Constructing a centrally primitive idempotent in the group algebra of the symmetric group Consider the group algebra of the symmetric group $ \mathbb{C} S_k$. 
Given some Young tableau $T$ of shape $\lambda$, let $a_{\lambda,T}$ and $b_{\lambda,T}$ be the row symmetrizer and column antisymmetrizer of the tableau respectively.
It is known that the Young symmetrizer $c_{\lambda,T} = a_{\lambda,T} b_{\lambda,T}$ is proportional to an idempotent. 
That is,
$c_{\lambda,T}^2 = m_\lambda c_{\lambda,T}$ with $m_\lambda \in \mathbb{R}$.
Using character theory, one can show that the element
$$\omega_\lambda = \sum_{\pi \in S_k} \pi c_{\lambda,T} \pi^{-1}$$ 
is proportional to a centrally primitive idempotent (see e.g. Proposition 2 in the notes by Graham Gill, representation theory of the symmetric group: basic elements). It therefore projects onto the isotypic component associated to $\lambda$.
Is there a more straightforward way (i.e. one that doesn't use character theory) to show that $\omega_\lambda$ is proportional to an idempotent, that is to show that $\omega_\lambda^2 = n_\lambda \omega_\lambda$ with $n_\lambda \in \mathbb{R}$?
edit: I feel one should be able to take advantage of the averaging operation / Reynolds operator $\alpha \mapsto \sum_{g \in G} g \alpha g^{-1}$. I don't quite know how to however.
edit2: Apologies, I have opened a question on mathoverflow just a moment before the first comment got in after a couple of days.
 A: Probably the easiest way to see this is using the isomorphism \begin{align}\mathbb{C}S_n&\cong\bigoplus_i M_{n_i}(\mathbb{C}),&(1)\end{align}
which is proved using Maschke's Theorem and the Wedderburn-Artin Theorem (I assume you know this). Throughout, I'll write $c_\lambda=c_{\lambda,T}$.
Starting with a fact about the right-hand side.

Lemma: Let $A=M_n(\mathbb{C})\oplus M_n(\mathbb{C})$. If $I$ is a left ideal of $M_n(\mathbb{C})$ and $I\oplus 0\cong 0\oplus I$ as a left $A$-module, then $I=(0)$.  

It follows that

Corollary: If $I$ is any nonzero ideal of $A=\bigoplus_i M_{n_i}(\mathbb{C})$ with support in $M_{n_i}(\mathbb{C})$ for some $i$, then $I$ is not is isomorphic (as a left $A$-module) to any left ideal $J$ with support in $M_{n_j}(\mathbb{C})$, $j\neq i$.

Now, turning to the right-hand side of the isomorphism, recall the minimal left ideal $\mathcal{I}_\lambda=\mathbb{C}S_n c_\lambda$. 

Proposition: If $w\in S_n$, then the map
  $$
\rho_w:\mathcal{I}_\lambda\to \mathcal{I}_\lambda w, \;\;\;\rho_w(x)=xw
$$
  is an isomorphism of left $\mathbb{C}S_n$-modules with inverse $\rho_{w^{-1}}$.

Now, since $\mathcal{I}_\lambda$ is a minimal left ideal, its image on the right-hand side of the isomorphism (1) has support in a single column of $M_{n_i}(\mathbb{C})$ for some $i$. For $w\in S_n$, $\mathcal{I}_\lambda w\cong \mathcal{I}_\lambda$. Since $\mathcal{I}_\lambda w$ is a minimal left ideal, its image under (1) has support in a single column of $M_{n_j}(\mathbb{C})$ for some $j$. Therefore, by the corollary, the image of $\mathcal{I}_\lambda w$ under isomorphism (1) has support in a single column of $M_{n_i}(\mathbb{C})$ for the same $i$ (it is worth noting at this point a standard linear algebra fact: right multiplication by an invertible matrix implements a combination of elementary column operations). In particular,

Proposition: The image of $\sum_{\pi\in S_n} \mathcal{I}_\lambda \pi^{-1}$ under isomorphism (1) has support in $M_{n_i}(\mathbb{C})$ for some $i$.

Next, observe that $w_\lambda=\sum_{\pi}\pi c_\lambda \pi^{-1}\in \sum_{\pi\in S_n} \mathcal{I}_\lambda \pi^{-1}$. Therefore, the image of $w_\lambda$ under (1) has support in a single $M_{n_i}(\mathbb{C})$. This element is clearly central since, for $\sigma\in S_n$,
\begin{align}
\sigma w_\lambda&=\sum_\pi \sigma\pi c_\lambda \pi^{-1}\\
&=\sum_\pi \sigma\pi c_\lambda \pi^{-1}\sigma^{-1}\sigma\\
&=\sum_\pi (\sigma\pi) c_\lambda (\sigma\pi)^{-1}\sigma\\
&=w_\lambda \sigma.
\end{align}
Since the center of $\bigoplus_i M_{n_i}(\mathbb{C})$ consists of matrices of the form $\oplus_i d_iI_{n_i}$, it follows that the image of $w_\lambda$ under (1) is of the form $0\oplus dI_{n_i}\oplus 0$. In particular, $$w_\lambda^2=d^2w_\lambda.$$
To show that $d\neq 0$, it is enough to show that $w_\lambda\neq 0$. But, this follows by expressing $c_\lambda=a_\lambda b_\lambda$ as a linear combination of elements of $S_n$. Indeed, if $\sigma$ is in the row stabilizer of $T$ (so $\sigma$ appears in $a_\lambda$), then $\sigma^{-1}$ is not in the column stabilizer of $T$ (so $\sigma^{-1}$ does not appear in $b_\lambda$). Therefore, the coefficient of the identity element in $c_\lambda$ is $1$. Since no non-identity element of $S_n$ is conjugate to the identity, it follows that the coefficient of the identity in $w_\lambda$ is $n!$, showing that $w_\lambda\neq0$.
