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

Theorem
Let $G$ be a group.

Let $H \triangleleft G$ where $\triangleleft$ denotes that $H$ is a normal subgroup of $G$.

Let $K \triangleleft G/H$ and $L = q_H^{-1} \left({K}\right)$, where $q_H$ is as defined in the Quotient Theorem for Epimorphisms.

Then:
 * $(1): \quad L \triangleleft G$
 * $(2): \quad$ There is a group isomorphism $\phi: \left({G / H}\right) / K \to G / L$ defined as:
 * $\phi \circ q_K \circ q_H = q_L$

Proof

 * By Quotient Mapping Canonical Epimorphism, both $q_K$ and $q_H$ are epimorphisms, that is, homomorphisms which are surjective.

From Composition of Homomorphisms we have that $q_K \circ q_H$ is a homomorphism.

From Composite of Surjections is a Surjection we have that $q_K \circ q_H$ is a surjection.

Therefore $q_K \circ q_H: G \to \left({G / H}\right) / K$ is an epimorphism.


 * Now:
 * $\forall x \in G: x \in \ker \left({q_K \circ q_H}\right) \iff q_K \left({q_H \left({x}\right)}\right) = K = e_{G/H}$

This means the same as:
 * $q_H \left({x}\right) \in \ker \left({q_K}\right) = K$

But:
 * $q_H \left({x}\right) \in K \iff x \in q_H^{-1} \left({K}\right) = L$

Thus:
 * $L = \ker \left({q_K \circ q_H}\right)$

By Kernel is Normal Subgroup of Domain, $L \triangleleft G$.


 * By Quotient Theorem for Group Epimorphisms, there is an group isomorphism $\psi$ from $G / L$ to $\left({G / H}\right) / K$ satisfying $\psi \circ q_L = q_K \circ q_L$.

Let $\phi = \psi^{-1}$.

Then $\phi$ is a group isomorphism from $\left({G / H}\right) / K$ to $G / L$:


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