# Group Epimorphism is Isomorphism iff Kernel is Trivial

## 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

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

## 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:

- $\map \ker \phi = \set {e_G}$

$\Box$

### Sufficient Condition

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

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

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.

$\blacksquare$

## Proof 2

From Kernel is Trivial iff Monomorphism, $\phi$ is a monomorphism if and only if $K = \left\{{e_G}\right\}$.

By definition, a group $G$ is an epimorphism is an isomorphism if and only if $G$ is also a monomorphism.

Hence the result.

$\blacksquare$