 Solution

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

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

G_{3}

G_{2} = G^{p}

is central because

[G_{2}, G] \subseteq G_{3}

, and has exponent dividing

p

because

G_{2}^{p} = G_{3}

. Thus the group

G / G_{3}

has nilpotency class at most two. Collection gives

(x y)^{p} = x^{p} y^{p} [y, x]^{(2 p)} \equiv x^{p} y^{p} (mod G_{3}),

since

[y, x] \in G_{2}

and

p ∣ (2 p)

. Also replacing

x

xu

with

u \in G_{2}

p

-th power unchanged modulo

G_{3}

. Hence the assignment is well-defined and is a group homomorphism. Its image contains all the generators

x^{p} G_{3}

G_{2} / G_{3}

, so it is onto:

G / G_{2} ↠ G_{2} / G_{3}, x G_{2} ⟼ x^{p} G_{3} .

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