 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2018/paper-2/10/b/solution

Every

D_{2} (g)

is invertible because both inversion and the conjugate transpose preserve invertibility. For

g, h \in G

, the group homomorphism property of

D_{1}

and reversal of multiplication under the adjoint give

D_{2} (g h) = [D_{1} ((g h)^{- 1})]^{†} = [D_{1} (h^{- 1}) D_{1} (g^{- 1})]^{†} = [D_{1} (g^{- 1})]^{†} [D_{1} (h^{- 1})]^{†} = D_{2} (g) D_{2} (h) .

Also

D_{2} (e) = I

. Hence $D_{2}$ is a group representation. It is the inverse-adjoint representation of

D_{1}

; the inverse and adjoint each reverse product order, so together they preserve it.

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