Finite Groups

Verify that

Ψ^{d} (g) := Ψ (g)^{- 1 *}

is a representation of

G

in the dual

E^{*}

E

For all $g \in G$ , we have $Ψ (g) \in G l (E)$ , so $Ψ (g)^{- 1} \in G l (E)$ and $Ψ^{d} (g) = Ψ (g)^{- 1 *} \in G l (E^{*})$ (this is indeed an invertible operator - its inverse is $Ψ (g)^{*}$ ). So we just have to show that $Ψ^{d}$ is a group homomorphism: For $g, h \in G$ , we have $Ψ^{d} (g h) = Ψ (g h)^{- 1 *} = (Ψ (g) Ψ (h))^{- 1 *} = (Ψ (h)^{- 1} Ψ (g)^{- 1})^{*} = Ψ (g)^{- 1 *} Ψ (h)^{- 1 *} = Ψ^{d} (g) Ψ^{d} (h) .$ Thus, $Ψ^{d}$ is a representation of $G$ in $E^{*}$ .