Quotient Theorem for Group Epimorphisms

Theorem
Let $\phi: \left({G, \oplus}\right) \to \left({H, \odot}\right)$ be a group epimorphism.

Let $e_G$ and $e_H$ be the identities of $G$ and $H$ respectively.

Let $K = \ker \left({\phi}\right)$.

There is one and only one group isomorphism $\psi: G / K \to H$ satisfying:


 * $\psi \circ q_K = \phi$

The epimorphism $\phi$ is an isomorphism iff $K = \left\{{e_G}\right\}$.

Proof

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

Thus $K = \left[\!\left[{e_G}\right]\!\right]_{\mathcal R_\phi}$.

From the Quotient Theorem for Epimorphisms, $\mathcal R_\phi$ is compatible with $\oplus$.

Thus from Kernel is Normal Subgroup of Domain, $K \triangleleft G$.

From Compatible Relation 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 $\psi \circ q_K = \phi$.


 * Now let $\phi$ be an isomorphism. Then $K = \left\{{e_G}\right\}$ as $\phi$ is injective.

Conversely, if $K = \left\{{e_G}\right\}$ and $\phi \left({x}\right) = \phi \left({y}\right)$, then $x \mathcal R_K y$ as $\mathcal R_\phi = \mathcal R_K$ from Congruence Modulo a Subgroup is an Equivalence.

Thus $x \oplus y^{-1} \in K$ by Congruence Class Modulo Subgroup is Coset.

Hence $x \oplus y^{-1} = e_G$ and so $x = y$.

Thus $\phi$ is injective, and an injective epimorphism is a isomorphism.