Schur's Lemma (Representation Theory)

Theorem
Let $(G,\cdot)$ be a finite group and let $V$ and $V^\prime$ be two irreducible $G$-modules.

Consider $f:V\to V^\prime$ an homomorphism of $G$-modules, then $f\equiv 0$ or $f$ is an isomorphism.

Proof
From Kernel is G-module, $\ker(f)$ is a $G$-submodule of $V$ and from Image is G-module, $\operatorname{Im}(f)$ is a $G$-submodule of $V^\prime$.

By the definition of irreducible, $\ker(f)=\{0\}$ or $\ker(f)=V$ from which follows that $f$ is inyective or $f=0$.

It also follow that $\operatorname{Im}(f)=\{0\}$ or $Im(f)=V^\prime$, thus $f$ is surjective or $f=0$.

In conclusion, $f=0$ or $f$ is inyective and surjective.

Corollary
If $V=V^\prime$ and is a $k$-vector space over an algebraically closed field, then $f=\lambda \operatorname{Id}_V$ where $\lambda\in k$.

In other words $\operatorname{End}(V)$ (endomorphisms of $V$) has the same structure as $k$; it's a field.

Proof of Corollary
If $f=0$, since $0\in k$ it can be written $f=0 Id_V$

Now if $f$ is an automorphism, the characteristic polynomial of $f$ is complete reducible in $k[x]$ because $k$ is algebraically closed; hence $f$ has all eigenvalue in $k$.

We take $\lambda\in k$ an eigenvalue of $f$ and consider the endomorphism $f-\lambda \operatorname{Id}_V:V\to V$.

Since $\ker(f-\lambda \operatorname{Id}_V)\ne\{0\}$ because $\lambda$ is an eigenvalue, it follows using Schur's Lemma that $f=\lambda \operatorname{Id}_V$