 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ia/paper-3/6d/i/solution

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

D_{8} = ⟨ r, s : r^{4} = s^{2} = e, srs = r^{- 1} ⟩

, one has

[r, s] = r^{- 2} = r^{2}

. The quotient by

⟨ r^{2} ⟩

C_{2} \times C_{2}

, so

D_{8}^{'} = {e, r^{2}} .

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