# How to complete this proof of the Schur-Weyl duality theorem using commutants?

I have a question on a proof from the following book :"Symmetry, representations and invariants by Roe Goodman and Nolan R. Wallach. Specifically on section 4.2.4 on Schur Weyl duality and the proof of the following statement

There are irreducible mutually inequivalent $$\mathfrak{S}_k$$-modules $$E^\lambda$$ and irreducible mutually inequivalent $$G :=GL(n,\mathbb{C})$$-modules $$F^\lambda$$ such that $$\bigotimes^k \mathbb{C}^n \cong \bigoplus_{\lambda} E^\lambda \otimes F^\lambda$$ as a representation of $$\mathfrak{S}_k \times G$$

In the book this is presented as a specific case of the the general duality theorem

I know from this theorem that there are irreducible, mutually inequivalent $$\mathcal{R}^G$$-modules $$E^\lambda$$ and irreducible, mutually inequivalent $$G := GL(n,\mathbb C)$$-modules $$F^\lambda$$ such that $$L := \bigotimes^k \mathbb C^n \cong \bigoplus_\lambda E^\lambda \otimes F^\lambda$$ as an $$\mathcal R^G \otimes \mathbb CG$$-module where $$\mathcal R^G := \{ T \in End(L) \vert \rho_k(g)T = T \rho_k(g) \quad \forall g \in G \}$$ is the commutant of $$\rho_k(G)$$ in $$End(L)$$ i.e. $$R^G = Comm(\rho_k(G))$$. Note that $$\rho_k$$ denotes the representation of $$G$$ on $$L$$

This is quite close to the final theorem. I need only prove that an $$R^G$$-module is the same as an $$\mathbb C\mathfrak{S}_k$$ module. Using the following theorem from the book

Theorem 4.2.10 Let $$\rho_k$$ and $$\sigma_k$$ denotes the representations of $$GL(n,\mathbb C)$$ and $$\mathfrak S_k$$ on $$\otimes^k \mathbb C^n$$ respectively. Then $$Comm(Span(\rho_k(G))) = Span(\sigma_k\mathfrak S_k)$$ and $$Comm(Span(\sigma_k \mathfrak S_k)) = Span \rho_k(G)$$

I can see that that $$\mathcal R^G = Span( \sigma_k \mathfrak S_k)$$ which is almost what I need. How can I conclude from here ?

Attempts:

A first idea would be to prove that $$Span \; \sigma_k \mathfrak S_k \cong \mathbb C \mathfrak S_k$$ as algebras. To do this I can consider the map

$$\sum_{s \in \mathfrak S_k} \mu_s s \mapsto \sum_s \mu_s \sigma_k(s)$$ and show that it is an algebra isomorphism. It is clearly surjective but I don't see how I could easily show injectivity. Suppose that $$\sum_s \mu_s \sigma_k(s) = 0$$ then to prove injectivity it suffices to show that the linear maps $$\sigma_k(s)$$ are linearly independent. But I don't see why this would be obvious.

A second idea would be to use the $$R^G$$ module structure on $$E^\lambda$$ to define a representation of $$\mathfrak S_k$$ on $$E^\lambda$$ by letting $$s$$ act on $$E^\lambda$$ by mapping $$v$$ to $$\sigma_k(s)v$$. This does turn $$E^\lambda$$ into an irreducible $$\mathbb C\mathfrak S_k$$-module but how can I be sure that these new $$\mathbb C \mathfrak S_k$$ modules are still mutually inequivalent ?

To complete the proof one must construct an $$\mathbb C\mathfrak S_k$$-module structure on $$E^\lambda$$. This can be done by using the $$Span\; \sigma_k(\mathfrak S_k)$$-module structure on $$E^\lambda$$.
$$\mathbb C \mathfrak S_k \rightarrow End(E^\lambda) : \sum_{s \in \mathfrak S_k} \lambda_s s \mapsto \sum_{s \in \mathfrak S_k} \lambda_s \sigma_k(s).$$
In this way the $$E^\lambda$$ are irreducible and non isomorphic as representation of $$\mathfrak S_k$$ since
$$E^\lambda \cong E^\mu \text{ as }\mathbb C \mathfrak S_k\text{-modules} \implies E^\lambda \cong E^\mu \text{ as Span } \sigma_k(\mathfrak{S}_k)\text{-modules.}$$