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

Theorem
Let $G$ be a group.

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

Let $K \lhd G/H$ and $L = q_H^{-1} \sqbrk K$, 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:
 * $L \lhd G$

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:
 * $L \lhd G$