MODEL ANSWERS TO THE SECOND HOMEWORK 1. (a) is surjective. Indeed, let t R be a non-negative real number. Then t has a square root s R. In this case f (s) = s2 = t, by definition of the square root. is not injective. For example, (1) = (-1) = 1.
MODEL ANSWERS TO THE SEVENTH HOMEWORK
1. For Chapter 2, Section 9: 1. Let : G1 G2 - G2 G1 be the homomorphism that sends (g1 , g2 ) to (g2 , g1 ). This is clearly a bijection. We check that it is a homomorphism. Suppose that (g1 , g2) and (h1 , h
MODEL ANSWERS TO THE FIFTH HOMEWORK 1. Chapter 3, Section 5: 1 (a) Yes. Given a and b Z, (ab) = [ab] = [a][b] = (a)(b). This map is clearly surjective but not injective. Indeed the kernel is easily seen to be nZ. (b) No. Suppose that G is not abelia
MODEL ANSWERS TO THE FOURTH HOMEWORK 2. Chapter 3, Section 1: 1 (a) 1 2 3 4 5 6 . 4 5 2 1 3 6 (b) 1 2 3 4 5 . 3 1 2 4 5 (c) 1 2 3 4 5 . 1 4 3 2 5 5. It suffices to find the cycle type and take the lowest common multiples of the individual lengths of
MODEL ANSWERS TO THE THIRD HOMEWORK 1. (b) Circles centre the origin. (c) The real line union , where the number m R {} represents the slope. 2. Chapter 2, Section 4: 9. [0] = 0 + H = {[0], [4], [8], [12]} [1] = 1 + H = {[1], [5], [9], [13]}
[2] =
MODEL ANSWERS TO THE SECOND HOMEWORK 1. Label the vertices of the square A, B, C, D, where we start at the top left hand corner and we go around the square clockwise. In particular A is opposite to C and B is opposite to D. There are three obvious ty
MODEL ANSWERS TO THE FIRST HOMEWORK 1. Chapter 1, 1: 1. Suppose that a and b are elements of S. By rule (1) a b = a. But by rule (2), a b = b a. Applying rule (1) we get a b = b a = b. Thus a = a b = b. As a and b are arbitrary, S can have at most on
MODEL ANSWERS TO THE FIFTH HOMEWORK 11. The solutions of the equation x2 = e are precisely the elements of G which are their own inverses. The function i : G - G is a bijection, since i is its own inverse. The elements which are not their own inverse
MODEL ANSWERS TO THE FOURTH HOMEWORK 1. (a) No, this is not a group. The rule for multiplication is not associative. For example, Also there is no identity. 0 is a right identity, as a - 0 = a, but 0 - 1 = -1 = 1, so that 0 is not a left identity. (b
MODEL ANSWERS TO THE THIRD HOMEWORK 1. 7. There are many ways to prove this. Perhaps the simplest proceeds as follows. Consider trying to construct a permutation f : S - S. Order the elements of S from 1 to n. We first have to decide where to send th