Show that the group of order 765 is abelian (Hint: let G act by conjugation on a normal Sylow p subgroup)
The Attempt at a Solution
By using Sylow`s third theorem, I have calculated that the number of Sylow-3 subgroups and Sylow 17 subgroups is both 1; so both of them are normal subgroups of G. I just don't really understand how to proceed from here; by G acting on either one of those subgroups; I get a group homomorphism G->S_Q (where Q is either the Sylow 17 subgroup or Sylow 3 subgroup). I believe that eventual goal is to show that G is cyclic.
Some random thoughts in my head: Aut(Q)=C_16 (Q being the 17-Sylow subgroup). Much thanks and any help is appreciated