 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2020/ii/paper-2/18g/c/i/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

G^{(0)} = G, G^{(r + 1)} = [G^{(r)}, G^{(r)}]

eventually reaches the trivial group; equivalently, it has a subnormal series with abelian quotients.

S_{4}

is solvable because

{1} ◃ V_{4} ◃ A_{4} ◃ S_{4}

has abelian quotients

V_{4}

C_{3}

, and

C_{2}

, where

V_{4}

S_{4} is solvable .

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