Questions: Modular Representation Theory

Question 1

Over a field of characteristic p dividing |G|, which fundamental theorem of ordinary representation theory fails?

• A. Schur's lemma
• B. Maschke's theorem (complete reducibility)
• C. The existence of a character map
• D. The fact that representations are group homomorphisms

Answer: (B)

Maschke's theorem requires dividing by |G|, which is impossible when char(k) divides |G| (since |G| = 0 in k). Without complete reducibility, the category of representations is no longer semisimple. Schur's lemma still holds (it uses only irreducibility and linear algebra). Characters still exist as traces, but they lose much of their power because non-isomorphic indecomposable modules can have the same composition factors.

Question 2

In characteristic p, an indecomposable module is always irreducible.

• T. True
• F. False

Answer: False

This is the key distinction in modular theory. A module is indecomposable if it cannot be written as M₁ ⊕ M₂ with both nonzero; it is irreducible if it has no proper nonzero submodules. In the semisimple case (Maschke), these coincide. In the modular case, there exist indecomposable modules with proper submodules that are not direct summands. For example, the 2-dimensional representation of ℤ/pℤ over 𝔽_p given by [[1,1],[0,1]] is indecomposable but not irreducible.

Question 3

The number of irreducible representations of G over an algebraically closed field of characteristic p equals:

• A. The number of conjugacy classes of G
• B. The number of p-regular conjugacy classes (classes of elements whose order is not divisible by p)
• C. The number of Sylow p-subgroups
• D. |G|/p

Answer: (B)

In characteristic 0, the number of irreducibles equals the total number of conjugacy classes. In characteristic p, it equals the number of p-regular (also called p'-) conjugacy classes. An element g is p-regular if its order is coprime to p. This is because the Brauer characters, which replace ordinary characters in modular theory, are only defined on p-regular elements and satisfy orthogonality relations on this restricted set.

Question 4

Brauer characters are defined only on p-regular elements of G. Why can't ordinary trace be used as a character in characteristic p?

Model answer: The trace of a matrix over a field of characteristic p takes values in that field, where distinct eigenvalue configurations can give the same trace (e.g., a matrix with p identical eigenvalues λ has trace pλ = 0). Brauer's solution is to lift eigenvalues to characteristic 0 via a ring of p-adic integers and take the trace there, but this only works for elements whose eigenvalues are roots of unity of order coprime to p — the p-regular elements.

In characteristic 0, the trace determines the multiset of eigenvalues (via Newton's identities). In characteristic p, information is lost: tr([[1,1],[0,1]]) = 2 = tr(I₂) over 𝔽₂, even though these matrices are not similar. Brauer characters bypass this by lifting to characteristic 0, recovering enough information to classify irreducible modules. The resulting theory is powerful but more intricate than ordinary character theory.