I want to show that no group of order $90=2 \cdot3^2 \cdot5 $ is simple. Part (iii) of Sylow's Theorem gives
$$n_2 \in\{1,3,5,9,15,45\} $$ $$n_3 \in \{1,10\} $$ $$n_5 \in \{1,6\} $$
I am aware of the solution that appears here, but I wonder if a more "brute force" approach is possible, as follows:
Assuming $G$ is a simple group of order $90$, we have
$$n_2 \in\{3,5,9,15,45\} $$ $$n_3=10 $$ $$n_5=6 $$
G then has $6(5-1)=24$ elements of order $5$, and $n_2$ elements of order $2$. The remaining $66-n_2$ elements must be exactly those in the union $$\bigcup_{i=1}^{10} P_i $$ of the Sylow 3-subgroups.
We now try to estimate the size of that union: The inclusion-exclusion principle gives
$$\left| \bigcup_{i=1}^{10} P_i \right|=\sum_{n=1}^{10} (-1)^{n+1} \sum_{i_1 <i_2<\dots <i_k} \left|P_{i_1} \cap P_{i_2} \cap \dots \cap P_{i_k} \right| $$
Obviously, the $n=1$ term in the sum above is $90$, and every intersection has possible order $1$ or $3$ (According to Lagrange's Theorem). Is is possible to find the minimal value of this expression?
I thought about splitting each inner sum into two, the sum over trivial intersections, and intersections of size $3$. But the calculations seem hard to handle.
Thank you!
