Questions: Group Algebras

Question 1

What is the dimension of the group algebra ℂ[S₃] as a vector space over ℂ?

• A. 3
• B. 6
• C. 9
• D. 36

Answer: (B)

dim(k[G]) = |G| as a k-vector space, since G itself is the basis. |S₃| = 6, so dim(ℂ[S₃]) = 6. As an algebra, ℂ[S₃] is 6-dimensional but non-commutative (since S₃ is non-abelian). By the Artin-Wedderburn theorem, ℂ[S₃] ≅ ℂ ⊕ ℂ ⊕ M₂(ℂ) as algebras, reflecting the three irreducible representations of dimensions 1, 1, and 2.

Question 2

A representation ρ: G → GL(V) is the same data as a left k[G]-module structure on V.

• T. True
• F. False

Answer: True

Given ρ, define the module action by (Σ aᵍeᵍ)·v = Σ aᵍρ(g)v. Conversely, given a k[G]-module V, define ρ(g)v = eᵍ·v. These constructions are inverse to each other and preserve all the relevant structure: subrepresentations correspond to submodules, G-equivariant maps correspond to module homomorphisms, and irreducibility corresponds to simplicity. This equivalence is the reason group algebras exist.

Question 3

The center Z(k[G]) of the group algebra consists of elements that commute with all of k[G]. What is the dimension of Z(ℂ[G]) for a finite group G?

Model answer: The dimension of Z(ℂ[G]) equals the number of conjugacy classes of G.

A basis for Z(ℂ[G]) is given by the class sums zC = Σ_{g∈C} eᵍ, one for each conjugacy class C. An element Σ aᵍeᵍ lies in the center if and only if it is constant on conjugacy classes (aᵍ = a_{hgh⁻¹} for all h), which means it is a linear combination of the class sums. Since the number of conjugacy classes also equals the number of irreducible representations, this connects the center of the group algebra to character theory.

Question 4

For an abelian group G, the group algebra k[G] is commutative. Over ℂ, what is ℂ[ℤ/nℤ] isomorphic to as an algebra?

• A. The polynomial ring ℂ[x]
• B. The matrix algebra Mₙ(ℂ)
• C. ℂ ⊕ ℂ ⊕ ··· ⊕ ℂ (n copies)
• D. The ring ℤ/nℤ tensored with ℂ

Answer: (C)

ℂ[ℤ/nℤ] ≅ ℂ[x]/(xⁿ − 1). Over ℂ, xⁿ − 1 factors into n distinct linear factors (the nth roots of unity), so by the Chinese Remainder Theorem, ℂ[x]/(xⁿ−1) ≅ ℂ ⊕ ℂ ⊕ ··· ⊕ ℂ. Each copy of ℂ corresponds to one irreducible representation (all 1-dimensional, since the group is abelian). This is the Artin-Wedderburn decomposition for cyclic groups.