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The Feit-Thompson conjecture states: If $p<q$ are primes, then $\frac{q^p-1}{q-1}$ does not divide $\frac{p^q-1}{p-1}$.

On page xiii of these proceedings of a conference at the University of Yamanashi (Japan) that took place in October of 2017, a proof of the Feit-Thompson conjecture was announced.

Questions:

  • I heard that this would greatly simplify the Feit–Thompson theorem on odd order groups. Can someone explain the simplification to someone who only is familiar with group theory on a basic algebra textbook level?

  • What other implications does the Feit-Thompson conjecture have?

José Hdz. Stgo.
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Mare
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    It would be helpful to state the conjecture: if $p<q$ are primes, then $(q^p-1)/(q-1)$ does not divide $(p^q-1)/(p-1)$. Feit and Thompson mention in their announcement "A solvability criterion for finite groups and some consequences" that this conjecture would simplify the proof of the odd order theorem by "rendering unnecessary the detailed use of generators and relations". I don't know what they mean by that. – Neil Strickland Sep 13 '17 at 09:45
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    @NeilStrickland You are right. I added the conjecture. – Mare Sep 13 '17 at 09:48
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    It would be very interesting what Georges Gonthier has to say on this. Seriously. You might contemplate asking him. (Though I cannot judge how relevant the Feit-Thompson conjecture is to Gonthier's project. Perhaps it is irrelevant.) – Peter Heinig Sep 13 '17 at 10:09
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    I would guess that the answer to your question "Can someone explain the simplification to someone who only is familiar with group theory on a basic algebra textbook level?" is almost certainly no. But it would be very interesting to have an explanation of the simplification that would be comprehensible to someone who was familiar with the basic structure of the proof of the odd order theorem. – Derek Holt Sep 13 '17 at 10:13
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    A small amount of information is available here: https://en.wikipedia.org/wiki/Feit–Thompson_theorem#Step_3._The_final_contradiction – David Roberts Sep 13 '17 at 11:35
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    @DavidRoberts's link: Feit–Thompson theorem: The final contradiction (which sounds like the name of a movie sequel). – LSpice Sep 19 '17 at 19:05
  • @Mare: The link to the proceedings is not working anymore. Can you fix it? – José Hdz. Stgo. Oct 21 '17 at 19:47
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    @JoséHdz.Stgo.

    http://www.cec.yamanashi.ac.jp/~ring/japan/abstract2017.pdf

    What's the status of the conjecture today?

     – Thibaut Dumont Dec 24 '17 at 13:19
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    @ThibautDumont: I really don't know. – José Hdz. Stgo. Aug 28 '18 at 20:06
  • Today I tried to write a strong form of this conjecture (thus if there aren't mistakes in the following statement or my experiments with a Pari/GP program, seems that Feit-Thompson conjecture should be a particular case of the following conjecture), but I don't know if it is in the literature or it has a good mathematical content (I don't know if it makes sense): There are no distinct square-free integers $p$ and $q$ (both greater than $1$ ) with $\gcd(p,q)=1$ such that $\frac{p^q-1}{p-1}$ divides $\frac{q^p-1}{q-1}$. Can you tell me anything about it professor @JoséHdz.Stgo. ? – user142929 Jan 30 '20 at 21:16

1 Answers1

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It is not true anymore that a proof of this conjecture would lead to significant simplifications. Peterfalvi proved in 1984 a weaker version of this conjecture, which suffices to get rid of the chapter involving generators and relations in the original paper. Bender and Glauberman reproduce this argument in their book in one of the appendices, and it takes only two or three pages. Surprisingly, although the statement is about the solvability of a certain equation in finite fields, the proof is group theoretic and not number theoretic in nature. Moreover, this statement does not follow in an obvious way from standard results in number theory such as Weil's estimates. Anyway, it is highly questionable whether replacing a short, elementary argument by a reference to a deep theorem should count as a simplification.

Jan-Christoph Schlage-Puchta
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