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8. Let Vreg = U1 ⊕. ⊕Us be the decomposition of the regular representation into irreps. Let W be any irrep of G. 5, we have that dim HomG (Vreg , W ) equals the number of Ui that are isomorphic to W . 12. Let U1 , . . , Ur be all the irreps of G, and let dim Ui = di . Then r d2i = |G| i=1 Proof. 8, Vreg = U1⊕d1 ⊕ . . ⊕ Ur⊕dr Now take dimensions of each side. 47 Notice this is consistent with out results on abelian groups. e. the number of irreps of G is the size of G. This is what we found. 13.

Then we know that V ∗ carries a representation of G, defined by ρHom(V,C) (g) : f → f ◦ ρV (g −1 ) We’ll denote this representation by (ρV )∗ , we call it the dual representation to ρV . 1. Let G = S3 = σ, τ | σ 3 = τ 2 = e, τ στ = σ −1 and let ρ be the 2-dimensional irrep of G. In the appropriate basis (see Problem Sheets) ρ(σ) = ω 0 0 ω −1 ρ(τ ) = 0 1 1 0 (where ω = e 2πi 3 ) The dual representation (in the dual basis) is ρ∗ (σ) = ρ(τ ) = ω −1 0 0 ω 0 1 1 0 This is equivalent to ρ under the change of basis P = 0 1 1 0 So in this case, ρ∗ and ρ are isomorphic.

This is the map ΦρV (g)(x) :V ∗ → C f → f (ρV (g)(x)) Now consider (ρV ∗ )∗ (g)(Φ(x)). By definition, this is the map Φx ◦ ρV ∗ (g −1 ) :V ∗ → C f → Φx ρV ∗ (g −1 )(f ) = Φx (f ◦ ρV (g)) = (f ◦ ρV (g)) (x) So Φ (ρV (g)(x)) and (ρV ∗ )∗ (g) (Φ(x)) are the same element of (V ∗ )∗ , so Φ is indeed G-linear. Therefore, (V ∗ )∗ and V are naturally isomorphic as representations. 3. Let V carry a representation of G. Then V is irreducible if and only if V ∗ is irreducible. Proof. e. it contains a non-trivial subrepresentation U ⊂ V .