Schur's Lemma

Statement

Let be any group (or a ring) and let and be irreducible representations (or irreducible modules). Then a morphism is either zero or an isomorphism

Proof

We can look at the . The kernel is always a subrepresentation (or module) of so it must be either or , so it is either injective or the zero map. If is an injection, then we can also look at which is a non-zero submodule, thus it must be all of and is both injective and surjective, thus an isomorphism.

Consequences of working over a field

Over then

1. Every morphism has the form for some (or other appropriate underlying field.)