   # In Example 3 of Section 3.1, find elements a and b of S ( A ) such that ( a b ) − 1 ≠ a − 1 b − 1 . From Example 3 of section 3.1: A = { 1 , 2 , 3 } and S ( A ) is a set of bijective functions defined on A . ### Elements Of Modern Algebra

8th Edition
Gilbert + 2 others
Publisher: Cengage Learning,
ISBN: 9781285463230

#### Solutions

Chapter
Section ### Elements Of Modern Algebra

8th Edition
Gilbert + 2 others
Publisher: Cengage Learning,
ISBN: 9781285463230
Chapter 3.2, Problem 5E
Textbook Problem
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## In Example 3 of Section 3.1, find elements a and b of S ( A ) such that ( a b ) − 1 ≠ a − 1 b − 1 .From Example 3 of section 3.1: A = { 1 , 2 , 3 } and S ( A ) is a set of bijective functions defined on A .

To determine

The elements a and b of S(A) such that (ab)1a1b1

### Explanation of Solution

Given information:

A={1,2,3} and S(A) set of functions defined from A to A.

Formula used:

1) Composition of function: Let f and g be the functions defined from A to A, then fg:AA, is defined as fg(x)=f[g(x)]

2) Definition of a group.

Suppose the binary operation is defined for element of a set G. The set G is a group with respect to , provided the following conditions hold.

1. G is closed under . That is xG and yG imply that xy is in G.

2. is associative. For all x,y,z in G, x(yz)=(xy)z.

3. G has an identity element e. There is an e in G such that xe=ex=x for all xG.

4. G contains inverses. For each aG, there exists bG such that ab=ba=e.

Calculation:

Let A={1,2,3}, S(A) be a set of bijective functions defined on A.

f(1) have three choices, f(2) have two choices and f(3) has only one choice.

So there are 3!=321=6 different mappings given by,

e={e(1)=1e(2)=2e(3)=3, σ={σ(1)=2σ(2)=1σ(3)=3, ρ={ρ(1)=2ρ(2)=3ρ(3)=1

γ={γ(1)=3γ(2)=2γ(3)=1, τ={τ(1)=3τ(2)=1τ(3)=2, δ={δ(1)=1δ(2)=3δ(3)=2

Thus, S(A)={e,ρ,τ,σ,γ,δ}.

In constructing the table for S(A), list the elements of S(A) in a column at the left

and in a row at the top.

Let e as an identity map, therefore

eρ=ρ,eτ=τ,eσ=σ,eγ=γ,eδ=δ

Similarly,

ρe=ρ,τe=τ,σe=σ,γe=γ,δe=δ.

When the product ρ2=ρρ is computed, then

ρ2(1)=ρ[ρ(1)]=ρ(2)=3ρ2(2)=ρ[ρ(2)]=ρ(3)=1ρ2(3)=ρ[ρ(3)]=ρ(1)=2ρρ=τ

Similarly, ρσ=γ, σρ=δ and so on.

Therefore, Cayley’s table for S(A) is given by,

eρτσγδeeρτσγδρρτeγδσττeρδσγσσδγeτργγσδρeτδδγστρe

Composition of e with other elements of S(A) satisfy the property,

(ab)1=a1b1

From the Cayley’s table,

a.

(ρρ)1=τ1=ρ,ρ1ρ1=ττ=ρ(ρρ)1=ρ1ρ1

b.

(ρτ)1=e1=e,ρ1τ1=τρ=e(ρτ)1=ρ1τ1

c.

(ρσ)1=γ1=γ,ρ1σ1=τσ=δ(ρσ)1ρ1σ1

d.

(ργ)1=δ1=δ,ρ1γ1=τγ=σ(ργ)1ρ1γ1

e.

(ρδ)1=σ1=σ,ρ1δ1=τδ=γ(ρδ)1ρ1δ1

f.

(τρ)1=e1=e,τ1ρ1=ρτ=e(τρ)1=τ1ρ1

g.

(ττ)1=ρ1=τ,τ1τ1=ρρ=τ(ττ)1=τ1τ1

h.

(τσ)1=δ1=δ,τ1σ1=ρσ=γ(τσ)1τ1σ1

i

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