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114 Groups and Rings 2008–09 Suggested solutions to exercise set 4 March 3, 2009 24.2 ([a]⊕[b])[c] = [a+b]c = [(a+b)c] = [ac+bc] = [ac]⊕[bc] = ([a][c])⊕([b][c]). 24.8 If a, b ∈ R with ab = ba then by Theorem 24.2, a(−b) = −(ab) = −(ba) = (−b)a and (−a)(−b) = ab = ba = (−b)(−a). 24.12 Let (G, +) be the additive group of R (that is, G is an Abelian group; as a set, G = R, and the group operation on G is the addition operation on R). Let a, b, c ∈ R. Since R is an Abelian group under addition, (a) the zero element of R is the identity element of the group G, so is unique (by Theorem 5.1(a)). (b) each element of G has a unique additive inverse by Theorem 5.1(b); this is what we mean by the negative of an element of R. (c) If a + b = a + c then b = c by Theorem 14.1(a). (d) If b + a = c + a then b = c by Theorem 14.1(b). (e) This is immediate from Theorem 14.1(c). (f) By Theorem 14.1(d), −(−a) = a. By Theorem 14.1(e) and the commutativity of addition, −(a + b) = (−b) + (−a) = (−a) + (−b). (g) We saw this at the top of page 77. 24.15 We know that E is an Abelian group under addition; and m ∗ (n ∗ k) = m ∗ ( 12 nk) = 12 m( 12 nk) = 14 mnk = 12 ( 12 mn)k = (m ∗ n) ∗ k 1 so ∗ is associative. Also, m ∗ (n + k) = 12 m(n + k) = 21 mn + 12 mk = (m ∗ n) + (m ∗ k) and ∗ is commutative (since m ∗ n = 21 mn = 12 nm = n ∗ m) so (n + k) ∗ m = m ∗ (n + k) = (m ∗ n) + (m ∗ k) = (n ∗ m) + (k ∗ m). So the distributive laws hold. So (E, +, ∗) is a ring. 2 is a unit for this ring since for any n ∈ E, n ∗ 2 = 2 ∗ n = 21 (2n) = n. 24.16 Using the distributive laws, for any a, b ∈ R we have (a + b)(a − b) = a(a − b) + b(a − b) = a2 − ab + ba − b2 = a2 − b2 + (ba − ab). So ∀a,b ∈ R : (a + b)(a − b) = a2 − b2 ⇐⇒ ∀a, b ∈ R : a2 − ab + ba − b2 = a2 − b2 ⇐⇒ ∀a, b ∈ R : −ab + ba = 0 ⇐⇒ ∀a, b ∈ R : ba = ab ⇐⇒ R is commutative. 24.18 We know that A is an Abelian group under addition; and a ∗ (b ∗ c) = 0 = (a ∗ b) ∗ c so ∗ is associative. Also, a ∗ (b + c) = 0 = 0 + 0 = (a ∗ b) + (a ∗ c) and (a + b) ∗ c = 0 = 0 + 0 = (a ∗ c) + (b ∗ c). So the distributive laws hold. So (A, +, ∗) is a ring. 24.22* First we check that (P(S), +) is an Abelian group. Since the definition of A + B is unchanged by swapping A and B, we see that + is a commutative operation. Note that ∅ ∈ P(S) and A + ∅ = (A ∪ ∅) \ (A ∩ ∅) = A \ ∅ = A = ∅ + A. 2 So ∅ is an identity for +. By drawing a Venn diagram, it’s easy to see that A + (B + C) = A \ (B ∪ C) ∪ B \ (C ∪ A) ∪ C \ (A ∪ B) . Since the right hand side is unchanged if we permute A, B, C, we have A + (B + C) = C + (A + B) = (A + B) + C. So + is associative. Also, A + A = A \ A = ∅ for each A ∈ P(S). So A is the inverse of A, for all A ∈ P(S). Since intersection is associative (again, this is easy to see from a Venn diagram if you don’t know it already), it only remains to check distributivity. Drawing a Venn diagram, we see that AB +AC = (A∩(B ∪C))\(A∩B ∩C) = A∩ (B ∪C)\(B ∩C) = A(B +C). Since intersection is commutative, BA + CA = AB + AC = A(B + C) = (B + C)A. So P(S) is a ring. 3