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Let $G=S_n$ be the permutation group on $n$ letters and $e\in\mathbb C[G]$ a central, primitive central idempotent. Let also $f\in\mathbb C[G]$ be central. I am reading this paper where the author concludes that $f\cdot e \in \mathbb C e$. Could someone tell me how this follows, and/or give a reference? I am sure this is a completely standard result.

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I am interested in the paper you are reading. Could you provide it? Thanks. –  awllower Jan 24 '13 at 14:24
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Maybe the following works: Any central elements could be written as a sum of central primitive idempotents, in which $e$ might appear or not. In any case, as $e$ is orthogonal to other central primitive idempotents, the result follows. –  awllower Jan 24 '13 at 14:57
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1 Answer

up vote 5 down vote accepted

Important grammar gripe: you mean "primitive central" rather than "central, primitive". You want $e$ to be primitive among the central idempotents, not among all idempotents. Otherwise the result would be trivial (over $\mathbb C$, that is, not over $\mathbb Q$, although for $G=S_n$ there is no difference), as the only central idempotents which are primitive among all idempotents come from $1$-dimensional irreps.

Now to the proof. Note that your statement (corrected) is a statement not so much about $\mathbb C\left[G\right]$ as about $Z\left(\mathbb C\left[G\right]\right)$. You want to prove that every primitive idempotent $e$ of $Z\left(\mathbb C\left[G\right]\right)$ and any $f\in Z\left(\mathbb C\left[G\right]\right)$ satisfy $fe\in\mathbb Ce$.

By Wedderburn's theorem, $\mathbb C\left[G\right]$ is a direct product of matrix rings over $\mathbb C$ (as a ring!). Hence, $Z\left(\mathbb C\left[G\right]\right)$ is a direct product of the centers of these matrix rings, i. e., a direct sum of $\mathbb C$'s. The primitive idempotents of $Z\left(\mathbb C\left[G\right]\right)$ thus are the $1$'s of these $\mathbb C$'s, and now everything is clear.

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Is it clear? I put my doubt into a solution, and I hope you can help dispel it. –  rschwieb Jan 24 '13 at 15:09
    
Looks like awllower ninja'd me. –  darij grinberg Jan 24 '13 at 15:28
    
Not sure if you can see the comment in the deleted solution, but here is the gist: thanks for pointing out my notation blindness! I am too used to talking about idempotents $e$ and $f$ :) –  rschwieb Jan 24 '13 at 15:29
    
Indeed, that wasn't so hard. Thanks! –  Jesko Hüttenhain Jan 24 '13 at 17:24
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