Quotient Epimorphism is Epimorphism/Group

From ProofWiki
Jump to navigation Jump to search

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:

\(\ds \forall x, y \in G: \ \ \) \(\ds \map {q_N} {x y}\) \(=\) \(\ds x y N\) Definition of Quotient Group Epimorphism
\(\ds \) \(=\) \(\ds \paren {x N} \paren {y N}\) Definition of Quotient Group
\(\ds \) \(=\) \(\ds \map {q_N} x \map {q_N} y\) 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 $x \in G$.

\(\ds x\) \(\in\) \(\ds \map \ker {q_N}\)
\(\ds \leadstoandfrom \ \ \) \(\ds \map {q_N} x\) \(=\) \(\ds e_{G/N}\) Definition of Kernel of Group Homomorphism
\(\ds \leadstoandfrom \ \ \) \(\ds x N\) \(=\) \(\ds e_{G/N} N\) Definition of Quotient Group Epimorphism
\(\ds \leadstoandfrom \ \ \) \(\ds x N\) \(=\) \(\ds N\) Coset by Identityā€ˇ
\(\ds \leadstoandfrom \ \ \) \(\ds x\) \(\in\) \(\ds N\) Left Coset Equals Subgroup iff Element in Subgroup

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

$\blacksquare$


Motivation

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 group operation 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 group operation.

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