Realization of irreducible $\mathfrak{S}_d$-modules and the representation theory of Lie algebra

Question

Let $n$ be a positive integer. It is well-known that a method to realize irreducible $\mathfrak{S}_d$-modules is to construct the so-called Specht modules $S^{\mu}$ which are submodules in the so-called $\textbf{permutation modules} \ \ M^{\mu}$ (see, e.g. Bruce E. Sagan's book ""The Symmetric group"")

Schur-Weyl duality supports such a point of view. Namely, the double centralizer property gives that $((\mathbb{C}^n)^{\otimes d})_{\mu}\cong M^{\mu}$ for all dominate weight $\mu$, i.e. $\mu$ is a partition, thus we may find all $\mathfrak{S}_d$-irreducible modules modules by decomposing $M^{\mu}$.

My question: Are there other reasons which explain why irreducible modules are in permutation modules? Thanks!

Answer 1

The irreducible modules of any finite group $G$ can be found in permutation modules, just because they can be found in the regular action of $G$ on itself.

Answer 2

As Dima Pasechnik says, every irreducible module for a finite group $G$ occurs as a composition factor of the regular representation of $G$, that is the permutation module coming from the regular permutation action of $G$ on itself. In this question Finite groups such that every irrep can be induced from trivial irrep of a subgroup ? , it is pointed out that the only finite groups which have complex irreducible modules not occurring in any permutation module on any non-trivial subgroup of $G$ are the so-called Frobenius complements.

Later note: In fact, over an algebraically closed field $F$ of characteristic $p$, every irreducible $FG$-module occurs as a composition factor of the permutation module on the cosets of a Sylow $p$-subgroup of $G.$ (In fact, each occurs both in the socle and the head of that permutation module).

Answer 3

This is a bit more specific than your question asked, but there's actually a beautiful relationship between the permutation modules $M^{\mu}$ and Specht modules $S^{\mu}$ (over $\mathbb C$): There is a unique (up to scalar) map $M^{\mu}\to S^{\mu}$ and the kernel of this map is spanned by the images of all maps $M^{\mu'}\to M^{\mu}$ for $\mu'> \mu$ in dominance order (or lexicographic order). In characteristic $p$, I don't think this is right any more, but it will be fixed if you consider the Schur algebra instead of $\mathbb{F}_p\mathfrak{S}_n$.

Alternatively, you can define signed permutation modules $\tilde{M}^{\nu}$, and there is a unique non-zero map up to scalar $M^{\mu}\to \tilde{M}^{\mu^t}$, and $S^{\mu}$ is the image of this map.