Questions: Permutation Representations

Question 1

S₃ acts on {1, 2, 3} by permutation. The resulting permutation representation has dimension 3. What is its decomposition into irreducible representations of S₃?

Model answer: The permutation representation decomposes as the trivial representation ⊕ the standard 2-dimensional representation.

The character of the permutation representation is χ(e) = 3, χ((12)) = 1, χ((123)) = 0 (counting fixed points). The trivial representation has character (1,1,1) and the standard has (2,0,−1). Taking inner products: ⟨χ, χ_triv⟩ = (3+3+0)/6 = 1, ⟨χ, χ_sign⟩ = (3−3+0)/6 = 0, ⟨χ, χ_std⟩ = (6+0+0)/6 = 1. So χ = χ_triv + χ_std. The trivial summand is the span of e₁+e₂+e₃.

Question 2

The character of a permutation representation counts fixed points: χ(g) = |Fix(g)| = |{x ∈ X : g·x = x}|.

• T. True
• F. False

Answer: True

In the permutation representation, g sends basis vector eₓ to e_{g·x}. The trace of this permutation matrix counts the number of basis vectors fixed by g, which is exactly |{x ∈ X : g·x = x}|. This makes permutation characters easy to compute without constructing matrices — just count fixed points for each group element.

Question 3

Every permutation representation on a set with |X| ≥ 2 contains the trivial representation as a subrepresentation.

• T. True
• F. False

Answer: True

The vector v = Σ_{x∈X} eₓ (the sum of all basis vectors) is fixed by every group element, since g permutes the basis vectors. So span{v} is a 1-dimensional G-invariant subspace on which G acts trivially. By Maschke's theorem (over ℂ), this splits off as a direct summand. The complementary (|X|−1)-dimensional subspace {Σ aₓeₓ : Σ aₓ = 0} is also G-invariant and may or may not be irreducible.

Question 4

Burnside's lemma states that the number of orbits of G acting on X equals:

• A. The dimension of the permutation representation
• B. (1/|G|) Σ_{g∈G} |Fix(g)|, the average number of fixed points
• C. The number of irreducible representations of G
• D. |X| / |G|

Answer: (B)

Burnside's lemma gives |X/G| = (1/|G|) Σ_{g∈G} |Fix(g)|. In representation-theoretic terms, this is ⟨χ_perm, χ_triv⟩ — the multiplicity of the trivial representation in the permutation representation. Since χ_triv = 1 everywhere, the inner product computes exactly the average number of fixed points. Each orbit contributes one copy of the trivial representation.

Question 5

If G acts transitively on X with stabilizer H = Stab(x₀), the permutation representation on X is isomorphic to:

• A. The regular representation of G
• B. The induced representation Ind_H^G(trivial)
• C. The trivial representation of dimension |X|
• D. The sign representation tensored with itself

Answer: (B)

When G acts transitively on X, we can identify X with G/H (the coset space). The permutation representation is then G acting on the left cosets of H by left multiplication. This is precisely the induced representation of the trivial representation of H to G: Ind_H^G(1_H). This connection between transitive permutation representations and induced representations is fundamental. When H = {e}, G/H = G and we recover the regular representation.