Preimage of Normal Subgroup of Quotient Group under Quotient Epimorphism is Normal
Theorem
Let $G$ be a group.
Let $H \lhd G$ where $\lhd$ denotes that $H$ is a normal subgroup of $G$.
Let $K \lhd G / H$.
Let $L = q_H^{-1} \sqbrk K$, where:
- $q_H: G \to G / H$ is the quotient epimorphism from $G$ to the quotient group $G / H$
- $q_H^{-1} \sqbrk K$ is the preimage of $K$ under $q_H$.
Then:
- $L \lhd G$
and there exists a group isomorphism $\phi: \paren {G / H} / K \to G / L$ defined as:
- $\phi \circ q_K \circ q_H = q_L$
Proof 1
By Quotient Mapping on Structure is Epimorphism, both $q_K$ and $q_H$ are epimorphisms.
From Composite of Group Epimorphisms is Epimorphism we have that $q_K \circ q_H: G \to \paren {G / H} / K$ is also an epimorphism.
Now:
- $\forall x \in G: x \in \map \ker {q_K \circ q_H} \iff \map {q_K} {\map {q_H} x} = K = e_{G / H}$
This means the same as:
- $\map {q_H} x \in \map \ker {q_K} = K$
But:
- $\map {q_H} x \in K \iff x \in \map {q_H^{-1} } K = L$
Thus:
- $L = \map \ker {q_K \circ q_H}$
By Kernel is Normal Subgroup of Domain:
- $L \lhd G$
$\Box$
By Quotient Mapping on Structure is Epimorphism, both $q_K$ and $q_H$ are epimorphisms.
From Composite of Group Epimorphisms is Epimorphism we have that $q_K \circ q_H: G \to \paren {G / H} / K$ is also an epimorphism.
Then a priori:
- $L \lhd G$
Hence by Quotient Theorem for Group Epimorphisms:
- there exists a group isomorphism $\psi: G / L \to \paren {G / H} / K$ satisfying:
- $\psi \circ q_L = q_K \circ q_L$
Let $\phi = \psi^{-1}$.
Then $\phi$ is a group isomorphism from $\paren {G / H} / K$ to $G / L$:
- $\phi \circ q_k \circ q_H = \phi \circ \psi \circ q_L = q_L$
$\blacksquare$
Proof 2
Let $e$ be the identity element of $G$.
Let $\RR$ be the congruence relation defined by $H$ in $G$.
Let $\SS$ be the congruence relation defined by $K$ in $G / H$.
Let $\TT$ be the relation on $G$ defined as:
- $\forall x, y \in G: x \mathrel \TT y \iff x H \mathrel \SS y H$
From Equivalence Relation induced by Congruence Relation on Quotient Structure is Congruence:
- $\TT$ is a congruence relation on $G$
Hence from Congruence Relation on Group induces Normal Subgroup:
- the equivalence class under $\TT$ of $e$, that is $\eqclass e \TT$, is a normal subgroup of $G$.
Then we have:
\(\ds L\) | \(=\) | \(\ds q_H^{-1} \sqbrk K\) | by hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds \set {x \in G: \map {q_H} x \in K}\) | Definition of Preimage of Subset under Mapping | |||||||||||
\(\ds \) | \(=\) | \(\ds \set {x \in G: x H \in K}\) | Definition of Quotient Mapping |
Recall that $H$ is the identity of $G / H$.
Then as $K$ is a subgroup of $G / H$:
- $H \in K$
from Identity of Subgroup.
Then:
\(\ds x\) | \(\in\) | \(\ds \eqclass e \TT\) | ||||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds x\) | \(\TT\) | \(\ds e\) | Definition of Equivalence Class | ||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds x H\) | \(\SS\) | \(\ds e H\) | |||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds x H\) | \(\in\) | \(\ds K\) | Definition of $\SS$, and a priori $H \in K$ |
That is:
- $\eqclass e \TT = L$
and so $L$ is a normal subgroup of $G$ a priori.
$\Box$
We can identify:
\(\ds q_\RR\) | \(\equiv\) | \(\ds q_H\) | as $\RR$ is the congruence relation defined by $H$ in $G$ | |||||||||||
\(\ds q_\SS\) | \(\equiv\) | \(\ds q_K\) | as $\SS$ is the congruence relation defined by $K$ in $G / H$ | |||||||||||
\(\ds q_\TT\) | \(\equiv\) | \(\ds q_L\) | Congruence Relation on Group induces Normal Subgroup |
From Equivalence Relation induced by Congruence Relation on Quotient Structure is Congruence we have:
- there exists a unique isomorphism $\phi$ from $\paren {G / H} / K$ to $G / L$ which satisfies:
- $\phi \circ q_\SS \circ q_\RR = q_\TT$
Thus using the above identfications:
- $\phi \circ q_K \circ q_H = q_L$
$\blacksquare$