Problem: I want to show that a simple group G of order 4*3*7*13 must be isomorphic to a subgroup of A_14 (the alternating group of 14 letters).
Attempt: It is easy to show that there are 14 sylow 13-subgroups. And so G act by conjugation on its sylow 7-subgroups with a trivial kernel (since G is simple). This shows that G is a subgroup of S_14 (the symmetric group of 14 letters). What's the next step? Thanks in advance.