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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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    Would it be correct to say that, of the two maps you speak of, *both* $f\circ g$ and $g\circ f$ are isomorphisms? Because that would imply $f$ and $g$ are themselves both isomorphisms, categorically even. (I ask because I am not familiar with the algebra $\mathbb{C}S_d$.)2012-02-29
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    Ah, that sounds right - I was about to write that right multiplication by $b_\lambda$ is not a well-defined map from $Aa_{\lambda}b_{\lambda}$ to from $Ab_{\lambda}a_{\lambda}$, but $Aa_{\lambda}b_{\lambda} \subset A$, so multiplication by $b_\lambda$ is indeed well-defined.2012-02-29
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    Hm, but then it's not immediately clear to me that multiplication by $b_\lambda a_\lambda$ is an isomorphism...2012-02-29
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    The first map must be injective, and since everything is finite-dimensional, must also be surjective. So the first map is an isomorphism (Similar reasoning works for the second map).2012-02-29
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    Also, out of curiosity, which book are you using? I am making a short list of nice books on rep. theory of symmetric groups.2012-02-29
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    I'm not sure about that logic, since that injectivity implies surjectivity only for linear maps on finite dimensional vector spaces. In this case, right multiplication by $a_\lambda$ does not appear to be a linear map. Also, the text is Fulton and Harris.2012-02-29
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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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