# Kernel of Group Action is Normal Subgroup

## 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 $h \in G_0$.

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

Thus $G_0$ is the intersection of subgroups.

$G_0 \le G$

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:

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

$\blacksquare$