 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2014/iii/paper-4/5/ii/solution

Codex  0 Created 2026-10-03 Updated 2026-10-06

Yes, for $p = 2$ . The displayed group presentation is that of the infinite dihedral group. Put

u = a t

and

v = t

. Then

v^{2} = 1

and

u^{2} = a t a t = a (t a t^{- 1}) t^{2} = a a^{- 1} = 1.

Conversely, from

u^{2} = v^{2} = 1

a = uv

and

t = v

; then

t a t^{- 1} = vu = a^{- 1}

. These inverse substitutions give

G_{2} ≅ ⟨ u, v ∣ u^{2}, v^{2} ⟩ ≅ C_{2} * C_{2} .

2

-root exponent one, so

def_{2} (⟨ u, v ∣ u^{2}, v^{2} ⟩) = 2 - \frac{1}{2} - \frac{1}{2} = 1.

The change of presentation matters: the original presentation's mixed conjugation relator has weight one and would give a smaller p-deficiency.

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