# Surjectivity of a homomorphism

AllRelative

## Homework Statement

let fx(z) = xz

let θ : Q* → Aut(Q)
with θ(x) = fx

Is θ an isomorphism?

## Homework Equations

Homomorphisms,
Surjectivity and injectivity

## The Attempt at a Solution

θ is a homomorphism because
θ(xy) = fxy = fx * fy = 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...
∀fx ∈ Aut(Q) , There is a unique x ∈ Q* for which
θ(x) = fx

Is the codomain = Aut(Q) ?

Mentor
2022 Award

## Homework Statement

let fx(z) = xz

let θ : Q* → Aut(Q)
with θ(x) = fx

Is θ an isomorphism?

## Homework Equations

Homomorphisms,
Surjectivity and injectivity

## The Attempt at a Solution

θ is a homomorphism because
θ(xy) = fxy = fx * fy = 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...
∀fx ∈ Aut(Q) , There is a unique x ∈ Q* for which
θ(x) = fx

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
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?

Mentor
2022 Award
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.

Mentor
2022 Award
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##.

Last edited:
AllRelative