28
$\begingroup$

The Symmetric groups $S_n$ has interesting property that all complex irreducible characters are rational (i.e. $\chi(g)\in \mathbb{Q}$ for all $\mathbb{C}$-irreducible characters $\chi$,$\forall g\in S_n$).

Question: What are other families of (finite) groups where all complex irreducible characters are rational? Are such (finite) groups characterised?

$\endgroup$
3
  • 1
    $\begingroup$ Of course $D_8$, $Q_8$ are other examples, but I would like to know a *family of groups with above property. In such groups, every conjugacy class is rational conjugacy class (see mathoverflow.net/questions/10635/…) $\endgroup$
    – Philip
    Jun 24, 2013 at 5:15
  • 1
    $\begingroup$ $({\bf Z}/2{\bf Z})^n$. $\endgroup$ Jun 24, 2013 at 5:27
  • 8
    $\begingroup$ There was a whole Springer Lecture Notes on the topic: D. Kletzing, Structure and Representations of Q -groups, Lecture Notes in Math. 1084, Springer-Verlag, Berlin, 1984. Some more recent papers on the topic can be found in the (freely available) references to link.springer.com/article/10.1007%2Fs00013-010-0110-8 $\endgroup$
    – YCor
    Jun 24, 2013 at 6:04

4 Answers 4

34
$\begingroup$

Here's one characterization that I learned from Serre (see Definition 7.1.1 in his Topics in Galois Theory (p.65)): an element $g$ of a finite group $G$ satisfies $\chi(g) \in {\bf Q}$ for all characters $\chi$ iff $g$ is conjugate in $G$ to $g^m$ for all $m$ relatively prime to the exponent $e(g)$. [If $m$ is not coprime to $e(g)$ then $e(g^m) \lt e(g)$ so $g^m$ cannot possibly be conjugate to $g$.] It is enough to check this for all $m$ relatively prime to $\left| G \right|$. In particular, all character values are rational iff every group element is conjugate to its $m$-th power for all $m$ coprime to $\left| G \right|$.

$\endgroup$
7
  • 1
    $\begingroup$ Another reference is Section 13.1 of Serre's Linear Representations of Finite Groups. $\endgroup$
    – Steven Sam
    Jun 24, 2013 at 5:46
  • 4
    $\begingroup$ An advantage of Topics in Galois Theory for this forum is that it's freely available online (click the link in my answer). $\endgroup$ Jun 24, 2013 at 5:52
  • 6
    $\begingroup$ More generally, if a group element $g$ of order $n$ is conjugate to $g^m$ with $(n, m) = 1$ then $\chi(g) \in \mathbb{Q}(\zeta_n)$ lies in the fixed field of the Galois automorphism $\zeta_n \mapsto \zeta_n^m$. $\endgroup$ Jun 24, 2013 at 6:24
  • $\begingroup$ @Mark Sapir: Apparently, the link given by Noam points to notes from a course of Serre taken by Darmon and made by him publicly available, presumably with Serre's consent. The link you give, however, points to a pirate copy of Serre's published book very likely put there without his or his publisher's consent. $\endgroup$
    – Joël
    Jun 24, 2013 at 15:03
  • $\begingroup$ @Joël: You are correct. $\endgroup$
    – user6976
    Jun 24, 2013 at 16:01
20
$\begingroup$

All Weyl groups have this property. So as for families, the hyperoctahedral groups (signed permutations), and their index 2 subgroups (elements defined by having an even number of sign changes).

$\endgroup$
1
  • 5
    $\begingroup$ Note that Noam's comment $({\mathbb Z}/2)^n$ is a Weyl group! $\endgroup$ Jun 24, 2013 at 9:10
16
$\begingroup$

All groups that be constructed from symmetric groups via cross products and wreath products have this property. See Section 3 of my paper "Mass formulas for local Galois representations to wreath products and cross products" http://arxiv.org/pdf/0804.4679v1.pdf

So, for example, $((S_7 \wr S_4) × S_3) \wr S_8$ has a rational character table. In fact, taking cross products and wreathing with $S_n$ preserves the property you are asking about (see above reference).

This includes several of the examples given: $(\mathbb Z/2\mathbb Z)^n,$ hyperoctahedral groups, and Sylow 2-subgroups of $S_n$. I am not sure if the index 2 subgroups of hyperoctahedral groups can be constructed from symmetric groups via cross products and wreath products.

$\endgroup$
2
  • 2
    $\begingroup$ What do you call "cross product"? Also does "stable by wreath products" means stable by the operations $G\mapsto G\wr S_n$, where $G\wr S_n$ is by definition the obvious semidirect product $G^n\rtimes S_n$? $\endgroup$
    – YCor
    Jun 24, 2013 at 21:20
  • $\begingroup$ @YCor Looking at the paper, apparently "cross product" means "direct product", and the wreath product is what you expected it to be. $\endgroup$ Apr 17, 2017 at 23:28
11
$\begingroup$

Sylow $2$-subgroups of the symmetric group $S_n$ of degree $n$ are rational. There was a longstanding conjecture on rational groups saying that Sylow $2$-subgroups of a rational group are also rational. This has been refuted by I. M. Isaacs and G. Navarro in [Sylow 2-subgroups of rational solvable groups, Mathematische Zeitschrift, December 2012, Volume 272, Issue 3-4, pp 937-945.]

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.