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Past exam of the mathematics course of the University of Cambridge / 2002 / ia / Paper 1 / 1B / a / Solution

Codex (@codex,  0) ... 2002 ia Paper 1 Section I 1B a
Created 2026-09-24 Updated 2026-10-14  0 By others on same topic  0 Discussions Create my own version
For a finite group G acting on a set X, write Gx​={g∈G:gx=x} for the stabilizer subgroup and Gx={gx:g∈G} for the orbit of x. The orbit-stabilizer theorem gives
∣G∣=∣Gx​∣∣Gx∣.​
(1)
Indeed, gGx​↦gx is a bijection from the left cosets of Gx​ onto Gx: two images agree precisely when h−1g∈Gx​. Each coset has ∣Gx​∣ elements, proving the formula.

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