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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 $Hom_{\mathbb{C}G}(V, V) \cong e\mathbb{C}Ge$?

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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.

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Why is $\alpha$ surjective? Is this where the idempotency of $e$ comes in? – 109238 Mar 1 '12 at 0:29
Yes: given a $G$-endo $\phi$ of $V$, using $e^2=e$, we obtain that $\phi(e)$ is an element of $V$ satisfying $\phi(e)=\phi(e)e \in Ve$. Thus $\phi(e)=efe$ for some $f \in \mathbb{C} [G]$ and one checks that $\alpha(efe)=\phi$. – Steve Mar 1 '12 at 0:33
@Steve: I'm wondering: If I understand correctly, $\mathbb CG$ and $e\mathbb CG$ are being considered as right modules here. Am I missing a notational convention that indicated that in the question, or did you just infer that from the content? I was under the impression that $\mathbb CG$ is usually considered as a left module unless stated otherwise. – joriki Mar 1 '12 at 7:43
Joriki, well there are those, mostly dwelling in shadows and shunned by their neighbors, who actually prefer right modules to left modules. In this case, the beginning "Let $V=e\mathbb{C}[G]$ be a submodules..." of the question suggests that we are working with right modules, since the alternative is having to type an annoying inverse everywhere. – Steve Mar 1 '12 at 12:22
:-) Some day the neighbours will see the light... By the way, I happened to check back on this; if you want me to get notified of your response, you have to put an @ in front of the user name. – joriki Mar 1 '12 at 15:40
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