We have a number of immediate consequences of the submodule theorem. First, and most important, the Specht modules form a complete list of irreducible modules for the symmetric group . We know that they’re irreducible, and that there’s one of them for each partition , which is the number of modules we’re looking for. But we need to show that the Specht modules corresponding to distinct partitions are themselves distinct. For this, we’ll use a lemma.
If is a nonzero intertwinor, then . Further, if , then must be multiplication by a scalar. Indeed, since there must be some polytabloid with . We decompose , and extent to all of by sending every vector in to . That is:
where the are -tableaux. Now, the can’t all be zero, so we must have at least one -tableau so that . But then our corollary of the sign lemma tells us that , as we asserted!
Further, if , then our other corollary shows us that for some scalar . We can thus calculate
and so multiplies every vector by .
As a consequence, the must be distinct for distinct permutations, since if then there is a nonzero homomorphism , and thus . But the same argument shows that , and thus .
More particularly, we have a decomposition
where the diagonal multiplicities are . The rest of these multiplicities will eventually have a nice interpretation.
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 .