Given the information on the Sylow numbers for a simple group of order found in the previous exercise, prove that there is no such group. [Hint: Work in the permutation representation of degree 819.]
Recall from the previous exercise that , so that . Now acts on by conjugation. Moreover, since has a unique Sylow 409-subgroup, this action has precisely one orbit of order 1, namely . By the Orbit Stabilizer Theorem and Lagrange, the remaining orbits have order 3, 409, or . Since , no orbit has this order. If some orbit has order 409, then there are remaining elements of , which are distributed among orbits of order three. However, mod 3, so there must be another orbit of order 1 – a contradiction. If no orbit has order 409, then there are elements of in orbits of order 3; however, mod 3, again a contradiction. Thus no simple group of order exists.