Orbit-Stabilizer Theorem/Proof 2

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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 $x \in X$.

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:

\(\ds h^{-1} * g * x\) \(=\) \(\ds h^{-1} * h * x\) applying $h^{-1}$ to both sides
\(\ds \paren{h^{-1} g } * x\) \(=\) \(\ds \paren{h^{-1} h } * x\) Group Action Axiom $\text {GA} 1$
\(\ds \) \(=\) \(\ds e * x\) Definition of Inverse Element
\(\ds \) \(=\) \(\ds x\) Group Action Axiom $\text {GA} 2$

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 $\Stab x$:

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

Thus:

\(\ds g_2 * x\) \(=\) \(\ds g_2 * \paren {g_2^{-1} g_1 * x}\) applying $g_2$ to both sides
\(\ds \) \(=\) \(\ds \paren {g_2 g_2^{-1} g_1} * x\) Group Action Axiom $\text {GA} 1$
\(\ds \) \(=\) \(\ds g_1 * x\)

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.

$\blacksquare$


Sources

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