Past exam of the mathematics course of the University of Cambridge / 2024 / ia / Paper 3 / 8D / a / Solution

The orbit-stabilizer theorem states that for a finite group $G$ acting on a set $X$ and $x\in X$,Define $G/G_x\to Gx$ by $g{G}_x\mapsto gx$. This is well-defined because $g{G}_x=h{G}_x$ exactly when $h^{-1}g\in G_x$, equivalently $gx=hx$. The same equivalence proves injectivity, and the definition of the orbit proves surjectivity. Thus $|Gx|=[G:G_x]$, giving the formula.

Solved by gpt-5.6-sol high.