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Math 546 Homework 1 Due Wednesday, January 25. This homework has two types of problems. • 546 Problems. All students (students enrolled in 546 and 701I) are required to turn these in. • 701I Problems. Only students enrolled in 701I are required to turn these in. Students not enrolled in 701I are welcome to turn these in as well. I especially welcome students looking for a challenge to attempt these. Note: Part 3 below contains solutions to some of the unassigned problems from Sections 1 and 2 of Saracino’s text. I am providing these as examples of how to write solutions. Please have a look at these, and make sure to read Sections 1 and 2 of the text. 1 546 Problems. 1.3 and 1.6. For each of the following sets S and functions ∗ on S×S, determine whether ∗ is a binary operation on S. If ∗ is a binary operation on S, determine whether it is commutative and whether it is associative. a) S = Z, a ∗ b = a + b2 . e) S = Z, a ∗ b = a + b − ab. f) S = R, a ∗ b = b. g) S = {1, −2, 3, 2, −4}, a ∗ b = |b|. 1.4. Is division a commutative operation on R+ ? Is it associative? Note: If S ⊆ R, then we define S + = {s ∈ S : s > 0}. 1.9. Let S = {a, b, c, d}. The following table defines a binary operation ∗ on S. Is ∗ commutative? Is it associative? * a b c d a a c b d b c a d b c b d a c d d b c a Note: For s1 , s2 ∈ S, we define s1 ∗ s2 to the be the element in the row that contains s1 and the column that contains s2 . For example, we have c ∗ b = d. 2.1. Which of the following are groups? Why or why not? d) The set {1, −1} under multiplication. f) R × R = {(x, y) : x, y ∈ R} under the operation (x, y) ∗ (z, w) = (x + z, y − w). g) R × R× = {(x, y) : x, y ∈ R and y 6= 0} under the operation (x, y) ∗ (z, w) = (x + z, yw). Recall that R× = R \ {0}. h) R \ {1} under the operation a ∗ b = a + b − ab. 2.5. Let S = {a, b, c}. The following table defines a binary operation ∗ on S. * a b c a a b c b c b c b c c c Is (S, ∗) a group? Why or why not? 2.8. Let G = {f : R → R : for all x ∈ R, f (x) 6= 0}. For all f , g ∈ G, define × by (f × g)(x) = f (x)g(x) for all x ∈ R. Is (G, ×) a group? Prove or disprove. 2.11. Let G = a 0 : a, b 6= 0 in R . Show that G forms a group under matrix multi0 b plication. 2 701I Problems. 1.10. How many binary operations are there on a set S with n elements? How many of these are commutative? 2.1 e) Let S = {q ∈ Q+ : √ q ∈ Q}, ∗ is ×. Is (S, ∗) a group? Why or why not? 2.6. Let S = {a, b, c}. The following table defines a binary operation ∗ on S. * a b c a a b c b c b c a c b a Is (S, ∗) a group? Why or why not? 2.10. Let a b 2 2 G= M = : a, b ∈ R, det M = a + b 6= 0 ⊂ GL(2, R). −b a Let × denote ordinary matrix multiplication. Show that (G, ×) is a group. 3 Examples. 1.3 and 1.6. For each of the following sets S and functions ∗ on S×S, determine whether ∗ is a binary operation on S. If ∗ is a binary operation on S, determine whether it is commutative and whether it is associative. b) S = Z, a ∗ b = a ∗ b = a2 b3 . Claim: ∗ is a binary operation on S. Proof: For all a, b ∈ Z, we have a ∗ b = a2 b3 ∈ Z, so ∗ is a binary operation. Claim: ∗ is not commutative on S. Proof: We note that b ∗ a = b2 a3 . Therefore, we have a ∗ b = b ∗ a if and only if a2 b3 = a3 b2 , which holds if and only if a = b or either of a, b = 0. We observe that 1 ∗ 2 = 12 23 = 8 6= 4 = 22 13 = 2 ∗ 1. Hence, ∗ is not commutative. Claim: ∗ is not associative on S. Proof: We compute (a ∗ b) ∗ c = (a2 b3 ) ∗ c = (a2 b3 )2 c3 = a4 b6 c3 , a ∗ (b ∗ c) = a ∗ (b2 c3 ) = a2 (b2 c3 )3 = a2 b6 c9 . Therefore, we have (a ∗ b) ∗ c = a ∗ (b ∗ c) if and only if a4 b6 c3 = a2 b6 c9 , which holds for a, b, c 6= 0 if only if a2 = c6 . Taking square roots, we see that associativity follows if and only if a = c3 . We note that (2 ∗ 1) ∗ 1 = 24 16 13 = 16 6= 4 = 22 16 19 = 2 ∗ (1 ∗ 1). Hence, ∗ is not associative. h) S = {1, 6, 3, 2, 18}, a ∗ b = ab. Claim: ∗ is not a binary operation on S. Proof: We note that 6, 2 ∈ S, but that 6 ∗ 2 = 12 6∈ S. Therefore, ∗ is not a binary operation. Since ∗ is not a binary operation, we do not consider whether or not ∗ is commutative or associative. 2.1. Which of the following are groups? Why or why not? b) S = 3Z = {3n : n ∈ Z}, the set of integers that are multiples of 3, ∗ is +. Claim: (S, ∗) is a group. Proof: It suffices to verify the axioms: (i) ∗ is a binary operation on S. To see this, let a = 3x, b = 3y ∈ 3Z. Then we have a + b = 3x + 3y = 3(x + y) ∈ 3Z, so ∗ is a binary operation on S. (ii) ∗ is associative on S. Since S = 3Z ⊂ Z, we see that 3Z inherits associativity under + from Z. (iii) (S, ∗) has an identity. We claim that 0 = 3 · 0 ∈ 3Z is an identity for (S, ∗). To see this, let a = 3x ∈ 3Z. Then we have a + 0 = 3x + 0 = 3x = 0 + 3x = 0 + a, so 0 is an identity for (S, ∗). (iv) (S, ∗) has inverses. To see this, let a = 3x ∈ 3Z. Then −a = −3x = 3(−x) ∈ 3Z has −a + a = 0 = a + (−a), so −a is an inverse for a. It follows that (S, ∗) is a group. Furthermore, (S, ∗) inherits commutativity from Z since S ⊆ Z, so (S, ∗) is an abelian group. i) S = Z, a ∗ b = a + b − 1. Claim: (S, ∗) is a group. Proof: It suffices to verify the axioms. (i) ∗ is a binary operation on S. To see this, let a, b ∈ Z. Then we have a ∗ b = a + b − 1 ∈ Z, so ∗ is a binary operation on S. (ii) ∗ is associative on S. To see this, let a, b, c ∈ Z. Then we have (a ∗ b) ∗ c = (a + b − 1) ∗ c = (a + b − 1) + c − 1 = a + (b + c − 1) − 1 = a ∗ (b + c − 1) = a ∗ (b ∗ c), so ∗ is associative on S. (iii) (S, ∗) has an identity. We claim that 1 ∈ Z is an identity for (S, ∗). To see this, let a ∈ Z. Then we have a ∗ 1 = a + 1 − 1 = a = 1 + a − 1 = 1 ∗ a, so 1 is an identity for (S, ∗). (iv) (S, ∗) has inverses. To see this, let a ∈ Z. Then we have 2 − a ∈ Z and a ∗ (2 − a) = a + (2 − a) − 1 = 1 = (2 − a) + a − 1 = (2 − a) ∗ a, so 2 − a is an inverse for a. It follows that (S, ∗) is a group. Furthermore, for all a, b ∈ Z, we have a∗b=a+b−1=b+a−1=b∗a implies that (S, ∗) is an abelian group.