 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ii/paper-2/19i/b/i/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

A^{2} = B^{2} = C^{2} = D^{2} = I, [A, B] = [C, D] = Z,

while every other pair among

A, B, C, D

commutes. Also

Z = - I

is central. Consequently every commutator of words in the generators lies in

⟨ Z ⟩

, so

G^{'} \subseteq ⟨ Z ⟩

. Since

Z = [A, B]

, the reverse inclusion holds and

G^{'} = ⟨ Z ⟩ .

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