Mat 311 Test 1, 2019 Solutions Question One 1.1 There exist integers x and y such that ax + cy =1sincegcd(a, c) =1. Thus (ax)b +(cy)b = b, implying a(xb)+(bc)y = b. Since a a(xb) and a (bc)y, thena [a(xb)+(bc)y]. Therefore a b. (3) 1.2 G = {z C : z n =1=e (2kπ)i,k Z} = {z =1 (1/n) = e (2kπ)i/n : k = 0, 1, 2, 3,,n 1} = {e (2πi/n)k : k =0, 1, 2, 3,,n 1} =< e (2πi)/n >. Thus G is cyclic. (3) 1.3 Consider a e= a, then4ae = a implies e =1/4. Thus e =1/4 isthe identity element of (Q +, ). Consider a b = 1/4. Then 4ab = 1/4 implies b = 1/16a is the inverse of a in (Q +, ). (3) 1.4 Let x, y G. Then x x = e = y y. This implies x 1 = x and y 1 = y. Therefore (xy) 2 = e implies xy(xy) =e = xx. Thus yxy = x by cancellation laws in G. Hence xy = y 1 x = yx, implying G is abelian. (3) 1.5 Let V = {e, a, b, c} be the Klein 4-group where e is the identity element. e a b c e e a b c a a e c b b b c e a c c b a e Now a 2 = e = b 2 = c 2 and V has order 4, but V has no element of order 4sincea, b and c have order 2. Thus V is not cyclic since no element of V can generate all its elenents. 1
Subgroups of V : V ; {e}; <a>= {e, a}; <b>= {e, b} and <c>= {e, c}. Lattice Diagram: V <a> <b> <c> (5) {e} 1.6 S = R { 1}. (a) We show that a b 1 for all a, b S. Ifa b = 1, then a + b + ab = 1, implying a(1 + b) = (1 + b). Since b 1, then 1 + b 0. Dividing by 1 + b, wehavea = 1, a contradiction. Hence a b 1 for all a, b S, implying is well defined. (b) Let a, b, c S. Then(a b) c =(a+b+ab) c =(a+b+ab)+c+(a+ b+ab)c = a+b+c+ab+ac+bc+(ab)c = a+b+c+ab+ac+bc+a(bc) (1). Also a (b c)=a (b + c + bc) =a +(b + c + bc) +a(b + c + bc) = a + b + c + ab + ac + bc + a(bc) (2). Thus (1) and (2) imply is associative. Consider a e = a, thena + e + ae = a implies e(1 + a) = 0. Since 1+a 0, we divide by it to obtain e = 0 is the identity element of S. Consider a b=0. Thena + b + ab = 0 implies b(1 + a) = a. Thus b = a is the inverse of a. If a = 1, then a =1+a implies 1 = 0, 1+a 1+a an absurdity. Hence a 1 S. Therefore (S, ) is a group. (c) (3 2) x = 9 implies (3+2+6) x =11 x =9. Thus11+x+11x =9, implying 11 + 12x = 9. Therefore 12x = 2, implying x = 1/6. (8) [25] Question Two 2.1 Suppose G has a finite number of subgroups. Each a G generates a cyclic subgroup of G. So G has a finite number of cyclic subgroups. 2
Let S = {G 1,G 2,,G n } be the set of all cyclic subgroups of G. Let a G, then<a> S, implying a G i for some i {1, 2,,n}. Therefore G n i=1 G i. Hence G = n i=1 G i. Suppose each G i is finite, then G = n i=1 G i <, i.e. G is finite. Suppose that G j is infinite for some j {1, 2,,n}. ThenG j = (Z, +). Z has an infinite number of subgroups of the form nz, n a positive integer. Thus G has an infinite number of subgroups, a contradiction! Hence each G i is finite, therefore G is finite. (5) 2.2 (a) Let G n = {n n non-singular matrices } with being the usual matrix multiplication. Then is an associative binary operation on G since matrix multiplication is associative. The identity element of G n is I n, the n x n identity matrix having 1 s on the main diagonal and zeros elsewhere. Since each matrix in G n is non-singular, it has an inverse A 1 and the inverse of A 1 is A, hence A 1 is non-singular and thus A 1 G n Therefore (G n, ) is a group. (3) (b) Let A, B S, then det(ab) = det(a)det(b) = 1 1 = 1, implying AB S. The identity matrix I n has determinant equal to 1, thus I n S. Let A S, thendet(a) =1anddet(A 1 )=1/det(A) =1/1 = 1, hence A 1 S. Therefore S is a subgroup of G n. (2) 2.3 Z 12 =< 1 >= {0; 1; 2; 3; 4; 5; 6; 7; 8; 9; 10; 11}. The other generators of Z 12 are elements 1r = r such that gcd(12,r) = 1. Thus these generators are r =5;7;11. So Z 12 =< 1 >=< 5 >=< 7 >=< 11 >. Subgroups of Z 12 are {0}; Z 12 ; < 2 >= {0; 2; 4; 6; 8; 10}. Other generators of < 2 > are elements 2m such that gcd(m, 6) = 1. Thus m =5, giving other generator of < 2 > as 10. Other subgroups: < 6 >= {0; 6} < 2 >, < 3 >. < 4 >= {0; 4; 8} =< 8 > < 2 >. < 3 >= {0; 3; 6; 9} =< 9 >. Lattice Diagram: Z 12 (5) < 3 > < 6 > < 2 > < 4 > 3 {0}
2.4 Let G =< a>= {a n : n Z} be a cyclic group and H a subgroup of G. Let b H, thenb = a k for some integer k. Let m be the least positive integer such that a m H. Claim: H =< a m > Let b H, thenb = a k for some integer k. By the Division Algorithm, there exist integers s and r such that 0 r<mand k = ms + r. Thus a k =(a m ) s a r, implying a r =(a m ) ( s) a k H since a m,a k H. But 0 r<m,sowemusthaver = 0 (otherwise the definition of m is contradicted). Therefore k = ms, implying b =(a m ) s <a m >. Thus H < a m > H, implying H =< a m >. Hence H is cyclic. (5) [20] Question Three 3.1 e a b c d e e a b c d a a b c d e b b c d e a c c d e a b d d e a b c The multiplication table is symmetric about the main diagonal, thus the group A is abelian. b 1 = c. (4) 3.2 TRUE: Z 6 =< 1 > and since the inverse of 1 is 5, then 5 is also a generator of Z 6. FALSE: The inverse of 2 is 4. FALSE: Matrix multiplication is not generally commutative. (3) 3.3 (a) S 3 is the group of permutations of 3 symbols, 1, 2 and 3 say, taken all at a time. Let A = {1, 2, 3}. Thus S 3 consists of bijective functions f : A A. The binary operation on S 3 is function composition. A permutation of A mapping1to1,2to3and3to2,isusu- 4
( ) ally denoted by α =. Thus we may list elements of S 1 3 2 3 ( ) ( ) ( ) as follows: ρ 1 = ; ρ 2 3 1 2 = ; ρ 3 1 2 0 = ; ( ) ( ) ( ) μ 1 = ; μ 1 3 2 2 = ; μ 3 2 1 3 =. So S 2 1 3 3 has 6 = 3! elements. The elements of S 3 may also be viewed as symmetries of an equilateral triangle, consisting of rotations and diagonal flips. Thus ρ 0, ρ 1 and ρ 2 are the rotations while μ 1 and μ 2 and μ 3 are the diagonal flips of an equilateral triangle. All the proper subgroups of S 3 are cyclic but S 3 is not cyclic. S 3 may be generated by two elements, for example S 3 =<ρ 1,μ 1 >. S 3 is NOT abelian since for example ρ 1 μ 1 = ( ) μ 1 ρ 1 = = μ 2 1 3 2 μ 3. ( 2 1 3 ) = μ 3 while (b) Subgroups of S 3 are S 3 ; {ρ 0 }; <ρ 1 >= {ρ 0 ; ρ 1 ; ρ 2 } = A 3 the alternating group on 3 symbols cosisting of all even permutations in S n ; <μ 1 >= {ρ 0 ; μ 1 }; <μ 2 >= {ρ 0 ; μ 2 }; <μ 3 >= {ρ 0 ; μ 3 }. (c) S 3 (8) <ρ 1 > < μ 1 >< μ 2 > < μ 3 > {ρ 0 } 5
( ) 4 5 6 3.4 σ = implies 6 5 4 ( ) 4 5 6 (a) σ 1 = 2 3 4 6 5 1 ( ) ( 4 5 6 4 5 6 (b) σ 2 = ; σ 4 6 1 2 5 3 3 = ( ) ( 3 4 ) 6 1 5 2 4 5 6 4 5 6 ; σ 2 3 4 6 5 1 5 = 4 5 6 permutation. Therefore O(σ) = 5. ) ; σ 4 = = σ 0,theidentity (c) ( 131 = 26 5 + ) 1 implies σ 131 = (σ 5 ) 26 σ 1 = (σ 0 ) 26 σ 1 = σ = 4 5 6. (6) 6 5 4 3.5 The identity permutation i = (1, 2)(1, 2) is even, hence i A n = {even permutations}. Let σ 1,σ 2 A n. Since each is even, then their product σ 1 σ 2 is a product of an even number of transpositions, hence it is even. Thus σ 1 σ 2 A n. Let σ A n,thenσ is a product of an even number of transpositions, say σ =(a 1,a 2 )(a 3,a 4 )(a 5,a 6 ) (a 2k+1,a 2k+2 ). Then σ 1 = (a 2k+1,a 2k+2 ) (a 5,a 6 )(a 3,a 4 )(a 1,a 2 ). Thus σ 1 isalsoaproductofan even number of transpositions and is therefore even. Since(a i,a j )(a i,a j )= i, thenσσ 1 = (a 1,a 2 )(a 3,a 4 )(a 5,a 6 ) (a 2k+1,a 2k+2 )(a 2k+1,a 2k+2 ) (a 5,a 6 )(a 3,a 4 )(a 1,a 2 )= i, showing that indeed σ 1 is the inverse of σ in A n. Hence A n S n. Since S n = n! and the number of even permutations equals the number of odd permutations, it follows that A n = 1 2 n!=n!. (4) 2 [25] END 6