 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2021/paper-2/10/b/solution

g g^{- 1} = g^{- 1} g = e

give

D (g) D (g^{- 1}) = D (e) = I, D (g^{- 1}) D (g) = I .

Thus

D (g^{- 1})

is both a left and right inverse of

D (g)

, proving

D (g^{- 1}) = [D (g)]^{- 1} .

If only multiplicativity is initially stated,

D (e)^{2} = D (e)

together with invertibility of representation matrices first implies

D (e) = I

. This explains why a group representation takes values in the general linear group rather than arbitrary square matrices.

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