (1) ⇒ (2). Since N is normal in G, gN = N g for all g ∈ G. Hence, for a given g ∈ G and n ∈ N, there exists an n in N such that gn = ng. Therefore, gng−1 = n ∈ N or gN g−1 ⊂ N. (2) ⇒ (3). Let g ∈ G. Since gN g−1 ⊂ N, we need only show N ⊂ gN g−1. For n ∈ N, g−1ng = g−1n(g−1)−1 ∈ N. Hence, g−1ng = n for some n ∈ N. Therefore, n = gng−1 is in gN g−1. (3) ⇒ (1). Suppose that gN g−1 = N for all g ∈ G. Then for any n ∈ N there exists an n ∈ N such that gng−1 = n. Consequently, gn = ng or
gN ⊂ N g. Similarly, N g ⊂ gN. Factor Groups If N is a normal subgroup of a group G, then the cosets of N in G form a group G/N under the operation (aN )(bN ) = abN. This group is called the factor or quotient group of G and N. Our ﬁrst task is to prove that G/N is indeed a group. Theorem 10.2 Let N be a normal subgroup of a group G. The cosets of N in G form a group G/N of order [G : N ]. Proof. The group operation on G/N is (aN )(bN ) = abN. This operation must be shown to be well-deﬁned; that is, group multiplication must be independent of the choice of coset representative. Let aN = bN and cN = dN. We must show that (aN )(cN ) = acN = bdN = (bN )(dN ). 10.1 FACTOR GROUPS AND NORMAL SUBGROUPS 161 Then a = bn1 and c = dn2 for some n1 and n2 in N. Hence, acN = bn1dn2N = bn1dN = bn1N d = bN d = bdN. The remainder of the theorem is easy: eN = N is the identity and g−1N is the inverse of gN. The order of G/N is, of course, the number of cosets of N in G. It is very important to remember that the elements in a factor group are sets of elements in the original group. Example 3. Consider the normal subgroup of S3, N = {(1), (123), (132)}. The cosets of N in S3 are N and (12)N. The factor group S3/N has the following multiplication table. N N N (12)N (12)N (12)N (12)N N This group is isomorphic to Z2. At ﬁrst, multiplying cosets seems both complicated and strange; however, notice that S3/N is a smaller group. The factor group displays a certain amount of information about S3. Actually, N = A3, the group of even permutations, and (12)N
= {(12), (13), (23)} is the set of odd permutations. The information captured in G/N is parity; that is, multiplying two even or two odd permutations results in an even permutation, whereas multiplying an odd permutation by an even permutation yields an odd permutation. Example 4. Consider the normal subgroup 3Z of Z. The cosets of 3Z in Z are 0 + 3Z = {..., −3, 0, 3, 6,...} 1 + 3Z = {..., −2, 1, 4, 7,...} 2 + 3Z = {..., −1, 2, 5, 8,...}. The group Z/3Z is given by the multiplication table below. 162 CHAPTER 10 NORMAL SUBGROUPS AND FACTOR GROUPS + 0 + 3Z 1 + 3Z 2 + 3Z 0 + 3Z 0 + 3Z 1 + 3Z 2 + 3Z 1 + 3Z 1 + 3Z 2 + 3Z 0 + 3Z 2 + 3Z 2 + 3Z 0 + 3Z 1 + 3Z In general, the subgroup nZ of Z is normal. The cosets of Z/nZ are nZ 1 + nZ 2 + nZ... (n − 1) + nZ. The sum of the cosets k + Z and l + Z is k + l + Z. Notice that we have written our cosets additively, because the group operation is integer addition. Example 5. Consider the dihedral group Dn, generated by the two elements r and s, satisfying the relations rn = id s2 = id srs = r−1. The element r actually generates the cyclic subgroup of rotations, Rn, of Dn. Since srs−1 = srs = r−1 ∈ Rn, the group of rotations is a normal subgroup of Dn; therefore, Dn/Rn is a group. Since there are exactly two elements in this group, it must be isomorphic to Z2. 10.2 The Simplicity of the Alternating Group Of special interest are groups with no nontrivial normal subgroups. Such groups are called simple groups. Of course, we already have a whole class of examples of simple groups, Zp, where p is prime. These groups are trivially simple since they
have no proper subgroups other than the subgroup consisting solely of the identity. Other examples of simple groups are not so easily found. We can, however, show that the alternating group, An, is simple for n ≥ 5. The proof of this result requires several lemmas. Lemma 10.3 The alternating group An is generated by 3-cycles for n ≥ 3. 10.2 THE SIMPLICITY OF THE ALTERNATING GROUP 163 Proof. To show that the 3-cycles generate An, we need only show that any pair of transpositions can be written as the product of 3-cycles. Since (ab) = (ba), every pair of transpositions must be one of the following: (ab)(ab) = id (ab)(cd) = (acb)(acd) (ab)(ac) = (acb). Lemma 10.4 Let N be a normal subgroup of An, where n ≥ 3. contains a 3-cycle, then N = An. If N Proof. We will ﬁrst show that An is generated by 3-cycles of the speciﬁc form (ijk), where i and j are ﬁxed in {1, 2,..., n} and we let k vary. Every 3-cycle is the product of 3-cycles of this form, since (iaj) = (ija)2 (iab) = (ijb)(ija)2 (jab) = (ijb)2(ija) (abc) = (ija)2(ijc)(ijb)2(ija). Now suppose that N is a nontrivial normal subgroup of An for n ≥ 3 such that N contains a 3-cycle of the form (ija). Using the normality of N, we see that [(ij)(ak)](ija)2[(ij)(ak)]−1 = (ijk) is in N. Hence, N must contain all of the 3-cycles (ijk) for 1 ≤ k ≤ n. By Lemma 10.3, these 3-cycles generate An; hence, N = An. Lemma 10.5 For n ≥ 5, every normal subgroup N of An contains a 3-cycle. Proof. Let σ be an arbitrary element in a normal subgroup N. There are several possible cycle structures for σ. • σ is a 3-cycle. • σ is the product of disj
oint cycles, σ = τ (a1a2 · · · ar) ∈ N, where r > 3. • σ is the product of disjoint cycles, σ = τ (a1a2a3)(a4a5a6). • σ = τ (a1a2a3), where τ is the product of disjoint 2-cycles. 164 CHAPTER 10 NORMAL SUBGROUPS AND FACTOR GROUPS • σ = τ (a1a2)(a3a4), where τ is the product of an even number of disjoint 2-cycles. If σ is a 3-cycle, then we are done. If N contains a product of disjoint cycles, σ, and at least one of these cycles has length greater than 3, say σ = τ (a1a2 · · · ar), then (a1a2a3)σ(a1a2a3)−1 is in N since N is normal; hence, σ−1(a1a2a3)σ(a1a2a3)−1 is also in N. Since σ−1(a1a2a3)σ(a1a2a3)−1 = σ−1(a1a2a3)σ(a1a3a2) = (a1a2 · · · ar)−1τ −1(a1a2a3)τ (a1a2 · · · ar)(a1a3a2) = (a1arar−1 · · · a2)(a1a2a3)(a1a2 · · · ar)(a1a3a2) = (a1a3ar), N must contain a 3-cycle; hence, N = An. Now suppose that N contains a disjoint product of the form σ = τ (a1a2a3)(a4a5a6). σ−1(a1a2a4)σ(a1a2a4)−1 ∈ N (a1a2a4)σ(a1a2a4)−1 ∈ N. Then since So σ−1(a1a2a4)σ(a1a2a4)−1 = [τ (a1a2a3
)(a4a5a6)]−1(a1a2a4)τ (a1a2a3)(a4a5a6)(a1a2a4)−1 = (a4a6a5)(a1a3a2)τ −1(a1a2a4)τ (a1a2a3)(a4a5a6)(a1a4a2) = (a4a6a5)(a1a3a2)(a1a2a4)(a1a2a3)(a4a5a6)(a1a4a2) = (a1a4a2a6a3). So N contains a disjoint cycle of length greater than 3, and we can apply the previous case. 10.2 THE SIMPLICITY OF THE ALTERNATING GROUP 165 Suppose N contains a disjoint product of the form σ = τ (a1a2a3), where τ is the product of disjoint 2-cycles. Since σ ∈ N, σ2 ∈ N, and σ2 = τ (a1a2a3)τ (a1a2a3) = (a1a3a2). So N contains a 3-cycle. The only remaining possible case is a disjoint product of the form σ = τ (a1a2)(a3a4), where τ is the product of an even number of disjoint 2-cycles. But σ−1(a1a2a3)σ(a1a2a3)−1 is in N since (a1a2a3)σ(a1a2a3)−1 is in N ; and so σ−1(a1a2a3)σ(a1a2a3)−1 = τ −1(a1a2)(a3a4)(a1a2a3)τ (a1a2)(a3a4)(a1a2a3)−1 = (a1a3)(a2a4). Since n ≥ 5, we can ﬁnd b ∈ {1, 2,..., n} such that b = a1, a2, a3, a4. Let µ = (a1a3b). Then µ−1(a
1a3)(a2a4)µ(a1a3)(a2a4) ∈ N and µ−1(a1a3)(a2a4)µ(a1a3)(a2a4) = (a1ba3)(a1a3)(a2a4)(a1a3b)(a1a3)(a2a4) = (a1a3b). Therefore, N contains a 3-cycle. This completes the proof of the lemma. Theorem 10.6 The alternating group, An, is simple for n ≥ 5. Proof. Let N be a normal subgroup of An. By Lemma 10.5, N contains a 3-cycle. By Lemma 10.4, N = An; therefore, An contains no proper nontrivial normal subgroups for n ≥ 5. Historical Note 166 CHAPTER 10 NORMAL SUBGROUPS AND FACTOR GROUPS One of the foremost problems of group theory has been to classify all simple ﬁnite groups. This problem is over a century old and has been solved only in the last few years. In a sense, ﬁnite simple groups are the building blocks of all ﬁnite groups. The ﬁrst nonabelian simple groups to be discovered were the alternating groups. Galois was the ﬁrst to prove that A5 was simple. Later mathematicians, such as C. Jordan and L. E. Dickson, found several inﬁnite families of matrix groups that were simple. Other families of simple groups were discovered in the 1950s. At the turn of the century, William Burnside conjectured that all nonabelian simple groups must have even order. In 1963, W. Feit and J. Thompson proved Burnside’s conjecture and published their results in the paper “Solvability of Groups of Odd Order,” which appeared in the Paciﬁc Journal of Mathematics. Their proof, running over 250 pages, gave impetus to a program in the 1960s and 1970s to classify all ﬁnite simple groups. Daniel Gorenstein was the organizer of this remarkable eﬀort. One of the last simple groups was the “Monster,” discovered by R. Greiss. The Monster, a 196,833 × 196,833 matrix group, is one of the 26 sporadic
, or special, simple groups. These sporadic simple groups are groups that ﬁt into no inﬁnite family of simple groups. Exercises 1. For each of the following groups G, determine whether H is a normal subgroup of G. If H is a normal subgroup, write out a Cayley table for the factor group G/H. (a) G = S4 and H = A4 (b) G = A5 and H = {(1), (123), (132)} (c) G = S4 and H = D4 (d) G = Q8 and H = {1, −1, I, −I} (e) G = Z and H = 5Z 2. Find all the subgroups of D4. Which subgroups are normal? What are all the factor groups of D4 up to isomorphism? 3. Find all the subgroups of the quaternion group, Q8. Which subgroups are normal? What are all the factor groups of Q8 up to isomorphism? 4. Let T be the group of nonsingular upper triangular 2 × 2 matrices with entries in R; that is, matrices of the form a b c 0, EXERCISES 167 where a, b, c ∈ R and ac = 0. Let U consist of matrices of the form 1 x 1 0, where x ∈ R. (a) Show that U is a subgroup of T. (b) Prove that U is abelian. (c) Prove that U is normal in T. (d) Show that T /U is abelian. (e) Is T normal in GL2(R)? 5. Show that the intersection of two normal subgroups is a normal subgroup. 6. If G is abelian, prove that G/H must also be abelian. 7. Prove or disprove: If H is a normal subgroup of G such that H and G/H are abelian, then G is abelian. 8. If G is cyclic, prove that G/H must also be cyclic. 9. Prove or disprove: If H and G/H are cyclic, then G is cyclic. 10. Let H be a subgroup of index 2 of a group G. Prove that H must be a normal subgroup of G. Conclude that Sn is
not simple. 11. Let G be a group of order p2, where p is a prime number. If H is a subgroup of G of order p, show that H is normal in G. Prove that G must be abelian. 12. If a group G has exactly one subgroup H of order k, prove that H is normal in G. 13. Deﬁne the centralizer of an element g in a group G to be the set C(g) = {x ∈ G : xg = gx}. Show that C(g) is a subgroup of G. If g generates a normal subgroup of G, prove that C(g) is normal in G. 14. Recall that the center of a group G is the set Z(G) = {x ∈ G : xg = gx for all g ∈ G }. (a) Calculate the center of S3. (b) Calculate the center of GL2(R). (c) Show that the center of any group G is a normal subgroup of G. (d) If G/Z(G) is cyclic, show that G is abelian. 168 CHAPTER 10 NORMAL SUBGROUPS AND FACTOR GROUPS 15. Let G be a group and let G = aba−1b−1; that is, G is the subgroup of all ﬁnite products of elements in G of the form aba−1b−1. The subgroup G is called the commutator subgroup of G. (a) Show that G is a normal subgroup of G. (b) Let N be a normal subgroup of G. Prove that G/N is abelian if and only if N contains the commutator subgroup of G. Sage Sage can esily determine if a subgroup is normal or not. If so, it can create the quotient group. However, the construction creates a new permuation group, isomorphic to the quotient group, so its utility is limited. 11 Homomorphisms 11.1 Group Homomorphisms One of the basic ideas of algebra is the concept of a homomorphism, a natural generalization of an isomorphism. If we relax the requirement that an isomorphism of groups be bijective, we have a homomorphism. A homomorphism between groups (G, ·) and (H
, ◦) is a map φ : G → H such that φ(g1 · g2) = φ(g1) ◦ φ(g2) for g1, g2 ∈ G. The range of φ in H is called the homomorphic image of φ. Two groups are related in the strongest possible way if they are isomorphic; however, a weaker relationship may exist between two groups. For example, the symmetric group Sn and the group Z2 are related by the fact that Sn can be divided into even and odd permutations that exhibit a group structure like that Z2, as shown in the following multiplication table. even even even odd odd odd odd even We use homomorphisms to study relationships such as the one we have just described. Example 1. Let G be a group and g ∈ G. Deﬁne a map φ : Z → G by φ(n) = gn. Then φ is a group homomorphism, since φ(m + n) = gm+n = gmgn = φ(m)φ(n). This homomorphism maps Z onto the cyclic subgroup of G generated by g. 169 170 CHAPTER 11 HOMOMORPHISMS Example 2. Let G = GL2(R). If A = a b c d is in G, then the determinant is nonzero; that is, det(A) = ad − bc = 0. Also, for any two elements A and B in G, det(AB) = det(A) det(B). Using the determinant, we can deﬁne a homomorphism φ : GL2(R) → R∗ by A → det(A). Example 3. Recall that the circle group T consists of all complex numbers z such that \|z\| = 1. We can deﬁne a homomorphism φ from the additive group of real numbers R to T by φ : θ → cos θ + i sin θ. Indeed, φ(α + β) = cos(α + β) + i sin(α + β) = (cos α cos β − sin α sin β) + i(sin α cos β + cos α sin β) = (cos α + i sin α) + (cos β + i sin β) = φ(α)φ(β
). Geometrically, we are simply wrapping the real line around the circle in a group-theoretic fashion. The following proposition lists some basic properties of group homomor- phisms. Proposition 11.1 Let φ : G1 → G2 be a homomorphism of groups. Then 1. If e is the identity of G1, then φ(e) is the identity of G2; 2. For any element g ∈ G1, φ(g−1) = [φ(g)]−1; 3. If H1 is a subgroup of G1, then φ(H1) is a subgroup of G2; 4. If H2 is a subgroup of G2, then φ−1(H2) = {g ∈ G1 : φ(g) ∈ H2} is a subgroup of G1. Furthermore, if H2 is normal in G2, then φ−1(H2) is normal in G1. Proof. (1) Suppose that e and e are the identities of G1 and G2, respectively; then eφ(e) = φ(e) = φ(ee) = φ(e)φ(e). By cancellation, φ(e) = e. 11.1 GROUP HOMOMORPHISMS 171 (2) This statement follows from the fact that φ(g−1)φ(g) = φ(g−1g) = φ(e) = e. (3) The set φ(H1) is nonempty since the identity of G2 is in φ(H1). Suppose that H1 is a subgroup of G1 and let x and y be in φ(H1). There exist elements a, b ∈ H1 such that φ(a) = x and φ(b) = y. Since xy−1 = φ(a)[φ(b)]−1 = φ(ab−1) ∈ φ(H1), φ(H1) is a subgroup of G2 by Proposition 3.10. (4) Let H2 be a subgroup of G2 and deﬁne H1 to be φ−1(H2); that is, H1 is the set of all g ∈ G1
such that φ(g) ∈ H2. The identity is in H1 since φ(e) = e. If a and b are in H1, then φ(ab−1) = φ(a)[φ(b)]−1 is in H2 since H2 is a subgroup of G2. Therefore, ab−1 ∈ H1 and H1 is a subgroup of G1. If H2 is normal in G2, we must show that g−1hg ∈ H1 for h ∈ H1 and g ∈ G1. But φ(g−1hg) = [φ(g)]−1φ(h)φ(g) ∈ H2, since H2 is a normal subgroup of G2. Therefore, g−1hg ∈ H1. Let φ : G → H be a group homomorphism and suppose that e is the identity of H. By Proposition 11.1, φ−1({e}) is a subgroup of G. This subgroup is called the kernel of φ and will be denoted by ker φ. In fact, this subgroup is a normal subgroup of G since the trivial subgroup is normal in H. We state this result in the following theorem, which says that with every homomorphism of groups we can naturally associate a normal subgroup. Theorem 11.2 Let φ : G → H be a group homomorphism. Then the kernel of φ is a normal subgroup of G. Example 4. Let us examine the homomorphism φ : GL2(R) → R∗ deﬁned by A → det(A). Since 1 is the identity of R∗, the kernel of this homomorphism is all 2 × 2 matrices having determinant one. That is, ker φ = SL2(R). Example 5. The kernel of the group homomorphism φ : R → C∗ deﬁned by φ(θ) = cos θ + i sin θ is {2πn : n ∈ Z}. Notice that ker φ ∼= Z. Example 6. Suppose that we wish to determine all possible homomorphisms φ from Z7 to Z12. Since the kernel of φ must be a subgroup of Z7, there are 172 CHAPTER 11
HOMOMORPHISMS only two possible kernels, {0} and all of Z7. The image of a subgroup of Z7 must be a subgroup of Z12. Hence, there is no injective homomorphism; otherwise, Z12 would have a subgroup of order 7, which is impossible. Consequently, the only possible homomorphism from Z7 to Z12 is the one mapping all elements to zero. Example 7. Let G be a group. Suppose that g ∈ G and φ is the homomorphism from Z to G given by φ(n) = gn. If the order of g is inﬁnite, then the kernel of this homomorphism is {0} since φ maps Z onto the cyclic subgroup of G generated by g. However, if the order of g is ﬁnite, say n, then the kernel of φ is nZ. 11.2 The Isomorphism Theorems Though at ﬁrst it is not evident that factor groups correspond exactly to homomorphic images, we can use factor groups to study homomorphisms. We already know that with every group homomorphism φ : G → H we can associate a normal subgroup of G, ker φ; the converse is also true. Every normal subgroup of a group G gives rise to homomorphism of groups. Let H be a normal subgroup of G. Deﬁne the natural or canonical homomorphism by φ : G → G/H φ(g) = gH. This is indeed a homomorphism, since φ(g1g2) = g1g2H = g1Hg2H = φ(g1)φ(g2). The kernel of this homomorphism is H. The following theorems describe the relationships among group homomorphisms, normal subgroups, and factor groups. Theorem 11.3 (First Isomorphism Theorem) If ψ : G → H is a group homomorphism with K = ker ψ, then K is normal in G. Let φ : G → G/K be the canonical homomorphism. Then there exists a unique isomorphism η : G/K → ψ(G) such that ψ = ηφ. 11.2 THE ISOMORPHISM THEOREMS 173 Proof. We already
know that K is normal in G. Deﬁne η : G/K → ψ(G) by η(gK) = ψ(g). We must ﬁrst show that this is a well-deﬁned map. Suppose that g1K = g2K. For some k ∈ K, g1k = g2; consequently, η(g1K) = ψ(g1) = ψ(g1)ψ(k) = ψ(g1k) = ψ(g2) = η(g2K). Since η(g1K) = η(g2K), η does not depend on the choice of coset representative. Clearly η is onto ψ(G). To show that η is one-to-one, suppose that η(g1K) = η(g2K). Then ψ(g1) = ψ(g2). This implies that ψ(g−1 1 g2) = e, or g−1 1 g2 is in the kernel of ψ; hence, g−1 1 g2K = K; that is, g1K = g2K. Finally, we must show that η is a homomorphism, but η(g1Kg2K) = η(g1g2K) = ψ(g1g2) = ψ(g1)ψ(g2) = η(g1K)η(g2K). Mathematicians often use diagrams called commutative diagrams to describe such theorems. The following diagram “commutes” since ψ = ηφ. ψ G H φ η G/K Example 8. Let G be a cyclic group with generator g. Deﬁne a map φ : Z → G by n → gn. This map is a surjective homomorphism since φ(m + n) = gm+n = gmgn = φ(m)φ(n). If \|g\| = m, then gm = e. Hence, ker φ = mZ and Clearly φ is onto. Z/ ker φ = Z/mZ ∼= G. On the other hand, if the order of g
is inﬁnite, then ker φ = 0 and φ is an isomorphism of G and Z. Hence, two cyclic groups are isomorphic exactly when they have the same order. Up to isomorphism, the only cyclic groups are Z and Zn. 174 CHAPTER 11 HOMOMORPHISMS Theorem 11.4 (Second Isomorphism Theorem) Let H be a subgroup of a group G (not necessarily normal in G) and N a normal subgroup of G. Then HN is a subgroup of G, H ∩ N is a normal subgroup of H, and H/H ∩ N ∼= HN/N. Proof. We will ﬁrst show that HN = {hn : h ∈ H, n ∈ N } is a subgroup of G. Suppose that h1n1, h2n2 ∈ HN. Since N is normal, (h2)−1n1h2 ∈ N. So (h1n1)(h2n2) = h1h2((h2)−1n1h2)n2 is in HN. The inverse of hn ∈ HN is in HN since (hn)−1 = n−1h−1 = h−1(hn−1h−1). Next, we prove that H ∩ N is normal in H. Let h ∈ H and n ∈ H ∩ N. Then h−1nh ∈ H since each element is in H. Also, h−1nh ∈ N since N is normal in G; therefore, h−1nh ∈ H ∩ N. Now deﬁne a map φ from H to HN/N by h → hN. The map φ is onto, since any coset hnN = hN is the image of h in H. We also know that φ is a homomorphism because φ(hh) = hhN = hN hN = φ(h)φ(h). By the First Isomorphism Theorem, the image of φ is isomorphic to H/ ker φ; that is, HN/N = φ(H) ∼= H/ ker φ. Since ker φ = {, HN/N = φ(H)
∼= H/H ∩ N. Theorem 11.5 (Correspondence Theorem) Let N be a normal subgroup of a group G. Then H → H/N is a one-to-one correspondence between the set of subgroups H containing N and the set of subgroups of G/N. Furthermore, the normal subgroups of H correspond to normal subgroups of G/N. Proof. Let H be a subgroup of G containing N. Since N is normal in H, H/N makes sense. Let aN and bN be elements of H/N. Then (aN )(b−1N ) = ab−1N ∈ H/N ; hence, H/N is a subgroup of G/N. EXERCISES 175 Let S be a subgroup of G/N. This subgroup is a set of cosets of N. If H = {g ∈ G : gN ∈ S}, then for h1, h2 ∈ H, we have that (h1N )(h2N ) = hhN ∈ S and h−1 1 N ∈ S. Therefore, H must be a subgroup of G. Clearly, H contains N. Therefore, S = H/N. Consequently, the map H → N/H is onto. Suppose that H1 and H2 are subgroups of G containing N such that H1/N = H2/N. If h1 ∈ H1, then h1N ∈ H1/N. Hence, h1N = h2N ⊂ H2 for some h2 in H2. However, since N is contained in H2, we know that h1 ∈ H2 or H1 ⊂ H2. Similarly, H2 ⊂ H1. Since H1 = H2, the map H → N/H is one-to-one. Suppose that H is normal in G and N is a subgroup of H. Then it is easy to verify that the map G/N → G/H deﬁned by gN → gH is a homomorphism. The kernel of this homomorphism is H/N, which proves that H/N is normal in G/N. Conversely, suppose that H/N is normal in G/N. The homomorphism given by G → G/N → G/N H
/N has kernel H. Hence, H must be normal in G. Notice that in the course of the proof of Theorem 11.5, we have also proved the following theorem. Theorem 11.6 (Third Isomorphism Theorem) Let G be a group and N and H be normal subgroups of G with N ⊂ H. Then G/H ∼= G/N H/N. Example 9. By the Third Isomorphism Theorem, Z/mZ ∼= (Z/mnZ)/(mZ/mnZ). Since \|Z/mnZ\| = mn and \|Z/mZ\| = m, we have \|mZ/mnZ\| = n. Exercises 1. Prove that det(AB) = det(A) det(B) for A, B ∈ GL2(R). This shows that the determinant is a homomorphism from GL2(R) to R∗. 176 CHAPTER 11 HOMOMORPHISMS 2. Which of the following maps are homomorphisms? If the map is a homomor- phism, what is the kernel? (a) φ : R∗ → GL2(R) deﬁned by (b) φ : R → GL2(R) deﬁned by φ(a) = φ(ac) φ : GL2(R) → R deﬁned by d) φ : GL2(R) → R∗ deﬁned by a b d c φ = ad − bc (e) φ : M2(R) → R deﬁned by a b d c φ = b, where M2(R) is the additive group of 2 × 2 matrices with entries in R. 3. Let A be an m × n matrix. Show that matrix multiplication, x → Ax, deﬁnes a homomorphism φ : Rn → Rm. 4. Let φ : Z → Z be given by φ(n) = 7n. Prove that φ is a group homomorphism. Find the kernel and the image of φ. 5. Describe all of the homomorphisms from Z24 to Z18. 6. Describe all of the homomorphisms from Z to Z12. 7. In the group Z24,
let H = 4 and N = 6. (a) List the elements in HN (we usually write H + N for these additive groups) and H ∩ N. (b) List the cosets in HN/N, showing the elements in each coset. (c) List the cosets in H/(H ∩ N ), showing the elements in each coset. (d) Give the correspondence between HN/N and H/(H ∩ N ) described in the proof of the Second Isomorphism Theorem. EXERCISES 177 8. If G is an abelian group and n ∈ N, show that φ : G → G deﬁned by g → gn is a group homomorphism. 9. If φ : G → H is a group homomorphism and G is abelian, prove that φ(G) is also abelian. 10. If φ : G → H is a group homomorphism and G is cyclic, prove that φ(G) is also cyclic. 11. Show that a homomorphism deﬁned on a cyclic group is completely determined by its action on the generator of the group. 12. Let G be a group of order p2, where p is a prime number. If H is a subgroup of G of order p, show that H is normal in G. Prove that G must be abelian. 13. If a group G has exactly one subgroup H of order k, prove that H is normal in G. 14. Prove or disprove: Q/Z ∼= Q. 15. Let G be a ﬁnite group and N a normal subgroup of G. If H is a subgroup of G/N, prove that φ−1(H) is a subgroup in G of order \|H\| · \|N \|, where φ : G → G/N is the canonical homomorphism. 16. Let G1 and G2 be groups, and let H1 and H2 be normal subgroups of G1 and G2 respectively. Let φ : G1 → G2 be a homomorphism. Show that φ induces a natural homomorphism φ : (G1/H1) → (G2/H2) if φ(H1) ⊆ H2. 17. If H and K are normal
subgroups of G and H ∩ K = {e}, prove that G is isomorphic to a subgroup of G/H × G/K. 18. Let φ : G1 → G2 be a surjective group homomorphism. Let H1 be a normal subgroup of G1 and suppose that φ(H1) = H2. Prove or disprove that G1/H1 ∼= G2/H2. 19. Let φ : G → H be a group homomorphism. Show that φ is one-to-one if and only if φ−1(e) = {e}. 20. Given a homomorphism φ : G → H deﬁne a relation ∼ on G by a ∼ b if φ(a) = φ(b) for a, b ∈ G. Show this relation is an equivalence relation and describe the equivalence classes. Additional Exercises: Automorphisms 1. Let Aut(G) be the set of all automorphisms of G; that is, isomorphisms from G to itself. Prove this set forms a group and is a subgroup of the group of permutations of G; that is, Aut(G) ≤ SG. 2. An inner automorphism of G, ig : G → G, 178 CHAPTER 11 HOMOMORPHISMS is deﬁned by the map ig(x) = gxg−1, for g ∈ G. Show that ig ∈ Aut(G). 3. The set of all inner automorphisms is denoted by Inn(G). Show that Inn(G) is a subgroup of Aut(G). 4. Find an automorphism of a group G that is not an inner automorphism. 5. Let G be a group and ig be an inner automorphism of G, and deﬁne a map by G → Aut(G) g → ig. Prove that this map is a homomorphism with image Inn(G) and kernel Z(G). Use this result to conclude that G/Z(G) ∼= Inn(G). 6. Compute Aut(S3) and Inn(S3). Do the same thing for D4. 7. Find all of the homomorphisms φ : Z → Z. What is Aut(Z)? 8. Find all of the autom
orphisms of Z8. Prove that Aut(Z8) ∼= U (8). 9. For k ∈ Zn, deﬁne a map φk : Zn → Zn by a → ka. Prove that φk is a homomorphism. 10. Prove that φk is an isomorphism if and only if k is a generator of Zn. 11. Show that every automorphism of Zn is of the form φk, where k is a generator of Zn. 12. Prove that ψ : U (n) → Aut(Zn) is an isomorphism, where ψ : k → φk. Sage Sage can create homomorphisms between groups, which can be used directly as functions, and then queried for their kernels and images. So there is great potential for exploring the many fundamental relationships between groups, normal subgroups, quotient groups and properties of homomorphisms. 12 Matrix Groups and Symmetry When Felix Klein (1849–1925) accepted a chair at the University of Erlangen, he outlined in his inaugural address a program to classify diﬀerent geometries. Central to Klein’s program was the theory of groups: he considered geometry to be the study of properties that are left invariant under transformation groups. Groups, especially matrix groups, have now become important in the study of symmetry and have found applications in such disciplines as chemistry and physics. In the ﬁrst part of this chapter, we will examine some of the classical matrix groups, such as the general linear group, the special linear group, and the orthogonal group. We will then use these matrix groups to investigate some of the ideas behind geometric symmetry. 12.1 Matrix Groups Some Facts from Linear Algebra Before we study matrix groups, we must recall some basic facts from linear algebra. One of the most fundamental ideas of linear algebra is that of a linear transformation. A linear transformation or linear map T : Rn → Rm is a map that preserves vector addition and scalar multiplication; that is, for vectors x and y in Rn and a scalar α ∈ R, T (x + y) = T (x) + T (y) T (αy) = αT (y). An m × n matrix with entries in R represents a linear transformation from Rn to Rm. If we write vectors x = (
x1,..., xn)t and y = (y1,..., yn)t in Rn 179 180 CHAPTER 12 MATRIX GROUPS AND SYMMETRY as column matrices, then an m × n matrix A =      a11 a12 a21 a22...... am1 am2 · · · · · ·... · · ·      a1n a2n... amn maps the vectors to Rm linearly by matrix multiplication. Observe that if α is a real number, A(x + y) = Ax + Ay and αAx = A(αx), where x =     .      x1 x2... xn We will often abbreviate the matrix A by writing (aij). Conversely, if T : Rn → Rm is a linear map, we can associate a matrix A with T by considering what T does to the vectors e1 = (1, 0,..., 0)t e2 = (0, 1,..., 0)t... en = (0, 0,..., 1)t. We can write any vector x = (x1,..., xn)t as x1e1 + x2e2 + · · · + xnen. Consequently, if T (e1) = (a11, a21,..., am1)t, T (e2) = (a12, a22,..., am2)t,... T (en) = (a1n, a2n,..., amn)t, 12.1 MATRIX GROUPS 181 then T (x) = T (x1e1 + x2e2 + · · · + xnen) = x1T (e1) + x2T (e2) + · · · + xnT (en) a1kxk,..., t amkxk n k=1 = n k=1 = Ax. Example 1. If we let T : R2 → R2 be the map given by
T (x1, x2) = (2x1 + 5x2, −4x1 + 3x2), the axioms that T must satisfy to be a linear transformation are easily veriﬁed. The column vectors T e1 = (2, −4)t and T e2 = (5, 3)t tell us that T is given by the matrix A = 2 5 −4 3. Since we are interested in groups of matrices, we need to know which matrices have multiplicative inverses. Recall that an n × n matrix A is invertible exactly when there exists another matrix A−1 such that AA−1 = A−1A = I, where · · ·......... 0 0 · · ·  0 0  ...   1 is the n × n identity matrix. From linear algebra we know that A is invertible if and only if the determinant of A is nonzero. Sometimes an invertible matrix is said to be nonsingular. Example 2. If A is the matrix then the inverse of A is 2 1 5 3, A−1 = 3 −1 2 −5. 182 CHAPTER 12 MATRIX GROUPS AND SYMMETRY We are guaranteed that A−1 exists, since det(A is nonzero. Some other facts about determinants will also prove useful in the course of this chapter. Let A and B be n × n matrices. From linear algebra we have the following properties of determinants. • The determinant is a homomorphism into the multiplicative group of real numbers; that is, det(AB) = (det A)(det B). • If A is an invertible matrix, then det(A−1) = 1/ det A. • If we deﬁne the transpose of a matrix A = (aij) to be At = (aji), then det(At) = det A. • Let T be the linear transformation associated with an n × n matrix A. Then T multiplies volumes by a factor of \| det A\|. In the case of R2, this means that T multiplies areas by \| det A\|. Linear maps, matrices, and determinants are covered in any elementary linear algebra text; however, if you have not had a course in linear algebra, it is a straightforward process to verify these
properties directly for 2 × 2 matrices, the case with which we are most concerned. The General and Special Linear Groups The set of all n × n invertible matrices forms a group called the general linear group. We will denote this group by GLn(R). The general linear group has several important subgroups. The multiplicative properties of the determinant imply that the set of matrices with determinant one is a subgroup of the general linear group. Stated another way, suppose that det(A) = 1 and det(B) = 1. Then det(AB) = det(A) det(B) = 1 and det(A−1) = 1/ det A = 1. This subgroup is called the special linear group and is denoted by SLn(R). Example 3. Given a 2 × 2 matrix A = a b c d, the determinant of A is ad − bc. The group GL2(R) consists of those matrices in which ad − bc = 0. The inverse of A is A−1 = 1 ad − bc d −b a −c. 12.1 MATRIX GROUPS 183 If A is in SL2(R), then −c Geometrically, SL2(R) is the group that preserves the areas of parallelograms. Let A−1 = d −b a. A = 1 1 0 1 be in SL2(R). In Figure 12.1, the unit square corresponding to the vectors x = (1, 0)t and y = (0, 1)t is taken by A to the parallelogram with sides (1, 0)t and (1, 1)t; that is, Ax = (1, 0)t and Ay = (1, 1)t. Notice that these two parallelograms have the same area. y (0, 1) y (1, 1) (1, 0) x (1, 0) x Figure 12.1. SL2(R) acting on the unit square The Orthogonal Group O(n) Another subgroup of GLn(R) is the orthogonal group. A matrix A is orthogonal if A−1 = At. The orthogonal group consists of the set of all orthogonal matrices. We write O(n) for the n × n orthogonal group. We leave as an exercise the proof that O(n) is
a subgroup of GLn(R). Example 4. The following matrices are orthogonal: 3/5 −4/5 3/5 4/5, 1/2 − 3/2 √ √ 3/2 1/2  , √ −1/ √ 1/ √ 1/ 0 2 √ 6 −2/ √ 1/ 3 1/ 6 1/ 1 184 CHAPTER 12 MATRIX GROUPS AND SYMMETRY There is a more geometric way of viewing the group O(n). The orthogonal matrices are exactly those matrices that preserve the length of vectors. We can deﬁne the length of a vector using the Euclidean inner product, or dot product, of two vectors. The Euclidean inner product of two vectors x = (x1,..., xn)t and y = (y1,..., yn)t is x, y = xty = (x1, x2,..., xn)      = x1y1 + · · · + xnyn.      y1 y2... yn We deﬁne the length of a vector x = (x1,..., xn)t to be x = x, x = x2 1 + · · · + x2 n. Associated with the notion of the length of a vector is the idea of the distance between two vectors. We deﬁne the distance between two vectors x and y to be x − y. We leave as an exercise the proof of the following proposition about the properties of Euclidean inner products. Proposition 12.1 Let x, y, and w be vectors in Rn and α ∈ R. Then 1. x, y = y, x. 2. x, y + w = x, y + x, w. 3. αx, y = x, αy = αx, y. 4. x, x ≥ 0 with equality exactly when x = 0. 5. If x, y = 0 for all x in Rn, then y = 0. Example 5. The vector x = (3, 4)t has length see that the orthogonal matrix √ 32 +
42 = 5. We can also A = 3/5 −4/5 3/5 4/5 preserves the length of this vector. The vector Ax = (−7/5, 24/5)t also has length 5. 12.1 MATRIX GROUPS 185 Since det(AAt) = det(I) = 1 and det(A) = det(At), the determinant of any orthogonal matrix is either 1 or −1. Consider the column vectors aj =           a1j a2j... anj of the orthogonal matrix A = (aij). Since AAt = I, ar, as = δrs, where δrs = 1 r = s 0 r = s is the Kronecker delta. Accordingly, column vectors of an orthogonal matrix all have length 1; and the Euclidean inner product of distinct column vectors is zero. Any set of vectors satisfying these properties is called an orthonormal set. Conversely, given an n×n matrix A whose columns form an orthonormal set, A−1 = At. We say that a matrix A is distance-preserving, length-preserving, or inner product-preserving when T x − T y = x − y, T x = x, or T x, T y = x, y, respectively. The following theorem, which characterizes the orthogonal group, says that these notions are the same. Theorem 12.2 Let A be an n × n matrix. The following statements are equivalent. 1. The columns of the matrix A form an orthonormal set. 2. A−1 = At. 3. For vectors x and y, Ax, Ay = x, y. 4. For vectors x and y, Ax − Ay = x − y. 5. For any vector x, Ax = x. Proof. We have already shown (1) and (2) to be equivalent. (2) ⇒ (3). Ax, Ay = (Ax)tAy = xtAtAy = xty = x, y. 186 CHAPTER 12 MATRIX GROUPS AND SYMMETRY (3) ⇒ (2). Since x, x = Ax, Ax = xtAtAy = x, AtAx, we know that x,
(AtA − I)x = 0 for all x. Therefore, AtA − I = 0 or A−1 = At. (3) ⇒ (4). If A is inner product-preserving, then A is distance-preserving, since Ax − Ay2 = A(x − y)2 = A(x − y), A(x − y) = x − y, x − y = x − y2. (4) ⇒ (5). If A is distance-preserving, then A is length-preserving. Letting y = 0, we have Ax = Ax − Ay = x − y = x. (5) ⇒ (3). We use the following identity to show that length-preserving implies inner product-preserving: x, y = 1 2 x + y2 − x2 − y2. Observe that Ax, Ay = Ax + Ay2 − Ax2 − Ay2 A(x + y)2 − Ax2 − Ay2 x + y2 − x2 − y2 = 1 2 1 2 1 2 = x, y. = Example 6. Let us examine the orthogonal group on R2 a bit more closely. An element T ∈ O(2) is determined by its action on e1 = (1, 0)t and 12.1 MATRIX GROUPS 187 y y (sin θ, − cos θ) (a, b) x (a, −b) (cos θ, sin θ) θ x Figure 12.2. O(2) acting on R2 e2 = (0, 1)t. If T (e1) = (a, b)t, then a2 + b2 = 1 and T (e2) = (−b, a)t. Hence, T can be represented by A = a −b a b cos θ − sin θ cos θ sin θ, = where 0 ≤ θ < 2π. A matrix T in O(2) either reﬂects or rotates a vector in R2 (Figure 12.2). A reﬂection about the horizontal axis is given by the matrix 1 0 0 −1, whereas a rotation by an angle θ in a counterclockwise direction must come from a matrix of the form cos θ sin θ sin θ − cos θ. A re�
�ection about a line is simply a reﬂection about the horizontal axis followed by a rotation. If det A = −1, then A gives a reﬂection. Two of the other matrix or matrix-related groups that we will consider are the special orthogonal group and the group of Euclidean motions. The special orthogonal group, SO(n), is just the intersection of O(n) and SLn(R); that is, those elements in O(n) with determinant one. The Euclidean group, E(n), can be written as ordered pairs (A, x), where A is in O(n) and x is in Rn. We deﬁne multiplication by (A, x)(B, y) = (AB, Ay + x). 188 CHAPTER 12 MATRIX GROUPS AND SYMMETRY The identity of the group is (I, 0); the inverse of (A, x) is (A−1, −A−1x). In Exercise 6, you are asked to check that E(n) is indeed a group under this operation. y y x + y x x x Figure 12.3. Translations in R2 12.2 Symmetry An isometry or rigid motion in Rn is a distance-preserving function f from Rn to Rn. This means that f must satisfy f (x) − f (y) = x − y for all x, y ∈ Rn. It is not diﬃcult to show that f must be a one-to-one map. By Theorem 12.2, any element in O(n) is an isometry on Rn; however, O(n) does not include all possible isometries on Rn. Translation by a vector x, Ty(x) = x + y is also an isometry (Figure 12.3); however, T cannot be in O(n) since it is not a linear map. We are mostly interested in isometries in R2. In fact, the only isometries in R2 are rotations and reﬂections about the origin, translations, and combinations of the two. For example, a glide reﬂection is a translation followed by a reﬂection (Figure 12.4). In Rn all isometries are given in the same manner. The proof
is very easy to generalize. Lemma 12.3 An isometry f that ﬁxes the origin in R2 is a linear transformation. In particular, f is given by an element in O(2). 12.2 SYMMETRY 189 y y x x x T (x) Figure 12.4. Glide reﬂections Proof. Let f be an isometry in R2 ﬁxing the origin. We will ﬁrst show that f preserves inner products. Since f (0) = 0, f (x) = x; therefore, x2 − 2f (x), f (y) + y2 = f (x)2 − 2f (x), f (y) + f (y)2 = f (x) − f (y), f (x) − f (y) = f (x) − f (y)2 = x − y2 = x − y, x − y = x2 − 2x, y + y2. Consequently, Now let e1 and e2 be (1, 0)t and (0, 1)t, respectively. If f (x), f (y) = x, y. x = (x1, x2) = x1e1 + x2e2, then f (x) = f (x), f (e1)f (e1) + f (x), f (e2)f (e2) = x1f (e1) + x2f (e2). The linearity of f easily follows. For any arbitrary isometry, f, Txf will ﬁx the origin for some vector x in R2; hence, Txf (y) = Ay for some matrix A ∈ O(2). Consequently, 190 CHAPTER 12 MATRIX GROUPS AND SYMMETRY f (y) = Ay + x. Given the isometries f (y) = Ay + x1 g(y) = By + x2, their composition is f (g(y)) = f (By + x2) = ABy + Ax2 + x1. This last computation allows us to identify the group of isometries on R2 with E(2). Theorem 12.4 The group of isometries on R2 is the Euclidean group, E(2). A symmetry group in Rn is a
subgroup of the group of isometries on Rn that ﬁxes a set of points X ⊂ R2. It is important to realize that the symmetry group of X depends both on Rn and on X. For example, the symmetry group of the origin in R1 is Z2, but the symmetry group of the origin in R2 is O(2). Theorem 12.5 The only ﬁnite symmetry groups in R2 are Zn and Dn. Proof. Any ﬁnite symmetry group G in R2 must be a ﬁnite subgroup of O(2); otherwise, G would have an element in E(2) of the form (A, x), where x = 0. Such an element must have inﬁnite order. By Example 6, elements in O(2) are either rotations of the form Rθ = cos θ − sin θ cos θ sin θ or reﬂections of the form Tφ = cos φ − sin φ cos φ sin φ 1 0 0 −1 cos φ sin φ sin φ − cos φ. = Notice that det(Rθ) = 1, det(Tφ) = −1, and T 2 φ = I. We can divide the proof up into two cases. In the ﬁrst case, all of the elements in G have determinant one. In the second case, there exists at least one element in G with determinant −1. Case 1. The determinant of every element in G is one. In this case every element in G must be a rotation. Since G is ﬁnite, there is a smallest angle, say θ0, such that the corresponding element Rθ0 is the smallest rotation in 12.2 SYMMETRY 191 the positive direction. We claim that Rθ0 generates G. If not, then for some positive integer n there is an angle θ1 between nθ0 and (n + 1)θ0. If so, then (n + 1)θ0 − θ1 corresponds to a rotation smaller than θ0, which contradicts the minimality of θ0. Case 2. The group G contains a reﬂection T. The kernel of the homomorphism φ : G → {−1, 1}
given by A → det(A) consists of elements whose determinant is 1. Therefore, \|G/ ker φ\| = 2. We know that the kernel is cyclic by the ﬁrst case and is a subgroup of G of, say, order n. Hence, \|G\| = 2n. The elements of G are Rθ,..., Rn−1 θ, T Rθ,..., T Rn−1 θ. These elements satisfy the relation T RθT = R−1 θ. Consequently, G must be isomorphic to Dn in this case. Figure 12.5. A wallpaper pattern in R2 The Wallpaper Groups Suppose that we wish to study wallpaper patterns in the plane or crystals in three dimensions. Wallpaper patterns are simply repeating patterns in the plane (Figure 12.5). The analogs of wallpaper patterns in R3 are crystals, which we can think of as repeating patterns of molecules in three dimensions (Figure 12.6). The mathematical equivalent of a wallpaper or crystal pattern is called a lattice. Let us examine wallpaper patterns in the plane a little more closely. Suppose that x and y are linearly independent vectors in R2; that is, one 192 CHAPTER 12 MATRIX GROUPS AND SYMMETRY Figure 12.6. A crystal structure in R3 vector cannot be a scalar multiple of the other. A lattice of x and y is the set of all linear combinations mx + ny, where m and n are integers. The vectors x and y are said to be a basis for the lattice. (−1, 1) (1, 1) (2, 0) (−1, −1) Figure 12.7. A lattice in R2 Notice that a lattice can have several bases. For example, the vectors (1, 1)t and (2, 0)t have the same lattice as the vectors (−1, 1)t and (−1, −1)t (Figure 12.7). However, any lattice is completely determined by a basis. Given two bases for the same lattice, say {x1, x2} and {y1, y2}, we can write y1 = α1x1 + α2x2 y2 = β1x1 + β2x2, 12.2 SYMMETRY 193 where α1, α2, β1
, and β2 are integers. The matrix corresponding to this transformation is α1 α2 β1 β2 U =. If we wish to give x1 and x2 in terms of y1 and y2, we need only calculate U −1; that is, U −1 y1 y2 =. x1 x2 Since U has integer entries, U −1 must also have integer entries; hence the determinants of both U and U −1 must be integers. Because U U −1 = I, det(U U −1) = det(U ) det(U −1) = 1; consequently, det(U ) = ±1. A matrix with determinant ±1 and integer entries is called unimodular. For example, the matrix 3 1 5 2 is unimodular. It should be clear that there is a minimum length for vectors in a lattice. We can classify lattices by studying their symmetry groups. The symmetry group of a lattice is the subgroup of E(2) that maps the lattice to itself. We consider two lattices in R2 to be equivalent if they have the same symmetry group. Similarly, classiﬁcation of crystals in R3 is accomplished by associating a symmetry group, called a space group, with each type of crystal. Two lattices are considered diﬀerent if their space groups are not the same. The natural question that now arises is how many space groups exist. A space group is composed of two parts: a translation subgroup and a point group. The translation subgroup is an inﬁnite abelian subgroup of the space group made up of the translational symmetries of the crystal; the point group is a ﬁnite group consisting of rotations and reﬂections of the crystal about a point. More speciﬁcally, a space group is a subgroup of G ⊂ E(2) whose translations are a set of the form {(I, t) : t ∈ L}, where L is a lattice. Space groups are, of course, inﬁnite. Using geometric arguments, we can prove the following theorem (see [5] or [6]). Theorem 12.6 Every translation group in R2 is isomorphic to Z × Z. The point group of G is G0 = {A : (A, b) ∈ G for some b
}. In particular, G0 must be a subgroup of O(2). Suppose that x is a vector in a lattice 194 CHAPTER 12 MATRIX GROUPS AND SYMMETRY Rectangular Square Rhombic Parallelogram Hexagonal Figure 12.8. Types of lattices in R2 L with space group G, translation group H, and point group G0. For any element (A, y) in G, (A, y)(I, x)(A, y)−1 = (A, Ax + y)(A−1, −A−1y) = (AA−1, −AA−1y + Ax + y) = (I, Ax); hence, (I, Ax) is in the translation group of G. More speciﬁcally, Ax must be in the lattice L. It is important to note that G0 is not usually a subgroup of the space group G; however, if T is the translation subgroup of G, then G/T ∼= G0. The proof of the following theorem can be found in [2], [5], or [6]. Theorem 12.7 The point group in the wallpaper groups is isomorphic to Zn or Dn, where n = 1, 2, 3, 4, 6. To answer the question of how the point groups and the translation groups can be combined, we must look at the diﬀerent types of lattices. Lattices can be classiﬁed by the structure of a single lattice cell. The possible cell shapes are parallelogram, rectangular, square, rhombic, and hexagonal (Figure 12.8). The wallpaper groups can now be classiﬁed according to the 12.2 SYMMETRY 195 types of reﬂections that occur in each group: these are ordinarily reﬂections, glide reﬂections, both, or none. Table 12.1. The 17 wallpaper groups Reﬂections or Glide Reﬂections? Notation and Space Groups Point Group Lattice Type p1 p2 p3 p4 p6 pm pg cm pmm pmg pgg c2mm p3m1, p31m p4m, p4g p6m parallelogram none parallelogram none none hexagonal none square none hexagonal reﬂ
ections rectangular glide reﬂections rectangular both rhombic reﬂections rectangular glide reﬂections rectangular both rectangular both rhombic both hexagonal both square both hexagonal Z1 Z2 Z3 Z4 Z6 D1 D1 D1 D2 D2 D2 D2 D3 D4 D6 Theorem 12.8 There are exactly 17 wallpaper groups. p4m p4g Figure 12.9. The wallpaper groups p4m and p4g The 17 wallpaper groups are listed in Table 12.1. The groups p3m1 and p31m can be distinguished by whether or not all of their threefold centers lie 196 CHAPTER 12 MATRIX GROUPS AND SYMMETRY on the reﬂection axes: those of p3m1 must, whereas those of p31m may not. Similarly, the fourfold centers of p4m must lie on the reﬂection axes whereas those of p4g need not (Figure 12.9). The complete proof of this theorem can be found in several of the references at the end of this chapter, including [5], [6], [10], and [11]. Historical Note Symmetry groups have intrigued mathematicians for a long time. Leonardo da Vinci was probably the ﬁrst person to know all of the point groups. At the International Congress of Mathematicians in 1900, David Hilbert gave a now-famous address outlining 23 problems to guide mathematics in the twentieth century. Hilbert’s eighteenth problem asked whether or not crystallographic groups in n dimensions were always ﬁnite. In 1910, L. Bieberbach proved that crystallographic groups are ﬁnite in every dimension. Finding out how many of these groups there are in each dimension is another matter. In R3 there are 230 diﬀerent space groups; in R4 there are 4783. No one has been able to compute the number of space groups for R5 and beyond. It is interesting to note that the crystallographic groups were found mathematically for R3 before the 230 diﬀerent types of crystals were actually discovered in nature. Exercises 1. Prove the identity x, y = 1 2 x + y2 − x2 − y2. 2. Show that O(n) is a group. 3. Prove that the following matrices are orthogonal. Are any of these
matrices in SO(n)? (a) (b) 1/ 1/ √ √ √ 2 −1/ √ 1/ 2 2 2 1/ −2/ √ 5 √ 5 √ √ 2/ 1/ 5 5 (c) (d) √ 5 √ 5   4/ −3/ 0   1/3 −2/3 −2/3   5 5 √ √ 3/ 4/ 0 0 0 −1 2/3 −2/3 1/3 2/3 2/3 1/3   EXERCISES 197 4. Determine the symmetry group of each of the ﬁgures in Figure 12.10. (a) (c) (b) Figure 12.10. 5. Let x, y, and w be vectors in Rn and α ∈ R. Prove each of the following properties of inner products. (a) x, y = y, x. (b) x, y + w = x, y + x, w. (c) αx, y = x, αy = αx, y. (d) x, x ≥ 0 with equality exactly when x = 0. (e) If x, y = 0 for all x in Rn, then y = 0. 6. Verify that is a group. E(n) = {(A, x) : A ∈ O(n) and x ∈ Rn} 7. Prove that {(2, 1), (1, 1)} and {(12, 5), (7, 3)} are bases for the same lattice. 8. Let G be a subgroup of E(2) and suppose that T is the translation subgroup of G. Prove that the point group of G is isomorphic to G/T. 9. Let A ∈ SL2(R) and suppose that the vectors x and y form two sides of a parallelogram in R2. Prove that the area of this parallelogram is the same as the area of the parallelogram with sides Ax and Ay. 10. Prove that SO(n) is a normal subgroup of O(n). 11. Show that any isometry f in Rn is a one-to-one map. 12. Show that an element
in E(2) of the form (A, x), where x = 0, has inﬁnite order. 13. Prove or disprove: There exists an inﬁnite abelian subgroup of O(n). 198 CHAPTER 12 MATRIX GROUPS AND SYMMETRY 14. Let x = (x1, x2) be a point on the unit circle in R2; that is, x2 A ∈ O(2), show that Ax is also a point on the unit circle. 1 + x2 2 = 1. If 15. Let G be a group with a subgroup H (not necessarily normal) and a normal subgroup N. Then G is a semidirect product of N by H if • H ∩ N = {id}; • HN = G. Show that each of the following is true. (a) S3 is the semidirect product of A3 by H = {(1), (12)}. (b) The quaternion group, Q8, cannot be written as a semidirect product. (c) E(2) is the semidirect product of O(2) by H, where H consists of all translations in R2. 16. Determine which of the 17 wallpaper groups preserves the symmetry of the pattern in Figure 12.5. Figure 12.11. 17. Determine which of the 17 wallpaper groups preserves the symmetry of the pattern in Figure 12.11. 18. Find the rotation group of a dodecahedron. 19. For each of the 17 wallpaper groups, draw a wallpaper pattern having that group as a symmetry group. References and Suggested Readings [1] Coxeter, H. M. and Moser, W. O. J. Generators and Relations for Discrete Groups, 3rd ed. Springer-Verlag, New York, 1972. [2] Grove, L. C. and Benson, C. T. Finite Reﬂection Groups. 2nd ed. Springer- Verlag, New York, 1985. EXERCISES 199 [3] Hiller, H. “Crystallography and Cohomology of Groups,” American Mathe- matical Monthly 93 (1986), 765–79. [4] Lockwood, E. H. and Macmillan, R. H. Geometric Symmetry. Cambridge University Press, Cambridge, 1978
. [5] Mackiw, G. Applications of Abstract Algebra. Wiley, New York, 1985. [6] Martin, G. Transformation Groups: An Introduction to Symmetry. Springer- Verlag, New York, 1982. [7] Milnor, J. “Hilbert’s Problem 18: On Crystallographic Groups, Fundamental Domains, and Sphere Packing,” Proceedings of Symposia in Pure Mathematics 18, American Mathematical Society, 1976. [8] Phillips, F. C. An Introduction to Crystallography. 4th ed. Wiley, New York, 1971. [9] Rose, B. I. and Staﬀord, R. D. “An Elementary Course in Mathematical Symmetry,” American Mathematical Monthly 88 (1980), 54–64. [10] Schattschneider, D. “The Plane Symmetry Groups: Their Recognition and Their Notation,” American Mathematical Monthly 85 (1978), 439–50. [11] Schwarzenberger, R. L. “The 17 Plane Symmetry Groups,” Mathematical Gazette 58 (1974), 123–31. [12] Weyl, H. Symmetry. Princeton University Press, Princeton, NJ, 1952. 13 The Structure of Groups The ultimate goal of group theory is to classify all groups up to isomorphism; that is, given a particular group, we should be able to match it up with a known group via an isomorphism. For example, we have already proved that any ﬁnite cyclic group of order n is isomorphic to Zn; hence, we “know” all ﬁnite cyclic groups. It is probably not reasonable to expect that we will ever know all groups; however, we can often classify certain types of groups or distinguish between groups in special cases. In this chapter we will characterize all ﬁnite abelian groups. We shall also investigate groups with sequences of subgroups. If a group has a sequence of subgroups, say G = Hn ⊃ Hn−1 ⊃ · · · ⊃ H1 ⊃ H0 = {e}, where each subgroup Hi is normal in Hi+1 and each of the factor groups Hi+1/Hi is abelian, then G is a solvable group. In addition to allowing us to
distinguish between certain classes of groups, solvable groups turn out to be central to the study of solutions to polynomial equations. 13.1 Finite Abelian Groups In our investigation of cyclic groups we found that every group of prime order was isomorphic to Zp, where p was a prime number. We also determined ∼= Zm × Zn when gcd(m, n) = 1. In fact, much more is true. Every that Zmn ﬁnite abelian group is isomorphic to a direct product of cyclic groups of prime power order; that is, every ﬁnite abelian group is isomorphic to a group of the type Z pα1 1 × · · · × Z n. pαn 200 13.1 FINITE ABELIAN GROUPS 201 First, let us examine a slight generalization of ﬁnite abelian groups. Suppose that G is a group and let {gi} be a set of elements in G, where i is in some index set I (not necessarily ﬁnite). The smallest subgroup of G containing all of the gi’s is the subgroup of G generated by the gi’s. If this subgroup of G is in fact all of G, then G is generated by the set {gi : i ∈ I}. In this case the gi’s are said to be the generators of G. If there is a ﬁnite set {gi : i ∈ I} that generates G, then G is ﬁnitely generated. Example 1. Obviously, all ﬁnite groups are ﬁnitely generated. For example, the group S3 is generated by the permutations (12) and (123). The group Z × Zn is an inﬁnite group but is ﬁnitely generated by {(1, 0), (0, 1)}. Example 2. Not all groups are ﬁnitely generated. Consider the rational numbers Q under the operation of addition. Suppose that Q is ﬁnitely generated with generators p1/q1,..., pn/qn, where each pi/qi is a fraction expressed in its lowest terms. Let p be some prime that does not divide any of the denominators q1,..., qn. We claim that 1/p cannot be
in the subgroup of Q that is generated by p1/q1,..., pn/qn, since p does not divide the denominator of any element in this subgroup. This fact is easy to see since the sum of any two generators is pi/qi + pj/qj = (piqj + pjqi)/(qiqj). Theorem 13.1 Let H be the subgroup of a group G that is generated by {gi ∈ G : i ∈ I}. Then h ∈ H exactly when it is a product of the form h = gα1 i1 · · · gαn in, where the gik ’s are not necessarily distinct. The reason that powers of a ﬁxed gi may occur several times in the product is that we may have a nonabelian group. However, if the group is abelian, then the gi’s need occur only once. For example, a product such as a−3b5a7 could always be simpliﬁed (in this case, to a4b5). Proof. Let K be the set of all products of the form gα1, where the i1 gik ’s are not necessarily distinct. Certainly K is a subset of H. We need only show that K is a subgroup of G. If this is the case, then K = H, since H is the smallest subgroup containing all the gi’s. · · · gαn in 202 CHAPTER 13 THE STRUCTURE OF GROUPS Clearly, the set K is closed under the group operation. Since g0 i = 1, It remains to show that the inverse of an element the identity is in K. g = gk1 1 · · · gkn in in K must also be in K. However, g−1 = (gk1 1 · · · gkn in )−1 = (g−kn 1 · · · g−k1 in ). Now let us restrict our attention to ﬁnite abelian groups. We can express any ﬁnite abelian group as a ﬁnite direct product of cyclic groups. More speciﬁcally, letting p be prime, we deﬁne a group G to be a p-group if every element in G has as its order a power of
p. For example, both Z2 × Z2 and Z4 are 2-groups, whereas Z27 is a 3-group. We shall prove that every ﬁnite abelian group is isomorphic to a direct product of cyclic p-groups. Before we state the main theorem concerning ﬁnite abelian groups, we shall consider a special case. Theorem 13.2 Every ﬁnite abelian group G is the direct product of p-groups. Proof. If \|G\| = 1, then the theorem is trivial. Suppose that the order of G is greater than 1, say \|G\| = pα1 1 · · · pαn n, where p1,..., pn are all prime, and deﬁne Gi to be the set of elements in G of order pk i for some integer k. Since G is an abelian group, we are guaranteed that Gi is a subgroup of G for i = 1,..., n. We must show that G = G1 × · · · × Gn. That is, we must be able to write every g ∈ G as a unique product gp1 · · · gpn where gpi is of the order of some power of pi. Since the order of g divides the order of G, we know that \|g\| = pβ1 1 pβ2 2 · · · pβn n for integers β1,..., βn. Letting ai = \|g\|/pβi i, the ai’s are relatively prime; hence, there exist integers b1,..., bn such that a1b1 + · · · + anbn = 1. Consequently, g = ga1b1+···+anbn = ga1b1 · · · ganbn. Since g(aibi)p βi i = gbi\|g\| = e, 13.1 FINITE ABELIAN GROUPS 203 it follows that gaibi must be in Gi. Let gi = gaibi. Then g = g1 · · · gn and Gi ∩ Gj = {e} for i = j. To show uniqueness, suppose that g = g1 · · · gn = h1 · · · hn, with hi ∈ Gi. Then e = (g1 · · · gn)(h1 ·
· · hn)−1 = g1h−1 1 · · · gnh−1 n. is a power of pi; hence, the order of g1h−1 1 The order of gih−1 least common multiple of the orders of the gih−1 order of the identity is 1. Therefore, \|gih−1 i i · · · gnh−1 n is the. This must be 1, since the \| = 1 or gi = hi for i = 1,..., n. i We shall now state the Fundamental Theorem of Finite Abelian Groups. Theorem 13.3 (Fundamental Theorem of Finite Abelian Groups) Every ﬁnite abelian group G is isomorphic to a direct product of cyclic groups of the form Z pα1 1 × Z pα2 2 × · · · × Z pαn n where the pi’s are primes (not necessarily distinct). Example 3. Suppose that we wish to classify all abelian groups of order 540 = 22 · 33 · 5. The Fundamental Theorem of Finite Abelian Groups tells us that we have the following six possibilities. • Z2 × Z2 × Z3 × Z3 × Z3 × Z5; • Z2 × Z2 × Z3 × Z9 × Z5; • Z2 × Z2 × Z27 × Z5; • Z4 × Z3 × Z3 × Z3 × Z5; • Z4 × Z3 × Z9 × Z5; • Z4 × Z27 × Z5. The proof of the Fundamental Theorem relies on the following lemma. Lemma 13.4 Let G be a ﬁnite abelian p-group and suppose that g ∈ G has maximal order. Then G can be written as g × H for some subgroup H of G. 204 CHAPTER 13 THE STRUCTURE OF GROUPS Proof. Suppose that the order of G is pn. We shall induct on n. If n = 1, then G is cyclic of order p and must be generated by g. Suppose now that the statement of the lemma holds for all integers k with 1 ≤ k < n and let g be of maximal order in G, say \|g\| = pm. Then apm = e for all a ∈ G. Now choose h in G such that h /∈ g, where
h has the smallest possible order. Certainly such an h exists; otherwise, G = g and we are done. Let H = h. We claim that g ∩ H = {e}. It suﬃces to show that \|H\| = p. Since \|hp\| = \|h\|/p, the order of hp is smaller than the order of h and must be in g by the minimality of h; that is, hp = gr for some number r. Hence, (gr)pm−1 = (hp)pm−1 = hpm = e, and the order of gr must be less than or equal to pm−1. Therefore, gr cannot generate g. Notice that p must occur as a factor of r, say r = ps, and hp = gr = gps. Deﬁne a to be g−sh. Then a cannot be in g; otherwise, h would also have to be in g. Also, ap = g−sphp = g−rhp = h−php = e. We have now formed an element a with order p such that a /∈ g. Since h was chosen to have the smallest order of all of the elements that are not in g, \|H\| = p. Now we will show that the order of gH in the factor group G/H must be the same as the order of g in G. If \|gH\| < \|g\| = pm, then H = (gH)pm−1 = gpm−1 H; hence, gpm−1 must be in g ∩ H = {e}, which contradicts the fact that the order of g is pm. Therefore, gH must have maximal order in G/H. By the Correspondence Theorem and our induction hypothesis, G/H ∼= gH × K/H for some subgroup K of G containing H. We claim that g ∩ K = {e}. If b ∈ g ∩ K, then bH ∈ gH ∩ K/H = {H} and b ∈ g ∩ H = {e}. It follows that G = gK implies that G ∼= g × K. The proof of the Fundamental Theorem of Finite Abelian Groups follows very quickly from Lemma 13.4. Suppose that G is a ﬁnite abelian group and let g be an element of maximal order in G. If g = G,
then we are done; otherwise, G ∼= Z\|g\| × H for some subgroup H contained in G by the lemma. Since \|H\| < \|G\|, we can apply mathematical induction. 13.2 SOLVABLE GROUPS 205 We now state the more general theorem for all ﬁnitely generated abelian groups. The proof of this theorem can be found in any of the references at the end of this chapter. Theorem 13.5 (The Fundamental Theorem of Finitely Generated Abelian Groups) Every ﬁnitely generated abelian group G is isomorphic to a direct product of cyclic groups of the form Z pα1 1 × Z pα2, pαn where the pi’s are primes (not necessarily distinct). 13.2 Solvable Groups A subnormal series of a group G is a ﬁnite sequence of subgroups G = Hn ⊃ Hn−1 ⊃ · · · ⊃ H1 ⊃ H0 = {e}, where Hi is a normal subgroup of Hi+1. If each subgroup Hi is normal in G, then the series is called a normal series. The length of a subnormal or normal series is the number of proper inclusions. Example 4. Any series of subgroups of an abelian group is a normal series. Consider the following series of groups: Z ⊃ 9Z ⊃ 45Z ⊃ 180Z ⊃ {0}, Z24 ⊃ 2 ⊃ 6 ⊃ 12 ⊃ {0}. Example 5. A subnormal series need not be a normal series. Consider the following subnormal series of the group D4: D4 ⊃ {(1), (12)(34), (13)(24), (14)(23)} ⊃ {(1), (12)(34)} ⊃ {(1)}. The subgroup {(1), (12)(34)} is not normal in D4; consequently, this series is not a normal series. A subnormal (normal) series {Kj} is a reﬁnement of a subnormal (normal) series {Hi} if {Hi} ⊂ {Kj}. That is, each Hi is one of the Kj. Example 6. The series Z ⊃ 3Z ⊃ 9Z ⊃ 45
Z ⊃ 90Z ⊃ 180Z ⊃ {0} 206 CHAPTER 13 THE STRUCTURE OF GROUPS is a reﬁnement of the series Z ⊃ 9Z ⊃ 45Z ⊃ 180Z ⊃ {0}. The correct way to study a subnormal or normal series of subgroups, {Hi} of G, is actually to study the factor groups Hi+1/Hi. We say that two subnormal (normal) series {Hi} and {Kj} of a group G are isomorphic if there is a one-to-one correspondence between the collections of factor groups {Hi+1/Hi} and {Kj+1/Kj}. Example 7. The two normal series Z60 ⊃ 3 ⊃ 15 ⊃ {0} Z60 ⊃ 4 ⊃ 20 ⊃ {0} of the group Z60 are isomorphic since Z60/3 ∼= 20/{0} ∼= Z3 3/15 ∼= 4/20 ∼= Z5 15/{0} ∼= Z60/4 ∼= Z4. A subnormal series {Hi} of a group G is a composition series if all the factor groups are simple; that is, if none of the factor groups of the series contains a normal subgroup. A normal series {Hi} of G is a principal series if all the factor groups are simple. Example 8. The group Z60 has a composition series Z60 ⊃ 3 ⊃ 15 ⊃ 30 ⊃ {0} with factor groups Z60/3 ∼= Z3 3/15 ∼= Z5 15/30 ∼= Z2 30/{0} ∼= Z2. 13.2 SOLVABLE GROUPS 207 Since Z60 is an abelian group, this series is automatically a principal series. Notice that a composition series need not be unique. The series Z60 ⊃ 2 ⊃ 4 ⊃ 20 ⊃ {0} is also a composition series. Example 9. For n ≥ 5, the series Sn ⊃ An ⊃ {(1)} is a composition series for Sn since Sn/An ∼= Z2 and An is simple. Example 10. Not every group has a composition series or a principal series. Suppose that {0} = H0 ⊂ H1 ⊂ ·
· · ⊂ Hn−1 ⊂ Hn = Z is a subnormal series for the integers under addition. Then H1 must be of ∼= nZ is an inﬁnite cyclic the form nZ for some n ∈ N. In this case H1/H0 group with many nontrivial proper normal subgroups. Although composition series need not be unique as in the case of Z60, it turns out that any two composition series are related. The factor groups of the two composition series for Z60 are Z2, Z2, Z3, and Z5; that is, the two composition series are isomorphic. The Jordan-H¨older Theorem says that this is always the case. Theorem 13.6 (Jordan-H¨older) Any two composition series of G are isomorphic. Proof. We shall employ mathematical induction on the length of the composition series. If the length of a composition series is 1, then G must be a simple group. In this case any two composition series are isomorphic. Suppose now that the theorem is true for all groups having a composition series of length k, where 1 ≤ k < n. Let G = Hn ⊃ Hn−1 ⊃ · · · ⊃ H1 ⊃ H0 = {e} G = Km ⊃ Km−1 ⊃ · · · ⊃ K1 ⊃ K0 = {e} 208 CHAPTER 13 THE STRUCTURE OF GROUPS be two composition series for G. We can form two new subnormal series for G since Hi ∩ Km−1 is normal in Hi+1 ∩ Km−1 and Kj ∩ Hn−1 is normal in Kj+1 ∩ Hn−1: G = Hn ⊃ Hn−1 ⊃ Hn−1 ∩ Km−1 ⊃ · · · ⊃ H0 ∩ Km−1 = {e} G = Km ⊃ Km−1 ⊃ Km−1 ∩ Hn−1 ⊃ · · · ⊃ K0 ∩ Hn−1 = {e}. Since Hi ∩Km−1 is normal in Hi+1 ∩Km−1, the Second Isomorphism Theorem (Theorem 11.4) implies that (Hi+
1 ∩ Km−1)/(Hi ∩ Km−1) = (Hi+1 ∩ Km−1)/(Hi ∩ (Hi+1 ∩ Km−1)) ∼= Hi(Hi+1 ∩ Km−1)/Hi, where Hi is normal in Hi(Hi+1 ∩ Km−1). Since {Hi} is a composition series, Hi+1/Hi must be simple; consequently, Hi(Hi+1 ∩ Km−1)/Hi is either Hi+1/Hi or Hi/Hi. That is, Hi(Hi+1 ∩ Km−1) must be either Hi or Hi+1. Removing any nonproper inclusions from the series Hn−1 ⊃ Hn−1 ∩ Km−1 ⊃ · · · ⊃ H0 ∩ Km−1 = {e}, we have a composition series for Hn−1. Our induction hypothesis says that this series must be equivalent to the composition series Hn−1 ⊃ · · · ⊃ H1 ⊃ H0 = {e}. Hence, the composition series G = Hn ⊃ Hn−1 ⊃ · · · ⊃ H1 ⊃ H0 = {e} and G = Hn ⊃ Hn−1 ⊃ Hn−1 ∩ Km−1 ⊃ · · · ⊃ H0 ∩ Km−1 = {e} are equivalent. If Hn−1 = Km−1, then the composition series {Hi} and {Kj} are equivalent and we are done; otherwise, Hn−1Km−1 is a normal subgroup of G properly containing Hn−1. In this case Hn−1Km−1 = G and we can apply the Second Isomorphism Theorem once again; that is, Km−1/(Km−1 ∩ Hn−1) ∼= (Hn−1Km−1)/Hn−1 = G/Hn−1. Therefore, G = Hn ⊃ Hn−1 ⊃ Hn−1 ∩ Km−1 ⊃ · · · ⊃ H0 ∩ Km−1 = {e} EXERCISES and G = Km �
� Km−1 ⊃ Km−1 ∩ Hn−1 ⊃ · · · ⊃ K0 ∩ Hn−1 = {e} are equivalent and the proof of the theorem is complete. 209 A group G is solvable if it has a composition series {Hi} such that all of the factor groups Hi+1/Hi are abelian. Solvable groups will play a fundamental role when we study Galois theory and the solution of polynomial equations. Example 11. The group S4 is solvable since S4 ⊃ A4 ⊃ {(1), (12)(34), (13)(24), (14)(23)} ⊃ {(1)} has abelian factor groups; however, for n ≥ 5 the series Sn ⊃ An ⊃ {(1)} is a composition series for Sn with a nonabelian factor group. Therefore, Sn is not a solvable group for n ≥ 5. Exercises 1. Find all of the abelian groups of order less than or equal to 40 up to isomor- phism. 2. Find all of the abelian groups of order 200 up to isomorphism. 3. Find all of the abelian groups of order 720 up to isomorphism. 4. Find all of the composition series for each of the following groups. (a) Z12 (b) Z48 (c) The quaternions, Q8 (d) D4 (e) S3 × Z4 (f) S4 (g) Sn, n ≥ 5 (h) Q 5. Show that the inﬁnite direct product G = Z2 × Z2 × · · · is not ﬁnitely generated. 6. Let G be an abelian group of order m. If n divides m, prove that G has a subgroup of order n. 210 CHAPTER 13 THE STRUCTURE OF GROUPS 7. A group G is a torsion group if every element of G has ﬁnite order. Prove that a ﬁnitely generated abelian torsion group must be ﬁnite. 8. Let G, H, and K be ﬁnitely generated abelian groups. Show that if G × H ∼= G × K, then H ∼= K. Give a counterexample
to show that this cannot be true in general. 9. Let G and H be solvable groups. Show that G × H is also solvable. 10. If G has a composition (principal) series and if N is a proper normal subgroup of G, show there exists a composition (principal) series containing N. 11. Prove or disprove: Let N be a normal subgroup of G. If N and G/N have composition series, then G must also have a composition series. 12. Let N be a normal subgroup of G. If N and G/N are solvable groups, show that G is also a solvable group. 13. Prove that G is a solvable group if and only if G has a series of subgroups G = Pn ⊃ Pn−1 ⊃ · · · ⊃ P1 ⊃ P0 = {e} where Pi is normal in Pi+1 and the order of Pi+1/Pi is prime. 14. Let G be a solvable group. Prove that any subgroup of G is also solvable. 15. Let G be a solvable group and N a normal subgroup of G. Prove that G/N is solvable. 16. Prove that Dn is solvable for all integers n. 17. Suppose that G has a composition series. If N is a normal subgroup of G, show that N and G/N also have composition series. 18. Let G be a cyclic p-group with subgroups H and K. Prove that either H is contained in K or K is contained in H. 19. Suppose that G is a solvable group with order n ≥ 2. Show that G contains a normal nontrivial abelian subgroup. 20. Recall that the commutator subgroup G of a group G is deﬁned as the subgroup of G generated by elements of the form a−1b−1ab for a, b ∈ G. We can deﬁne a series of subgroups of G by G(0) = G, G(1) = G, and G(i+1) = (G(i)). (a) Prove that G(i+1) is normal in (G(i)). The series of subgroups G(0) = G ⊃ G(1) ⊃ G(2)
⊃ · · · is called the derived series of G. (b) Show that G is solvable if and only if G(n) = {e} for some integer n. EXERCISES 211 21. Suppose that G is a solvable group with order n ≥ 2. Show that G contains a normal nontrivial abelian factor group. 22. Zassenhaus Lemma. Let H and K be subgroups of a group G. Suppose also that H ∗ and K ∗ are normal subgroups of H and K respectively. Then (a) H ∗(H ∩ K ∗) is a normal subgroup of H ∗(H ∩ K). (b) K ∗(H ∗ ∩ K) is a normal subgroup of K ∗(H ∩ K). (c) H ∗(H ∩ K)/H ∗(H ∩ K ∗) ∼= K ∗(H ∩ K)/K ∗(H ∗ ∩ K) ∼= (H ∩ K)/(H ∗ ∩ K)(H ∩ K ∗). [Hint: Use the diagram in Figure 13.1. The Zassenhaus Lemma is often referred to as the Butterﬂy Lemma because of this diagram.] H K H ∗(H ∩ K) H ∩ K K∗(H ∩ K) H ∗(H ∩ K∗) K∗(H ∗ ∩ K) H ∗ (H ∗ ∩ K)(H ∩ K∗) K∗ H ∗ ∩ K H ∩ K∗ Figure 13.1. The Zassenhaus Lemma 23. Schreier’s Theorem. Use the Zassenhaus Lemma to prove that two sub- normal (normal) series of a group G have isomorphic reﬁnements. 24. Use Schreier’s Theorem to prove the Jordan-H¨older Theorem. Programming Exercises Write a program that will compute all possible abelian groups of order n. What is the largest n for which your program will work? 212 CHAPTER 13 THE STRUCTURE OF GROUPS References and Suggested Readings Each of the following references contains a proof of the Fundamental Theorem of Finitely Generated Abelian Groups.
[1] Hungerford, T. W. Algebra. Springer, New York, 1974.. [2] Lang, S. Algebra. 3rd ed. Springer, New York, 2002. [3] Rotman, J. J. An Introduction to the Theory of Groups. 4th ed. Springer, New York, 1995. Sage Sage is able to create direct products of cyclic groups, though they are realized as permutation groups. This is a situation that should improve. However, with a classiﬁcation of ﬁnite abelian groups, we can describe how to construct in Sage every group of order less than 16. 14 Group Actions Group actions generalize group multiplication. If G is a group and X is an arbitrary set, a group action of an element g ∈ G and x ∈ X is a product, gx, living in X. Many problems in algebra may best be attacked via group actions. For example, the proofs of the Sylow theorems and of Burnside’s Counting Theorem are most easily understood when they are formulated in terms of group actions. 14.1 Groups Acting on Sets Let X be a set and G be a group. A (left) action of G on X is a map G × X → X given by (g, x) → gx, where 1. ex = x for all x ∈ X; 2. (g1g2)x = g1(g2x) for all x ∈ X and all g1, g2 ∈ G. Under these considerations X is called a G-set. Notice that we are not requiring X to be related to G in any way. It is true that every group G acts on every set X by the trivial action (g, x) → x; however, group actions are more interesting if the set X is somehow related to the group G. Example 1. Let G = GL2(R) and X = R2. Then G acts on X by left multiplication. If v ∈ R2 and I is the identity matrix, then Iv = v. If A and B are 2 × 2 invertible matrices, then (AB)v = A(Bv) since matrix multiplication is associative. Example 2. Let G = D4 be the symmetry group of a square. If X = {1, 2, 3, 4} is the set of vertices of the square
, then we can consider D4 to 213 214 CHAPTER 14 GROUP ACTIONS consist of the following permutations: {(1), (13), (24), (1432), (1234), (12)(34), (14)(23), (13)(24)}. The elements of D4 act on X as functions. The permutation (13)(24) acts on vertex 1 by sending it to vertex 3, on vertex 2 by sending it to vertex 4, and so on. It is easy to see that the axioms of a group action are satisﬁed. In general, if X is any set and G is a subgroup of SX, the group of all permutations acting on X, then X is a G-set under the group action for σ ∈ G and x ∈ X. (σ, x) → σ(x) Example 3. If we let X = G, then every group G acts on itself by the left regular representation; that is, (g, x) → λg(x) = gx, where λg is left multiplication: e · x = λex = ex = x (gh) · x = λghx = λgλhx = λg(hx) = g · (h · x). If H is a subgroup of G, then G is an H-set under left multiplication by elements of H. Example 4. Let G be a group and suppose that X = G. If H is a subgroup of G, then G is an H-set under conjugation; that is, we can deﬁne an action of H on G, via H × G → G, (h, g) → hgh−1 for h ∈ H and g ∈ G. Clearly, the ﬁrst axiom for a group action holds. Observing that (h1h2, g) = h1h2g(h1h2)−1 2 )h−1 1 = h1(h2gh−1 = (h1, (h2, g)), we see that the second condition is also satisﬁed. 14.1 GROUPS ACTING ON SETS 215 Example 5. Let H be a subgroup of G and LH the set of left cosets of H. The set LH is a G-set under the action (g
, xH) → gxH. Again, it is easy to see that the ﬁrst axiom is true. Since (gg)xH = g(gxH), the second axiom is also true. If G acts on a set X and x, y ∈ X, then x is said to be G-equivalent to y if there exists a g ∈ G such that gx = y. We write x ∼G y or x ∼ y if two elements are G-equivalent. Proposition 14.1 Let X be a G-set. Then G-equivalence is an equivalence relation on X. Proof. The relation ∼ is reﬂexive since ex = x. Suppose that x ∼ y for x, y ∈ X. Then there exists a g such that gx = y. In this case g−1y = x; hence, y ∼ x. To show that the relation is transitive, suppose that x ∼ y and y ∼ z. Then there must exist group elements g and h such that gx = y and hy = z. So z = hy = (hg)x, and x is equivalent to z. If X is a G-set, then each partition of X associated with G-equivalence is called an orbit of X under G. We will denote the orbit that contains an element x of X by Ox. Example 6. Let G be the permutation group deﬁned by G = {(1), (123), (132), (45), (123)(45), (132)(45)} and X = {1, 2, 3, 4, 5}. Then X is a G-set. The orbits are O1 = O2 = O3 = {1, 2, 3} and O4 = O5 = {4, 5}. Now suppose that G is a group acting on a set X and let g be an element of G. The ﬁxed point set of g in X, denoted by Xg, is the set of all x ∈ X such that gx = x. We can also study the group elements g that ﬁx a given x ∈ X. This set is more than a subset of G, it is a subgroup. This subgroup is called the stabilizer subgroup or isotropy subgroup of x. We will denote the stabilizer subgroup of x by G
x. Remark. It is important to remember that Xg ⊂ X and Gx ⊂ G. Example 7. Let X = {1, 2, 3, 4, 5, 6} and suppose that G is the permutation group given by the permutations {(1), (12)(3456), (35)(46), (12)(3654)}. 216 CHAPTER 14 GROUP ACTIONS Then the ﬁxed point sets of X under the action of G are X(1) = X, X(35)(46) = {1, 2}, X(12)(3456) = X(12)(3654) = ∅, and the stabilizer subgroups are G1 = G2 = {(1), (35)(46)}, G3 = G4 = G5 = G6 = {(1)}. It is easily seen that Gx is a subgroup of G for each x ∈ X. Proposition 14.2 Let G be a group acting on a set X and x ∈ X. The stabilizer group, Gx, of x is a subgroup of G. Proof. Clearly, e ∈ Gx since the identity ﬁxes every element in the set X. Let g, h ∈ Gx. Then gx = x and hx = x. So (gh)x = g(hx) = gx = x; hence, the product of two elements in Gx is also in Gx. Finally, if g ∈ Gx, then x = ex = (g−1g)x = (g−1)gx = g−1x. So g−1 is in Gx. We will denote the number of elements in the ﬁxed point set of an element g ∈ G by \|Xg\| and denote the number of elements in the orbit of x ∈ X by \|Ox\|. The next theorem demonstrates the relationship between orbits of an element x ∈ X and the left cosets of Gx in G. Theorem 14.3 Let G be a ﬁnite group and X a ﬁnite G-set. If x ∈ X, then \|Ox\| = [G : Gx]. Proof. We know that \|G\|/\|Gx\| is the number of left cosets of Gx in G by Lagrange’s Theorem (Theorem
6.5). We will deﬁne a bijective map φ between the orbit Ox of X and the set of left cosets LGx of Gx in G. Let y ∈ Ox. Then there exists a g in G such that gx = y. Deﬁne φ by φ(y) = gGx. First we must show that this map is well-deﬁned and does not depend on our selection of g. Suppose that h is another element in G such that hx = y. Then gx = hx or x = g−1hx; hence, g−1h is in the stabilizer subgroup of x. Therefore, h ∈ gGx or gGx = hGx. Thus, y gets mapped to the same coset regardless of the choice of the representative from that coset. To show that φ is one-to-one, assume that φ(x1) = φ(x2). Then there exist g1, g2 ∈ G such that x1 = g1x and x2 = g2x. Since there exists a g ∈ Gx such that g2 = g1g, x2 = g2x = g1gx = g1x = x1; 14.2 THE CLASS EQUATION 217 consequently, the map φ is one-to-one. Finally, we must show that the map φ is onto. Let gGx be a left coset. If gx = y, then φ(y) = gGx. 14.2 The Class Equation Let X be a ﬁnite G-set and XG be the set of ﬁxed points in X; that is, XG = {x ∈ X : gx = x for all g ∈ G}. Since the orbits of the action partition X, \|X\| = \|XG\| + n i=k \|Oxi\|, where xk,..., xn are representatives from the distinct nontrivial orbits of X. Now consider the special case in which G acts on itself by conjugation, (g, x) → gxg−1. The center of G, Z(G) = {x : xg = gx for all g ∈ G}, is the set of points that are ﬁx
ed by conjugation. The nontrivial orbits of the action are called the conjugacy classes of G. If x1,..., xk are representatives from each of the nontrivial conjugacy classes of G and \|Ox1\| = n1,..., \|Oxk \| = nk, then \|G\| = \|Z(G)\| + n1 + · · · + nk. The stabilizer subgroups of each of the xi’s, C(xi) = {g ∈ G : gxi = xig}, are called the centralizer subgroups of the xi’s. From Theorem 14.3, we obtain the class equation: \|G\| = \|Z(G)\| + [G : C(x1)] + · · · + [G : C(xk)]. One of the consequences of the class equation is that the order of each conjugacy class must divide the order of \|G\|. Example 8. It is easy to check that the conjugacy classes in S3 are the following: {(1)}, {(123), (132)}, {(12), (13), (23)}. The class equation is 6 = 1 + 2 + 3. 218 CHAPTER 14 GROUP ACTIONS Example 9. The conjugacy classes for D4 are {(1)}, {(13)(24)}, {(13), (24)}, {(1432), (1234)}, {(12)(34), (14)(23)}. The class equation is. Example 10. For Sn it takes a bit of work to ﬁnd the conjugacy classes. We begin with cycles. Suppose that σ = (a1,..., ak) is a cycle and let τ ∈ Sn. By Theorem 6.10, τ στ −1 = (τ (a1),..., τ (ak)). Consequently, any two cycles of the same length are conjugate. Now let σ = σ1σ2 · · · σr be a cycle decomposition, where the length of each cycle σi is ri. Then σ is conjugate to every other τ ∈ Sn whose cycle decomposition has the same lengths. The number of conjugate classes in Sn is the number of ways in which n can be partitioned into sums of positive integers. For
example, we can partition the integer 3 into the following three sums; therefore, there are three conjugacy classes. The problem of ﬁnding the number of such partitions for any positive integer n is what computer scientists call NP-complete. This eﬀectively means that the problem cannot be solved for a large n because the computations would be too time-consuming for even the largest computer. Theorem 14.4 Let G be a group of order pn where p is prime. Then G has a nontrivial center. Proof. We apply the class equation \|G\| = \|Z(G)\| + n1 + · · · + nk. Since each ni > 1 and ni \| G, p must divide each ni. Also, p \| \|G\|; hence, p must divide \|Z(G)\|. Since the identity is always in the center of G, \|Z(G)\| ≥ 1. Therefore, \|Z(G)\| ≥ p and there exists some g ∈ Z(G) such that g = 1. 14.3 BURNSIDE’S COUNTING THEOREM 219 Corollary 14.5 Let G be a group of order p2 where p is prime. Then G is abelian. Proof. By Theorem 14.4, \|Z(G)\| = p or p2. If \|Z(G)\| = p2, then we are done. Suppose that \|Z(G)\| = p. Then Z(G) and G/Z(G) both have order p and must both be cyclic groups. Choosing a generator aZ(G) for G/Z(G), we can write any element gZ(G) in the quotient group as amZ(G) for some integer m; hence, g = amx for some x in the center of G. Similarly, if hZ(G) ∈ G/Z(G), there exists a y in Z(G) such that h = any for some integer n. Since x and y are in the center of G, they commute with all other elements of G; therefore, gh = amxany = am+nxy = anyamx = hg, and G must be abelian. 14.3 Burnside’s Counting Theorem Suppose that we are to color the vertices of a square with two diﬀerent colors, say black and white. We
might suspect that there would be 24 = 16 diﬀerent colorings. However, some of these colorings are equivalent. If we color the ﬁrst vertex black and the remaining vertices white, it is the same as coloring the second vertex black and the remaining ones white since we could obtain the second coloring simply by rotating the square 90◦ (Figure 14.1). Figure 14.1. Equivalent colorings of square 220 CHAPTER 14 GROUP ACTIONS Burnside’s Counting Theorem oﬀers a method of computing the number of distinguishable ways in which something can be done. In addition to its geometric applications, the theorem has interesting applications to areas in switching theory and chemistry. The proof of Burnside’s Counting Theorem depends on the following lemma. Lemma 14.6 Let X be a G-set and suppose that x ∼ y. Then Gx is isomorphic to Gy. In particular, \|Gx\| = \|Gy\|. Proof. Let G act on X by (g, x) → g · x. Since x ∼ y, there exists a g ∈ G such that g · x = y. Let a ∈ Gx. Since gag−1 · y = ga · g−1y = ga · x = g · x = y, we can deﬁne a map φ : Gx → Gy by φ(a) = gag−1. The map φ is a homomorphism since φ(ab) = gabg−1 = gag−1gbg−1 = φ(a)φ(b). Suppose that φ(a) = φ(b). Then gag−1 = gbg−1 or a = b; hence, the map is injective. To show that φ is onto, let b be in Gy; then g−1bg is in Gx since g−1bg · x = g−1b · gx = g−1b · y = g−1 · y = x; and φ(g−1bg) = b. Theorem 14.7 (Burnside) Let G be a ﬁnite group acting on a set X and let k denote the number of orbits of X. Then k = 1 \|G\| g∈G \|Xg\|. Proof. We look at all the ﬁxed points x of
all the elements in g ∈ G; that is, we look at all g’s and all x’s such that gx = x. If viewed in terms of ﬁxed point sets, the number of all g’s ﬁxing x’s is g∈G \|Xg\|. However, if viewed in terms of the stabilizer subgroups, this number is x∈X \|Gx\|; 14.3 BURNSIDE’S COUNTING THEOREM 221 hence, g∈G \|Xg\| = x∈X \|Gx\|. By Lemma 14.6, y∈Ox \|Gy\| = \|Ox\| · \|Gx\|. By Theorem 14.3 and Lagrange’s Theorem, this expression is equal to \|G\|. Summing over all of the k distinct orbits, we conclude that g∈G \|Xg\| = x∈X \|Gx\| = k · \|G\|. Example 11. Let X = {1, 2, 3, 4, 5} and suppose that G is the permutation group G = {(1), (13), (13)(25), (25)}. The orbits of X are {1, 3}, {2, 5}, and {4}. The ﬁxed point sets are X(1) = X X(13) = {2, 4, 5} X(13)(25) = {4} X(25) = {1, 3, 4}. Burnside’s Theorem says that k = 1 \|G\| g∈G \|Xg\| = 1 4 (5 + 3 + 1 + 3) = 3. A Geometric Example Before we apply Burnside’s Theorem to switching-theory problems, let us examine the number of ways in which the vertices of a square can be colored black or white. Notice that we can sometimes obtain equivalent colorings by simply applying a rigid motion to the square. For instance, as we have pointed out, if we color one of the vertices black and the remaining three white, it does not matter which vertex was colored black since a rotation will give an equivalent coloring. The symmetry group of a square, D4, is given by the following permuta- tions: (1) (1234) (13) (12)(34) (24
) (14)(23) (1432) (13)(24) 222 CHAPTER 14 GROUP ACTIONS The group G acts on the set of vertices {1, 2, 3, 4} in the usual manner. We can describe the diﬀerent colorings by mappings from X into Y = {B, W } where B and W represent the colors black and white, respectively. Each map f : X → Y describes a way to color the corners of the square. Every σ ∈ D4 induces a permutation σ of the possible colorings given by σ(f ) = f ◦ σ for f : X → Y. For example, suppose that f is deﬁned by f (1) = B f (2) = W f (3) = W f (4) = W and σ = (12)(34). Then σ(f ) = f ◦ σ sends vertex 2 to B and the remaining vertices to W. The set of all such σ is a permutation group G on the set of possible colorings. Let X denote the set of all possible colorings; that is, X is the set of all possible maps from X to Y. Now we must compute the number of G-equivalence classes. 1. X(1) = X since the identity ﬁxes every possible coloring. 24 = 16. \| X\| = 2. X(1234) consists of all f ∈ X such that f is unchanged by the permutation (1234). In this case f (1) = f (2) = f (3) = f (4), so that all values of f must be the same; that is, either f (x) = B or f (x) = W for every vertex x of the square. So \| X(1234)\| = 2. 3. \| X(1432)\| = 2. 4. For X(13)(24), f (1) = f (3) and f (2) = f (4). Thus, \| X(13)(24)\| = 22 = 4. 5. \| X(12)(34)\| = 4. 6. \| X(14)(23)\| = 4. 7. For X(13), f (1) = f (3) and the other corners can be of any color; hence, \| X(13)\| = 23 = 8. 8. \|
X(24)\| = 8. By Burnside’s Theorem, we can conclude that there are exactly (24 + 21 + 22 + 21 + 22 + 22 + 23 + 23) = 6 1 8 ways to color the vertices of the square. 14.3 BURNSIDE’S COUNTING THEOREM 223 Proposition 14.8 Let G be a permutation group of X and X the set of functions from X to Y. Then there exists a permutation group G acting on X, where σ ∈ G is deﬁned by σ(f ) = f ◦ σ for σ ∈ G and f ∈ X. Furthermore, if n is the number of cycles in the cycle decomposition of σ, then \| Xσ\| = \|Y \|n. Proof. Let σ ∈ G and f ∈ X. Clearly, f ◦ σ is also in X. Suppose that g is another function from X to Y such that σ(f ) = σ(g). Then for each x ∈ X, f (σ(x)) = σ(f )(x) = σ(g)(x) = g(σ(x)). Since σ is a permutation of X, every element x in X is the image of some x in X under σ; hence, f and g agree on all elements of X. Therefore, f = g and σ is injective. The map σ → σ is onto, since the two sets are the same size. Suppose that σ is a permutation of X with cycle decomposition σ = σ1σ2 · · · σn. Any f in Xσ must have the same value on each cycle of σ. Since there are n cycles and \|Y \| possible values for each cycle, \| Xσ\| = \|Y \|n. Example 12. Let X = {1, 2,..., 7} and suppose that Y = {A, B, C}. If g is the permutation of X given by (13)(245) = (13)(245)(6)(7), then n = 4. Any f ∈ Fg must have the same value on each cycle in g. There are \|Y \| = 3 such choices for any value, so \|Fg\| = 34 = 81. Example 13. Suppose that we wish to color the vertices of
a square using four diﬀerent colors. By Proposition 14.8, we can immediately decide that there are (44 + 41 + 42 + 41 + 42 + 42 + 43 + 43) = 55 1 8 possible ways. Switching Functions In switching theory we are concerned with the design of electronic circuits with binary inputs and outputs. The simplest of these circuits is a switching function that has n inputs and a single output (Figure 14.2). Large electronic circuits can often be constructed by combining smaller modules of this kind. The inherent problem here is that even for a simple circuit a large number of diﬀerent switching functions can be constructed. With only four inputs 224 CHAPTER 14 GROUP ACTIONS x1 x2... xn f f (x1, x2,..., xn) Figure 14.2. A switching function of n variables and a single output, we can construct 65, 536 diﬀerent switching functions. However, we can often replace one switching function with another merely by permuting the input leads to the circuit (Figure 14.3). a b f f (a, b) a b f f (b, a) = g(a, b) Figure 14.3. A switching function of two variables We deﬁne a switching or Boolean function of n variables to be a function from Zn 2 to Z2. Since any switching function can have two possible values for each binary n-tuple and there are 2n binary n-tuples, 22n switching functions are possible for n variables. In general, allowing permutations of the inputs greatly reduces the number of diﬀerent kinds of modules that are needed to build a large circuit. The possible switching functions with two input variables a and b are listed in Table 14.1. Two switching functions f and g are equivalent if g can be obtained from f by a permutation of the input variables. For example, g(a, b, c) = f (b, c, a). In this case g ∼ f via the permutation (acb). In the case of switching functions of two variables, the permutation (ab) reduces 16 possible switching functions to 12 equivalent functions since f2 ∼ f4 f3 ∼ f5 f10 ∼ f12 f11 ∼ f13. For three input variables there are 223 = 256 possible switching functions; in the case of four variables there are 224 = 65,536. The number of equivalence 14
.3 BURNSIDE’S COUNTING THEOREM 225 Table 14.1. Switching functions in two variables Inputs 0 0 1 0 0 1 1 1 Inputs 0 0 1 1 0 1 0 1 f0 0 0 0 0 f8 1 0 0 0 f1 0 0 0 1 f9 1 0 0 1 f2 0 0 1 0 f10 1 0 1 0 Outputs f4 f3 0 0 1 0 0 1 1 0 Outputs f12 f11 1 1 1 0 0 1 0 1 f5 0 1 0 1 f13 1 1 0 1 f6 0 1 1 0 f14 1 1 1 0 f7 0 1 1 1 f15 1 1 1 1 classes is too large to reasonably calculate directly. It is necessary to employ Burnside’s Theorem. Consider a switching function with three possible inputs, a, b, and c. As we have mentioned, two switching functions f and g are equivalent if a permutation of the input variables of f gives g. It is important to notice that a permutation of the switching functions is not simply a permutation of the input values {a, b, c}. A switching function is a set of output values for the inputs a, b, and c, so when we consider equivalent switching functions, we are permuting 23 possible outputs, not just three input values. For example, each binary triple (a, b, c) has a speciﬁc output associated with it. The permutation (acb) changes outputs as follows: (0, 0, 0) → (0, 0, 0) (0, 0, 1) → (0, 1, 0) (0, 1, 0) → (1, 0, 0)... (1, 1, 0) → (1, 0, 1) (1, 1, 1) → (1, 1, 1). 226 CHAPTER 14 GROUP ACTIONS Let X be the set of output values for a switching function in n variables. Then \|X\| = 2n. We can enumerate these values as follows: (0,..., 0, 1) → 0 (0,..., 1, 0) → 1 (0,..., 1, 1) → 2... (1,..., 1, 1) → 2n − 1. Now let us consider a circuit with four input variables and a single output. Suppose that we can perm
ute the leads of any circuit according to the following permutation group: (a) (ac) (bd) (adcb) (abcd) (ab)(cd) (ad)(bc) (ac)(bd). The permutations of the four possible input variables induce the permutations of the output values in Table 14.2. Hence, there are (216 + 2 · 212 + 2 · 26 + 3 · 210) = 9616 1 8 possible switching functions of four variables under this group of permutations. This number will be even smaller if we consider the full symmetric group on four letters. Table 14.2. Permutations of switching functions in four variables Group Permutation (a) (ac) (bd) (adcb) (abcd) (ab)(cd) (ad)(bc) (ac)(bd) Switching Function Permutation (0) (2, 8)(3, 9)(6, 12)(7, 13) (1, 4)(3, 6)(9, 12)(11, 14) (1, 2, 4, 8)(3, 6.12, 9)(5, 10)(7, 14, 13, 11) (1, 8, 4, 2)(3, 9, 12, 6)(5, 10)(7, 11, 13, 14) (1, 2)(4, 8)(5, 10)(6, 9)(7, 11)(13, 14) (1, 8)(2, 4)(3, 12)(5, 10)(7, 14)(11, 13) (1, 4)(2, 8)(3, 12)(6, 9)(7, 13)(11, 14) Number of Cycles 16 12 12 6 6 10 10 10 Historical Note EXERCISES 227 William Burnside was born in London in 1852. He attended Cambridge University from 1871 to 1875 and won the Smith’s Prize in his last year. After his graduation he lectured at Cambridge. He was made a member of the Royal Society in 1893. Burnside wrote approximately 150 papers on topics in applied mathematics, diﬀerential geometry, and probability, but his most famous contributions were in group theory. Several of Burnside’s conjectures have stimulated research to this day. One such conjecture was that every group of odd order is solvable; that is, for a group G of odd order, there exists a sequence of subgroups G = Hn ⊃
Hn−1 ⊃ · · · ⊃ H1 ⊃ H0 = {e} such that Hi is normal in Hi+1 and Hi+1/Hi is abelian. This conjecture was ﬁnally proven by W. Feit and J. Thompson in 1963. Burnside’s The Theory of Groups of Finite Order, published in 1897, was one of the ﬁrst books to treat groups in a modern context as opposed to permutation groups. The second edition, published in 1911, is still a classic. Exercises 1. Examples 1–5 in the ﬁrst section each describe an action of a group G on a set X, which will give rise to the equivalence relation deﬁned by G-equivalence. For each example, compute the equivalence classes of the equivalence relation, the G-equivalence classes. 2. Compute all Xg and all Gx for each of the following permutation groups. (a) X = {1, 2, 3}, G = S3 = {(1), (12), (13), (23), (123), (132)} (b) X = {1, 2, 3, 4, 5, 6}, G = {(1), (12), (345), (354), (12)(345), (12)(354)} 3. Compute the G-equivalence classes of X for each of the G-sets in Exercise 2. For each x ∈ X verify that \|G\| = \|Ox\| · \|Gx\|. 4. Let G be the additive group of real numbers. Let the action of θ ∈ G on the real plane R2 be given by rotating the plane counterclockwise about the origin through θ radians. Let P be a point on the plane other than the origin. (a) Show that R2 is a G-set. (b) Describe geometrically the orbit containing P. (c) Find the group GP. 5. Let G = A4 and suppose that G acts on itself by conjugation; that is, (g, h) → ghg−1. (a) Determine the conjugacy classes (orbits) of each element of G. 228 CHAPTER 14 GROUP ACTIONS (b) Determine all of the isotropy subgroups for each element of G. 6. Find
the conjugacy classes and the class equation for each of the following groups. (a) S4 (b) D5 (c) Z9 (d) Q8 7. Write the class equation for S5 and for A5. 8. If a square remains ﬁxed in the plane, how many diﬀerent ways can the corners of the square be colored if three colors are used? 9. How many ways can the vertices of an equilateral triangle be colored using three diﬀerent colors? 10. Find the number of ways a six-sided die can be constructed if each side is marked diﬀerently with 1,..., 6 dots. 11. Up to a rotation, how many ways can the faces of a cube be colored with three diﬀerent colors? 12. Consider 12 straight wires of equal lengths with their ends soldered together to form the edges of a cube. Either silver or copper wire can be used for each edge. How many diﬀerent ways can the cube be constructed? 13. Suppose that we color each of the eight corners of a cube. Using three diﬀerent colors, how many ways can the corners be colored up to a rotation of the cube? 14. Each of the faces of a regular tetrahedron can be painted either red or white. Up to a rotation, how many diﬀerent ways can the tetrahedron be painted? 15. Suppose that the vertices of a regular hexagon are to be colored either red or white. How many ways can this be done up to a symmetry of the hexagon? 16. A molecule of benzene is made up of six carbon atoms and six hydrogen atoms, linked together in a hexagonal shape as in Figure 14.4. (a) How many diﬀerent compounds can be formed by replacing one or more of the hydrogen atoms with a chlorine atom? (b) Find the number of diﬀerent chemical compounds that can be formed by replacing three of the six hydrogen atoms in a benzene ring with a CH3 radical. 17. How many equivalence classes of switching functions are there if the input variables x1, x2, and x3 can be permuted by any permutation in S3? What if the input variables x1, x2, x3, and x4 can be permuted by any permutation in S4? EXERCIS
ES 229 H H H H H H Figure 14.4. A benzene ring 18. How many equivalence classes of switching functions are there if the input variables x1, x2, x3, and x4 can be permuted by any permutation in the subgroup of S4 generated by the permutation (x1x2x3x4)? 19. A striped necktie has 12 bands of color. Each band can be colored by one of four possible colors. How many possible diﬀerent-colored neckties are there? 20. A group acts faithfully on a G-set X if the identity is the only element of G that leaves every element of X ﬁxed. Show that G acts faithfully on X if and only if no two distinct elements of G have the same action on each element of X. 21. Let p be prime. Show that the number of diﬀerent abelian groups of order pn (up to isomorphism) is the same as the number of conjugacy classes in Sn. 22. Let a ∈ G. Show that for any g ∈ G, gC(a)g−1 = C(gag−1). 23. Let \|G\| = pn and suppose that \|Z(G)\| = pn−1 for p prime. Prove that G is abelian. 24. Let G be a group with order pn where p is prime and X a ﬁnite G-set. If XG = {x ∈ X : gx = x for all g ∈ G} is the set of elements in X ﬁxed by the group action, then prove that \|X\| ≡ \|XG\| (mod p). 25. If G is a group of order pn, where p is prime and n ≥ 2, show that G must have a proper subgroup of order p. If n ≥ 3, is it true that G will have a proper subgroup of order p2? Programming Exercise Write a program to compute the number of conjugacy classes in Sn. What is the largest n for which your program will work? 230 CHAPTER 14 GROUP ACTIONS References and Suggested Reading [1] De Bruijin, N. G. “P´olya’s Theory of Counting,” in Applied Combinatorial Mathematics, Beckenbach, E. F., ed. Wiley,
New York, 1964. [2] Eidswick, J. A. “Cubelike Puzzles—What Are They and How Do You Solve Them?” American Mathematical Monthly 93 (1986), 157–76. [3] Harary, F., Palmer, E. M., and Robinson, R. W. “P´olya’s Contributions to Chemical Enumeration,” in Chemical Applications of Graph Theory, Balaban, A. T., ed. Academic Press, London, 1976. [4] G˚ading, L. and Tambour, T. Algebra for Computer Science. Springer-Verlag, New York, 1988. [5] Laufer, H. B. Discrete Mathematics and Applied Modern Algebra. PWS-Kent, Boston, 1984. [6] P´olya, G. and Read, R. C. Combinatorial Enumeration of Groups, Graphs, and Chemical Compounds. Springer-Verlag, New York, 1985. [7] Shapiro, L. W. “Finite Groups Acting on Sets with Applications,” Mathematics Magazine, May–June 1973, 136–47. Sage Sage has many commands related to conjugacy, which is a group action. It also has commands for orbits and stabilizers of permutation groups. In the supplement, we illustrate the automorphism group of a (combinatorial) graph as another example of a group action on the vertex set of the graph. 15 The Sylow Theorems We already know that the converse of Lagrange’s Theorem is false. If G is a group of order m and n divides m, then G does not necessarily possess a subgroup of order n. For example, A4 has order 12 but does not possess a subgroup of order 6. However, the Sylow Theorems do provide a partial converse for Lagrange’s Theorem: in certain cases they guarantee us subgroups of speciﬁc orders. These theorems yield a powerful set of tools for the classiﬁcation of all ﬁnite nonabelian groups. 15.1 The Sylow Theorems We will use the idea of group actions to prove the Sylow Theorems. Recall for a moment what it means for G to act on itself by conjugation and how conjugacy classes are distributed in the group according to the class
equation, discussed in Chapter 14. A group G acts on itself by conjugation via the map (g, x) → gxg−1. Let x1,..., xk be representatives from each of the distinct conjugacy classes of G that consist of more than one element. Then the class equation can be written as \|G\| = \|Z(G)\| + [G : C(x1)] + · · · + [G : C(xk)], where Z(G) = {g ∈ G : gx = xg for all x ∈ G} is the center of G and C(xi) = {g ∈ G : gxi = xig} is the centralizer subgroup of xi. We now begin our investigation of the Sylow Theorems by examining subgroups of order p, where p is prime. A group G is a p-group if every element in G has as its order a power of p, where p is a prime number. A subgroup of a group G is a p-subgroup if it is a p-group. Theorem 15.1 (Cauchy) Let G be a ﬁnite group and p a prime such that p divides the order of G. Then G contains a subgroup of order p. 231 232 CHAPTER 15 THE SYLOW THEOREMS Proof. We will use induction on the order of G. If \|G\| = p, then clearly G must have an element of order p. Now assume that every group of order k, where p ≤ k < n and p divides k, has an element of order p. Assume that \|G\| = n and p \| n and consider the class equation of G: \|G\| = \|Z(G)\| + [G : C(x1)] + · · · + [G : C(xk)]. We have two cases. Case 1. The order of one of the centralizer subgroups, C(xi), is divisible by p for some i, i = 1,..., k. In this case, by our induction hypothesis, we are done. Since C(xi) is a proper subgroup of G and p divides \|C(xi)\|, C(xi) must contain an element of order p. Hence, G must contain an element of order p. Case 2. The order of no centralizer subgroup is divisible
by p. Then p divides [G : C(xi)], the order of each conjugacy class in the class equation; hence, p must divide the center of G, Z(G). Since Z(G) is abelian, it must have a subgroup of order p by the Fundamental Theorem of Finite Abelian Groups. Therefore, the center of G contains an element of order p. Corollary 15.2 Let G be a ﬁnite group. Then G is a p-group if and only if \|G\| = pn. Example 1. Let us consider the group A5. We know that \|A5\| = 60 = 22·3·5. By Cauchy’s Theorem, we are guaranteed that A5 has subgroups of orders 2, 3 and 5. The Sylow Theorems give us even more information about the possible subgroups of A5. We are now ready to state and prove the ﬁrst of the Sylow Theorems. The proof is very similar to the proof of Cauchy’s Theorem. Theorem 15.3 (First Sylow Theorem) Let G be a ﬁnite group and p a prime such that pr divides \|G\|. Then G contains a subgroup of order pr. Proof. We induct on the order of G once again. If \|G\| = p, then we are done. Now suppose that the order of G is n with n > p and that the theorem is true for all groups of order less than n. We shall apply the class equation once again: \|G\| = \|Z(G)\| + [G : C(x1)] + · · · + [G : C(xk)]. First suppose that p does not divide [G : C(xi)] for some i. Then pr \| \|C(xi)\|, since pr divides \|G\| = \|C(xi)\| · [G : C(xi)]. Now we can apply the induction hypothesis to C(xi). 15.1 THE SYLOW THEOREMS 233 Hence, we may assume that p divides [G : C(xi)] for all i. Since p divides \|G\|, the class equation says that p must divide \|Z(G)\|; hence, by Cauchy’s Theorem, Z(G) has an element of order p, say g. Let N be the group
generated by g. Clearly, N is a normal subgroup of Z(G) since Z(G) is abelian; therefore, N is normal in G since every element in Z(G) commutes with every element in G. Now consider the factor group G/N of order \|G\|/p. By the induction hypothesis, G/N contains a subgroup H of order pr−1. The inverse image of H under the canonical homomorphism φ : G → G/N is a subgroup of order pr in G. A Sylow p-subgroup P of a group G is a maximal p-subgroup of G. To prove the other two Sylow Theorems, we need to consider conjugate subgroups as opposed to conjugate elements in a group. For a group G, let S be the collection of all subgroups of G. For any subgroup H, S is a H-set, where H acts on S by conjugation. That is, we have an action H × S → S h · K → hKh−1 deﬁned by for K in S. The set N (H) = {g ∈ G : gHg−1 = H} is a subgroup of G. Notice that H is a normal subgroup of N (H). In fact, N (H) is the largest subgroup of G in which H is normal. We call N (H) the normalizer of H in G. Lemma 15.4 Let P be a Sylow p-subgroup of a ﬁnite group G and let x have as its order a power of p. If x−1P x = P, then x ∈ P. Proof. Certainly x ∈ N (P ), and the cyclic subgroup, xP ⊂ N (P )/P, has as its order a power of p. By the Correspondence Theorem there exists a subgroup H of N (P ) such that H/P = xP. Since \|H\| = \|P \| · \|xP \|, the order of H must be a power of p. However, P is a Sylow p-subgroup contained in H. Since the order of P is the largest power of p dividing \|G\|, H = P. Therefore, H/P is the trivial subgroup and xP = P, or x ∈ P. Lem
ma 15.5 Let H and K be subgroups of G. The number of distinct H-conjugates of K is [H : N (K) ∩ H]. 234 CHAPTER 15 THE SYLOW THEOREMS Proof. We deﬁne a bijection between the conjugacy classes of K and the right cosets of N (K) ∩ H by h−1Kh → (N (K) ∩ H)h. To show that this map is a bijection, let h1, h2 ∈ H and suppose that (N (K)∩H)h1 = (N (K)∩H)h2. Then h2h−1 2 Kh2, and the map is an injection. It is easy to see that this map is surjective; hence, we have a one-to-one and onto map between the H-conjugates of K and the right cosets of N (K) ∩ H in H. 1 ∈ N (K). Therefore, K = h2h−1 1 Kh1 = h−1 1 Kh1h−1 2 or h−1 Theorem 15.6 (Second Sylow Theorem) Let G be a ﬁnite group and p a prime dividing \|G\|. Then all Sylow p-subgroups of G are conjugate. That is, if P1 and P2 are two Sylow p-subgroups, there exists a g ∈ G such that gP1g−1 = P2. Proof. Let P be a Sylow p-subgroup of G and suppose that \|G\| = prm and \|P \| = pr. Let P = {P = P1, P2,..., Pk} consist of the distinct conjugates of P in G. By Lemma 15.5, k = [G : N (P )]. Notice that \|G\| = prm = \|N (P )\| · [G : N (P )] = \|N (P )\| · k. Since pr divides \|N (P )\|, p cannot divide k. Given any other Sylow p-subgroup Q, we must show that Q ∈ P. Consider the Q-conjugacy classes of each Pi. Clearly, these conjugacy classes partition P. The size of the partition containing Pi is [Q : N (Pi)
∩ Q]. Lagrange’s Theorem tells us that this number is a divisor of \|Q\| = pr. Hence, the number of conjugates in every equivalence class of the partition is a power of p. However, since p does not divide k, one of these equivalence classes must contain only a single Sylow p-subgroup, say Pj. Therefore, for some Pj, x−1Pjx = Pj for all x ∈ Q. By Lemma 15.4, Pj = Q. Theorem 15.7 (Third Sylow Theorem) Let G be a ﬁnite group and let p be a prime dividing the order of G. Then the number of Sylow p-subgroups is congruent to 1 (mod p) and divides \|G\|. Proof. Let P be a Sylow p-subgroup acting on the set of Sylow p-subgroups, P = {P = P1, P2,..., Pk}, by conjugation. From the proof of the Second Sylow Theorem, the only P -conjugate of P is itself and the order of the other P -conjugacy classes is a 15.2 EXAMPLES AND APPLICATIONS 235 power of p. Each P -conjugacy class contributes a positive power of p toward \|P\| except the equivalence class {P }. Since \|P\| is the sum of positive powers of p and 1, \|P\| ≡ 1 (mod p). Now suppose that G acts on P by conjugation. Since all Sylow psubgroups are conjugate, there can be only one orbit under this action. For P ∈ P, \|P\| = \|orbit of P\| = [G : N (P )]. But [G : N (P )] is a divisor of \|G\|; consequently, the number of Sylow p-subgroups of a ﬁnite group must divide the order of the group. Historical Note Peter Ludvig Mejdell Sylow was born in 1832 in Christiania, Norway (now Oslo). After attending Christiania University, Sylow taught high school. In 1862 he obtained a temporary appointment at Christiania University. Even though his appointment was relatively brief, he inﬂuenced students such as Sophus Lie (1842–1899). Sylow had a chance at
a permanent chair in 1869, but failed to obtain the appointment. In 1872, he published a 10-page paper presenting the theorems that now bear his name. Later Lie and Sylow collaborated on a new edition of Abel’s works. In 1898, a chair at Christiania University was ﬁnally created for Sylow through the eﬀorts of his student and colleague Lie. Sylow died in 1918. 15.2 Examples and Applications Example 2. Using the Sylow Theorems, we can determine that A5 has subgroups of orders 2, 3, 4, and 5. The Sylow p-subgroups of A5 have orders 3, 4, and 5. The Third Sylow Theorem tells us exactly how many Sylow p-subgroups A5 has. Since the number of Sylow 5-subgroups must divide 60 and also be congruent to 1 (mod 5), there are either one or six Sylow 5-subgroups in A5. All Sylow 5-subgroups are conjugate. If there were only a single Sylow 5-subgroup, it would be conjugate to itself; that is, it would be a normal subgroup of A5. Since A5 has no normal subgroups, this is impossible; hence, we have determined that there are exactly six distinct Sylow 5-subgroups of A5. The Sylow Theorems allow us to prove many useful results about ﬁnite groups. By using them, we can often conclude a great deal about groups of a particular order if certain hypotheses are satisﬁed. 236 CHAPTER 15 THE SYLOW THEOREMS Theorem 15.8 If p and q are distinct primes with p < q, then every group G of order pq has a single subgroup of order q and this subgroup is normal in G. Hence, G cannot be simple. Furthermore, if q ≡ 1 (mod p), then G is cyclic. Proof. We know that G contains a subgroup H of order q. The number of conjugates of H divides pq and is equal to 1 + kq for k = 0, 1,.... However, 1 + q is already too large to divide the order of the group; hence, H can only be conjugate to itself. That is, H must be normal in G. The group G also has a Sylow p
-subgroup, say K. The number of conjugates of K must divide q and be equal to 1 + kp for k = 0, 1,.... Since q is prime, either 1 + kp = q or 1 + kp = 1. If 1 + kp = 1, then K is normal in G. In this case, we can easily show that G satisﬁes the criteria, given in Chapter 8, for the internal direct product of H and K. Since H is isomorphic to Zq and K is isomorphic to Zp, G ∼= Zp × Zq ∼= Zpq by Theorem 9.10. Example 3. Every group of order 15 is cyclic. This is true because 15 = 5 · 3 and 5 ≡ 1 (mod 3). Example 4. Let us classify all of the groups of order 99 = 32 · 11 up to isomorphism. First we will show that every group G of order 99 is abelian. By the Third Sylow Theorem, there are 1 + 3k Sylow 3-subgroups, each of order 9, for some k = 0, 1, 2,.... Also, 1 + 3k must divide 11; hence, there can only be a single normal Sylow 3-subgroup H in G. Similarly, there are 1 + 11k Sylow 11-subgroups and 1 + 11k must divide 9. Consequently, there is only one Sylow 11-subgroup K in G. By Corollary 14.5, any group of order p2 is abelian for p prime; hence, H is isomorphic either to Z3 × Z3 or to Z9. Since K has order 11, it must be isomorphic to Z11. Therefore, the only possible groups of order 99 are Z3 × Z3 × Z11 or Z9 × Z11 up to isomorphism. To determine all of the groups of order 5 · 7 · 47 = 1645, we need the following theorem. Theorem 15.9 Let G = aba−1b−1 : a, b ∈ G be the subgroup consisting of all ﬁnite products of elements of the form aba−1b−1 in a group G. Then G is a normal subgroup of G and G/G is abelian. 15.2 EXAMPLES AND APPLICATIONS 237 The subgroup G of
G is called the commutator subgroup of G. We leave the proof of this theorem as an exercise. Example 5. We will now show that every group of order 5 · 7 · 47 = 1645 is abelian, and cyclic by Corollary 9.11. By the Third Sylow Theorem, G has only one subgroup H1 of order 47. So G/H1 has order 35 and must be abelian by Theorem 15.8. Hence, the commutator subgroup of G is contained in H which tells us that \|G\| is either 1 or 47. If \|G\| = 1, we are done. Suppose that \|G\| = 47. The Third Sylow Theorem tells us that G has only one subgroup of order 5 and one subgroup of order 7. So there exist normal subgroups H2 and H3 in G, where \|H2\| = 5 and \|H3\| = 7. In either case the quotient group is abelian; hence, G must be a subgroup of Hi, i = 1, 2. Therefore, the order of G is 1, 5, or 7. However, we already have determined that \|G\| = 1 or 47. So the commutator subgroup of G is trivial, and consequently G is abelian. Finite Simple Groups Given a ﬁnite group, one can ask whether or not that group has any normal subgroups. Recall that a simple group is one with no proper nontrivial normal subgroups. As in the case of A5, proving a group to be simple can be a very diﬃcult task; however, the Sylow Theorems are useful tools for proving that a group is not simple. Usually some sort of counting argument is involved. Example 6. Let us show that no group G of order 20 can be simple. By the Third Sylow Theorem, G contains one or more Sylow 5-subgroups. The number of such subgroups is congruent to 1 (mod 5) and must also divide 20. The only possible such number is 1. Since there is only a single Sylow 5-subgroup and all Sylow 5-subgroups are conjugate, this subgroup must be normal. Example 7. Let G be a ﬁnite group of order pn, n > 1 and p prime. By Theorem 14.4, G has a
nontrivial center. Since the center of any group G is a normal subgroup, G cannot be a simple group. Therefore, groups of orders 4, 8, 9, 16, 25, 27, 32, 49, 64, and 81 are not simple. In fact, the groups of order 4, 9, 25, and 49 are abelian by Corollary 14.5. Example 8. No group of order 56 = 23 · 7 is simple. We have seen that if we can show that there is only one Sylow p-subgroup for some prime p 238 CHAPTER 15 THE SYLOW THEOREMS dividing 56, then this must be a normal subgroup and we are done. By the Third Sylow Theorem, there are either one or eight Sylow 7-subgroups. If there is only a single Sylow 7-subgroup, then it must be normal. On the other hand, suppose that there are eight Sylow 7-subgroups. Then each of these subgroups must be cyclic; hence, the intersection of any two of these subgroups contains only the identity of the group. This leaves 8 · 6 = 48 distinct elements in the group, each of order 7. Now let us count Sylow 2-subgroups. There are either one or seven Sylow 2-subgroups. Any element of a Sylow 2-subgroup other than the identity must have as its order a power of 2; and therefore cannot be one of the 48 elements of order 7 in the Sylow 7-subgroups. Since a Sylow 2-subgroup has order 8, there is only enough room for a single Sylow 2-subgroup in a group of order 56. If there is only one Sylow 2-subgroup, it must be normal. For other groups G it is more diﬃcult to prove that G is not simple. Suppose G has order 48. In this case the technique that we employed in the last example will not work. We need the following lemma to prove that no group of order 48 is simple. Lemma 15.10 Let H and K be ﬁnite subgroups of a group G. Then Proof. Recall that \|HK\| = \|H\| · \|K\| \|H ∩ K\|. HK = {hk : h ∈ H, k ∈ K}. Certainly, \|HK\| ≤ \|H\| · \|K\| since some element in HK could be written
as the product of diﬀerent elements in H and K. It is quite possible that h1k1 = h2k2 for h1, h2 ∈ H and k1, k2 ∈ K. If this is the case, let a = (h1)−1h2 = k1(k2)−1. Notice that a ∈ H ∩ K, since (h1)−1h2 is in H and k2(k1)−1 is in K; consequently, h2 = h1a−1 k2 = ak1. Conversely, let h = h1b−1 and k = bk1 for b ∈ H ∩ K. Then hk = h1k1, where h ∈ H and k ∈ K. Hence, any element hk ∈ HK can be written in EXERCISES 239 the form hiki for hi ∈ H and ki ∈ K, as many times as there are elements in H ∩ K; that is, \|H ∩ K\| times. Therefore, \|HK\| = (\|H\| · \|K\|)/\|H ∩ K\|. Example 9. To demonstrate that a group G of order 48 is not simple, we will show that G contains either a normal subgroup of order 8 or a normal subgroup of order 16. By the Third Sylow Theorem, G has either one or three Sylow 2-subgroups of order 16. If there is only one subgroup, then it must be a normal subgroup. Suppose that the other case is true, and two of the three Sylow 2subgroups are H and K. We claim that \|H ∩ K\| = 8. If \|H ∩ K\| ≤ 4, then by Lemma 15.10, \|HK\| = = 64, 16 · 16 4 which is impossible. Notice that H ∩ K has index two in both of H and K, so is normal in both, and thus H and K are each in the normalizer of H ∩ K. Because H is a subgroup of N (H ∩ K) and because N (H ∩ K) has strictly more than 16 elements, \|N (H ∩ K)\| must be a multiple of 16 greater than 1, as well as dividing 48. The only possibility is that \|N (H ∩ K)\| = 48. Hence, N (
H ∩ K) = G. The following famous conjecture of Burnside was proved in a long and diﬃcult paper by Feit and Thompson [2]. Theorem 15.11 (Odd Order Theorem) Every ﬁnite simple group of nonprime order must be of even order. The proof of this theorem laid the groundwork for a program in the 1960s and 1970s that classiﬁed all ﬁnite simple groups. The success of this program is one of the outstanding achievements of modern mathematics. Exercises 1. What are the orders of all Sylow p-subgroups where G has order 18, 24, 54, 72, and 80? 2. Find all the Sylow 3-subgroups of S4 and show that they are all conjugate. 3. Show that every group of order 45 has a normal subgroup of order 9. 4. Let H be a Sylow p-subgroup of G. Prove that H is the only Sylow p-subgroup of G contained in N (H). 240 CHAPTER 15 THE SYLOW THEOREMS 5. Prove that no group of order 96 is simple. 6. Prove that no group of order 160 is simple. 7. If H is a normal subgroup of a ﬁnite group G and \|H\| = pk for some prime p, show that H is contained in every Sylow p-subgroup of G. 8. Let G be a group of order p2q2, where p and q are distinct primes such that q \|p2 − 1 and p \|q2 − 1. Prove that G must be abelian. List three pairs of primes satisfying these conditions. 9. Show directly that a group of order 33 has only one Sylow 3-subgroup. 10. Let H be a subgroup of a group G. Prove or disprove that the normalizer of H is normal in G. 11. Let G be a ﬁnite group divisible by a prime p. Prove that if there is only one Sylow p-subgroup in G, it must be a normal subgroup of G. 12. Let G be a group of order pr, p prime. Prove that G contains a normal subgroup of order pr−1. 13. Suppose that G is a ﬁnite group of order pnk, where k <
p. Show that G must contain a normal subgroup. 14. Let H be a subgroup of a ﬁnite group G. Prove that gN (H)g−1 = N (gHg−1) for any g ∈ G. 15. Prove that a group of order 108 must have a normal subgroup. 16. Classify all the groups of order 175 up to isomorphism. 17. Show that every group of order 255 is cyclic. 18. Let G have order pe1 1 · · · pen P1,..., Pn where \|Pi\| = pei n and suppose that G has n Sylow p-subgroups i. Prove that G is isomorphic to P1 × · · · × Pn. 19. Let P be a normal Sylow p-subgroup of G. Prove that every inner automor- phism of G ﬁxes P. 20. What is the smallest possible order of a group G such that G is nonabelian and \|G\| is odd? Can you ﬁnd such a group? 21. The Frattini Lemma. If H is a normal subgroup of a ﬁnite group G and P is a Sylow p-subgroup of H, for each g ∈ G show that there is an h in H such that gP g−1 = hP h−1. Also, show that if N is the normalizer of P, then G = HN. 22. Show that if the order of G is pnq, where p and q are primes and p > q, then G contains a normal subgroup. 23. Prove that the number of distinct conjugates of a subgroup H of a ﬁnite group G is [G : N (H)]. 24. Prove that a Sylow 2-subgroup of S5 is isomorphic to D4. EXERCISES 241 25. Another Proof of the Sylow Theorems. (a) Suppose p is prime and p does not divide m. Show that p \| pkm pk. (b) Let S denote the set of all pk element subsets of G. Show that p does not divide \|S\|. (c) Deﬁne an action of G on S by left multiplication, aT = {at : t ∈ T } for a ∈

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