The Submodule Theorem
Sorry for the delay.
Let be a submodule of one of the Young tabloid modules. Then I say that either contains the Specht module , or it is contained in the orthogonal complement . In particular, each Specht module is irreducible, since any nontrivial submodule cannot be contained in the orthogonal complement, and so it must also contain — and thus be equal to — .
To see this, let be any vector, and let be a -tableau. Our last result showed us that for some . First, we’ll assume that there is some pair of and so that . In this case, . Since generates all of — the Specht modules are cyclic — we must have .
On the other hand, what if there is no such pair of and ? That is, for all vectors and -tableaux . We can calculate
So each vector is orthogonal to each polytabloid . Since the polytabloids span , we must have .