Orbit-Stabilizer Theorem

Theorem

Let $G$ be a group which acts on a finite set $X$.

Let $\operatorname{Orb} \left({x}\right)$ be the orbit of $x$.

Let $\operatorname{Stab} \left({x}\right)$ be the stabilizer of $x$ by $G$.

Let $\left[{G : \operatorname{Stab} \left({x}\right)}\right]$ be the index of $\operatorname{Stab} \left({x}\right)$ in $G$.

Then:

$\displaystyle \left|{\operatorname{Orb} \left({x}\right)}\right| = \left[{G : \operatorname{Stab} \left({x}\right)}\right] = \frac {\left|{G}\right|} {\left|{\operatorname{Stab} \left({x}\right)}\right|}$

Proof 1

Let us define the mapping:

$\phi: G \to \operatorname{Orb} \left({x}\right)$

such that:

$\phi \left({g}\right) = g * x$

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 2:

$\phi \left({g}\right) = \phi \left({h}\right) \iff g^{-1} h \in \operatorname{Stab} \left({x}\right)$

This means:

$g \equiv h \left({\bmod\, \operatorname{Stab} \left({x}\right)}\right)$

Thus there is a well-defined bijection:

$G / \operatorname{Stab} \left({x}\right) \to \operatorname{Orb} \left({x}\right)$

given by:

$g \, \operatorname{Stab} \left({x}\right) \mapsto g * x$

So $\operatorname{Orb} \left({x}\right)$ has the same number of elements as $G / \operatorname{Stab} \left({x}\right)$.

That is:

$\left|{\operatorname{Orb} \left({x}\right)}\right| = \left[{G : \operatorname{Stab} \left({x}\right)}\right]$

The result follows.

$\blacksquare$

Proof 2

For $K$ subgroup of $G$:

$g * x = h * x \iff g K = h K$

where $gK$ and $hK$ are left cosets of $K$.

By definition, $g * x$ and $h * x$ are elements of of the orbit of $x$.

We conclude that there is a bijection between the elements of the orbit and the left cosets of any subgroup $K$.

Therefore:

$\left|{G : K}\right| = \left|{\operatorname {Orb} \left({x}\right)}\right|$

From Lagrange's theorem:

$\left|{G}\right| = \left|{K}\right| * \left|{H : K}\right|$

Because $\operatorname{Stab} \left({x}\right)$ is a subset of $G$, setting $\operatorname{Stab} \left({x}\right) = K$ yields:

$\left|{G}\right| = \left|{\operatorname{Stab} \left({x}\right)}\right| * \left|{\operatorname {Orb} \left({x}\right)}\right|$

$\blacksquare$

Sources

• J.A. Green: Sets and Groups (1965)... (previous)... (next): $\S 6.5$
• Allan Clark: Elements of Abstract Algebra (1971)... (previous)... (next): $\S 54$
• John F. Humphreys: A Course in Group Theory (1996): $\S 10$: Example $10.16$