 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ia/paper-3/5d/c/ii/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

Every element satisfies

x^{2} = e

, including the identity. For any

x, y \in G

e = (x y)^{2} = x y x y .

Since

x^{- 1} = x

and

y^{- 1} = y

, this implies

x y = y x

, so

G

is abelian. It is therefore a vector space over the two-element field, with group operation as vector addition and scalar multiplication by zero or one. A basis of this finite vector space gives

G ≅ C_{2}^{r} = r factors C_{2} \times \dots \times C_{2}

for some

r \geq 0

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