Quotient Theorem for Group Epimorphisms

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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 is 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 $\mathcal R_\phi$ be the equivalence on $G$ defined by $\phi$.

\(\displaystyle \forall x \in G: e_G\) \(\mathcal R_\phi\) \(\displaystyle x\)
\(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle \map \phi x\) \(=\) \(\displaystyle \map \phi {e_G}\) Definition of $\mathcal R_\phi$
\(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle \map \phi x\) \(=\) \(\displaystyle e_H\) Homomorphism to Group Preserves Identity


Thus:

$K = \eqclass {e_G} {\mathcal R_\phi}$


From the Quotient Theorem for Epimorphisms:

$\mathcal R_\phi$ is compatible with $\oplus$

Thus from Kernel is Normal Subgroup of Domain:

$K \lhd G$

From Congruence Relation induces Normal Subgroup, $\mathcal R_\phi$ 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