Questions: Regular Representation

Question 1

What is the character of the regular representation evaluated at a non-identity element g ≠ e?

• A. |G|
• B. 1
• C. 0
• D. −1

Answer: (C)

For g ≠ e, left multiplication by g sends every basis vector eₕ to e_{gh} ≠ eₕ (since gh ≠ h when g ≠ e). So the permutation matrix for ρ(g) has all zeros on the diagonal, giving tr(ρ(g)) = 0. At the identity, ρ(e) = I_{|G|}, so χ_reg(e) = |G|. This character — |G| at e and 0 elsewhere — encodes the fact that the regular representation contains each irreducible dᵢ times.

Question 2

The regular representation of any nontrivial group is irreducible.

• T. True
• F. False

Answer: False

The regular representation has dimension |G| and always decomposes into irreducible summands. For example, the trivial representation (the span of Σ eᵍ) is always a one-dimensional subrepresentation. By Maschke's theorem (over ℂ), the regular representation decomposes as ⊕ᵢ Vᵢ^{dᵢ}, where the sum runs over all irreducible representations and dᵢ = dim(Vᵢ). Only for the trivial group G = {e} is the regular representation irreducible.

Question 3

The formula |G| = Σᵢ dᵢ² (sum of squares of irreducible dimensions) follows from which property of the regular representation?

Model answer: The regular representation has dimension |G|, and it decomposes as ⊕ᵢ Vᵢ^{dᵢ} where each irreducible Vᵢ appears with multiplicity dᵢ = dim(Vᵢ). So |G| = dim(ℂ[G]) = Σᵢ dᵢ · dim(Vᵢ) = Σᵢ dᵢ².

The multiplicity of Vᵢ in the regular representation can be computed via characters: nᵢ = ⟨χ_reg, χᵢ⟩ = (1/|G|)·|G|·χᵢ(e) = dᵢ, since χ_reg is zero off the identity. So each irreducible appears exactly dim(Vᵢ) times, and the dimension count gives the sum-of-squares formula.