Definition:Quotient Epimorphism

Theorem
Let $$G$$ be a group.

Let $$N$$ be a normal subgroup of $$G$$.

Let $$G / N$$ be the quotient group of $$G$$ by $$N$$.

Then the mapping $$\phi: G \to G / N$$, defined as:


 * $$\phi: G \to G / N: \phi \left({x}\right) = x N$$

is a group epimorphism, and its kernel is $$N$$.

This epimorphism is known as: from $$G$$ to $$G / N$$.
 * the natural (epi)morphism,
 * the quotient (epi)morphism;
 * the canonical (epi)morphism

Proof
The proof follows from Quotient Mapping Canonical Epimorphism.


 * When $$N \triangleleft G$$, we have:

$$ $$ $$

Therefore $$\phi$$ is a homomorphism.


 * $$\forall x \in G: x N \in G / N = \phi \left({x}\right)$$, so $$\phi$$ is surjective.

Therefore $$\phi$$ is an epimorphism.


 * Let $$x \in G$$.

$$ $$ $$ $$ $$ $$

Conversely, $$ n \in N $$

$$ $$ $$ $$ $$

Combining (Eq. a) and (Eq. b) we get $$ ker\left(\phi\right) = N $$.

Comment
In Kernel is Normal Subgroup of Domain it was shown that the kernel of a group homomorphism is a normal subgroup of its domain. In this result it has been shown that every normal subgroup is a kernel of at least one group homomorphism of the group of which it is the subgroup.

We see that when a subgroup is normal, its cosets make a group using the product rule defined as in this result. However, it is not possible to make the left or right cosets of a non-normal subgroup into a group using the same sort of product rule. Otherwise there would be a group homomorphism with that subgroup as the kernel, and we have seen that this can not be done unless the subgroup is normal.