MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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…) – Philip Jun 24 '13 at 5:15
$({\bf Z}/2{\bf Z})^n$. – Noam D. Elkies Jun 24 '13 at 5:27
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 – YCor Jun 24 '13 at 6:04
up vote 29 down vote accepted

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 $\left| G \right|$.

share|cite|improve this answer
Another reference is Section 13.1 of Serre's Linear Representations of Finite Groups. – Steven Sam Jun 24 '13 at 5:46
An advantage of Topics in Galois Theory for this forum is that it's freely available online (click the link in my answer). – Noam D. Elkies Jun 24 '13 at 5:52
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 $\zeta_n \mapsto \zeta_n^m$. – Qiaochu Yuan Jun 24 '13 at 6:24
@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. – Joël Jun 24 '13 at 15:03
@Joël: You are correct. – Mark Sapir Jun 24 '13 at 16:01

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).

share|cite|improve this answer
Note that Noam's comment $({\mathbb Z}/2)^n$ is a Weyl group! – Allen Knutson Jun 24 '13 at 9:10

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"

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.

share|cite|improve this answer
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 $G^n\rtimes S_n$? – YCor Jun 24 '13 at 21:20

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.]

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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