2
$\begingroup$

Let $V = e\mathbb{C}G$ be a submodule of $\mathbb{C}G$ and $e \in \mathbb{C}G$ is such that $e^{2} = e$. Why is it that $Hom_{\mathbb{C}G}(V, V) \cong e\mathbb{C}Ge$?

1 Answers 1

2

There is a natural map $\alpha: e \mathbb{C}[G] e \rightarrow \mathrm{End}_G(V)$ defined by $\alpha(efe)(h)=efeh \quad \hbox{for $f \in \mathbb{C}[G]$ and $h \in e \mathbb{C}[G]$.}$

Using the fact that a $G$-endomorphism $\phi:V \rightarrow V$ is determined by the image of $e$ it is straightforward to check that $\alpha$ is an isomorphism.

  • 0
    Waving @joriki, thanks for the tip!2012-03-01