# 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.

$\blacksquare$

## Sources

- 1971: Allan Clark:
*Elements of Abstract Algebra*... (previous) ... (next): Chapter $2$: The Sylow Theorems: $\S 53 \beta$