Feit–Thompson theorem

It is quite mysterious that every finite group of odd order is solvable. You see that odd factorial order is even number. Prime number is included in this theory.

3!/2=3

5!/2=60

7!/2=2520

9!/2=181440

11!/2=19958400

13!/2=3113510400

15!/2=653837184000

17!/2=177843714048000

19!/2=60822550204416000

21!/2=25545471085854720000


3,5,7,11,13,17,19 are prime numbers.



G is a group, and P and Q are odd primes. (P < Q)


(1+KQ) is the subgroups of G, so it is a divisor of ｜G｜ . K is an integer.
Therefore, P^2 is divided by (1+KQ).


∴

(1+KQ)=1,P or P^2







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