Quotient Theorem for Group Epimorphisms
Theorem
Let $\struct {G, \oplus}$ and $\struct {H, \odot}$ be groups.
Let $\phi: \struct {G, \oplus} \to \struct {H, \odot}$ be a group epimorphism.
Let $e_G$ and $e_H$ be the identities of $G$ and $H$ respectively.
Let $K = \map \ker \phi$ be the kernel of $\phi$.
There exists one and only one group isomorphism $\psi: G / K \to H$ satisfying:
- $\psi \circ q_K = \phi$
where $q_K$ is the quotient epimorphism from $G$ to $G / K$.
Proof
Let $\RR_\phi$ be the equivalence on $G$ defined by $\phi$.
\(\ds \forall x \in G: \, \) | \(\ds e_G\) | \(\RR_\phi\) | \(\ds x\) | |||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \map \phi x\) | \(=\) | \(\ds \map \phi {e_G}\) | Definition of $\RR_\phi$ | ||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \map \phi x\) | \(=\) | \(\ds e_H\) | Homomorphism to Group Preserves Identity |
Thus:
- $K = \eqclass {e_G} {\RR_\phi}$
From the Quotient Theorem for Epimorphisms:
- $\RR_\phi$ is compatible with $\oplus$
Thus from Kernel is Normal Subgroup of Domain:
- $K \lhd G$
From Congruence Relation induces Normal Subgroup, $\RR_\phi$ is the equivalence defined by $K$.
Thus, again by Quotient Theorem for Epimorphisms, there is a unique epimorphism $\psi: G / K \to H$ satisfying $\psi \circ q_K = \phi$.
$\blacksquare$
Also known as
Some sources call this the Factor Theorem for Group Epimorphisms.
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {II}$: New Structures from Old: $\S 12$: Homomorphisms: Theorem $12.6$