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

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

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

Proof. Since the ground field is algebraically closed, the linear transformation $f: V\to V$ has an eigenvalue; call it $\lambda$ . By definition, $f - \lambda 1$ is not invertible, and hence equal to zero by Schur's lemma. In other words, $f = \lambda$ , a scalar.