Brian Bi
\[ \DeclareMathOperator{\End}{End} \DeclareMathOperator{\char}{char} \DeclareMathOperator{\tr}{tr} \DeclareMathOperator{\ker}{ker} \DeclareMathOperator{\im}{im} \DeclareMathOperator{\sgn}{sgn} \DeclareMathOperator{\Hom}{Hom} \DeclareMathOperator{\span}{span} \DeclareMathOperator{\diag}{diag} \DeclareMathOperator{\Id}{Id} \DeclareMathOperator{\ad}{ad} \newcommand\d{\mathrm{d}} \newcommand\pref[1]{(\ref{#1})} \]

Remark: I did not understand the proof of Lemma 5.13.3 given in the text, so I consulted Fulton & Harris, which cited Curtis & Reiner. It appears that the key is to consider the linear operator \(T\) on \(\mathbb{C}[S_n]\) that acts by right-multiplication by \(c_\lambda\). This was not clear in Etingof's proof.

Explicitly, let \(\gamma\) denote the scalar such that \(c_\lambda^2 = \gamma c_\lambda\). Since the coefficient of the identity element in \(c_\lambda\) is 1, computing the trace of \(T\) in the basis consisting of the elements of \(S_n\) yields \(n!\). We now compute \(\tr T\) in a second way. Write \(\mathbb{C}[S_n] = V_\lambda \oplus W\), a direct sum of vector spaces. (Note: the argument used in the proof of Maschke's theorem implies that it's possible to choose \(W\) to be a subrepresentation of \(\mathbb{C}[S_n]\), but we won't need that fact here.) Now \(n! = \tr T = \tr(T|_{V_\lambda}) + \tr(T|_W)\). For \(x \in V_\lambda\), we have \(x = yc_\lambda\) for some \(y \in \mathbb{C}[S_n]\), so that \(T(x) = xc_\lambda = yc_\lambda^2 = \gamma yc_\lambda = \gamma x\). Thus, \(T|_{V_\lambda} = \gamma \Id\) and the trace on this block is \(\gamma \dim V_\lambda\). On the other hand, for arbitrary \(x \in \mathbb{C}[S_n]\), we have \(T(x) \in V_\lambda\), so \(\tr(T|_W) = 0\). This implies that \(n! = \gamma \dim V_\lambda\), or \(\gamma = \frac{n!}{\dim V_\lambda}\) as required.