Show injectivity, surjectivity and kernel of groups

[AllRelative]

Homework Statement

I am translating so bear with me.
We have two group homomorphisms:
α : G → G'
β : G' → G

Let β(α(x)) = x ∀x ∈ G

Show that
1)β is a surjection
2)α an injection
3) ker(β) = ker(α ο β) (Here ο is the composition of functions.)

Homework Equations

This is from a introductory Group Theory course.
I know:
the definition of homomorphisms
Let f: G → G' and e and e' are the two neutral elements.
f is injective if and only if Ker(f) = e
f is surjective if and only if image(f) = G'

The Attempt at a Solution

My initial reflex would have been writing α = β^-1 but I know that we don't know that yet.

I started with let x,y ∈ G so that α(x) = x' and α(y) = y'

Showing β as a surjection means that image(β) = G. I am unsure of where to start proving that.
Showing that α is injective means I need to show that α^-1(e') = e
Again I am unsure of where to start.

Thank you

 

[fresh_42]

Homework Statement

I am translating so bear with me.
We have two group homomorphisms:
α : G → G'
β : G' → G

Let β(α(x)) = x ∀x ∈ G

Show that
1)β is a surjection
2)α an injection
3) ker(β) = ker(α ο β) (Here ο is the composition of functions.)

Homework Equations

This is from a introductory Group Theory course.
I know:
the definition of homomorphisms
Let f: G → G' and e and e' are the two neutral elements.
f is injective if and only if Ker(f) = e
f is surjective if and only if image(f) = G'

The Attempt at a Solution

My initial reflex would have been writing α = β^-1 but I know that we don't know that yet.

I started with let x,y ∈ G so that α(x) = x' and α(y) = y'

Showing β as a surjection means that image(β) = G. I am unsure of where to start proving that.
Showing that α is injective means I need to show that α^-1(e') = e
Again I am unsure of where to start.

Thank you

a) Let's assume we have an $x\in G$. Then $\beta (\alpha (x)) = x.$ Does this $x$ have a pre-image under $\beta\,$?
b) Let's assume an $x\in G$ with $\alpha(x)=1.$ What does $\beta \circ \alpha = \operatorname{id}_G$ tell you?
c) Show both inclusions: $\operatorname{ker}\beta \subseteq \operatorname{ker}(\alpha \beta)$ and $\operatorname{ker}(\alpha \beta) \subseteq \operatorname{ker}\beta$ by first assuming an element $y \in \operatorname{ker}\beta$ and next an element $x \in \operatorname{ker}(\alpha \beta)\,.$