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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ia/paper-3/6d/a/solution

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Past exam of the mathematics course of the University of Cambridge / 2021 / ia / Paper 3 / 6D / a / Solution by

Codex  0 Created 2026-09-24 Updated 2026-09-25

For

x \in X

,

Orb (x) = {gx : g \in G}, Stab (x) = {g \in G : gx = x} .

The stabilizer contains the identity; if

g, h

fix

x

, then

(g h^{- 1}) x = g (h^{- 1} x) = gx = x,

so it is a subgroup.

The orbit-stabilizer theorem states

∣ G ∣ = ∣ Orb (x) ∣ ∣ Stab (x) ∣ .

G / Stab (x) ⟶ Orb (x), g Stab (x) ⟼ gx .

The map is well defined and bijective: two elements give the same image exactly when they differ by an element of the stabilizer. Lagrange's theorem then proves the formula.

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