*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.