Surjectivity of a homomorphism

[AllRelative]

Homework Statement

let f_x(z) = xz

let θ : Q^* → Aut(Q)
with θ(x) = f_x

Is θ an isomorphism?

Homework Equations

Homomorphisms,
Surjectivity and injectivity

The Attempt at a Solution

θ is a homomorphism because
θ(xy) = f_xy = f_x * f_y = θ(x)θ(y)

θ is injective because the neutral element of Aut(Q) is f₁ and
θ^-1(f₁) = 1 the neutral element of Q^*

Prooving the surjectivity is where I am uncertain...
∀f_x ∈ Aut(Q) , There is a unique x ∈ Q^* for which
θ(x) = f_x

Is the codomain = Aut(Q) ?

 

[fresh_42]

Homework Statement

let f_x(z) = xz

let θ : Q^* → Aut(Q)
with θ(x) = f_x

Is θ an isomorphism?

Homework Equations

Homomorphisms,
Surjectivity and injectivity

The Attempt at a Solution

θ is a homomorphism because
θ(xy) = f_xy = f_x * f_y = θ(x)θ(y)

θ is injective because the neutral element of Aut(Q) is f₁ and
θ^-1(f₁) = 1 the neutral element of Q^*

Prooving the surjectivity is where I am uncertain...
∀f_x ∈ Aut(Q) , There is a unique x ∈ Q^* for which
θ(x) = f_x

Is the codomain = Aut(Q) ?

You cannot write ##\forall f_x \in \operatorname{Aut}(\mathbb{Q})## since this already suggest, there is a preimage which you are looking for. You have to start with an automorphism ##\sigma \in \operatorname{Aut}(\mathbb{Q})## and show, that there is an ##x\in \mathbb{Q}^* ## such that ##\sigma(z)=x\cdot z##.

To do this, you also have to know what ##\operatorname{Aut}(\mathbb{Q})## here means. It cannot be the field ##\mathbb{Q}## nor the multiplicative group ##\mathbb{Q}^*##, nor a ring. Here we have ##\operatorname{Aut}(\mathbb{Q})=\operatorname{Aut}(\mathbb{Q},+)## and you can operate with ##\sigma(1)##.

 

[AllRelative]

I knew what Surjectivity was but I now see that I was always trying to prove it the wrong way.

So in general to prove the surjectivity of a group homomorphism f: G → G', I need to show that for any element y of the codomain(G') there exists at least an element x of the domain(G) such that f(x) = y.

Doing this shows that the image of f is G' entirely.Am I seeing that right?

 

[fresh_42]

Yes, as long as you don't impose any conditions on ##y##. It must be an arbitrary element, so only conditions are allowed, which makes it an element of ##G'##.

 

[AllRelative]

Yes, as long as you don't impose any conditions on ##y##. It must be an arbitrary element, so only conditions are allowed, which makes it an element of ##G'##.

Thanks! It suddenly makes a lot more sense.

 

[fresh_42]

Thanks! It suddenly makes a lot more sense.

By the way, ##G\,'## is an unfortunate notation for a second group, better choose ##\varphi\, : \,G \longrightarrow H##. The reason is that many authors abbreviate the commutator ##G\,'=[G,G]=\{\,aba^{-1}b^{-1}\,|\,a,b \in G\,\}## which is an important normal subgroup of ##G##.