Orbit-Stabilizer Theorem/Proof 1
Jump to navigation
Jump to search
Theorem
Let $G$ be a group which acts on a finite set $X$.
Let $x \in X$.
Let $\Orb x$ denote the orbit of $x$.
Let $\Stab x$ denote the stabilizer of $x$ by $G$.
Let $\index G {\Stab x}$ denote the index of $\Stab x$ in $G$.
Then:
- $\order {\Orb x} = \index G {\Stab x} = \dfrac {\order G} {\order {\Stab x} }$
Proof
Let us define the mapping:
- $\phi: G \to \Orb x$
such that:
- $\map \phi g = g * x$
where $*$ denotes the group action.
It is clear that $\phi$ is surjective, because from the definition $x$ was acted on by all the elements of $G$.
Next, from Stabilizer is Subgroup: Corollary:
- $\map \phi g = \map \phi h \iff g^{-1} h \in \Stab x$
This means:
- $g \equiv h \pmod {\Stab x}$
Thus there is a well-defined bijection:
- $G \mathbin / \Stab x \to \Orb x$
given by:
- $g \, \Stab x \mapsto g * x$
So $\Orb x$ has the same number of elements as $G \mathbin / \Stab x$.
That is:
- $\order {\Orb x} = \index G {\Stab x}$
The result follows.
$\blacksquare$
Sources
- 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $2$: The Sylow Theorems: $\S 54$