- State the orbit-stabilizer theorem
Solution
Let a group \(G\) act on a set \(X\), and let \(x\in X\). Write \[ G_x=\{g\in G:g\cdot x=x\} \] for the stabilizer and \(G\cdot x\) for the orbit.
1. The map\[ \Phi:G/G_x\longrightarrow G\cdot x, \qquad gG_x\longmapsto g\cdot x \]is a bijection.
It is well-defined: if \(gG_x=hG_x\), then \(h^{-1}g\in G_x\), so \[ g\cdot x=h(h^{-1}g)\cdot x=h\cdot x. \] It is surjective by the definition of the orbit. If \(g\cdot x=h\cdot x\), then \[ h^{-1}g\cdot x=x, \] so \(h^{-1}g\in G_x\) and therefore \(gG_x=hG_x\). Hence it is injective.
2. Therefore\[ |G\cdot x|=[G:G_x] \]whenever these cardinalities are finite.
By 1, the orbit is in bijection with the left coset space \(G/G_x\).
3. If \(G\) is finite, then\[ |G|=|G\cdot x|\,|G_x|. \]
2. This is the orbit–stabilizer theorem.