Show that every group homomorphism can be written as a composition $ι ◦ π$
of an injective group homomorphism $ι$ with a surjective group homomorphism $π$.
HINT: Use the first isomorphism theorem, aka the fundamental homomorphism
theorem.
I don't know how to start this proof.
If we have a group homomorphism $\phi : G \to H$, then surely $π : G \to X$ and $ι : X \to H$ ?
Best Answer
Let $\phi: G \to H$ be a group morphism. The (proof of the) first isomorphism theorem tells us that
$$\iota: G/\ker \phi \to H: g \ker \phi \mapsto \phi(g)$$
is a well-defined group morphism that is injective.
Consider the natural projection $$\pi: G \to G/\ker \phi: g \mapsto g \ker \phi$$ which is a surjection. Then
$$\phi= \iota \circ \pi$$
and we are done.