Stabilizer is Subgroup
From ProofWiki
Contents |
Theorem
Let $G$ be a group which acts on a set $X$.
Let $\operatorname{Stab} \left({x}\right)$ be the stabilizer of $x$ by $G$.
Then for each $x \in X$, $\operatorname{Stab} \left({x}\right)$ is a subgroup of $G$.
Corollary 1
Let $G$ be a group whose identity is $e$, which acts on a set $X$.
Then:
- $\forall x \in X: e \in \operatorname{Stab} \left({x}\right)$
Corollary 2
Let $G$ be a group whose identity is $e$, which acts on a set $X$.
Then:
- $\forall g, h \in G: g * x = h * x \iff g^{-1} h \in \operatorname{Stab} \left({x}\right)$
Proof
- $\operatorname{Stab} \left({x}\right)$ can not be empty, because $e * x = x \implies e \in \operatorname{Stab} \left({x}\right)$.
- Let $g, h \in \operatorname{Stab} \left({x}\right)$.
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle g, h\) | \(\in\) | \(\displaystyle \operatorname{Stab} \left({x}\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \implies\) | \(\displaystyle \) | \(\displaystyle g * x\) | \(=\) | \(\displaystyle x, h * x = x\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \implies\) | \(\displaystyle \) | \(\displaystyle g * \left({h * x}\right)\) | \(=\) | \(\displaystyle x\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \implies\) | \(\displaystyle \) | \(\displaystyle \left({g h}\right) * x\) | \(=\) | \(\displaystyle x\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \implies\) | \(\displaystyle \) | \(\displaystyle g h\) | \(\in\) | \(\displaystyle \operatorname{Stab} \left({x}\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) |
- Let $g \in \operatorname{Stab} \left({x}\right)$.
Then $x = \left({g^{-1} g}\right) * x = g^{-1} * \left({g * x}\right) = g^{-1} * x$.
Hence $g^{-1} \in \operatorname{Stab} \left({x}\right)$.
- Thus the conditions for the Two-step Subgroup Test are fulfilled, and $\operatorname{Stab} \left({x}\right) \le G$.
$\blacksquare$
Sources
- J.A. Green: Sets and Groups (1965)... (previous)... (next): $\S 5.6$
- Thomas A. Whitelaw: An Introduction to Abstract Algebra (1978)... (previous)... (next): Exercise $6.5$
- John F. Humphreys: A Course in Group Theory (1996): $\S 10$: Proposition $10.9$