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 $\phi = \psi \circ q_N$
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
- 1967: George McCarty: Topology: An Introduction with Application to Topological Groups ... (previous) ... (next): Chapter $\text{II}$: Groups: Factoring Morphisms: Corollary