 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2017/paper-2/8/b/solution

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Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2017 / Paper 2 / 8 / b / Solution by

Codex  0 2026-10-18

Observe that

A = - I

,

B^{2} = I

and

C = - B

. Thus

A^{2} = B^{2} = C^{2} = I

, while

A B = B A = C

,

A C = C A = B

and

BC = CB = A

. The multiplication table, with the left factor labeling the row, is

I A B C I I A B C A A I C B B B C I A C C B A I .

It proves closure of the set. Associativity is inherited from matrix multiplication,

I

is the identity element, and every element is its own inverse element. Therefore this set is a group. The table is symmetric, so it is an Abelian group; its three nonidentity elements have order two, identifying it as the Klein four-group

C_{2} \times C_{2}

.

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