Kernel of Group Action is Normal Subgroup

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Theorem

Let $G$ be a group whose identity is $e$.

Let $X$ be a set.

Let $\phi: G \times X \to X$ be a group action.


Let $G_0$ denote the kernel of $\phi$.


Then $G_0$ is a normal subgroup of $G$.


Proof

Let $h \in G_0$.

\(\displaystyle h\) \(\in\) \(\displaystyle G_0\)
\(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle h\) \(\in\) \(\displaystyle \set {g \in G: \forall x \in X: g \cdot x = x}\) Definition of Kernel of Group Action
\(\displaystyle \leadstoandfrom \ \ \) \(\, \displaystyle \forall x \in X: \, \) \(\displaystyle h\) \(\in\) \(\displaystyle \set {g \in G: g \cdot x = x}\)
\(\displaystyle \leadstoandfrom \ \ \) \(\, \displaystyle \forall x \in X: \, \) \(\displaystyle h\) \(\in\) \(\displaystyle \Stab x\)
\(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle h\) \(\in\) \(\displaystyle \bigcap_{x \mathop \in X} \Stab x\)
\(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle G_0\) \(=\) \(\displaystyle \bigcap_{x \mathop \in X} \Stab x\)

From Stabilizer is Subgroup:

$\Stab x \le G$

Thus $G_0$ is the intersection of subgroups.

By Intersection of Subgroups is Subgroup:

$G_0 \le G$


To prove normality it is sufficient to show:

$\forall g \in G: g G_0 g^{-1} = G_0$

Let $h \in G_0, g \in G$ be arbitrary.

Then:

\(\displaystyle \paren {g h g^{-1} } \cdot x\) \(=\) \(\displaystyle g \cdot \paren {h \cdot \paren { g^{-1} \cdot x } }\) Group Action Associates with Group Operation: Group Action Axiom $\text {GA} 1$
\(\displaystyle \) \(=\) \(\displaystyle g \cdot \paren {g^{-1} \cdot x}\) because $h \in \Stab { g^{-1} \cdot x}$
\(\displaystyle \) \(=\) \(\displaystyle \paren {g \cdot g^{-1} } \cdot x\) Group Action Associates with Group Operation: Group Action Axiom $\text {GA} 1$
\(\displaystyle \) \(=\) \(\displaystyle x\) $e \cdot x = x$: Group Action Axiom $\text {GA} 2$


Therefore:

$g h g^{-1} \in G_0$

so:

$g G_0 g^{-1} \subseteq G_0$

Conversely suppose that $h \in G_0$.

Then by the above:

$h' = g^{-1} h g \in G_0$

Therefore:

$h = g h' g^{-1} \in g G_0 g^{-1}$

and so:

$G_0 \subseteq g G_0 g^{-1}$

This concludes the proof.

$\blacksquare$


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