# Feit-Thompson conjecture

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?

• 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 9:45
• 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
• A small amount of information is available here: en.wikipedia.org/wiki/Feit–Thompson_theorem#Step_3._The_final_contradiction – theHigherGeometer Sep 13 '17 at 11:35
• @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
• @JoséHdz.Stgo. cec.yamanashi.ac.jp/~ring/japan/abstract2017.pdf What's the status of the conjecture today? – Thibaut Dumont Dec 24 '17 at 13:19