EXAM 3 MAT 23 Modern Algebra I Fall 201 Name: Section: I All answers must include either supporting work or an explanation of your reasoning. MPORTANT: These elements are considered main part of the answer and will be graded. 1. (10 pts) Recall that M 2 (R) denotes the set of all 2 ( 2 matrices ) of real entries. Let H be the a b subset of M 2 (R) consisting of all matrices of the form for a, b R. Is H closed under (a) matrix addition? Solution: ( ) a b + So H is closed under matrix addition. (b) matrix multiplication? Solution: ( a b So, H is closed under matrix multiplication. ( ) ( ) c d a + c (b + d) = H. d c b + d a + c ) ( ) ( ) c d ac bd (ad + bc) = H. d c ad + bc ac bd 2. (10 pts) Let n be a positive integer and let nz = {nm : m Z}. (a) Show that (nz, +) is a group. Solution: Let a, b nz. Then a = nk and b = nl for some k, l Z. a+b = nk+nl = n(k+l) nz so nz is closed under +. The addition + is associative on Z and nz Z, so it is also associative on nz. 0 nz and it is an identity element. For each nk nz, nk = n( k) nz is an inverse element of nk. Therefore, (nz, +) is a group. (It is indeed an abelian group.) (b) Show that (nz, +) = (Z, +). Hint: You need to find an isomorphism ϕ : Z nz. Solution: Define ϕ : Z nz by ϕ(k) = nk for each k Z. Then ϕ is clearly well-defined, one-to-one and onto. ϕ is a homomorphism since ϕ(k + l) = n(k + l) = nk + nl = ϕ(k) + ϕ(l). Therefore, ϕ is an isomorphism and (nz, +) ϕ = (Z, +).
EXAM 3 MAT 23 Modern Algebra I 2 3. (20 pts) Let S be the set of all real numbers except 1. Define on S by a b = a + b + ab. (a) Show that gives a binary operation on S. Solution: Suppose that a b = a+b+ab = 1. Then we obtain that a = 1 or b = 1. Therefore, S is closed under. (b) Show that (S, ) is a group. Solution: Using the fact that + and are associative on R, one can easily show that is also associative on S = R \ { 1}. Suppose that e is an identity element. Then for any a S, we have a e = a + e + ae = a. Since a 1, this implies that e = 0. Let a S and let a denote an inverse element of a. Then we have a a = a + a + aa = 0. Since a 1, this implies that a = a abelian group.) a+1 (c) Find the solution of the equation 2 x 3 = 7 in S. Solution: By the associative law, 2 x 3 = (2 x) 3 1. Therefore, (S, is a group. (It is indeed an = (2 + x + 2x) 3 = (2 + 3x) 3 = 2 + 3x + 3 + (2 + 3x)3 = 12x + 11. So, the equation tunrs into a linear equation 12x + 11 = 7 whose solution is x = 1 3 S.
EXAM 3 MAT 23 Modern Algebra I 3. (5 pts) Show that if H and K are subgroup of an abelian group G, then is a subgroup of G. HK = {hk : h H and k K} Solution: Let a, b HK. Then a = hk and b = h k for some h, h H and k, k K. ab 1 = (hk)(h k ) 1 = (hk)(k 1 h 1 ) = (hh 1 )(kk 1 ) HK, since H and K are subgroups of an abelian group G. Hence, HK is a subgroup of G. 5. (5 pts) Show that a group with no proper nontrivial subgroup is cyclic. Solution: Let G be a group with no proper nontrivial subgroup. If G = {e}, clearly G is cyclic. Suppose that G {e}. Then there exists a G such that a e and a G. Since G has no proper nontrivial subgroup, a = G. Therefore, G is cyclic. 6. (10 pts) List elements of Z 2 Z. Find the order of each of the elements. Is this group cyclic? Solution: Z 2 Z = {(0, 0), (0, 1), (0, 2), (0, 3), (1, 0), (1, 1), (1, 2), (1, 3)}. The order of each element is given by: (0, 0) = 1, (0, 1) =, (0, 2) = 2, (0, 3) =, (1, 0) = 2, (1, 1) =, (1, 2) = 2, (1, 3) =. Since none of the elements has order 8, Z 2 Z is not cyclic. Alternatively, 2 and are not relatively prime, so Z 2 Z is not cyclic. 7. (10 pts) List elements of Z 3 Z. Find the order of each of the elements. Is this group cyclic? Solution: Z 3 Z = {(0, 0), (0, 1), (0, 2), (0, 3), (1, 0), (1, 1), (1, 2), (1, 3), (2, 0), (2, 1), (2, 2), (2, 3)}. The order of each element is given by: (0, 0) = 1, (0, 1) =, (0, 2) = 2, (0, 3) =, (1, 0) = 3, (1, 1) = 12, (1, 2) = 6, (1, 3) = 12, (2, 0) = 3, (2, 1) = 12, (2, 2) = 6, (2, 3) = 12. Since there are elements of order 12, Z 3 Z is cyclic. Alternatively, 3 and are relatively prime, Z 3 Z is cyclic.
EXAM 3 MAT 23 Modern Algebra I 8. (10 pts) Find the order of the given element of the direct product. (a) (3, 10, 9) in Z Z 12 Z 15 Solution: The order of 3 in Z is =, the order of 10 in Z 12 (3,) 12 is = 6, and the order of 9 (10,12) 15 in Z 15 is = 5. Hence the order of (3, 10, 9) in Z (9,15) Z 12 Z 15 is the least common multiple of, 6 and 5 which is 60. (b) (3, 6, 12, 16) in Z Z 12 Z 20 Z 2 Solution: The order of 3 in Z is (as seen in the previous problem), the order of 6 in Z 12 is 12 = 2, the order of 12 in Z 20 (6,12) 20 is = 5, and the order of 16 in Z 2 (12,20) 2 is = 3. Hence, the (16,2) order of (3, 6, 12, 16) in Z Z 12 Z 20 Z 2 is the least common multiple of, 2, 5 and 3 which is 60. 9. (5 pts) Prove that a direct product of abelian groups is abelian. Solution: Let G i, i = 1, 2,, n be abelian groups. Let (a 1, a 2,, a n ), (b 1, b 2,, b n ) n i=1 G i. Then (a 1, a 2,, a n )(b 1, b 2,, b n ) = (a 1 b 1, a 2 b 2,, a n b n ) Therefore, the direct product n i=1 G i is abelian. = (b 1 a 1, b 2 a 2,, b n a n ) = (b 1, b 2,, b n )(a 1, a 2,, a n ). 10. (5 pts) Find all abelian groups, up to isomorphism, of order 8. Solution: 8 = 2 3 = 2 2 2 = 2 2 2, so up to isomorphism are all abelian groups of order 8. Z 2 Z 2 Z 2, Z 2 Z, Z 8
EXAM 3 MAT 23 Modern Algebra I 5 11. (10 pts) Let G be the group of all real-valued functions on [0, 1], where we define, for f, g G, addition by (f + g)(x) = f(x) + g(x) for every x [0, 1]. If N = {f G : f ( 1 ) = 0}, prove that G/N = (R, +). Hint: Find an onto homomorphism ϕ : G R such that ker ϕ = N and then use the Fundamental Homomorphism Theorem (the First Isomorphism Theorem). Solution: Define a map ϕ : G R by ϕ(f) = f for each f G. Then ϕ is 1. well-defined: If f = g G then ϕ(f) = f ( 1 ) = g ( 1 ) = ϕ(g). 2. onto: Let r R. Define f : [0, 1] R to be the constant function f(x) = r for all x [0, 1]. Then f G and f ( 1 ) = r. 3. a homomorphism: ϕ(f + g) = (f + g) = f + g = ϕ(f) + ϕ(g). The kernel of ϕ is N = {f G : f ( 1 ) = 0}. Therefore, by the Fundamental Homomorphism Theorem, G/N = (R, +).