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][1]). 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 $n_\lambda \in \mathbb{R}$?

Alternatively, is there a (published) reference for the construction of $\omega_\lambda$ by such averaging operation?

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.


  [1]: http://www.math.toronto.edu/murnaghan/courses/mat445/Symmetric.pdf