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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?

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    $\begingroup$ 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
    Commented Jun 24, 2013 at 5:15
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    $\begingroup$ $({\bf Z}/2{\bf Z})^n$. $\endgroup$
    – Noam D. Elkies
    Commented Jun 24, 2013 at 5:27
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    $\begingroup$ 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
    Commented Jun 24, 2013 at 6:04

4 Answers 4

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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 $\left| G \right|$.

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    $\begingroup$ Another reference is Section 13.1 of Serre's Linear Representations of Finite Groups. $\endgroup$
    – Steven Sam
    Commented Jun 24, 2013 at 5:46
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    $\begingroup$ An advantage of Topics in Galois Theory for this forum is that it's freely available online (click the link in my answer). $\endgroup$
    – Noam D. Elkies
    Commented Jun 24, 2013 at 5:52
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    $\begingroup$ 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 $\zeta_n \mapsto \zeta_n^m$. $\endgroup$
    – Qiaochu Yuan
    Commented 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
    Commented Jun 24, 2013 at 15:03
  • $\begingroup$ @Joël: You are correct. $\endgroup$
    – user6976
    Commented Jun 24, 2013 at 16:01
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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).

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    $\begingroup$ Note that Noam's comment $({\mathbb Z}/2)^n$ is a Weyl group! $\endgroup$
    – Allen Knutson
    Commented Jun 24, 2013 at 9:10
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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.

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    $\begingroup$ 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
    Commented 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. $\endgroup$
    – zibadawa timmy
    Commented Apr 17, 2017 at 23:28
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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.]

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