Preimage of Normal Subgroup of Quotient Group under Quotient Epimorphism is Normal/Proof 2

Proof
Let $e$ be the identity element of $G$.

Let $\RR$ be the congruence relation defined by $H$ in $G$.

Let $\SS$ be the congruence relation defined by $K$ in $G / H$.

Let $\TT$ be the relation on $G$ defined as:
 * $\forall x, y \in G: x \mathrel \TT y \iff x H \mathrel \SS y H$

From Equivalence Relation induced by Congruence Relation on Quotient Structure is Congruence:
 * $\TT$ is a congruence relation on $G$

Hence from Congruence Relation on Group induces Normal Subgroup:
 * the equivalence class under $\TT$ of $e$, that is $\eqclass e \TT$, is a normal subgroup of $G$.

Then we have:

Recall that $H$ is the identity of $G / H$.

Then as $K$ is a subgroup of $G / H$:
 * $H \in K$

from Identity of Subgroup.

Then:

That is:
 * $\eqclass e \TT = L$

and so $L$ is a normal subgroup of $G$.

We can identify:

From Equivalence Relation induced by Congruence Relation on Quotient Structure is Congruence we have:


 * there exists a unique isomorphism $\phi$ from $\paren {G / H} / K$ to $G / L$ which satisfies:
 * $\phi \circ q_\SS \circ q_\RR = q_\TT$

Thus using the above identfications:
 * $\phi \circ q_K \circ q_H = q_L$