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This might be a silly question, but I don't understand why the solution to the following problem implies the result:

Let $A = \mathbb{C}S_d$ and let $c_{\lambda}$ denote the Young symmetrizer (with $c_{\lambda} = a_{\lambda}b_{\lambda}$). The problem asks to show that $Aa_{\lambda}b_{\lambda} \cong Ab_{\lambda}a_{\lambda}$. The solution says to consider the maps from $Aa_{\lambda}b_{\lambda} \to Ab_{\lambda}a_{\lambda} \to Aa_{\lambda}b_{\lambda}$ given by multiplication by $a_\lambda$ and $b_\lambda$ respectively. The composite is a scalar multiplication on a vector space (since $c_\lambda$ is idempotent) and hence an isomorphism.

I understand all this argument, but how does it follow from this that implies the needed isomorphism? A composite map on an vector space that is an isomorphism certainly does not imply that the component maps are isomorphisms...

edit: Also, it seems that every element $Aa_\lambda b_\lambda$ can be written as $e_g e_p e_q$ where $e_g \in A$, $e_p$ and $e_q$ are components of the sums $a_\lambda$ and $b_\lambda$. So can I just write an explicit map $e_g e_p e_q \mapsto e_g e_q e_p$? Is it right to sa that this doesn't work because this is a bijection but not an isomorphism?

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    Josef, multiplication by $a_\lambda$ (like by any element of the group algebra) _is_ a linear map of finite dimensional vector spaces.2012-03-02

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