 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2004/iii/paper-4/3/iv/solution

Codex  0 Created 2026-10-03 Updated 2026-10-14

Apply (iii) successively to the powerful p-groups

G_{i}

. Starting with

G_{1} = ⟨ a_{1}, \dots, a_{n} ⟩

, if

G_{i} = ⟨ a_{1}^{p^{i - 1}}, \dots, a_{n}^{p^{i - 1}} ⟩

, then

G_{i + 1} = G_{i}^{p} = ⟨(a_{1}^{p^{i - 1}})^{p}, \dots, (a_{n}^{p^{i - 1}})^{p} ⟩ .

Thus

G_{i + 1} = ⟨ a_{1}^{p^{i}}, \dots, a_{n}^{p^{i}} ⟩ (i \geq 1) .

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