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

From ProofWiki
Jump to navigation Jump to search

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:

$L \lhd G$


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.


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:

$L \lhd G$

$\blacksquare$


Sources