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Let $G$ be a finite group, and let $r$ - the number of it's conjugate classes, so our group has $r$ classes of irreducible representations. $\varkappa_1,..,\varkappa_r$ - irreduciible characters, and $n_1,..,n_r$ - the degrees of corresponding linear spaces. Let $C[G]$ be group algebra over the field of complex numbers. Assume, that $I_i=\frac{n_i}{|G|}\sum\limits_g\overline{\varkappa_i(g)}g$.

My question is the following: how to prove that every irreducible $C[G]$ submodule of group algebra (it will be an irreducible space-representation of our group at the sime) which corresponds the character $\varkappa_i$ belongs to $I_iC[G]$? I read (Kostrikin,"Introduction into algebra",book 3, chapter 4) that it can be somehow deduced from the next computation: $\varkappa_j(I_i)=\frac{n_i}{|G|}\sum\limits_g\overline{\varkappa_i(g)}\varkappa_j(g)=n_i\delta_{ij}$ But i have no ideas about the next step. (Excuse me for my english, I am not a native speaker).

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migrated from mathoverflow.net Jul 25 '13 at 20:47

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    $\begingroup$ I_i is central, so Schur's lemma says it acts by a scalar. But we know the trace of this scalar by the computation. $\endgroup$ – John Wiltshire-Gordon Jul 25 '13 at 13:52

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