I want to show that the only subgroups of Sn (the symmetric group of n elements) containing Sn-1 are Sn and Sn-1. So essentially, all that's needed to be checked is that there is no subgroup of order greater than (n-1)!, the order of Sn-1 and less than n!, the order of Sn. I was first thinking that if there did exist such a subgroup of Sn, then it would contradict Lagrange's theorem, as then (n-1)! would not divide the order of said subgroup. However, that fails for large enough n, as the said subgroup may have order (n-1)*k, where 2<=k<n. Any ideas? Or should I just try a different approach? Can induction work here?