# 114 Groups and Rings 2008–09 Suggested solutions to exercise set 4 ```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
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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.
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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.
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