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 onetoone. 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 5cycle 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! 5cycles. 5 5 (b) How many permutations of order 5 are there in S 9? Answer: There are ( ) 9 5! 5cycles Order of 5cycle 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 5cycles. (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 11cycle 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 nonnegative 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 onetoone. f : Z 4 Z 2 where f(x) = x(mod2) (Check!) 2. Give an example of a group homomorphism which is onetoone 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 onetoone 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 nonzero 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 psubgroup 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 psubgroup 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 psubgroup of G is contained in one of the P i C. Theorem 2 All psylow subgroups of a finite group G are conjugate to each other. Theorem 3 Let G be a finite group. Then #{psylow subgroups of G} divides G. Theorem 4 Let G be a finite group. Then #{psylow subgroups of G} 1(mod p). Theorem 5 Let G be a finite group. Then #{psylow subgroups of G} = 1 if and only if the psylow subgroup is normal. 1. Let G be a group. Let X be a psubgroup of G and a G. Prove that axa 1 is a psubgroup 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 psubgroup of G and a G. Prove that ap a 1 is a Sylow psubgroup 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 #{5Sylow subgroups of G} using Theorem 3? r 5 = 1, 5, 7, 35 (b) What are the possible #{5Sylow subgroups of G} using Theorem 4? r 5 = 1, 6, 11, 16, 21, 26, 31 (c) How many 5Sylow subgroups does G have? Exactly one. (Explain!) (d) Let P 5 be a 5Sylow subgroup of G. How many elements does P 5 have? P 5 = 5 (Explain!) (e) What are the possible #{7Sylow subgroups of G} using Theorem 3? r 7 = 1, 5, 7, 35 (f) What are the possible #{7Sylow subgroups of G} using Theorem 4? r 7 = 1, 8, 15, 22, 29 (g) How many 7Sylow subgroups does G have? Exactly one. (Explain!) (h) Let P 7 be a 7Sylow 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. 2Sylow subgroup is normal, therefore there is only one 2Sylow 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 2Sylow subgroup has order 2 2 (12)(34), (13)(24), (14)(23) is a subgroup of order 4, hence a 2Sylow subgroup. (ij) / A 4 since (ij) are odd permutations. 4cycles 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 2sylow subgroup P 2 = {(1), (12)(34), (13)(24), (14)(23)}. 3Sylow 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 3Sylow 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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