I've been studying for my final exams, and I came across the following question:
If possible, give an example of a simple group $G$ with $n>1$ Sylow $p$-subgroups such that the order of $G$ does not divide $n!$. If not possible, briefly explain why.
Now, it didn't take me long to think of the following "numerical" example: a group of order $60=2^2\cdot3\cdot5$ where there are $3$ Sylow $2$-subgroups. My question is whether or not such a group exists, and if it does, is there any simple way to describe these groups based on knowing how the Sylow structure is built?
In trying to describe the above group, I think I showed that it can't exist. This is because there either have to be $4$ Sylow 3-subgroups or $10$ Sylow 3-subgroups. The second case leads to a contradiction since then there would have to be $k$ Sylow 5-subgroups such that $k(5-1)=30$, and likewise, in the first case, we find we'd have to have 13 Sylow 5-subgroups, also a contradiction. Is all of this correct and is there an easy way to determine whether such a group exists or not?