No group of order 400 is simple

Solution 1:

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 $G$.