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Solutions to assigned problems from Sections 3.1, page 142, and 3.2, page 150 In Exercises 3.1.2, 3.1.4, and 3.1.10, decide whether each of the given sets is a group with respect to the indicated operation. If it is not a group, state a condition in Definition 3.1 that fails to hold. 3.1.2 The set of all irrational numbers with operation addition. This is not a group. Note that π and −π are irrational numbers, but π + (−π) = 0, which is not an irrational number. Thus condition 1, of Definition 3.1 fails to hold. This means that multiplication is not a binary operation on the set of all irrational numbers. 3.1.4 The set of all positive rational numbers with operation multiplication. This is a group. 3.1.10 The set E of all even integers with operation multiplication. This is not a group. Condition 3, of Definition 3.1 fails to hold. 3.1.14 The table below defines an operation of multiplication on the set S = {e, a, b, c}. Find a condition in Definition 3.1 that fails to hold, and thereby show that S is not a group. × e a b c e e e e e a a a a a b b b b b c c c c c Condition 3 of Definition 3.1 fails to hold. a × e = e 6= a shows that e is not an identity element. a × b = b 6= a shows that b is not an identity element. a × c = c 6= a shows that c is not an identity element. b × a = a 6= b shows that a is not an identity element. In Exercises 3.1.22 and 3.1.26, decide whether each of the given sets is a group with respect to the indicated operation. If it is not a group, state all of the conditions in Definintion 3.1 that fail to hold. If it is a group, state its order. 3.1.22 The set {[1], [2], [3], [4]} ⊆ Z5 with operation multiplication. This is a group. 3.1.26 The set {[0], [2], [4], [6]} ⊆ Z8 with operation addition. This is a group. 3.1.28 Let G be the set of eight elements G = {1, i, j, k, −1, −i, − j, −k} with identity element 1 and non-commutative multiplication given by (−1)2 = 1, i2 = j2 = k2 = −1, i j = − ji = k, jk = −kg = i, ki = −ik = j, −x = (−1)x = x(−1) for all x in G. Given that G is a group of order 8, write out the multiplication table for G. This group is known as the quaternion group. + 1 i j k −1 −i −j −k 1 i j k 1 i j k i −1 k − j j −k −1 i k j −i −1 −1 −i − j −k −i 1 −k j −j k 1 −i −k − j i 1 −1 −i −1 −i −i 1 −j k −k − j 1 i i −1 j −k k j −j −j −k 1 i j k −1 −i −k −k j −i 1 k −j i −1 Sections 3.1 and 3.2 Solutions 2 Section 3.2 3.2.4 An element x in a multiplicative group G is called idempotent if x2 = x. Prove that the identity element e is the only idempotent element in a group G. Proof. Clearly e is an idempotent element by the definition of the identity element. So assume that x2 = x for some x ∈ G. Then x·x = x −1 =⇒ x (x · x) = x−1 x multiplying both sides by x−1 on the left =⇒ (x−1 x)x = x−1 x using associativity =⇒ ex = e as x−1 x = e =⇒ x = e Thus e is the only idempotent element in a group G. 3.2.5 Let A = {1, 2, 3} and let S (A) be the group of permutations on A as in Example 3 of Section 3.1. Find elements a and b of S (A) such that (ab)−1 6= a−1 b−1 . Let a = ρ and b = σ. Then (ρ ◦ σ)−1 = γ−1 = γ 6= δ = ρ2 ◦ γ = ρ−1 ◦ γ−1 . (There are other answers.) 3.2.14 Prove that a group G is abelian if and only if (ab)−1 = a−1 b−1 for all a and b in G. Proof. Assume first that (ab)−1 = a−1 b−1 for all a and b in G. Let a, b ∈ G. Then ab = ((ab)−1 )−1 as (x−1 )−1 = x for any x ∈ G by Theorem 3.4, part c. = (b−1 a−1 )−1 by the reverse order law = (b−1 )−1 (a−1 )−1 by applying the hypothesis of the statement to a−1 , b−1 ∈ G = ba as (x−1 )−1 = x for any x ∈ G by Theorem 3.4, part c. Thus ab = ba for any a, b ∈ G. Therefore G is abelian. Conversely assume that G is abelian. Then (ab)−1 = b−1 a−1 by the reverse order law = a−1 b−1 as G is abelian. This shows the desired property for all elements a and b of G.