Rigorous proof of isomorphism

  • #1

Homework Statement


*This is from a Group Theory class
**My secondary aim is to practice writing the math perfectly because I tend to loose a lot of points for not doing so in exams...

Let λ ∈ Q*
fλ : Q → Q defined as fλ(x) = λx

a) Show that fλ is and automorphism of the group of rationals with the sum operation (Q,+)

Homework Equations



Let f: (G,*) → (G', ⋅)
e is the neutral element of G
e' is the neutral element of G'

The definition of homomorphisms:
f is a homomorphism if
f(x*y) = f(x) ⋅ f(y) , ∀x,y ∈ G

f is injective ⇔ ker(f) = {e}
f is surjective ⇔ Im(f) = G'

The Attempt at a Solution



Let's show that fλ is a homomorphism.

Let x,y ∈ Q
fλ(x+y) = λ(x+y) = λx + λy = fλ(x) + fλ(y)
which proves that fλ is a homomorphism.


Let's show that fλ is injective.

The neutral element of Q is 0.
0 = λx ⇒ x = 0 because λ ≠ 0
We can therefore write Ker(fλ ) = {0} which proves that fλ is injective.


Let's show that fλ is surjective.

Every element y of Q can be written with the form λx where x∈Q.
This means that Im(fλ) = Q.
This proves that fλ is surjective.


Finally, we proved that fλ is a endomorphism that is injective and surjective. This is the definition of an automorphism.
 

Answers and Replies

  • #2
13,228
10,167

Homework Statement


*This is from a Group Theory class
**My secondary aim is to practice writing the math perfectly because I tend to loose a lot of points for not doing so in exams...

Let λ ∈ Q*
fλ : Q → Q defined as fλ(x) = λx

a) Show that fλ is and automorphism of the group of rationals with the sum operation (Q,+)

Homework Equations



Let f: (G,*) → (G', ⋅)
e is the neutral element of G
e' is the neutral element of G'

The definition of homomorphisms:
f is a homomorphism if
f(x*y) = f(x) ⋅ f(y) , ∀x,y ∈ G

f is injective ⇔ ker(f) = {e}
f is surjective ⇔ Im(f) = G'

The Attempt at a Solution



Let's show that fλ is a homomorphism.

Let x,y ∈ Q
fλ(x+y) = λ(x+y) = λx + λy = fλ(x) + fλ(y)
which proves that fλ is a homomorphism.


Let's show that fλ is injective.

The neutral element of Q is 0.
0 = λx ⇒ x = 0 because λ ≠ 0
We can therefore write Ker(fλ ) = {0} which proves that fλ is injective.


Let's show that fλ is surjective.

Every element y of Q can be written with the form λx where x∈Q.
This means that Im(fλ) = Q.
This proves that fλ is surjective.


Finally, we proved that fλ is a endomorphism that is injective and surjective. This is the definition of an automorphism.
Right so far. I would have written in the surjective part: Every element ##y## can be written ##y=1\cdot y= \lambda \cdot \lambda^{-1} \cdot y = f_\lambda(\lambda^{-1}y)## to actually show the preimage.

In general, you will have to consider more points in a proof:
  1. well definition
  2. domain
  3. codomain
  4. homomorphy
  5. injectivity
  6. surjectivity
The first point comes into play when there are more than one possible images. E.g. if we have ##f: \mathbb{Q} \longrightarrow \mathbb{R}## defined as ##f(x)=\,##decimal representation of ##x##, then we want to be sure that ##f(1/2)=f(2/4)##. This is no problem in your example, and also the points 2. and 3. are obvious, but in general should be considered, because if only a rule for a function is given, it's sometimes not automatically clear that the image belongs to the claimed domain.
 
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  • #3
Right so far. I would have written in the surjective part: Every element ##y## can be written ##y=1\cdot y= \lambda \cdot \lambda^{-1} \cdot y = f_\lambda(\lambda^{-1}y)## to actually show the preimage.

In general, you will have to consider more points in a proof:
  1. well definition
  2. domain
  3. codomain
  4. homomorphy
  5. injectivity
  6. surjectivity
The first point comes into play when there are more than one possible images. E.g. if we have ##f: \mathbb{Q} \longrightarrow \mathbb{R}## defined as ##f(x)=\,##decimal representation of ##x##, then we want to be sure that ##f(1/2)=f(2/4)##. This is no problem in your example, and also the points 2. and 3. are obvious, but in general should be considered, because if only a rule for a function is given, it's sometimes not automatically clear that the image belongs to the claimed domain.
Thanks a lot man! This is exactly the kind of insights I was hoping for.
 

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