Consider the representation of ℤ/2ℤ on ℝ² where the generator acts by ρ(1) = [[1, 1], [0, 1]]. The subspace W = span{e₁} is invariant. Is this representation reducible?
ANo — the representation is irreducible because W is only one-dimensional
BYes — it is reducible because W is a proper invariant subspace, but it is not completely reducible because there is no invariant complement to W
CYes — it is reducible and completely reducible because every subspace of ℝ² has a complement
DNo — the representation is irreducible because ρ(1) is not diagonalizable
The representation is reducible since W = span{e₁} is a proper G-invariant subspace (ρ(1)e₁ = e₁ ∈ W). However, it is NOT completely reducible: there is no G-invariant complement to W. If U were such a complement, it would be spanned by some vector v = ae₁ + be₂ with b ≠ 0, and ρ(1)v = ae₁ + be₂ + be₁ = (a+b)e₁ + be₂. For U to be invariant, (a+b)e₁ + be₂ must be a scalar multiple of ae₁ + be₂, which forces b = 0 — contradiction. This is an example of a representation that is reducible but indecomposable.
Question 2 True / False
Every one-dimensional representation is irreducible.
TTrue
FFalse
Answer: True
A one-dimensional vector space V has only two subspaces: {0} and V itself. Both are trivially G-invariant. Since there are no proper nontrivial subspaces, the definition of irreducibility is vacuously satisfied. One-dimensional representations are always irreducible, regardless of the group or the field.
Question 3 Short Answer
Why are irreducible representations considered the 'atoms' of representation theory?
Think about your answer, then reveal below.
Model answer: Under Maschke's theorem conditions (finite group, characteristic zero or coprime to |G|), every representation decomposes uniquely as a direct sum of irreducible representations. So irreducibles are the building blocks from which all representations are constructed, analogous to prime numbers in arithmetic.
The analogy to prime factorization is precise: just as every positive integer factors uniquely into primes, every (finite-dimensional, semisimple) representation decomposes uniquely into irreducibles. Classifying all representations of a group therefore reduces to two problems: finding all irreducible representations, and understanding how they combine via direct sums. The first problem is solved by character theory; the second by tools like induced representations and tensor products.
Question 4 Multiple Choice
A representation is called completely reducible if it is a direct sum of irreducible subrepresentations. Which of the following is NOT a condition that guarantees complete reducibility?
AG is a finite group and the field has characteristic zero
BG is a finite group and char(F) does not divide |G|
CThe representation is unitary (preserves an inner product)
DThe representation has prime dimension
Having prime dimension does not guarantee complete reducibility — a representation of prime dimension p could be indecomposable but reducible (containing a proper invariant subspace with no invariant complement). The other three conditions do guarantee complete reducibility: the first two are forms of Maschke's theorem, and unitarity allows you to take orthogonal complements of invariant subspaces, which are automatically invariant.