Preimage of Normal Subgroup of Quotient Group under Quotient Epimorphism is Normal

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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:

$\phi \circ q_K \circ q_H = q_L$

$\blacksquare$