I have to show that there is no simple group of order $96$ using the sylow theorems.
I know that $96 = 2^5\cdot 3$, and from the third sylow theorem $n_2 = 1$ or $3$ and $n_3 = 1$ or $4$ or $16$.
I have seen some proofs of this statement that use a so called index factorial theorem but I haven't learned about this. Is there a way to prove this using just the sylow theorems?