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If a group $G$ acts on a set $X$, then we can speak of permutation representation on $K[X]$. now set of all $k$-subsets $X_k$ of $n={1,2,...,n}$ is a $S_n$ - set and we can speak about permutation representation on $K[X_k]$. by decomposing it we get $[k/2]$ irreducible representations $V_0, . . .,V_{[k/2]} $ for k less than or equal to $n/2$. Similarly we can do this for any partition of n instead of (k,n-k) which we consider in the above case and we obtain all irreducible representation of $S_n$ with the assumption that $K[X]$ is completely reducible. I am looking for any other method (instead of permutation representation) which give as all irreducible representations of $S_n$ and its character values in simpler way. Is there any method like this? thank you.

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What is $K$?${}$ – Gerry Myerson Apr 14 '13 at 11:05
any field which is algebraically closed. – GA316 Apr 14 '13 at 11:11
we have assumed that K[X] is completely reducible. – GA316 Apr 14 '13 at 11:11
thanks. Is there any way to get simple representations? – GA316 Apr 14 '13 at 12:09
Sorry, but your method is not clear from me... how do you get the non-hook irreps? Anyway, you are aware of the fact that if we fix a faithful rep $V$ of the finite group $G$, then every irrep of $G$ is a direct summand of some tensor power of $V$ ? – darij grinberg Apr 14 '13 at 16:04

Since you only care about the completely reducible case, I'll assume $K=\mathbb{C}$.

The easiest way that I know of to construct irreducible $S_n$ representations is a special case of the construction here. In particular, symmetric group is a quotient of the (degenerate) affine Hecke algebra appearing there (obtained by setting $x_1=0$). If you take $\mu=0$ in the link, then the skew shape $\lambda/\mu$ is just $\lambda$ and the module that is constructed is the irreducible representation $S_\lambda$.

This recovers the character formula $s_\lambda=\sum_T x^T$, where the sum is over all standard tableaux of shape $\lambda$ (as described in Macdonald's text, for example).

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There is a general study of representations of symmetric groups via young tableaux. In particular there is a combinatorial expression for the characters. I think that's as simple as it gets currently.

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This is precisely the construction I linked to above. – David Hill Apr 27 '13 at 16:09

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