Schur's lemmaSchursLemma2002-11-04 09:54:152006-06-12 22:01:25Theorem\usepackage{amsmath,amsfonts,amssymb,amsthm}
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Schur's lemma is a fundamental result in representation theory,
an elementary observation about irreducible modules, which is nonetheless
noteworthy because of its profound applications.
\begin{lemma}[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.
\end{lemma}
\begin{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.
\end{proof}
One of the most important consequences of Schur's lemma is the following.
\begin{corollary}
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.
\end{corollary}
\begin{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.
\end{proof}