Questions: Maschke's Theorem

Question 1

Maschke's theorem fails when the characteristic of the field divides |G|. Give a concrete example.

Model answer: Take G = ℤ/pℤ over the field 𝔽_p. The representation ρ(1) = [[1, 1], [0, 1]] has the invariant subspace span{e₁}, but no G-invariant complement exists — the representation is reducible but not completely reducible.

The matrix [[1,1],[0,1]] has order p in GL₂(𝔽_p), giving a valid representation of ℤ/pℤ. The subspace span{e₁} is invariant (fixed by ρ(1)), but any candidate complement must contain a vector with nonzero second component, and applying ρ(1) shifts it by e₁, pushing it outside the complement. The averaging trick fails because dividing by |G| = p is impossible in characteristic p.

Question 2

The key step in the proof of Maschke's theorem involves 'averaging' a projection over the group. What does this averaging accomplish?

• A. It makes the projection have norm 1
• B. It turns an arbitrary linear projection onto an invariant subspace into a G-equivariant projection, whose kernel is then also G-invariant
• C. It diagonalizes every group element simultaneously
• D. It constructs the irreducible decomposition directly

Answer: (B)

Given a G-invariant subspace W ⊆ V, pick any linear projection π: V → W. Define π̃ = (1/|G|) Σ_{g∈G} ρ(g)∘π∘ρ(g)⁻¹. This averaged projection is G-equivariant: ρ(h)π̃ = π̃ρ(h). Its kernel ker(π̃) is therefore G-invariant and provides the required complement V = W ⊕ ker(π̃). The averaging uses the group's finiteness (to sum) and the characteristic condition (to divide by |G|).

Question 3

Maschke's theorem applies to all representations of all groups.

• T. True
• F. False

Answer: False

Maschke's theorem has two crucial hypotheses: the group must be finite, and the characteristic of the field must not divide the group's order. It fails for infinite groups in general (though unitary representations of compact groups satisfy an analogue via Haar measure). It also fails in modular representation theory, where char(F) divides |G| — this is a rich and much harder subject.

Question 4

Maschke's theorem implies that every representation of S₃ over ℂ is a direct sum of irreducible representations. How many non-isomorphic irreducible representations does S₃ have?

• A. 2
• B. 3
• C. 4
• D. 6

Answer: (B)

The number of non-isomorphic irreducible representations of a finite group over ℂ equals the number of conjugacy classes. S₃ has three conjugacy classes: {e}, {(12),(13),(23)}, {(123),(132)}. So it has three irreducible representations: the trivial (degree 1), the sign (degree 1), and the standard (degree 2). Their degrees satisfy 1² + 1² + 2² = 6 = |S₃|, confirming the dimension formula.