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.