Group Epimorphism is Isomorphism iff Kernel is Trivial/Proof 1

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 $\map \phi x = e_H$.

From Epimorphism Preserves Identity, that element $x$ is $e_G$:
 * $\map \phi {e_G} = e_H$

Thus by definition of kernel:
 * $\map \ker \phi = \set {e_G}$

Sufficient Condition
Let $K := \map \ker \phi = \set {e_G}$.

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

and thus from Kernel is Normal Subgroup of Domain:
 * $K \lhd G$

From Congruence Relation induces Normal Subgroup, $\RR_\phi$ is the equivalence defined by $K$.

Let $\RR_K$ be the congruence modulo $K$ induced by $K$.

Suppose $\map \phi x = \map \phi y$.

Then:
 * $x \mathop {\RR_K} y$

as $\RR_\phi = \RR_K$ 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.