Quotient Theorem for Group Homomorphisms/Corollary 2

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

Then:
 * $N \subseteq K$


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

Proof
From Quotient Theorem for Group Homomorphisms: Corollary 1:


 * $N \subseteq K$


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