Group Epimorphism is Isomorphism iff Kernel is Trivial/Proof 1

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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 $\phi$.


the epimorphism $\phi$ is an isomorphism

if and only if

$K = \set {e_G}$


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


$x \mathop {\RR_K} y$

as $\RR_\phi = \RR_K$ from Congruence Modulo Subgroup is Equivalence Relation.

Thus by Congruence Class Modulo Subgroup is Coset:

$x \oplus y^{-1} \in K$


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

and so:

$x = y$

Thus $\phi$ is injective.

By definition, an injective epimorphism is a isomorphism.