I want to show that a simple group of order 60 is isomorphic to $A_5$. In the process, I am stuck at the part in which I have to show that the number of Sylow 2 subgroups (whose orders are 4) cannot be 1 or 3; it therefore has to be 5 or 15.
I searched for all contents of Hungerford chapter 2, but am not able to prove this fact... Could anyone please help me?
If there is only one Sylow $p$-subgroup, then it is normal.
If there are $r$ of them, then $G$ acts by conjugation transitively on the set of $p$-Sylow subgroups. If $G$ is simple, this action must be faithful, and $G$ is isomorphic to a transitive subgroup of $S_r$.