Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 :)

share|improve this question
    
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. –  Jim Humphreys Jun 8 '11 at 13:48
2  
@Jim: The Sylow 2-subgroups of the symmetric groups are iterated wreath products, and these are rational groups. –  Frieder Ladisch Jun 8 '11 at 14:41
    
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. –  Jim Humphreys Jun 8 '11 at 15:36
1  
An almost-simple rational group has rational Sylow 2-subgroups by Feit–Seitz and a quick check. –  Jack Schmidt Jun 9 '11 at 13:54
add comment

2 Answers 2

up vote 4 down vote accepted

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

share|improve this answer
1  
@Mark Lewis: Thanks Mark for the comment. May I ask you to inform me the related paper? :-) –  Babak Sorouh Sep 24 '11 at 17:36
    
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. –  Mark Lewis Nov 21 '11 at 19:34
    
The paper is online, but not in print yet, I think: springerlink.com/content/e6428q435750326l –  Steve D Dec 4 '11 at 18:43
add comment

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).

share|improve this answer
    
@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". –  Babak Sorouh Jun 8 '11 at 16:39
    
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. –  Geoff Robinson Jun 8 '11 at 17:23
    
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. –  Geoff Robinson Jun 8 '11 at 19:37
    
@Geoff:Thanks for sharing thoughts. :) –  Babak Sorouh Jun 9 '11 at 16:58
add comment

Your Answer

 
discard

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.