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

Table of contents
- Solution a

Solution

Direct matrix multiplication gives

P^{2} = P - I

and hence

P^{3} = - I

. It follows that

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.

 Ancestors (12)

 View article source

 Discussion (0)

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.

  See all articles in the same topic Create my own version