Schur's Lemma (Representation Theory)

Theorem
Let $\left({G, \cdot}\right)$ be a finite group.

Let $V$ and $V'$ be two irreducible $G$-modules.

Let $f: V \to V'$ be a homomorphism of $G$-modules.

Then either:
 * $f \left({v}\right) = 0$ for all $v \in V$

or:
 * $f$ is an isomorphism.

Proof
From Kernel is G-Module, $\ker \left({f}\right)$ is a $G$-submodule of $V$.

From Image is G-Module, $\operatorname{Im} \left({f}\right)$ is a $G$-submodule of $V'$.

By the definition of irreducible:
 * $\ker \left({f}\right) = \left\{{0}\right\}$

or:
 * $\ker \left({f}\right) = V$

If $\ker \left({f}\right) = V$ then by definition:
 * $f \left({v}\right) = 0$ for all $v \in V$

Let $\ker \left({f}\right) = \left\{{0}\right\}$.

Then from Linear Transformation is Injective iff Kernel Contains Only Zero:
 * $f$ is injective.

It also follows that:
 * $\operatorname{Im} \left({f}\right) = V'$

Thus $f$ is surjective and injective.

Thus by definition $f$ is a bijection and thence an isomorphism.