Quotient Theorem for Group Homomorphisms/Corollary 2/Proof 1

Theorem

Let $\struct {G, \odot}$ and $\struct {H, *}$ be groups whose identities are $e_G$ and $e_H$ respectively.

Let $\phi: G \to H$ be a group epimorphism.

Let $K$ be the kernel of $\phi$.

Let $N$ be a normal subgroup of $G$.

Let $q_N: G \to G / N$ denote the quotient epimorphism from $G$ to the quotient group $G / N$.

Then:

$N \subseteq K$

if and only if:

there exists a group epimorphism $\psi: G / N \to H$ such that $\phi = \psi \circ q_N$

Proof

From Quotient Theorem for Group Homomorphisms: Corollary 1:

$N \subseteq K$

if and only if:

there exists a group homomorphism $\psi: G / N \to H$ such that $\phi = \psi \circ q_N$

From Surjection if Composite is Surjection, it follows that the group homomorphism $\psi$ is a surjection.

Hence by definition, $\psi$ is an epimorphism.

$\blacksquare$