 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/a/solution

P^{2} = P - I

and hence

P^{3} = - I

P^{0} P^{3} = (1001), = (- 1 0 0 - 1), P P^{4} = (1 - 1 10), = (- 1 1 - 1 0), P^{2} P^{5} = (0 - 1 1 - 1), = (01 - 1 1),

and

P^{6} = I

. These six matrices are distinct; consequently no smaller positive power is the identity element. Thus

⟨ P ⟩ = {I, P, P^{2}, P^{3}, P^{4}, P^{5}} ≅ C_{6}, ord (P) = 6.

Products add exponents modulo six and

P^{- 1} = P^{5}

. This verifies the cyclic group structure explicitly.

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