# Quotient Group of Quotient Group is Isomorphic to Quotient Group by Preimage under Quotient Mapping

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## 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 $L = q_H^{-1} \left[{K}\right]$, where:

- $q_H: G \to G/H$ is the quotient epimorphism from $G$ to the quotient group $G/H$
- $q_H^{-1} \left[{K}\right]$ is the preimage of $K$ under $q_H$.

Then there exists 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 on Structure is Canonical Epimorphism, both $q_K$ and $q_H$ are epimorphisms.

From Composite of Group Epimorphisms is Epimorphism we have that $q_K \circ q_H: G \to \left({G / H}\right) / K$ is also an epimorphism.

From Preimage of Normal Subgroup of Quotient Group under Quotient Mapping is Normal:

- $L \lhd G$

By Quotient Theorem for Group Epimorphisms:

- there exists a group isomorphism $\psi: G / L \to \left({G / H}\right) / K$ 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 $G / L$:

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

$\blacksquare$

## Sources

- 1965: Seth Warner:
*Modern Algebra*... (previous) ... (next): $\S 12$: Example $12.4$