Name: Solutions - AI FINAL EXAM

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1 total Name: Solutions - AI FINAL EXAM The first 7 problems will each count 10 points. The best 3 of # 8-13 will count 10 points each. Total is 100 points. A 4th problem from # 8-13 may count up to 5 pts extra credit. You may use a calculator (but it is not needed, you may retain factorials, powers, in your answer). Closed notes. There are some formulas at the end. Good luck! I. Each of # 1-7 counts ten points each. 1. (a) Describe all Abelian groups of order 3 5, up to isomorphism. Explain the connection with the set of partitions of 5. Ans. These correspond 1-1 with the seven partitions of 5: Z 3 5 Z 3 4 Z 3 Z 3 3 Z 3 2 Z 3 3 Z 3 Z 3 Z 3 2 Z 3 2 Z 3 Z 3 2 Z 3 Z 3 Z 3 Z 3 Z 3 Z 3 Z 3 Z 3 (b) Which of these have elements of order 27? Ans The order of an element in a direct sum divides LCM(orders of factors). So the first four have elements of order 27: Z 3 5, Z 3 4 Z 3, Z 3 3 Z 3 2, Z 3 3 Z 3 Z 3 (c) Determine all Abelian groups of order 75, up to isomorphism, writing each as a direct sum of cyclic groups of prime power order. Ans Since 75 = 5 2 3, and GCD(25, 3) = 1 an Abelian group of order 75 is the product of a 5-group of order 25, and Z 3. There are two such 5- groups, Z 25 and Z 5 Z 5, so we have two Abelian groups of order 75. Recalling that Z mn = Z m Z n if GCD(m, n) = 1, we have Z 25 Z 3 = Z75, and Z 5 Z 5 Z 3 = Z5 Z 15 as the two groups, up to isomorphism. 1

2 2. Let G = (Z 80, +) = {0, 1, 2,..., 79} under (+ mod 80). (a) How many elements of order 80 are there? Ans. Since 80 = 2 5 5, φ(80) = 80(1 1/2)(1 1/5) = 32. Explanation: there are 80/2 multiples of 2, 80/5 multioles of 5, but that double counts 80/10 multiples of 10, so =32 elements from {0, 1,..., 79} are relatively prime to 80. (b) Write down 5 different elements of order 80. Ans 1, 3, 7, 9, 11, 13, 17, 19, 21,.... (c) List all elements of order 10 in G. There is one subgroup H = 80/10 = 8 of order 10. Note that φ(10) = φ(5) φ(2) = 4 1, The generators of H are the multiples of 8 by integers in U(10), so they are 8(1, 3, 7, 9) = {8, 24, 56, 72}. These are the four elements of G having order 10. (d) Determine all subgroups of G having order 5. Ans Since Z 80 is cyclic, there is a unique subgroup W = 80/5 = 16 of order 5. (Its generators are the multiples {16a GCD(a, 5) = 1}, so 16 = 32 = 64 = 96 ). 2

3 3. Let S 7 be the group of permutations of the set {1, 2, 3, 4, 5, 6, 7}. (a) Find one element of order 6, one of order 10 and one of order 12 in S 7. Ans Order 6: any element with disjoint cycle decomposition (6,1), (3,2,2), or (3,2,1,1), so g = (1, 2, 3, 4, 5, 6) suffices. Order 10: Disjoint cycle decomposition (5,2), so g = (1, 2, 3, 4, 5) (6, 7). Order 12: Disjoint cycle decomposition (4,3), so g = (1, 2, 3, 4) (5, 6, 7) is one. (b) How many elements of order 6 are there in S 7? Ans There are 3 cycle decompositions: (6,1): ( 7 6) 5! = 7 5! = 840. (3,2,1,1): ( ( 7 3) 2! 1! 4 ) 2 = = (3,2,2,1): ( 7 3) 2! 1! ( 4 2) 1 2! = 210. Total: = 1470 elements of order 6. (c) Prove that there are no elements of order 9 in S 7. Ans Since the order of a product of disjoint cycles, is the LCM of the lengths, we would need a cycle of length 9 in the disjoint cycle decomposition of g. This is possible only in S n, n 9. 3

4 4. Consider the group S 8 of permutations of the set {1, 2, 3, 4, 5, 6, 7, 8}. (a) Write the disjoint cycle decomposition of g = (256) (125) (13542). Ans (1, 3)(2, 5, 4, 6). (b) Let h = (1235)(46)(78). Give the disjoint cycle decompositions of h 2 and of h 3. Ans h 2 = (13)(25), h 3 = (1, 5, 3, 2)(46)(78). (c) How many conjugates are there of h in S 8? Ans ( ( 8 4) 3! 4 ) 2 1 = ! 6 = !

5 5. Let ϕ : Z 10 Z 30 be the group homomorphism defined as ϕ(x) = 6x (mod 30). You may assume that ϕ is a homomorphism. (a) Find K = Ker(ϕ). Ans K = {0, 5} = 5. (b) Find the image Im(ϕ). Ans (0, 6, 12, 18, 24). (c) Determine the quotient Z 10 /K. List all the elements of the factor group. Write each coset as a list of specific elements. Ans The cosets are 0 = {0, 5}, 1 = {1, 6}, 2 = {2, 8}, 3 = {3, 8}, 4 = {4, 9}. (d) Give the isomorphism ϕ : Z 10 /K Im(ϕ). Ans ϕ(i) = 6i mod 30, i = 0, 1, 2, 3, 4. 5

6 6. Let S 5 be the group of all permutations on 5 elements {1, 2, 3, 4, 5}. (a) Prove that the subgroup A 5 of all even permutations of the same 5 elements {1, 2, 3, 4, 5}, is a normal subgroup of S 5. Ans. Method i: Since A 5 = S 5 /2, then ga 5 = A 5 = A 5 g for g A 5. And ga 5 = S 5 A 5 = A 5 g if g / A 5. If follows that A 5 is normal. This proof shows that if H has index 2 in G, ( G = 2 H ), then H is normal in G. Method ii. If g is even, and h A 5, then ghg 1 is the product of three evens, so is even. If g is odd, then so is g 1, since g 1 has the same lengths of disjoint cycles as g, and these lengths determine whether g 1 is even or odd. 1 But then ghg 1 is the product of an (odd even odd) is itself even. In each case ghg 1 A 5 so A 5 is normal in S 5. (b) Prove that the subgroup H = (254) generated by the permutation (254) is not a normal subgroup of S 5. Ans. Consider g = (12). Then g(254)g 1 = (154) / H = {(2, 5, 4), (2, 4, 5), e}. This shows that H is not normal in S 5. (c) Determine which subgroups H of S 5 are conjugates of H. Your answer should be either a list, or an intrinsic property of H that explains simply how to identify a conjugate, (not just the definition that g S 5 H = ghg 1.) Ans.. Since H is generated by a 3-cycle, the conjugates are all subgroups H generated by a 3-cycle, so all of the form (a, b, c). In fact, (a, 2)(b, 5)(c, 4) (254) ((a, 2)(b, 5)(c, 4)) 1 = (a, b, c), so taking g = (a, 2)(b, 5)(c, 4) we have ghg 1 = (a, b, c). There are ( 5 3) /2 = 10 conjugates (divide by 2 since (a, b, c) and (a, c, b) generate the same subgroup ). 1 An odd length cycle is the product of an even number of transpositions so is even, and an even length cycle is the product of an odd number of transpositions, so is odd. (Worthy of Charles L. Dodgson, alias Lewis Carroll, mathematician and author of Alice in Wonederland ) 6

7 7. Let G be a group of order G = 75. (a) Find the possible values for r 5, the number of 5-Sylow subgroups. Ans. Since 75 = 5 2 3, r 5 1 mod 5 and divides 3, so the only possibility is r 5 = 1. (b) Find the possible values for r 3, the number of 3-Sylow subgroups.. Ans. Since 75 = 25 3, r 3 1 mod 3 and divides 25. The divisors of 25 are 1, 5, 25, and 5 2 mod 3 so r 3 = 1 or 25. (c) Prove that there are no simple groups of order 75. That is, prove that a group of order 75 must contain a proper normal subgroup other than the identity subgroup. Ans. Since r 5 = 1, the 5-Sylow subgroup H, of order 25, has as conjugate only itself, so is normal in G. 7

8 II. The best 3 of #8-13 count ten points each. A 4-th may count 5 pts EC. 8. Let G = A B be the external direct product. Assume that A = 25 and B = 10, and let (e A, e B ) be the identity element of G. (a) What is G? Ans A B = A B = 250. (The direct sum has as base set all pairs (a, b) a A, b B) (b) What are the possible orders of elements of G? Ans. The order of (a, b) G, is (a, b) = LCM( a, b ) so must divide LCM ( A, B ) = LCM (25, 10) = 50. So the possible orders are 1,2,5,10,25,50. (c) Prove that the subgroup H = {(a, e B ) a A} is a normal subgroup of G. You may assume that H is a subgroup of G. Ans First, easier proof : We must show ghg 1 H for all g G. Let g = (a 1, b). (Note: We must choose a 1 possibly different from a). Since (a 1, b) 1 = (a 1 1, b 1 ) we have (a 1, b) (a, e B ) (a 1, b) 1 = (a 1 a a 1 1, b e B b 1 ) = (a 1 a a 1 1, e B ) H proving that H is normal in G. Second proof. We must show that g G, gh = Hg as sets. Let g = (a 1, b), then gh = (a 1, b) {a, e B ) a H} = {(a 1 a, b e B ) a A} = (A, b) since {a 1 a a A} = A, as, given a 2 A take a = (a 1 ) 1 a 2, and a 1 ((a 1 ) 1 a 2 ) = a 2. Similarly, Hg = {a, e B ) a H} (a 1, b) = {(a a 1, e B b) a A} = (A, b), so the left coset gh = Hg, the right coset. Note: Often in proving a subgroup is normal, the criterion ghg 1 H in the first proof is simpler to use than the equality of cosets in the second proof. But you may wish to consider which definition is more useful, when showing that N is normal has some partiuclar consequence. Also, in general a 1 a aa 1, since A is not assumed commutative. So key to the second proof is that a 1 A = Aa 1 = A as sets, since A is a group. 8

9 9. (a) Determine the order U(75) of the group U(75) of those integers {i 0 i 74 GCD (i, 75) = 1} which have multiplicative inverses mod 75. Ans. We have U(75) = U(25) U(3) = Z 20 Z 2 = Z4 Z 5 Z 2, of order 40. Or, U(75) = 75(1 1/5)(1 1/3) = 40 (One eliminates the multiples of 5, then the multiples of 3, but one has double counted the multiples of 15: this is inclusion-exclusion in combinatorics. For m = p n 1 1 p nr r we have, similarly φ(m) = m(1 1/p 1 ) (1 1/p r ). (b) Prove that U(9) and D 3 have the same number of elements, but are not isomorphic. (D 3 is the dihedral group.) Ans U 9 = Z9 3 = Z 6 so U 9 = 6 = S 3. But U(9) is Abelian (commutative), and S 3 is not since (12)(123) (123) (12). Or U(9) is cyclic with a generator (2) of order 6, and S 3 has no element of order 6. (Students cited other properties where they are different, for example U(9) has a unique element of order 2, but S 3 has 3 order two cycles (12), (13), (23). (c) Prove that Z 15 and Z 5 Z 3 are isomorphic. Ans. (Complete). Define a homomorphism φ : Z 15 Z 5 Z 3, φ(k) = (k mod 5, k mod 3). Then φ is a homomorphism since φ(a + b) = (a + b mod 5, a + b mod 3) = (a mod 5, a mod 3) + (b mod 5, b mod 3) = φ(a) + φ(b), and φ( a) = φ(a). Let k Ker(φ), then 5 k and 3 k so GCD(5, 3) = 15 k and k = 0 in Z 15. Thus φ is 1-1, but Z 15 = Z 5 Z 3 = 15 so φ is an isomorphism. Or We accepted the less complete answer that both Z 15 and Z 5 Z 3 are cyclic groups, the latter since GCD(5, 3) = 1, hence they are isomorphic since there is a unique cyclic group of given order, up to isomorphism. What underpins this second answer is the above complete proof. 9

10 10. Let G be a group satisfying G = 75. Assume that G has a cyclic normal subgroup N of order 25, generated by n. (a) Denote by Aut(N) the group of all isomorphisms β: N N. Show that Aut(N) = Z 20. Ans. By theorem Aut(Z 25 ) = U(25) acting by multiplication, and we have U(25) = Z 20 = Z4 Z 5. (b) Let H be a subgroup of G having order 3, generated by an element h. Assume that hnh 1 = n. Show that this implies that G = H N is a direct sum. Ans Also H N = e, since H N divides H = 3 and N = 25 and GCD( H, N ) = 1. Since N is assumed cyclic, and H must be cyclic, having order 3, we have h n k h 1 = (h n h 1 ) k = n k, and h 2 n k h 2 = h (h n k h 1 ) h 1 = hn k h 1 = n k, showing that H commutes with N, implying HN = G. This shows G = H N. (c) Let H =< h > be a subgroup of G having order 3. Show that, indeed, hnh 1 = n (this was assumed in part (b). Now you are asked to prove it). Steps: First show that hnh 1 N, so hnh 1 = n k for some k. Then show k = 1, by considering the homomorphism θ : H Aut(N) given by θ(h t ) (n i ) = h t n i (h t ) 1. Hint: Consider the orders of θ(h), H, and Aut(N). Ans. Since N is normal, hnh 1 N and θ defines an automorphism of N. Since θ(h) is a quotient group of H and also a subgroup of Aut(N), the order θ(h) divides both H and Aut(N), so divides GCD (3,20)=1, so θ(h) = e (see also problem # 13B). This means hnh 1 = n. 10

11 11. Let G be a group of order G = 125 acting on a set X with X = 100. (a) What are the possible sizes of the orbits of this action? Ans. The size of an orbit divides G = 125, and is no larger than X = 100 so can be 1,5, or 25. (we accepted also 125 here). (b) Prove: if the action of G on X has one fixed point then it must have at least 5 fixed points. Ans We have X = X G + X i where X G is the fixed points and the X i are the larger orbits. But X is a multiple of 5, and the orders of the larger orbits are 5 or 25. So 5 divides X G. If X G it has at least 5 elements. (c) Prove that any action of G on X has at least 2 orbits. Ans Since the sizes of orbits are 1,5, or 25 and 100 is a multiple of 25, we obtain the smallest number of orbits when the action of G on X has 4 orbits of 25 elements each. 11

12 12. Recall that the group G = D 4 of symmetries of the square X with vertices A, B, C, D acts on the set S = {AB, BC, CD, DA} of sides of X, as well as on the set D = {AC, BD} of diagonals of X. (a) Determine the stabilizer H = Stab G (AB) of AB, and the orbit of AB in S. Ans H = Stab G (AB) = (e, (AB)(CD)). The powers of τ = (ABCD) take AB to the other sides so the orbit of AB is S. These satisfy G = Stab G (AB) #{orbit of AB}. (b) Write the stabilizer Stab G (DA) as a conjugate of H. Ans. Since τ 1 (AB) = DA, we have Stab G (DA) = τ 1 Hτ. (c) Determine whether Stab G (AB) is a normal subgroup of D 4. Explain your answer. Ans. No, since the conjugate Stab G (DA) = τ 1 Hτ H, as (AD)(BC) Stab G (DA) but is not in H. (d) Determine the stabilizer W = Stab G (AC) of AC in D. Is it a normal subgroup of G. Why or why not? Ans W = (e, τ 2, (AC), (BD)) satisfies D 4 = 2 W, so is normal in D 4 (see # 6a, proof (i)). Or: Stab G (BD) = W (by inspection). Since all conjugates of W are W itself, W is normal in G = D 4. Note: When G acts on a set S, the conjugates of a stabilizer H of s S are exactly the stabilizers of the elements in the orbit of s. 12

13 13. You will be credited with either (13A) or (13B) but not both. 13A. The dihedral group D 6 acts on the set X = K 6 of colorings of the edges of the regular hexagon with a set K of n colors. Determine how many different-colored hexagons can be made with (up to) n given colors. Two colorings are considered the same if the first colored hexagon can be moved in space so that it is congruent to the second. 13B. Let ϕ : G W be a homomorphism of groups and denote by ϕ(g) the image of ϕ. i. Let g G. Show that the order ϕ(g) divides g. ii. Show that ϕ(g) divides GCD ( G, W ). Ans 13A. Let the Hexagon have vertices {A, B, C, D, E.F }, and let τ = (ABCDEF ). Then X e = n 6 ; and X τ = X (τ 5) = n as all sides must be the same color. But X (τ 2) = X (τ 4) = n 2, since every other side must be colored the same. X (τ 3) = n 3 since sides AB, BC, CD may each have one of n different colors. For a flip as f = f AD = (CE)(BF ) about the diagonal (AD), X f = n 3 as again sides AB, BC, CD each may have one of n colors. Finally the the three flips like f v = (CD)(BE)(AF ) with axes the perpendicular to two opposite sides, have X fv = n 4, since for f v the sides CD, AF, AB, BC may each have one of n different colors, By Burnside theorem we have # orbits v = (n 6 + 3n 4 + 4n 3 + 2n 2 + 2n) /12 Ans 13B. First proof of (i). Let k = g, then g k = e. This implies (ϕ(g)) k = ϕ(v k ) = ϕ(e G ) = e W. This implies ϕ(v) divides k. Second Proof of i. By restricting ϕ to H = g, we have ϕ(g) = ϕ(h) and H = K H ϕ(h) where K H is the kernel of ϕ on H (since ϕ(h) = H/K H ). Thus ϕ(g) divides g = H. Proof of (ii). Since ϕ(g) W is a subgroup of W, ϕ(g) W. Let K be the kernel of ϕ: since G/K = ϕ(g) we have G = K ϕ(g) so ϕ(g) G. Thus ϕ(g) divides GCD ( G, W ). Sylow: Let G = p n m with GCD(p, m) = 1 and r p = #{Sylow p-subgroups}. Then r p 1(mod p), and r p m. ( ) Burnside: # orbits v = g G Sg / G, where S g = F ix(g). U(p k ) = Z p k p k 1 for p an odd prime. U(2k ) = Z 2 k 2 Z 2 for k > 1. 13