Preimage of Normal Subgroup under Epimorphism is Normal Subgroup

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Theorem

Let $\struct {G_1, \circ}$ and $\struct {G_2, *}$ be groups.

Let $\phi: \struct {G_1, \circ} \to \struct {G_2, *}$ be a group epimorphism.

Let $H$ be a normal subgroup of $\struct {G_2, *}$.

Then:

$\phi^{-1} \sqbrk H$ is a normal subgroup of $\struct {G_1, \circ}$

where $\phi^{-1} \sqbrk H$ denotes the preimage of $H$ under $\phi$.


Proof

Recall that as $\phi$ is an group epimorphism, it is also a fortiori a group Homomorphism.

Hence:

$\forall a, b \in G_1: \map \phi {a \circ b} = \map \phi a * \map \phi b$


Let $H$ be a normal subgroup of $\struct {G_2, *}$.

From Preimage of Subgroup under Epimorphism is Subgroup:

$\phi^{-1} \sqbrk H$ is a subgroup of $\struct {G_1, \circ}$.


It remains to be shown that $\phi^{-1} \sqbrk H$ is normal.

Let $K = \phi^{-1} \sqbrk H$.

That is:

$H = \map \phi K$


To prove that $K$ is normal, we are to show that:

$\forall g \in G: k \in K \iff g \circ k \circ g^{-1} \in K$

and:

$\forall g \in G: k \in K \iff g^{-1} \circ k \circ g \in K$


So, let $g \in G$ and $k \in K$ be arbitrary.

By definition of preimage of $H$ under $\phi$:

$\exists h \in H: \map \phi k = h$

Then:

\(\ds \map \phi {g \circ k \circ g^{-1} }\) \(=\) \(\ds \map \phi g * \map \phi k * \map \phi {g^{-1} }\) Definition of Group Homomorphism
\(\ds \) \(=\) \(\ds \map \phi g * h * \map \phi {g^{-1} }\) as $h = \map \phi k$
\(\ds \) \(\in\) \(\ds H\) as $H$ is normal
\(\ds \leadsto \ \ \) \(\ds g \circ k \circ g^{-1}\) \(\in\) \(\ds K\) Definition of Preimage of Subset under Mapping: $K = \phi^{-1} \sqbrk H$


In the same way:

\(\ds \map \phi {g^{-1} \circ k \circ g}\) \(=\) \(\ds \map \phi {g^{-1} } * \map \phi k * \map \phi g\) Definition of Group Homomorphism
\(\ds \) \(=\) \(\ds \map \phi {g^{-1} } * h * \map \phi q\) as $h = \map \phi k$
\(\ds \) \(\in\) \(\ds H\) as $H$ is normal
\(\ds \leadsto \ \ \) \(\ds g^{-1} \circ k \circ g\) \(\in\) \(\ds K\) Definition of Preimage of Subset under Mapping: $K = \phi^{-1} \sqbrk H$

Thus:

$\forall g \in G: k \in K \implies g \circ k \circ g^{-1} \in K$

and:

$\forall g \in G: k \in K \implies g^{-1} \circ k \circ g \in K$

$\Box$


Then:

\(\ds g \circ k \circ g^{-1}\) \(\in\) \(\ds K\)
\(\ds \leadsto \ \ \) \(\ds \map \phi {g \circ k \circ g^{-1} }\) \(\in\) \(\ds H\) Definition of $K$
\(\ds \leadsto \ \ \) \(\ds \map \phi g * \map \phi k * \map \phi {g^{-1} }\) \(\in\) \(\ds H\) Definition of Group Homomorphism
\(\ds \leadsto \ \ \) \(\ds \map \phi k\) \(\in\) \(\ds H\) as $H$ is normal
\(\ds \leadsto \ \ \) \(\ds k\) \(\in\) \(\ds K\) Definition of Preimage of Subset under Mapping: $K = \phi^{-1} \sqbrk H$

and:

\(\ds g^{-1} \circ k \circ g\) \(\in\) \(\ds K\)
\(\ds \leadsto \ \ \) \(\ds \map \phi {g^{-1} \circ k \circ g}\) \(\in\) \(\ds H\) Definition of $K$
\(\ds \leadsto \ \ \) \(\ds \map \phi {g^{-1} } * \map \phi k * \map \phi g\) \(\in\) \(\ds H\) Definition of Group Homomorphism
\(\ds \leadsto \ \ \) \(\ds \map \phi k\) \(\in\) \(\ds H\) as $H$ is normal
\(\ds \leadsto \ \ \) \(\ds k\) \(\in\) \(\ds K\) Definition of Preimage of Subset under Mapping: $K = \phi^{-1} \sqbrk H$

Thus:

$\forall g \in G: k \in K \impliedby g \circ k \circ g^{-1} \in K$

and:

$\forall g \in G: k \in K \impliedby g^{-1} \circ k \circ g \in K$

$\Box$


Putting this together:

$\forall g \in G: k \in K \iff g \circ k \circ g^{-1} \in K$

and:

$\forall g \in G: k \in K \iff g^{-1} \circ k \circ g \in K$

and the result follows.

$\blacksquare$


Sources