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) $$L \triangleleft G$$;
 * 2) 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 $$G / L$$:


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