Quotient Theorem for Group Homomorphisms/Corollary 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 homomorphism.

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 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 $\phi = \psi \circ q_N$

This can be illustrated by means of the following commutative diagram:


 * $\begin{xy}\xymatrix@L+2mu@+1em{

G \ar[dr]^*{\phi} \ar[d]_*{q} & \\ G / N \ar@{-->}[r]_*{\psi} & H }\end{xy}$

Necessary Condition
Suppose $\psi$ exists as defined.

Then:
 * $K = \phi^{-1} \sqbrk {\set {e_H} } = q^{-1} \sqbrk {\psi^{-1} \sqbrk {\set {e_H} } }$

That is: $\psi^{-1} \sqbrk {\set {e_H} }$ is the kernel of $\psi$.

So by Kernel is Normal Subgroup of Domain, $\psi^{-1} \sqbrk {\set {e_H} }$ is a normal subgroup of $G / N$.

This corresponds via $q^{-1}$ to $K$, which must then be a normal subgroup of $G$ which contains $N$.

Sufficient Condition
Suppose $N \subseteq K$.

Let $\psi: G / N \to H$ be defined as:
 * $\forall g \odot N \in G / N: \map \psi {g \odot N} = \map \phi g \in H$

We have that:
 * $n \in g \odot N \implies n^{-1} \odot g \in N \subseteq K$

So:
 * $\map \phi {n^{-1} \odot g} = \map \phi n^{-1} * \map \phi g = e_H$

and so $\map \phi n = \map \phi g$.

Thus $\psi$ is well-defined.

Hence $\psi$ is the required group homomorphism.