Solutions - AI Some solutions are reworded/rewritten in order to be more in the style that I was teaching (with GT in front of it). FINAL EXAM Problems #1-8 will each count ten points. The best 3 of # 9-14 will count ten points each. (*) denotes extra credit. There are several formulas at the end. Good luck! 1. (a) Determine all Abelian groups (up to isomorphism) of order 3 4 and give a connection to a set of partitions. Ans. These correspond 1-1 with the five partitions of 4: 4 3 + 1 2 + 2 2 + 1 + 1 1 + 1 + 1 + 1 Z 3 4 Z 3 3 Z 3 Z 3 2 Z 3 2 Z 3 2 Z 3 Z 3 Z 3 Z 3 Z 3 Z 3 (b) Determine all Abelian groups of order 200, up to isomorphism. Ans Each such group factors into A B with A = 25 and B = 8. The Abelian groups A correspond to partitions of 2, and B to partitions of 3. Thus we have A = Z 5 2 or Z 5 Z 5 ; and B = Z 2 3 or Z 2 2 Z 2 or Z 2 Z 2 Z 2, so there are 6 = 2 3 Abelian groups of order 200, up to isomorphism, direct sums A B. 2. (a) Prove that in a group aba 1 = b ab = ba. Ans. We first prove. Assume first that aba 1 = b. Multiplying by a on the right we have (aba 1 ) a = ba. By associativity the left side is (ab) (a 1 a) = (ab) e = ab, and we have ab = ba We have shown. Assume now that ab = ba. Multiply on the right by a 1 to obtain aba 1 = (ba)a 1 = b(aa 1 ) = be = b, reversing the above steps. This shows and completes the proof that aba 1 = b ab = ba. (b) Let G be a finite group such that each element satisfies g 2 = e. a. Show that G is Abelian. Ans. Since (ab) 2 = (ab)(ab) = e, we have a((ab)(ab))b = ab (multiply on the left by a and on right by b. By associativity this is a 2 (ba)(b 2 ) = ab, implying e(ba)e = ab and ab = ba. b. Show that G is isomorphic to a direct product of cyclic groups of order 2. Ans. Every element has order 2 or 1, so G is an Abelian 2-group. However since there are no elements of order 4, the structure theorem for Abelian p-groups implies that G = Z 2 Z 2. 1
3. Consider the map: θ : Z 9 Z 24 : θ(a) = 8a mod 24. (a) Indicate why θ is a homomorphism. Ans. We need to show that θ is well defined, and is a homomorphism. Here are two approaches. Method a. There is a homomorphism θ : Z, + Z 24, θ (k) = 8k mod 24, since a homorphism from Z, + is determined uniquely by θ (1) = 8. Since θ (9) = 0, this homomorphism descends to a homomorphism θ : Z 9 Z 24. OR Method b. Since θ(9) = 8 9 mod 24=0 mod 24, the map θ is well-defined. We need to show that θ respects sums and inverses. We have θ(a + b) = 8(a + b) mod 24. By distributivity of in Z 24 this is 8a + 8b mod 24 = θ(a) + θ(b). Also θ( a) = 8( a) = (8a) mod 24, again by distributivity. (GT) Claim 1: θ : Z 9 Z 24 defined by θ(a) = 8a (mod 24) is well defined. Since the group operation in Z 9 is addition (mod 9), we have to check that θ(9) = θ(0 9 ) = 0 24 Z 24, which is true since θ(9) = 8 9 (mod 24) = 72 (mod 24) = 0 Claim 2: θ(a + b) = θ(a) + θ(b), i.e. θ is a group homomorphism (respects group operations). θ(a + b) = 8(a + b) (mod 24). By distributivity of in Z 24 this is 8a + 8b (mod 24) = θ(a) + θ(b) (b) Determine the kernel K and the image Im(θ) = θ(z 9 ) of θ. Ans. K = {0, 3, 6}. The image θ(z 9 ) = {0, 8, 16}. (c) List the left cosets of K in Z 9. Ans: 0 = K = {0, 3, 6}, 1 = 1 + K = {1, 4, 7}, 2 = 2 + K = {2, 5, 8}. (d) Give the isomorphism θ from Z 9 /K to the image Im(θ) = θ(z 9 ). Ans. θ(0)=θ(k)=0(mod 24), θ(1)=θ(1+k)=8(mod 24), θ(2)=θ(2+k)=16(mod 24). You may wish to include the corresponding Cayley tables. 2
4. (a) Let H be a subgroup of G. Prove that the left cosets satisfy ah bh ah bh =. Ans We show the contrapositive. Suppose that ah bh and let c ah bh. Then c = ah 1 = bh 2, whence a = (bh 2 )(h 1 ) 1. Let h H then ah = b(h 2 h 1 ) 1 h bh by the closure property of the subgroup H. We have shown ah bh. By a similar argument, interchanging the roles of ah and bh by symmetry we have bh ah. We have shown ah bh ah = bh. This is the contrapositive to the statement ah bh ah bh =, so is equivalent to it. (b) Under what condition on H do the left cosets of H in G form a group under the composition inherited from G, i.e. under the operation defined as: (ah)(bh) := (ab)h? Ans. H must be normal in G: ghg 1 H g G. 3
5. Consider the group S 8 of permutations of {1, 2, 3, 4, 5, 6, 7, 8}. (a) Determine the disjoint cycle decomposition of h = (1 2 3 7)(3 4)(1 2 5 6). Ans. (1347)(256) (b) Write h as a product of 2-cycles. Is h in the alternating group A 8? Ans. h = (17)(14)(13)(26)(25) (There are many correct answers). No, h is a produc of an odd number of 2-cycles, so is not in A 8. (c) Determine the largest order of an element of S 8 and specify such an element. Ans. We look at possible disjoint cycle lengths so partitions of 8, and ask for maximum LCM, so wish to avoid common factors in the lengths. We have (7, 1), (5, 3) and none with three parts having no common factor. So 3 5 = 15 is greatest order, occurring for (12437)(586). (And many other permutations given by (5,3) disjoint cycle decomposition. You know how to determine how many such permutations are there! - Don t need it here.) (d) Determine the orbit of g = (1 2 3)(4 5) under the conjugation action of S 8 on S 8. Ans All elements having the same partition (3,2,1,1,1) as disjoint cycle lengths. So all of the form (a, b, c)(d, e) with a, b, c, d, e distinct integers 1 a, b, c, d, e 8. (e) (*) How many elements are in the orbit of g = (1 2 3)(4 5)? Ans ( 8 3) (2!) ( 5 2) 1! = 1120. Note: be prepared for a question with repeated disjoint cycle lengths as 3,3,2 in S 8 or 4, 4, 3, 3 in S 14. 6. (a) Prove that in a group G the centralizer C(a) = {g G gag 1 = a} of the element a G is a subgroup of G. Ans. Using the two step process, we need to show C(a) is closed under products and inverses. Suppose b, c C(a). Then (bc)a(bc) 1 = bcac 1 b 1 = b(cac 1 )b 1. Since c C(a) this is bab 1 = a since b C(a). So C(a) is closed under product. Now assume b C(a) so bab 1 = a. Multiply by b 1 on the left and b on he right to obtain b 1 (bab 1 )b = b 1 ab, which simplifies to a = b 1 ab, implying b 1 C(a). (b) (*) Show that the only groups of order 6 are D 3 = S 3 and Z 6. Let G be a group of order 6. By the Cauchy theorem (or Sylow), G has a subgroup N of order 3, so N = Z 3 as there is only one group of order 3. And G contains H = Z 2. Since N = G /2, N is normal in G and H acts by conjugation on N. Let g N be a generator of N and h a generator of H. We have hnh 1 = n or hnh 1 = n 2, as conjugation preserves order. In the former case G = H N = Z 2 Z 3 = Z6 since GCD(3,2)=1. In the latter case G = S 3. 4
7. Let G be a group of order 2450 = 7 2 5 2 2. (a) Determine the possible numbers r 7 of Sylow 7-subgroups. What is the order of a Sylow 7-subgroup of G? Ans. r 7 1 mod 7 and divides 5 2 2. The divisors of 50 are 1, 2, 5, 10, 25, 50. Of these r 7 = 1 or r 7 = 50 are the only that satisfy r 7 1 mod 7. (b) Determine the possible numbers r 5 of Sylow 5-subgroups, Ans. r 5 1 mod 5 and divides 7 2 2. Divisors of 7 2 2 are 1, 2, 7, 14, 49, 98, so r 5 = 1 is the only possibility. (c) Determine the possible numbers r 2 of Sylow 2-subgroups. Ans. r 2 1 mod 2 and divides 7 2 5 2, so ostensibly can be any divisor of 7 2 5 2 (nine possibilities). (d) Show that G must have a proper non-trivial normal subgroup. Ans. Since r 5 = 1, there is a unique Sylow 5-subgroup of order 25, that is normal. (e) (*) What else can you say about G? Ans. (i). Groups of order p 2 are Abelian (see text) so the unique 5-Sylow subgroup N = Z 5 Z 5 or N = Z 25. Since the 7-Sylow subgroups are all conjugate to one of them, they are all isomorphic to either Z 7 Z 7 or to Z 49, one or the other. (ii). If G is Abelian, then G = H N Z 2, where H is the Sylow 7-subgroup and N the Sylow 5-subgroup, so there are 4 possibilities. (iii). Some non-abelian groups of order 2450 are D 49 Z 25, D 49 Z 5 Z 5, D 25 Z 49, D 25 Z 7 Z 7. D 25 49 where D n is the dihedral group, of order 2n. (iv). With a little more work we can show that always r 7 = 1 and that G contains an Abelian subgroup W = H N of order (7 2 5 2 ), so W has index 2 in G and is normal. We will show that the order of Aut(N) is relatively prime to H: then the action θ of H by conjugation on N is the identity the image θ(h) Aut(N) is e Aut(N). We know Aut(Z 25 ) = Z 25 5 = Z 20. Also, Aut(Z 5 Z 5 ) = Gl 2 (Z 5 ) and Gl 2 (Z 5 ) = (24)(20) 1 Thus in each case Aut(N) is relatively prime to 49. So the action of any 7 Sylow subgroup H on N is trivial: that is, the elements of H commute with those of N, and W = H N is a subgroup of G. It follows that r 7 = 1: there is not room for another 7-Sylow subgroup, much less 50 of them! (iv). Let m generate a 2-Sylow subgroup of G, so m has order 2. The conjugation action of m on G takes W to W and determines τ m Aut(W ) of order 2 (τ m is an involution). We have shown that specifying the structure of G comes down to choosing Abelian groups N, H or order 25 and 49, respectively, and an involution τ m on W. Because the orders of H, N are relatively prime, Aut(W ) = Aut(N) Aut(N), and we need only study the order 2 elements in Aut(W) to determine G. (Is (ii), (iii) a complete list?) 1 This may not have been discussed in the course. In general Gl 2 (Z p ) = (p 2 1)(p 2 p). 5
8. (a) Let a group H of order 49 act on a set with 23 elements. Show that the action has a fixed point. Ans Let H act on S. Since Orbit(s) Stab H (s) = H = 49, the possible sizes of the orbits are 1, 7 and 49. The fixed points s satisfy Orbit(s) = 1, and Stab H (s) = H. Since S = 23 there can be 1,2, or 3 orbits of 7 elements, with the rest of S being fixed points. Unless p divides S there must be fixed points under the action of a p-group. (b) Let a group H of order 49 act on itself by conjugation. Show that the center Z(H) must have 7 or 49 elements. Ans. The center is comprised of the fixed points under the conjugation action of H on H; it is noempty since e Z(H). We can write H = Z(H) O 1 O 2., where each of O 1, O 2, is an orbit with 7 elements. It folllows that Z(H) = H O 1 O 2 is divisible by 7. II. The best 3 out of the following 6 problems will count ten points each. 9. Suppose G has two normal subgroups H, N such that HN = G and GCD( H, N ) = 1. Show that G = H N is a direct product (or direct sum). Ans. (i). We have H N H, N so H N divides H, N so H N divides GCD( H, N ) = 1. So H N = e. These are the conditions for G = H N. (text definition). (ii) (More detail) We show that H commutes with N (even if H, N are themselves non- Abeliean). Suppose that h 1 n 1 = h 2 n 2 with h 1, h 2 H, n 1, n 2 N. Then (h 2 ) 1 h 1 = n 2 (n 1 ) 1 H N = e, so (h 2 ) 1 h 1 = e, h 1 = h 2 and likewise n 1 = n 2. Now let h H, n N then hnh 1 N, since N is normal, so hnh 1 = n 2, hn = hn 2, but we have just shown that this implies n 2 = n. So we have hnh 1 = n for all h H, n N. It follows that hn = nh for all h H, n N and G = H N. 10. Let H and N be groups and H N the external direct product. (a) Show that the set H = {(h, e N ) h H, e N identity in N} is a normal subgroup of H N. Ans. Let g = (h 1, n) G = H N. Then g(h, e N )g 1 = (h 1 hh 1 1, ne N n 1 by definition of the composition in H N. But h 1 hh 1 1 H, since H is a subgroup, and ne N n 1 = e N, if follows that gh g 1 H so H is normal. (b) Let N = {(e H, n) n N, e H identity in H}. Show that every element z H N can be written as z = xy with x N and y H. Ans. Let z = (h, n) H N. Then z = (h, e N ) (e H, n) where x = (h, e N ) H and y = (e H, n) N. 6
11. The dihedral group D 4 acts on the set X = K 4 of colorings of the edges of the square with a set K of n colors. Determine how many different-colored squares can be made with (up to) n given colors. Two colorings are considered the same if the first colored square can be moved in space so that it is the same as the second. Ans. Labeling the vertices of the square counter clockwise A, B, C, D we have D 4 = {τ, τ 2, τ 3, e, f h, f v, (AC), (BD)} where τ = (A, B, C, D), f h = (AB)(CD), f v = (AD)(BC). We have X τ = { all sides colored the same }, so X τ = X τ 3 = n. Also X τ 2 = X (AC) = X BD = n 2, while X f h = X f h = n 3, and X e = X = n 4. By Burnside theorem, the number w of orbits of X under D 4 satisfies, since D 4 = 8, w = n4 +2n 3 +3n 2 +2n 8. 12. Consider the group G = Z 6 Z 4. (a) What are the possible orders of elements in G? Ans. (g, h) = LCM ( g, h ). Since g divides 6, h divides 4, (g, h) can be any divisor of LCM (6,4)=12. (b) Give an example of at least one element of each of the possible orders. Ans. (0, 0) = 1, (3, 2) = 2, (2, 0) = 3, (0, 1) = 4, (2, 2) = 6, (1, 1) = 12. (c) Prove that G = Z 6 Z 4 is not cyclic. G = 6 4 = 24, but G has no elements of order 24 by (a) so cannot be cyclic. 13. Recall that the group D 6 is the group of symmetries of the regular hexagon with vertices A, B, C, D, E, F. Let D 6 act on the set T = {ACE.BDF } of two inscribed triangles. (a) Determine the stabilizer Stab G (ACE) of ACE, and the orbit of ACE. Ans. The stabilizer W = Stab G (ACE) is {g D 6 g {A, C, E} = {A, C, E} (permutes the vertices A,C,E). This is W = {τ 2, τ 4, e, f AD = (BF )(CE), f BE = (AC)(DF ), f CF = (BD)(AE)}. The orbit of ACE = {ACE, BDF } = T. (b) Write the stabilizer of Stab G (BDF ) as a conjugate of Stab G (ACE). Ans Stab G (BDF ) = τ Stab G (ACE) τ 1 = α Stab G (BDF ) α 1, α = (AB)(F C)(DE) or α = any flip about an axis through two opposite sides. (c) Determine whether Stab G (ACE) is a normal subgroup of D 6. Explain your answer. Ans. Yes, for either of two reasons: (i). Stab G (BDF ) = Stab G (ACE), as is seen by observing that each element w of W takes the set {B, D, F } to itself: as each element permutes the set {A, C, E}. Thus every conjugate gw g 1 of W is W. 7
(ii). W has 6 elements, so W = D 6 /2. It follows that W is normal, since for g / W, gw = G W = W g (the two left cosets partition G, as well as the two right coeets, and ew = W e). 14. Let S 6 be the permutation group on 6 elements {1, 2, 3, 4, 5, 6}. Let the group G = S 6 act on the set X = S 6 by left multiplication, i.e. ϕ : S 6 S 6 S 6 is given by ϕ(g, x) := gx for each g S 6 and each x S 6. (a) Write the definition of the orbit of x X under the action of a group G, i.e. O x. Ans. The orbit O x = G x of x X is the subset {gx g G} of X. (b) Let x = (135) S 6. Find the orbit of x, i.e. O x. Ans Let g S 6. Then (135) y, y = (135) 1 g X and y satisfies (135) y = g. So the orbit of (135) is all of G. Sylow: Let G = p n m with GCD(p, m) = 1. Then ( # Sylow p groups r p 1 mod p, and r p m. ) Burnside: # orbits v = g G F ix(g) / G 8