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The following is a lemma from Fulton and Harris' book -Representation theory,a first course (page 53):
Lemma: For all $x\in \mathbb{C}\mathfrak{S}_r$, $c_{\lambda}\cdot x\cdot c_{\lambda}= scalar \;multiple\; of \;c_{\lambda} $. In particualr, $c_{\lambda}\cdot c_{\lambda}=n_{\lambda}c_{\lambda}.$
In the Lemma, $c_{\lambda}$ denotes the Young symmetriser corresponding to one standard tableau on partition $\lambda$.

The authors give a clear formula for $n_{\lambda}$. My question is that how to deduce a general formula for the scalar multiple of $c_{\lambda}$ for general $x\in \mathfrak{S}_r$? Until now I have no good ideas, so could anyone give me some suggestions on this problem. Thanks a lot!

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  • $\begingroup$ The $n_\lambda$ is the product of the hook lengths of $\lambda$. Combine Lemma 4.42 with Theorem 4.53 in arxiv.org/abs/0901.0827v5. (That said, I hope Fulton/Harris also have a proof of this somewhere.) $\endgroup$
    – darij grinberg
    Commented May 4, 2015 at 1:17
  • $\begingroup$ You may misread my question. From the lemma, $c_{\lambda}xc_{\lambda}=n_{x}c_{\lambda}$, where $n_{x}$ is the scalar multiple depending on $x$. The authors have determined $n_{(1)}=n_{\lambda}$, but what is $n_x$ for $x≠(1)$? $\endgroup$
    – yang
    Commented May 4, 2015 at 1:27
  • $\begingroup$ Ah! You want the general $n_x$. Sorry. $\endgroup$
    – darij grinberg
    Commented May 4, 2015 at 1:31

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