Is fλ an Automorphism of the Rational Numbers Group?

[AllRelative]

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^*
f_λ : Q → Q defined as f_λ(x) = λx

a) Show that f_λ 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 f_λ is a homomorphism.

Let x,y ∈ Q
f_λ(x+y) = λ(x+y) = λx + λy = f_λ(x) + f_λ(y)
which proves that f_λ is a homomorphism.Let's show that f_λ is injective.

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

Every element y of Q can be written with the form λx where x∈Q.
This means that Im(f_λ) = Q.
This proves that f_λ is surjective.Finally, we proved that f_λ is a endomorphism that is injective and surjective. This is the definition of an automorphism.

 

[fresh_42]

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^*
f_λ : Q → Q defined as f_λ(x) = λx

a) Show that f_λ 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 f_λ is a homomorphism.

Let x,y ∈ Q
f_λ(x+y) = λ(x+y) = λx + λy = f_λ(x) + f_λ(y)
which proves that f_λ is a homomorphism.Let's show that f_λ is injective.

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

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

 

[AllRelative]

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.