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 $Hom_{\mathbb{C}G}(V, V) \cong e\mathbb{C}Ge$?
Question about the group algebra $\mathbb{C}G$
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abstract-algebra
1 Answers
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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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0Waving @joriki, thanks for the tip! – 2012-03-01