Q:

No group of order 400 is simple

I've been given the question of showing that no group of order $400$ is simple. I've tried to attack it via the Sylow theorems for about a week now, but all the tricks and methods I know seem to be failing horribly.
Things I've tried:
Trying to produce a contradiction by giving a map into $S_n$ by elements acting by conjugation on Sylow 5-subgroups doesn't work, since there are 16 such Sylow 5-subgroups, and 400 divides $16!$, so it might very well be an injection and therefore we can't obviously find a nontrivial kernel.
Trying element counting is messy and I can't get it to come out the way I want- for instance, we can show that each of the Sylow 5-subgroups is isomorphic to $\mathbb{Z}_5\times \mathbb{Z}_5$, so there should be at least 125 elements of order divisible by only 5, but I can't see this producing a contradiction with any of the  things I can find out about Sylow 2-subgroups.
Anyways, I'm probably missing something fairly obvious, and I would appreciate any hints, solutions, or other help that you could give.

A:

Here is an outline for a solution. 
First of all, $|G| = 400 = 2^4 \cdot 5^2\ $. By Sylow's theorem we know that the number of Sylow 5-subgroups must be a divisor of $2^4$ and that it is $1$ modulo $5$. Thus it is either $1$ or $2^4$. If there is only one Sylow 5-subgroup, it must be normal. 
For the other case, suppose first that the intersections of different Sylow 5-subgroups are always trivial. By counting elements you can conclude that $G$ has exactly one Sylow 2-subgroup, which is then normal. 
If we have Sylow 5-subgroups $P$ and $Q$ such that $P \cap Q \neq \{1\}$, then $|P \cap Q| = 5$. Therefore $P \cap Q$ is normal in $P$ and $Q$, and thus is normal in the subgroup $\langle P, Q \rangle$ generated by $P$ and $Q$. Finally, show that either $\langle P, Q \rangle$ is normal in $G$ or equals $G$.

