S11MTH 3175 Group Theory (Prof.Todorov) Final (Practice Some Solutions) 2 BASIC PROPERTIES

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1 S11MTH 3175 Group Theory (Prof.Todorov) Final (Practice Some Solutions) 2 BASIC PROPERTIES 1 Some Definitions For your convenience, we recall some of the definitions: A group G is called simple if it has no proper normal subgroups, i.e. the only normal subgroups are {e} and G. Let x G be an element of a group G. A conjugate of x by g G is the element gxg 1. Let x G be an element of a group G. A conjugacy class of x is C(x) = {gxg 1 g G}. Let H < G be a subgroup of a group G. A conjugate of H by g G is the subgroup ghg 1. Let H < G be a subgroup of a group G. A conjugacy class of H is C(H) = {ghg 1 g G}. 2 Basic properties 1. Let x G be an element of a group G. Prove that x = gxg 1 for any g G. Proof: Let x = n. Then x n = e (by the definition of order of an element). (gxg 1 ) n = (gxg 1 )(gxg 1 )... (gxg 1 ) = gx(g 1 g)x(g 1 g)... (g 1 g)xg 1 = gxexe... exg 1 = gx n g 1 = geg 1 = gg 1 = e Since (gxg 1 ) n = e it follows that gxg 1 divides n. Let gxg 1 = k. Then ks = n for some integer s. From gxg 1 = k it follows that e = (gxg 1 ) k = (gxg 1 )(gxg 1 )... (gxg 1 ) = gx k g 1 and therefore e = gx k g 1. Therefore g 1 eg = g 1 gx k g 1 g after multiplying the above equation by g 1 and by g on the right. Therefore e = x k after using associative law, and properties of identity e and inverses. Therefore n = x divides k. Hence k = nt. So we have: n = ks = nts where k, n, s, t are all positive integers. Therefore ts = 1 and since all numbers are positive integers it follows that s = t = 1 and therefore k = n. Hence gxg 1 = k = n = x for all g G. 2. Let H < G be a subgroup of a finite group G. Prove that H = ghg 1 for any g G. Proof: Consider the map f : H ghg 1 given by f(x) = gxg 1. - f is a map H ghg 1 since f(x) ghg 1 for every x H. - f is one to one: Suppose f(x 1 ) = f(x 2 ). Then gx 1 g 1 = gx 2 g 1. - Multiply by g 1 on left and g on the right, use associative law, inverse and identity, and get x 1 = x 2. Therefore f is one-to-one. 1

2 S11MTH 3175 Group Theory (Prof.Todorov) Final (Practice Some Solutions) 3 PERMUTATIONS - f is onto: Any element gxg 1 ghg 1 is f(x) for some x H. - Therefore f is a bijection between finite sets. (No need to check group homomorphism for this, since the question was only about the number of elements.) - So H = f(h) = ghg 1. 3 Permutations 1. Let α = (24367) S 8, be a permutation in the permutation group on 8 letters: {1, 2, 3, 4, 5, 6, 7, 8}. (a) Find (28)(24367) (28)(24367) = (1)(243678) (b) Find (24367)(28) (24367)(28) = (1)(284367) (c) Find (28)(24367)(28) 1 (28)(24367)(28) 1 = (1)(243678)(28) 1 = (243678)(28) = (84367) (d) Find (28)(24367)(28) (84367) (e) Find (281)(24367) (f) Find (281) 1 (182) (g) Find (281)(24367)(281) 1 (84367) 2. Prove that a 5-cycle in any permutation group has order Let S 9 be the permutation group on 9 letters: {1, 2, 3, 4, 5, 6, 7, 8, 9}. (a) How many cycles of order 5 are there in S 9? Answer: There are ( ) 9 5! 5-cycles. 5 5 (b) How many permutations of order 5 are there in S 9? Answer: There are ( ) 9 5! 5-cycles Order of 5-cycle is 5. - Order of a product of disjoint cycles is least common multiple of the orders of the cycles. - Therefore the only possibility for order 5 in S 9 is 5 = lcm(5, 1, 1, 1, 1), hence 5-cycles. (c) How many permutations of order 10 are there in S 9? Answer: - There are ( ) 9 5! (4) 2! (2) 2! 1 permutations with cycle decomposition (5, 2, 2) ! 2

3 S11MTH 3175 Group Theory (Prof.Todorov) Final (Practice Some Solutions) 4 CYCLIC GROUPS ) 5! (4) 2! 5 - There are ( 9 permutations with cycle decomposition (5, 2, 1, 1) The only possibilities to get permutations of order 10, are the cycle decompositions: (5, 2, 2) and (5, 2, 1, 1) since lcm(5, 2, 2) = 10 and lcm(5, 2, 1, 1) = 10. (d) How many permutations of order 11 are there in S 9? Answer: None. Reason: - Since 11 is prime, need to get 11-cycle in order to get permutation of order There are no partitions of 9, which have 11 elements (e) How many permutations of order 12 are there in S 9? 4. Let α = (abc) and let β = (ad). Prove that the conjugate of α by β is (dbc). 5. Let α = (abcdefg) and let β = (bmf). Prove that the conjugate of α by β is (amcdebg). 6. Let α = (abcdefg) and let β = (arbsct). What is the conjugate of α by β? Prove it. 7. Let α = (236). Find β such that βαβ 1 = (143). β = (63)(34)(21). βαβ 1 = (63)(34)(21)(236)((63)(34)(21)) 1 = (63)(34)(21)(236)(21)(34)(63) = (143) 8. Let α = (1236). Find β such that βαβ 1 = (1435). β = (1)(24)(3)(65) You should check that βαβ 1 = (1435). 9. Let α = (1236). Prove that there is no β such that βαβ 1 = (14)(36). No β exists since conjugation preserves disjoint cycle structure. 10. Let α = (1236). Prove that there is no β such that βαβ 1 = (14). No β exists since conjugation preserves disjoint cycle structure. 4 Cyclic Groups 1. Write the definition of a cyclic group. Group is cyclic if it can be generated by one element. 2. Prove that a subgroup of a finite cyclic group is cyclic group. Proof: Let G be a finite cyclic group. Then there exists an element a G such that G = a. - Let n be the smallest positive integer such that a n = e, where e is the identity of G. - Then G = {e, a, a 2, a 3,..., a n 1 }. - Let H < G be a subgroup of G. - Let i be the smallest positive integer such that a i H. 3

4 S11MTH 3175 Group Theory (Prof.Todorov) Final (Practice Some Solutions) 4 CYCLIC GROUPS - Claim: H = a i, i.e. every element of H is of the form (a i ) j for some j. - Proof of the Claim: Let x H. Since x H G we have x G. - Then x = a m for some non-negative integer m, since G = a. - Let d = gcd(i, m). Then there exist integers α, β Z such that d = αi + βm. - So a d = a αi+βm = (a i ) α (a m ) β. - Since a i H it follows that (a i ) α H by closure of subgroup under group operation. - Since a m H it follows that (a m ) β H by closure of subgroup under group operation. - Since (a i ) α H and (a m ) β H it follows that a αi+βm H again by closure of subgroup under group operation. - Therefore a d H. - Since i is the smallest positive such integer, it follows that d = i. - Therefore i m. So m = ij for some integer j. - Hence x = a m = a ij = (a i ) j for some j, i.e. every element x H can be expressed in terms of a i. 3. Prove that a quotient of a finite cyclic group is cyclic group. 4. Prove that a cyclic group of order 60 is isomorphic to Z Consider a cyclic group of order 16, i.e. G = a where a = 16. (a) Write all generators of G. Solution: If G = a and a = n, then generators of G are all a i such that gcd(i, n) = 1. - Generators of G = a where a = 16 are: {a = a 1, a 3, a 5, a 7, a 9, a 11, a 13, a 15 } (b) Write the formula for the number of generators of a cyclic group of order n and check that you got the correct number of generators according to the formula. Solution: The number of positive integers smaller then n and relatively prime to n is given by Euler phi function ϕ(n). - We have ϕ(p k ) = p k 1 (p 1) - Therefore ϕ(16) = ϕ(2 4 ) = 2 3 (2 1) = 8 (agrees with the above 8 generators). (c) Write all elements of order 8 in G. Solution: If G = a, a = n and k n, then a n/k = k and a i = a j if and only if gcd(i, n) = gcd(j, n). - Therefore a 16/8 = 8, i.e. a 2 = 8. - To find the other elements of order 8, need to find other integers j < 16, so that gcd(j, 16) = gcd(2, 16) = 2. So j = 2, 6, 10, 14 - Therefore the elements of order 8 are: a 2, a 6, a 10, a 14. (d) Write the formula for the number of elements of order 8 in a cyclic group of order 16 and check that you got the correct number of elements according to the formula. Solution: If G = a, a = n, k n, then the number of elements of order k in G is ϕ(k). - So the number of elements of order k = 8 is ϕ(8) = ϕ(2 3 ) = 2 2 (2 1) = 4, which is exactly how many we got in the previous part. 4

5 S11MTH 3175 Group Theory (Prof.Todorov) Final (Practice Some Solutions) 4 CYCLIC GROUPS (e) Write all elements of order 4 in G. Solution: a 16/4 = 4, i.e. a 4 = 4. - To find the other elements of order 4, need to find other integers j < 16, so that gcd(j, 16) = gcd(4, 16) = 4. So j = 4, 12 - Therefore the elements of order 4 are: a 4, a 12. (f) Write the formula for the number of elements of order 4 in a cyclic group of order 16 and check that you got the correct number of elements according to the formula. Solution: The number of elements of order k = 4 is ϕ(4) = ϕ(2 2 ) = 2 1 (2 1) = 2, which is exactly how many we got in the previous part. (g) Write all elements of order 2 in G. Solution: a 16/2 = 2, i.e. a 8 = 2. - To find the other elements of order 2, need to find other integers j < 16, so that gcd(j, 16) = gcd(8, 16) = 8. So j = 8 - Therefore the elements of order 2 are: a 8. (h) Write the formula for the number of elements of order 2 in a cyclic group of order 16 and check that you got the correct number of elements according to the formula. Solution: The number of elements of order k = 2 is ϕ(2) = ϕ(2 1 ) = 2 0 (2 1) = 1, which is exactly how many we got in the previous part. 6. Consider the group Z 16. (Remember, operation is addition modulo 16). (a) Write all generators of Z 16. (b) Write the formula for the number of generators of the group Z 16 and check that you got the correct number of generators according to the formula. (c) Write all elements of order 8 in Z 16. (d) Write the formula for the number of elements of order 8 in the group Z 16 and check that you got the correct number of elements according to the formula. (e) Write all elements of order 4 in Z 16. (f) Write the formula for the number of elements of order 4 in the group Z 16 and check that you got the correct number of elements according to the formula. (g) Write all elements of order 2 in Z 16. (h) Write the formula for the number of elements of order 2 in the group Z 16 and check that you got the correct number of elements according to the formula. 7. Consider the group Z 49. (Remember, operation is addition modulo 49). (a) What are the possible orders of elements? Solution: Since Z 49 is cyclic group of order 49, the only possible orders of elements are: 1, 7, 49. 5

6 S11MTH 3175 Group Theory (Prof.Todorov) Final (Practice 5 EXTERNAL Some Solutions) PRODUCTS OF GROUP (b) List all the elements and their orders. Which formula are you using? Does it agree with your computations? Solution: Since Z n is cyclic group of order n, we can use everything we know about cyclic groups, but use additive notation!!! - Use additive version of the two statements: a n/k = k, which is (n/k)a = k and a i = a j if and only if gcd(i, n) = gcd(j, n), which is ia = ja if and only if gcd(i, n) = gcd(j, n) - Notice that Z n is generated by 1 (additively!!!) - Elements of order 49 : (49/49)1 = 1 1 = 1 - Other elements of order 49 are j < 49 such that gcd(j, 49) = gcd(1, 49) = 1 - Elements of order 49: 1,2,3,4,5,6,8,9,10,11,12,13,15,16,17,18,19,20,22,23,24,25,26,27,29,30,31,32,33,34,3 - Notice ϕ(49) = ϕ(7 2 ) = (7 1) = 7 6 = 42 - Elements of order 7: 7,14,21,28,35,42 - Notice ϕ(7) = 7 1 = 6 - Elements of order 1: 0 5 External Products of Group 1. Consider G = Z 4 Z 3. (a) How many elements does G have? Solution: 4 3 = 12 (b) What is the order of the element (0, 1) Z 4 Z 3? Solution: 3 (c) What is the order of the element (0, 2) Z 4 Z 3? Solution: 3 (d) What is the order of the element (1, 1) Z 4 Z 3? Solution: (1, 1) = lcm( 1, 1 ) = lcm(4, 3) = 12 (e) What is the order of the element (3, 2) Z 4 Z 3? (f) What are all the possible orders of elements (a, b) Z 4 Z 3? (a, b) = lcm( a, b ), a = 1, 2, 4 and b = 1, 3 (g) Describe all elements (a, b) Z 4 Z 3 of order 3? (h) Describe all elements (a, b) Z 4 Z 3 of order 4? (i) Describe all elements (a, b) Z 4 Z 3 of order 12? 2. Prove that Z 4 Z 3 = Z12. 6

7 S11MTH 3175 Group Theory (Prof.Todorov) Final (Practice7Some GROUP Solutions) HOMOMORPHISMS 6 Abelian Groups 1. Describe all Abelian groups (up to isomorphism) of order Describe all Abelian groups (up to isomorphism) of order 100 which are not cyclic. Answer: 100 = Abelian groups of order 2 2, up to isomorphism are: Z 2 2 (which corresponds to the partition (2) of 2 Z 2 Z 2 (which corresponds to the partition (1,1) of 2. - Abelian groups of order 5 2, up to isomorphism are: Z 5 2 (which corresponds to the partition (2) of 2 Z 5 Z 5 (which corresponds to the partition (1,1) of 2. - Abelian groups of order 100 = , up to isomorphism are the 4 possible combinations: Z 2 2 Z 5 2 Z 2 2 Z 5 Z 5 Z 2 Z 2 Z 5 2 Z 2 Z 2 Z 5 Z 5 - Use that Z m Z n = Zmn when gcd(m, n) = 1. So Z 2 2 Z 5 2 = Z 100 which is cyclic!!! Z 2 2 Z 5 Z 5 = Z20 Z 5, which is NOT cyclic Z 2 Z 2 Z 5 2 = Z 2 Z 50, which is NOT cyclic Z 2 Z 2 Z 5 Z 5 = Z10 Z 10 again NOT cyclic. 7 Group Homomorphisms 1. Give an example of a group homomorphism which is onto but not one-to-one. f : Z 4 Z 2 where f(x) = x(mod2) (Check!) 2. Give an example of a group homomorphism which is one-to-one but not onto. f : Z 4 Z 12 where f(x) = 3x(mod12) (Check!) 3. Give an example of a group homomorphism which is neither one-to-one not onto. f : Z 4 Z 12 where f(x) = 6x(mod12) (Check!) 4. Write all isomorphisms Z 12 Z 4 Z 3. Z 12 is a cyclic group, generated by 1, so need to determine image of 1. In order to have isomorphism, need to find all elements of order 12 in Z 4 Z 3. (a, b) = lcm( a, b ) Need elements a Z 4 with a = 4 and b Z 3 with b = 3. 7

8 S11MTH 3175 Group Theory (Prof.Todorov) Final (Practice7Some GROUP Solutions) HOMOMORPHISMS a = 1, 3 and b = 1, 2 (a, b) = (1, 1), or (a, b) = (1, 2), or (a, b) = (3, 1), or (a, b) = (3, 2) There are 4 isomorphisms: f 1 : Z 12 Z 4 Z 3 where f 1 (1) = (1, 1) and therefore: f(x) = (x(mod4), x(mod3)): f 1 (1) = (1, 1), f 1 (2) = (2, 2), f 1 (3) = (3, 0), f 1 (4) = (0, 1), f 1 (5) = (1, 2), f 1 (6) = (2, 0) f 1 (7) = (3, 1), f 1 (8) = (0, 2), f 1 (9) = (1, 0), f 1 (10) = (2, 1), f 1 (11) = (3, 2), f 1 (0) = f 1 (12) = (0, 0). f 2 : Z 12 Z 4 Z 3 where f 2 (1) = (1, 2) and therefore: f(x) = (x(mod4), 2x(mod3)) f 3 : Z 12 Z 4 Z 3 where f 3 (1) = (3, 1) and therefore: f(x) = (3x(mod4), x(mod3)) f 4 : Z 12 Z 4 Z 3 where f 2 (1) = (3, 2) and therefore: f(x) = (3x(mod4), 2x(mod3)) 5. Consider group homomorphism f : Z 12 Z 4 Z 3 given by f(a) = (a, 0). (a) Describe Ker(f). Write down all the elements. Ker(f) = {4, 8, 0} (b) Describe Im(f). Write down all the elements. Im(f) = {(0, 0), (1, 0), (2, 0), (3, 0)} 6. Consider group homomorphism g : Z 12 Z 24 given by g(a) = 2a. (a) Describe Ker(g). Write down all the elements of Ker(g). Ker(g) = {0} (b) Describe Im(g). Write down all the elements of Im(g). Im(g) = {2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 0} 7. Consider group homomorphism ϕ : Z 12 Z 24 given by ϕ(a) = 4a. (a) Describe Ker(ϕ). Write down all the elements of Ker(ϕ). Ker(ϕ) = {0, 6} (b) Describe Im(ϕ). Write down all the elements of Im(ϕ). Im(ϕ) = {4, 8, 12, 16, 20, 0} 8. Consider group homomorphism φ : Z 12 Z 24 given by φ(a) = 6a. (a) Describe Ker(φ). Write down all the elements of Ker(φ). (b) Describe Im(φ). Write down all the elements of Im(φ). 9. Describe all group homomorphisms Z 25 Z 24? f(x) = 0 is the only group homomorphism Z 25 Z 24. (Prove this!) 8

9 S11MTH 3175 Group Theory (Prof.Todorov) Final (Practice Some Solutions) 9 SYLOW THEORY (a) Describe Ker(f). Write down all the elements of Ker(f). Ker(f) = Z 25 (b) Describe Im(f). Write down all the elements of Im(f). Im(f) = {0} Z Prove that there is no group homomorphism ω : Z 12 Z 24 such that ω(a) = a. If ω(a) = a then ω(1 12 ) = = 12 for 1 Z 12 but ω(1 12 ) = 1 24 = 24 for 1 Z 24. If ω is a group homomorphism then ω(a) must divide a which is not true if ω(1 12 ) = Group Actions 1. Let G be group of order 25, i.e. G = 25. Prove that the center of G, Z(G) {e}. Proof: Let G act on itself (i.e. X = G) by conjugation. - The action ϕ : G G G is given by ϕ(g, x) = gxg 1. - X = G = O x =1 ( O x ) + O x >1 ( O x ) = Z(G) + O x >1 ( O x ). - O x G = 25. So if O x 1 then 5 O x. - Also 5 G. - Therefore 5 Z(G). - Since e Z(G), we know that Z(G) 0, so Z(G) is a non-zero multiple of 5. - Therefore Z(G) has at least 5 elements. 2. Let G be group of order 81, i.e. G = 81. Prove that G has a normal subgroup. 3. Let G be group of order p k, p a prime and k 1, k Z. Prove that G has a normal subgroup. 9 Sylow Theory We ve defined and proved the following in class, so you may use this: Definition 1 Let G be a group such that G = p k m where p is prime and p does not divide m. A subgroup H of a group G is called a p-subgroup if H = p i. Definition 2 Let G be a group such that G = p k m where p is prime and p does not divide m. A subgroup H of a group G is called a Sylow p-subgroup if H = p k. Theorem 1 Let G be a group such that G = p k m where p is prime and p does not divide m and 1 k. Then there exists a subgroup H of order H = p i for each 1 i k. 9

10 S11MTH 3175 Group Theory (Prof.Todorov) Final (Practice Some Solutions) 9 SYLOW THEORY Proposition Let G be a group such that G = p k m where p is prime and p does not divide m. Let P be a subgroup of G of order p k. Let C = {gp g 1 g G} = {P = P 1, P 2,..., P n } be all conjugates of P. Then any p-subgroup of G is contained in one of the P i C. Theorem 2 All p-sylow subgroups of a finite group G are conjugate to each other. Theorem 3 Let G be a finite group. Then #{p-sylow subgroups of G} divides G. Theorem 4 Let G be a finite group. Then #{p-sylow subgroups of G} 1(mod p). Theorem 5 Let G be a finite group. Then #{p-sylow subgroups of G} = 1 if and only if the p-sylow subgroup is normal. 1. Let G be a group. Let X be a p-subgroup of G and a G. Prove that axa 1 is a p-subgroup of G. This follows from Problem 2. in Basic properties. (Give details of the proof!) 2. Let G be a group. Let P be a Sylow p-subgroup of G and a G. Prove that ap a 1 is a Sylow p-subgroup of G. This follows from Problem 2. in Basic properties. (Give details of the proof!) 3. Let G be a group of order G = 35. (a) What are the possible #{5-Sylow subgroups of G} using Theorem 3? r 5 = 1, 5, 7, 35 (b) What are the possible #{5-Sylow subgroups of G} using Theorem 4? r 5 = 1, 6, 11, 16, 21, 26, 31 (c) How many 5-Sylow subgroups does G have? Exactly one. (Explain!) (d) Let P 5 be a 5-Sylow subgroup of G. How many elements does P 5 have? P 5 = 5 (Explain!) (e) What are the possible #{7-Sylow subgroups of G} using Theorem 3? r 7 = 1, 5, 7, 35 (f) What are the possible #{7-Sylow subgroups of G} using Theorem 4? r 7 = 1, 8, 15, 22, 29 (g) How many 7-Sylow subgroups does G have? Exactly one. (Explain!) (h) Let P 7 be a 7-Sylow subgroup of G. Find P 7. 10

11 S11MTH 3175 Group Theory (Prof.Todorov) Final (Practice Some Solutions) 10 X P 7 = 7 (Explain!) 4. Find ALL Sylow subgroups of Z 12. Prove that these are all. 12 = Since Z 12 is abelian, all of it s subgroups are normal. 2-Sylow subgroup is normal, therefore there is only one 2-Sylow subgroup and it has order 2 2. The only subgroup of order 4 is P 2 = 3 = {3, 6, 9, 0} < Z Sylow subgroup has order 3 and it is P 3 = 4 = {4, 8, 0} < Z Determine the Sylow subgroups of the alternating group A 4 (the even permutations of {1, 2, 3, 4}. A 4 = 12 = r 2 12 implies r 2 = 1, 2, 3, 4, 6, 12 r 2 = 1(mod2) implies r 2 = 1, 3, 5, 7, 9, 11 r 2 = 1 or 3 2-Sylow subgroup has order 2 2 (12)(34), (13)(24), (14)(23) is a subgroup of order 4, hence a 2-Sylow subgroup. (ij) / A 4 since (ij) are odd permutations. 4-cycles are also odd permutations, hence not in A 4 Therefore there are no other elements of order 2 or 4 in A 4, hence there is only one 2-sylow subgroup P 2 = {(1), (12)(34), (13)(24), (14)(23)}. 3-Sylow subgroups have order 3. r 3 12 implies r 3 = 1, 2, 3, 4, 6, 12 r 3 = 1(mod3) implies r 3 = 1, 4, 7, 10 r 2 = 1 or 4 P 3,1 = {(123), (132), (1)}, P 3,2 = {(124), (142), (1)}, P 3,3 = {(134), (143), (1)}, P 3,4 = {(234), (243), (1)} These are the 4 different 3-Sylow subgroups. 10 x 1. Prove that the conjugacy class of (2314) in S 4 has exactly 6 elements. 2. How many elements are there in the conjugacy class of (13)(26) in S 6? 11

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