# Quotient Epimorphism is Epimorphism/Group

## Theorem

Let $G$ be a group.

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

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

Let $q_N: G \to G / N$ be the quotient epimorphism from $G$ to $G / N$:

- $\forall x \in G: \map {q_N} x = x N$

Then $q_N$ is a group epimorphism whose kernel is $N$.

## Proof

The proof follows from Quotient Mapping on Structure is Canonical Epimorphism.

When $N \lhd G$, we have:

\(\displaystyle \forall x, y \in G: \ \ \) | \(\displaystyle \map {q_N} {x y}\) | \(=\) | \(\displaystyle x y N\) | Definition of Quotient Group Epimorphism | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \paren {x N} \paren {y N}\) | Definition of Quotient Group | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \map {q_N} x \map {q_N} y\) | Definition of Quotient Group Epimorphism |

Therefore $q$ is a homomorphism.

We have that:

- $\forall x \in G: x N \in G / N = \map {q_N} x$

so $q$ is surjective.

Therefore $q$ is an epimorphism.

Let $x \in G$.

\(\displaystyle x\) | \(\in\) | \(\displaystyle \map \ker {q_N}\) | |||||||||||

\(\displaystyle \leadstoandfrom \ \ \) | \(\displaystyle \map {q_N} x\) | \(=\) | \(\displaystyle e_{G/N}\) | Definition of Kernel of Group Homomorphism | |||||||||

\(\displaystyle \leadstoandfrom \ \ \) | \(\displaystyle x N\) | \(=\) | \(\displaystyle e_{G/N} N\) | Definition of Quotient Group Epimorphism | |||||||||

\(\displaystyle \leadstoandfrom \ \ \) | \(\displaystyle x N\) | \(=\) | \(\displaystyle N\) | Coset by Identity | |||||||||

\(\displaystyle \leadstoandfrom \ \ \) | \(\displaystyle x\) | \(\in\) | \(\displaystyle N\) | Left Coset Equals Subgroup iff Element in Subgroup |

thus proving that $\map \ker {q_N} = N$ from definition of subset.

$\blacksquare$

## 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.

## Sources

- 1965: J.A. Green:
*Sets and Groups*... (previous) ... (next): Chapter $7$: Homomorphisms: Exercise $6$ - 1966: Richard A. Dean:
*Elements of Abstract Algebra*... (previous) ... (next): $\S 1.10$: Theorem $24$ - 1967: George McCarty:
*Topology: An Introduction with Application to Topological Groups*... (previous) ... (next): $\text{II}$: Quotient Groups: Theorem $3$ - 1970: B. Hartley and T.O. Hawkes:
*Rings, Modules and Linear Algebra*... (previous) ... (next): $\S 2.2$: Homomorphisms - 1971: Allan Clark:
*Elements of Abstract Algebra*... (previous) ... (next): Chapter $2$: Group Homomorphism and Isomorphism: $\S 60 \beta$ - 1978: Thomas A. Whitelaw:
*An Introduction to Abstract Algebra*... (previous) ... (next): $\S 50.5$ Quotient groups - 1996: John F. Humphreys:
*A Course in Group Theory*... (previous) ... (next): Chapter $8$: The Homomorphism Theorem: Remark