Feit and Thompson, in regards to the Feit-Thompson theorem, have said that proving this conjecture would simplify their proof of their theorem, “rendering unnecessary the detailed use of generators and relations.” In 1971, Stephens strengthened the conjecture to state that
always, and then found the counterexample , . The numbers and do have 112643 as their greatest common divisor, but dividing the former by the latter leaves a remainder of 149073454345008273252753518779212742886488244343395482423. No other counterexamples have been found to Stephen’s stronger version of the conjecture.
- 1 N. M. Stephens, “On the Feit-Thompson Conjecture” Math. of Computation 25 115 (1971): 625
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