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As we know, a finite group $G$ is a rational group if $\chi (g)\in\mathbb{Q}$, where $\chi$ is every irreducible charahter and $g\in G$. I have an interesting question that is "Is 2-Sylow subgroup of a rational group also a rational group?"

Any hints will be appreciated :)

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  • $\begingroup$ The answer seems to be no. Keep in mind that symmetric groups are rational groups, so that's the first case to investigate starting with $S_4$ and its $2$-Sylow subgroups. $\endgroup$
    – Jim Humphreys
    Commented Jun 8, 2011 at 13:48
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    $\begingroup$ @Jim: The Sylow 2-subgroups of the symmetric groups are iterated wreath products, and these are rational groups. $\endgroup$
    – Frieder Ladisch
    Commented Jun 8, 2011 at 14:41
  • $\begingroup$ Yes, that sounds familiar now that you mention it. But I still wonder whether the questioner has begun by looking at cases where the given group is rational. In any case, Geoff's answer is quite useful. $\endgroup$
    – Jim Humphreys
    Commented Jun 8, 2011 at 15:36
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    $\begingroup$ An almost-simple rational group has rational Sylow 2-subgroups by Feit–Seitz and a quick check. $\endgroup$
    – Jack Schmidt
    Commented Jun 9, 2011 at 13:54

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This had been a long standing conjecture, but it has now been answered negatively. Isaacs and Navarro have found a counterexample.

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    $\begingroup$ @Mark Lewis: Thanks Mark for the comment. May I ask you to inform me the related paper? :-) $\endgroup$
    – Mikasa
    Commented Sep 24, 2011 at 17:36
  • $\begingroup$ Sorry for taking so long to reply. I don't think the paper of Isaacs and Navarro has appeared any where yet. (I have a copy of the preprint.) I will try to post a link when it does appear. $\endgroup$
    – Mark Lewis
    Commented Nov 21, 2011 at 19:34
  • $\begingroup$ The paper is online, but not in print yet, I think: springerlink.com/content/e6428q435750326l $\endgroup$
    – Steve D
    Commented Dec 4, 2011 at 18:43
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This is a fairly long-standing question in certain quarters, though I would need to check who was the first to ask it (if such a person is well-defined). Isaacs and Karagueuzian answered a somewhat related question in the negative (around 2002), disproving a conjecture of Kirillov. They proved that a Sylow $2$-subgroup of ${\rm GL}(13,2)$ is not a real group. (Recall that the definition of a real group is analagous to the definition of a rational group given in the question. A finite group is real if and only if all its complex irreducible characters are real-valued, which is equivalent to all its elements being conjugate to their inverses). However, I should point out that in my original post, I had mis-remembered the content of the Isaacs-Karagueuzian result. Contrary to my earlier statement, the group ${\rm GL}(13,2)$ is not itself a real group. For example, it contains an element of order $127$ which is not conjugate to its inverse. As far as I am aware, the given question about rational groups is open. One of the difficulties with the question is that rational groups are relatively rare (a loose statement, I know, but justifiable).

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  • $\begingroup$ @Geoff: @Jim: I point out that, we know some properties about rational group , for example every quotient of a rational group is rational group and Sylow 2-subgroup of symmetric groups are rational group. Also, we know that the question is a conjecture and is a very hard problem. W. Feit said that, "Probably this problem is wrong but still there is not counterexample for it". $\endgroup$
    – Mikasa
    Commented Jun 8, 2011 at 16:39
  • $\begingroup$ Yes, what you say is not inconsistent with what I said, I think. The composition factors of rational finite groups are almost completely understood ( see the MR review by A. Turull of the 2005 Proc LMS paper by P. Hegedus, for example). I almost agree with Feit, except that I could not predict that the conjecture is wrong: nevertheless, rational groups are few and far between, and it is difficult to make uniform statements about them, as their structure varies considerably. $\endgroup$
    – Geoff Robinson
    Commented Jun 8, 2011 at 17:23
  • $\begingroup$ Regarding Feit's statement, I just mean that I would not go as far. Until the conjecture is proved or disproved, I would feel unable to say that it is probably wrong-it may well be, but I don't know and can't say. $\endgroup$
    – Geoff Robinson
    Commented Jun 8, 2011 at 19:37
  • $\begingroup$ @Geoff:Thanks for sharing thoughts. :) $\endgroup$
    – Mikasa
    Commented Jun 9, 2011 at 16:58

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