let us consider a homomorphism phi: R^2 --> R^2 transforming a point (x,y) € R^2

Question

let us consider a homomorphism phi: R^2 --> R^2 transforming a point (x,y) € R^2 into a point (2x+4y, 3x+6y).

a) prove that the groups ker of phi and (R^2/ ker of phi ) both are isomorphic to R

b) is phi an automorphism ?
here R stands for the group of real numbers with respect to addition and R^2 denotes a direct sum of two copies of R

4. Let us consider a homomorphism :R2 Ra transforming a paint (z, e R into a point (2z 4w.3z 6) a) rove that the graups Kers snd R /Kero both are isomarphic to R b) Is an aut tomorph Here R stands for the group of real numbers with respect to addition and Ra denotes a direct sum of two copies of R

Explanation / Answer

phi:R^2 --- R^2

by phi(x,y)=(2x+4y,3x+6y)

Kern phi = {(x,y) belongs to R^2 :phi(x,y)=(0,0)}

{(x,y) in R^2:(2x+4y,3x+6y)=0}

2x+4y=0,3x+6y=0

x+2y=0,x+2y=0

x=-2y

i.e

Kern(phi)={(x,y) in R^2:x=-2y}

Now define a map g:R to kern(phi)

g(x)= (-2x,x)

Verify that g is an isomorphism

For one one onto ,

ker(g)={x belongs to R:g(x)=(0,0)}

={x in R: (-2x,x)=(0,0)}

{(0)}

Which impkies that g is one onto ,hence isomorphism

And by group isomorphism theorem

R^2/kern(phi) isomorphic to Im(phi)

By rank nullity theorem ,rank(phi)=1 since kernal (phi ) has dim 1

Which implies that img(phi) has dim 1 that is R

Hence R^2/kern(phi) isimorphic to R

B) phi is not automorphism since from part we get kern(phi) not equal to {0} that is not one one onto hence not automorphism

Related Questions

Navigate

• Browse (All)
• Browse L
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.