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

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:
 * $L \lhd G$

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 :
 * $L \lhd G$

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 $G / L$:


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