Orbit-Stabilizer Theorem

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 1

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.

$\blacksquare$

Proof 2

Let $x \in X$.

Let $\phi: \Orb x \to G \mathbin / \Stab x$ be a mapping from the orbit of $x$ to the left coset space of $\Stab x$ defined as:

$\forall g \in G: \map \phi {g * x} = g \, \Stab x$

where $*$ is the group action.

Note: this is not a homomorphism because $\Orb x$ is not a group.

Suppose $g * x = h * x$ for some $g, h \in G$.

Then:

$h^{-1} g * x = h^{-1} h * x$

and so:

$h^{-1} g * x = x$

Thus:

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

so by Left Coset Space forms Partition:

$g \, \Stab x = h \, \Stab x$

demonstrating that $\phi$ is well-defined.

Let $\map \phi {g_1 * x} = \map \phi {g_2 * x}$ for some $g_1, g_2 \in G$.

Then:

$g_1 \, \Stab x = g_2 \, \Stab x$

and so by Left Coset Space forms Partition:

$g_2^{-1} g_1 \in \Stab x$

So by definition of $\Stab x$:

$x = g_2^{-1} g_1 * x$

Thus:

\(\ds g_2 * x\)

\(=\)

\(\ds g_2 * \paren {g_2^{-1} g_1 * x}\)

applying $g_2$ to both sides

\(\ds \)

\(=\)

\(\ds \paren {g_2 g_2^{-1} g_1} x\)

Definition of Group Action

\(\ds \)

\(=\)

\(\ds g_1 * x\)

thus demonstrating that $\phi$ is injective.

As the left coset $g \, \Stab x$ is $\map \phi {g * x}$ by definition of $\phi$, it follows that $\phi$ is a surjection.

The result follows.

$\blacksquare$