   # Let H be the group given in Exercise 17 of Section 3.3 , and let S ( A ) be as given in Example 2 of this section. Find an isomorphism from H to S ( A ) . Sec. 3.3 , # 17 ≫ Consider the set of matrices H = { I 2 , M 1 , M 2 , M 3 , M 4 , M 5 } , where I 2 = [ 1 0 0 1 ] , M 1 = [ 1 0 − 1 − 1 ] , M 2 = [ 0 1 − 1 − 1 ] M 3 = [ − 1 − 1 1 0 ] , M 4 = [ − 1 − 1 0 1 ] , M 5 = [ 0 1 1 0 ] Show that H is a subgroup of G L ​ ( 2 , ℝ ) , the general linear group of order 2 over ℝ . ### 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 3.5, Problem 5E
Textbook Problem
1 views

## Let H be the group given in Exercise 17 of Section 3.3 , and let S ( A ) be as given in Example 2 of this section. Find an isomorphism from H to S ( A ) .Sec. 3.3 , # 17 ≫ Consider the set of matrices H = { I 2 , M 1 , M 2 , M 3 , M 4 , M 5 } , where I 2 = [ 1 0 0 1 ] , M 1 = [ 1 0 − 1 − 1 ] , M 2 = [ 0 1 − 1 − 1 ] M 3 = [ − 1 − 1 1 0 ] , M 4 = [ − 1 − 1 0 1 ] , M 5 = [ 0 1 1 0 ] Show that H is a subgroup of G L ​ ( 2 , ℝ ) , the general linear group of order 2 over ℝ .

To determine

To find: An isomorphism from H to S(A).

### Explanation of Solution

Given information:

H={I2,M1,M2,M3,M4,M5} where I2=,M1=,M2=,M3=,M4=,M5= under multiplication.

S(A)={IA,ρ,τ,σ,γ,δ} where

IA={IA(1)=1IA(2)=2IA(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

under multiplication.

Formula used:

A mapping ϕ:GG' is isomorphism if

1) ϕ is one-to-one correspondence from G to G'.

2) ϕ(xy)=ϕ(x)ϕ(y) for all x,y in G.

Calculation:

Let the first group H={I2,M1,M2,M3,M4,M5} under multiplication and second group S(A)={IA,ρ,τ,σ,γ,δ} under multiplication.

A mapping ϕ:HS(A) defined by ϕ(I2)=IA,ϕ(M1)=σ,ϕ(M2)=ρ,ϕ(M3)=τ,ϕ(M4)=γ,ϕ(M5)=δ.

For every element of H, there is a different image in S(A), so ϕ:HS(A) is one-one. That is,

abϕ(a)ϕ(b), so ϕ:HS(A) is one-one.

For every element of S(A), there is preimage in H, so ϕ:HS(A) is onto. That is,

xS(A),aH such that ϕ(a)=x

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