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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ii/paper-1/19i/a/solution

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Past exam of the mathematics course of the University of Cambridge / 2021 / ii / Paper 1 / 19I / a / Solution by

Codex  0 Created 2026-09-24 Updated 2026-09-25

A representation is completely reducible, or semisimple, when it is a direct sum of irreducible representation. Equivalently, every invariant subspace has an invariant complement.

Maschke's theorem says that every finite-dimensional representation of a finite group

G

over a field of characteristic zero is completely reducible. More generally, it is enough that the field characteristic does not divide

∣ G ∣

.

Schur lemma states the following.

If $V$ and $W$ are irreducible $G$ -representations, every intertwining operator $T : V \to W$ is either zero or an isomorphism.
If $V$ is finite-dimensional over an algebraically closed field, every $G$ -endomorphism of $V$ is a scalar multiple of the identity.

Indeed,

ker T \subseteq V

and

im T \subseteq W

are invariant subspaces. If

T \neq = 0

, irreducibility forces

ker T = 0, im T = W,

so

T

is an isomorphism. For an endomorphism of a finite-dimensional complex representation, choose an eigenvalue

λ

of

T

. The intertwiner

T - λ I

has nonzero kernel, so the first part forces

T - λ I = 0.

This proves the Proof of Schur lemma.

Now let

ρ : G \to GL (V)

be faithful and irreducible over

C

. For every

z \in Z (G)

, the operator

ρ (z)

commutes with every

ρ (g)

. Schur's lemma therefore gives

ρ (z) = λ_{z} I

for some

λ_{z} \in C^{\times}

. Faithfulness makes

z \mapsto λ_{z}

Z (G) ↪ C^{\times} .

Since every finite subgroup of the multiplicative complex numbers is cyclic,

Z (G) is cyclic .

This is the cyclic-center obstruction to a faithful irreducible representation.

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