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I'm trying to understand the following decomposition of $M^{(n-2,2)}$

here is the part of the article with the relevant decomposition

I don't understand why the last representation (which is $\frac{n(n-3)}{2}$-dimensional) is irreducible.

help very appreciated.

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    Well, $M^{(n-2,2)}$ has simple summand $S^{(n-2,2)}$ which has dimension $n(n-3)/2$. Having identified the compliment, the author is done. – 2017-02-14
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    But why is it irreducible? – 2017-02-14

1 Answers 1

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Using general theory, we know that $M^{(n-2,2)}$ contains a copy of the $n(n-3)/2$-dimensional representation $S^{(n-2,2)}$. Now, the author finds an $n$-dimensional submodule isomorphic to $M^{(n-1,1)}=S^{(n)}\oplus S^{(n-1,1)}$.

Using general theory again, the compliment of $M^{(n-1,1)}$ is an $S_n$-submodule. Since $\dim M^{(n-2,2)}={n\choose 2}$, this compliment has dimension ${n\choose 2}-n=\frac{n(n-3)}{2}$. Since there is no room left for anything else, this compliment must be the irreducible representation $S^{(n-2,2)}$.