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 view of point. 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!

  • $\begingroup$ I guess you mean: $\mathbb{C}\mathfrak{S}_d$-modules (it's important to specify the field). $\endgroup$ – YCor May 29 '16 at 21:24

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.

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  • $\begingroup$ Thank you very much for your comments. However, I am still confused about it. It is not clear to me that every irreducible module can be found in permutation modules, namely, the induction from Young subgroups (Bruce E. Sagan's book ""The Symmetric group""). Anyway, thanks! $\endgroup$ – Steven May 29 '16 at 19:18
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    $\begingroup$ The regular representation is a permutation module in the sense you mean, it is induced from the trivial representation of the the trivial subgroup, which is the Young subgroup associated to the partition (1,1,1,...,1). $\endgroup$ – Nate May 29 '16 at 21:20
  • $\begingroup$ Oh! I understand it now. Thank you very much! $\endgroup$ – Steven May 30 '16 at 4:40

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).

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  • $\begingroup$ Thank you very much for your comments. However, I am still confused about it. Here by permutation module I mean inductions from Young subgroups. I don't know why all irreducible modules can be found in this manner. $\endgroup$ – Steven May 29 '16 at 19:16
  • $\begingroup$ You asked for other examples. However, regarding Young modules explicitly, it was know to Frobenius that the character ring of the symmetric group is spanned by the characters of the permutation characters induced from Young subgroups, so all complex irreducible characters do show up. This also implies that all irreducible characteristic $p$-modules show up as composition factors of such permutation modules. $\endgroup$ – Geoff Robinson May 29 '16 at 19:55
  • $\begingroup$ This is interesting. Thank you very much! $\endgroup$ – Steven May 30 '16 at 4:41

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