I need to show that there are no simple groups of order $945$.
I've tried the regular method using the Sylow theorems.
$|G|=945=3^3\cdot5\cdot7 $
If $G$ is simple then there should be 7 Sylow-3 groups ; 21 Sylow-5 groups and 15 Sylow-7 groups. Even if they would all intersect trivially, there will still be no contradiction.
Any ideas?