Math 546 Homework Problems. Due Wednesday, January 25. This homework has two types of problems.

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1 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 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 + b 2. e) S = Z, a b = a + b ab. f) S = R, a b = b. g) S = {1, 2, 3, 2, 4}, a b = b Is division a commutative operation on R +? Is it associative? Note: If S R, then we define S + = {s S : s > 0} 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 s 1, s 2 S, we define s 1 s 2 to the be the element in the row that contains s 1 and the column that contains s 2. For example, we have c b = d.

2 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 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 Let S = {a, b, c}. The following table defines a binary operation on S. Is (S, ) a group? Why or why not? * a b c a a b c b b b c c c c c 2.8. Let G = {f : R R : for all x R, f(x) 0}. For all f, g G, define by Is (G, ) a group? Prove or disprove Let G = {( ) } a 0 : a, b 0 in R. Show that G forms a group under matrix multi- 0 b plication. (f g)(x) = f(x)g(x) for all x R I Problems 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?

3 2.6. Let S = {a, b, c}. The following table defines a binary operation on S. Is (S, ) a group? Why or why not? Let G = { M = ( a ) b b a * a b c a a b c b b a c c c b a } : a, b R, det M = a 2 + b 2 0 GL(2, R). 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 = a 2 b 3. Claim: is a binary operation on S. Proof: For all a, b Z, we have a b = a 2 b 3 Z, so is a binary operation. Claim: is not commutative on S. Proof: We note that b a = b 2 a 3. Therefore, we have a b = b a if and only if a 2 b 3 = a 3 b 2, which holds if and only if a = b or either of a, b = 0. We observe that Hence, is not commutative. Claim: is not associative on S. Proof: We compute 1 2 = = 8 4 = = 2 1. (a b) c = (a 2 b 3 ) c = (a 2 b 3 ) 2 c 3 = a 4 b 6 c 3, a (b c) = a (b 2 c 3 ) = a 2 (b 2 c 3 ) 3 = a 2 b 6 c 9. Therefore, we have (a b) c = a (b c) if and only if a 4 b 6 c 3 = a 2 b 6 c 9, which holds for a, b, c 0 if only if a 2 = c 6. Taking square roots, we see that associativity follows if and only if a = c 3. We note that Hence, is not associative. (2 1) 1 = = 16 4 = = 2 (1 1).

4 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 S. Therefore, is not a binary operation. Since is not a binary operation, we do not consider whether or not is commutative or associative 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 so is associative on S. = a (b + c 1) = a (b c),

5 (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 = 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 implies that (S, ) is an abelian group. a b = a + b 1 = b + a 1 = b a