Group Epimorphism is Isomorphism iff Kernel is Trivial/Proof 1

Theorem
Let $\left({G, \oplus}\right)$ and $\left({H, \odot}\right)$ be groups.

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 $K = \ker \left({\phi}\right)$ be the kernel of $\phi$.

Then:
 * the epimorphism $\phi$ is an isomorphism


 * $K = \left\{{e_G}\right\}$

Necessary Condition
Let $\phi$ be an isomorphism.

Then by definition $\phi$ is a bijective homomorphism.

Thus by definition of bijection, $\phi$ is an injection.

By definition of injection, there exists exactly one element $x$ of $G$ such that $\phi \left({x}\right) = e_H$.

From Epimorphism Preserves Identity, that element $x$ is $e_G$:
 * $\phi \left({e_G}\right) = e_H$

Thus by definition of kernel:
 * $\ker \left({\phi}\right) = \left\{{e_G}\right\}$

Sufficient Condition
Let $K := \ker \left({\phi}\right) = \left\{{e_G}\right\}$.

From the Quotient Theorem for Epimorphisms:
 * $\mathcal R_\phi$ is compatible with $\oplus$

and 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 $K$.

Let $\mathcal R_K$ be the congruence modulo $K$ induced by $K$.

Suppose $\phi \left({x}\right) = \phi \left({y}\right)$.

Then:
 * $x \mathop{\mathcal R_K} y$

as $\mathcal R_\phi = \mathcal R_K$ from Congruence Modulo Subgroup is Equivalence Relation.

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

Hence:
 * $x \oplus y^{-1} = e_G$

and so:
 * $x = y$

Thus $\phi$ is injective.

By definition, an injective epimorphism is a isomorphism.