No Simple Group of Order 144 I see the proof here: http://crazyproject.wordpress.com/2010/07/17/no-simple-groups-of-order-144-525-2025-or-3159-exist/, but I can't follow it, so could someone please explain to me how this proof works?  Or maybe offer an alternative proof?
 A: Both of the things you're asking about could be classified as part of the game of eliminating possibilities. No doubt you already knew that, but maybe the specific strategies they applied were things you hadn't thought about before.
The $n!$ trick eliminates subgroups of low index:
Suppose $G$ had a subgroup $H$ of index $k$ with $1<k\leq 5$. Then $G$ can act on the cosets of $H$ in the obvious way. If the group were simple, then the kernel of the action would be trivial, and $G$ would have to embed into $Sym(k)$. But the proof is pointing out that $G$ can't embed in $Sym(k)$ since $|G|\not||Sym(k)|=k!$
In this way, they are able to eliminate the $3$ and $4$ case. The reason he said five and not six is that $2^3|6$. S/He could just as well have said "$4!$" and eliminated the $3$ and $4$ case in the same way.
Looking at intersections of $p$-Sylows can be a good way to force a large normalizer
The index of $N_G(P_3\cap Q_3)$, must be in $\{1,2,4,8\}$ considering the order of $G$ and that the normalizer contains $3$-Sylow subgroups. We eliminated $2,4$ in the last paragraph. $1$ doesn't work because if that were the case, you would have a normal subgroup of order $3$, which when multiplied by any $2$-sylow would result in a subgroup of index $3$ (not allowed.) (The author of that page doesn't seem to have eliminated this case.) So the only option left is $8$.
Applying Sylow theory again to the normalizer, there can be only one $3$-Sylow: but this contradicts the fact that both $Q_3$ and $P_3$ are there. So all the alternatives are exhausted, and $G$ must have a normal $2$-Sylow subgroup or a normal $3$-Sylow subgroup.
So again: what's useful about $p$-Sylow intersections? At least in small cases like this ($p^2$) it's an easy way to pull those two Sylows into a single subgroup of $G$ which share a normal subgroup. This generally produces a group of smaller order which you can reanalyze. If it is in fact the whole group, then you have a normal subgroup with which to produce other subgroups, hopefully of low index (because you hopefully have already ruled out low index subgroups.)
In case you haven't read it before, there is a refined version of the basic Sylow theorems that takes into consideration intersections of Sylows. You can find the statement here on page 59. I didn't see that it was applicable in this particular case, but it's still worth knowing.
