Lemma 1 (Schur's lemma)   Let be a finite group and let and be irreducible -modules. Then, every -module homomorphism is either invertible or the trivial zero map.

Proof. Note that both the kernel, , and the image, , are -submodules of and , respectively. Since is irreducible, is either trivial or all of . In the former case, is all of -- also because is irreducible -- and hence is invertible. In the latter case, is the zero map.

Corollary 1   Let be a finite-dimensional, irreducible -module taken over an algebraically closed field. Then, every -module homomorphism is equal to a scalar multiplication.

Proof. Since the ground field is algebraically closed, the linear transformation has an eigenvalue; call it . By definition, is not invertible, and hence equal to zero by Schur's lemma. In other words, , a scalar.