Kernel of Group Action is Normal Subgroup

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Theorem

Let $\struct{G, \cdot}$ be a group whose identity is $e$.

Let $X$ be a set.

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

$\map \phi x = g * x$


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


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


Proof 1

Let $h \in G_0$.

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

From Stabilizer is Subgroup:

$\Stab x \le G$

Thus $G_0$ is the intersection of subgroups.

By Intersection of Subgroups is Subgroup:

$G_0 \le G$



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Let $g \circ G_0$ denote the subset product of $g$ with $G_0$.

To prove normality it is sufficient to show:

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

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

We have that:

$h \in \Stab { g^{-1} * x}$


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Then:

\(\ds \paren {g \cdot h \cdot g^{-1} } * x\) \(=\) \(\ds \paren {g \cdot h} * \paren { g^{-1} * x }\) Group Action Associates with Group Operation: Group Action Axiom $\text {GA} 1$
\(\ds \) \(=\) \(\ds g * \paren {h * \paren { g^{-1} * x } }\) Group Action Associates with Group Operation: Group Action Axiom $\text {GA} 1$
\(\ds \) \(=\) \(\ds g * \paren {g^{-1} * x}\) because $h \in \Stab { g^{-1} * x}$
\(\ds \) \(=\) \(\ds \paren {g \cdot g^{-1} } * x\) Group Action Associates with Group Operation: Group Action Axiom $\text {GA} 1$
\(\ds \) \(=\) \(\ds e * x\) Group Axiom $\text G 4$:
\(\ds \) \(=\) \(\ds x\) Group Action Axiom $\text {GA} 2$



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Therefore:

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

so:

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

Conversely suppose that:

$h \in G_0$

Then by the above:

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

Therefore:

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

and so:

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

This concludes the proof.

$\blacksquare$

Proof 2

We are given $\phi: G \times X \to X$ is a group action such that:

$\map \phi x = g * x$

We are given $G_0$ is the kernel of $\phi$.

Let $\phi_g : X \to X$ be the mapping:

$\map {\phi_g} x = \map \phi {g, x}$

for any $g \in G$

By Definition 2 of Kernel of Group Action, the kernel of the group action is the kernel of its permutation representation.

By definition, the permutation representation of $G$ associated to the group action is the group homomorphism $\rho : G \to \struct {\map \Gamma X, \circ}$ which sends $g$ to $\phi_g$.

By Correspondence Between Group Actions and Permutation Representations, we have that $\rho$ is the permutation representation associated to $\phi$

Let $\tilde \phi: G \to \map \Gamma X$ be the permutation representation associated to $\phi$, defined by:

$\map {\tilde \phi} g := \phi_g$

By Group Action defines Permutation Representation $\map {\tilde \phi} g$ is a group homomorphism.

By Group Homomorphism Preserves Identity, $\map {\tilde \phi} e$ is the identity element for the kernel of the permutation representation of $G$ associated to the group action.



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By Inner Automorphism Maps Subgroup to Itself iff Normal we have that:

$\forall x \in G_0: \kappa_x \sqbrk {G_0} = G_0$

Hence the result.

$\blacksquare$

Sources

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