Quotient Theorem for Group Homomorphisms/Corollary 1

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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:

$N \subseteq K$

if and only if:

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:


$\begin{xy}\xymatrix@L+2mu@+1em{ G \ar[dr]^*{\phi} \ar[d]_*{q} & \\ G / N \ar@{-->}[r]_*{\psi} & H }\end{xy}$


Proof

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 $N$.

$\Box$


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.

$\blacksquare$


Sources