Download Solutions to assigned problems from Sections 3.1, page 142, and

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.