Preimage of Normal Subgroup under Epimorphism is Normal Subgroup
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
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {II}$: New Structures from Old: $\S 12$: Homomorphisms: Exercise $12.20 \ \text {(b)}$