According to PlanetMath Schur's lemma is this:

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.


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.


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.


In the proof above, why can Schur's lemma be applied to $\,f - \lambda 1$? Is it assumed that $\,f - \lambda 1$ is also a $G$-module homomorphism?

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    $\begingroup$ The set of all $G$-module homomorphisms from $V$ to $W$ is a vector space, and it contains $1$. $\endgroup$ – Qiaochu Yuan Sep 10 '12 at 22:47

$f$ is a $G$-module homomorphisms by assumption. Let $\rho : G \rightarrow GL_k(V)$ by the representation, where $k$ is the field. Then you have

$(f - \lambda I)(\rho(g)(v)) = f(\rho(g)(v)) - \lambda I(\rho(g)( v)) = \rho(g)(f(v)) - \lambda \rho(g)(v)$

by using the fact that $f$ is a $G$-module homomorphism. Since $\rho(g) \in GL_k(V)$, this means that $\rho(g)$ is a k-linear transformation, hence

$= \rho(g)(f(v)) - \rho(g)(\lambda v) = \rho(g)(f(v) - \lambda I(v)) = \rho(g)((f - \lambda I))(v)$

So $f - \lambda I$ has been shown to be $G$-linear.


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