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Suppose $\lambda$ is a partition of the natural number $n$ and $T$ is a standard Young Tableaux of shape $\lambda$. Let $$P_{\lambda}:=\lbrace g\in S_n:g\text{ preserves the rows of }T\rbrace$$ and $$Q_{\lambda}:=\lbrace g\in S_n:g\text{ preserves the columns of }T\rbrace.$$ If $$a_{\lambda}:=\sum_{g\in P_{\lambda}}e_g \in \mathbb{C}S_n,$$ $$b_{\lambda}:=\sum_{g\in Q_{\lambda}}sgn(g)e_g \in \mathbb{C}S_n,$$ then the Young symmetrizer is given by $$c_{\lambda}:=a_{\lambda}b_{\lambda}.$$ Basically these elements encode the entire unitary representation theory of $S_n$, which, by the way, I am rather familiar with, but I've recently been asked why these elements should be non-zero. Thinking about it, I could not come up with an elementary answer and looking at different proofs of the classification theorem did not help either. Hence, I decided to ask it here: Why is $c_{\lambda}$ non-zero?

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We can write $$c_\lambda =\sum_{g\in P_\lambda ,g^\prime\in Q_\lambda}sgn(g^\prime )\,e_{gg^\prime}$$

It is enough to show that coefficient of $e_{id}$ (where $id$ is the identity) in the above sum is non zero; in fact, we will show that the coefficient is $1$. This will follow from the fact that $gg^\prime =id$, for $g\in P_\lambda , g^\prime\in Q_\lambda$, has a unique solution; namely $g=id, g^\prime =id$. Otherwise, if $gg^\prime =hh^\prime=id$ for some $h$ and $h^\prime $, then $g/h=g^\prime /h^\prime$. But $g/h\in P_\lambda$ and $g^\prime /h^\prime\in Q_\lambda$ and therefore both belong to $P_\lambda\cap Q_\lambda$. But clearly $P_\lambda\cap Q_\lambda =\{id\}$, and this implies $g=g^\prime$ and $h=h^\prime$.

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  • $\begingroup$ Ah, yes, wonderful. Thank you very much! $\endgroup$ – Nephry Feb 2 '16 at 23:49

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