Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 151 / 4 / c / i

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- Solution i

Solution

The trivial action gives

H^{1} (D_{n}, Z) = Hom (D_{n}, Z) = 0

because

D_{n}

is finite. By part b(ii),

H^{2} (D_{n}, Z) ≅ Hom (D_{n}, Q / Z) ≅ Hom (D_{n}^{ab}, Q / Z) .

In the abelianization the relation

aba = b^{- 1}

becomes

b = b^{- 1}

. Hence

D_{n}^{ab} ≅ {C_{2}, C_{2} \times C_{2}, n odd, n even .

Each finite cyclic group is naturally isomorphic to its character group in

Q / Z

, so

H^{2} (D_{n}, Z) ≅ {Z /2 Z, (Z /2 Z)^{2}, n odd, n even .

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