Definition of the Young Symmetrizer

Solution 1:

Let $A = \Bbb{C}[S_n]$, we wish to prove that $Aa_Tb_T \cong Ab_Ta_T$ as left $A$ - modules. To do this we need the crucial fact that $a_Tb_Ta_Tb_T = n_T (a_Tb_T)$ for some constant $n_T \neq 0$. This comes upon knowing that $Aa_Tb_T$ is an irreducible representation of $S_n$ and using Schur's Lemma. Now define maps

$$f : Aa_Tb_T \stackrel{\cdot \frac{a_T}{\sqrt{n_T}}}{\longrightarrow} Ab_Ta_T \hspace{5mm} \text{and}\hspace{5mm} g: Ab_Ta_T \stackrel{\cdot \frac{b_T}{\sqrt{n_T}}}{\longrightarrow} Aa_Tb_T$$

which are simply just right multiplication by $a_T/\sqrt{n_T}$ and $b_T/\sqrt{n_T}$ respectively. Note that division by $n_T$ is ok because we are just dividing by a number. Then for any $x \in Aa_Tb_T$, write $x = y a_Tb_T$ for some $y \in A$. Then $$g(f(x)) = \frac{ya_Tb_Ta_Tb_T}{n_T} = \frac{n_T \cdot (ya_Tb_T)}{n_T} = x$$ and similarly for any $z \in Ab_Ta_T$, we find $f(g(z)) = z$. It follows that $f$ and $g$ define mutual left $A$ - module inverses and so $Aa_Tb_T \cong Ab_Ta_T$.