Preimage of Normal Subgroup of Quotient Group under Quotient Epimorphism is Normal/Proof 1

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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

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 $G / L$:

$\phi \circ q_k \circ q_H = \phi \circ \psi \circ q_L = q_L$

$\blacksquare$


Sources

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