How do I show that every group of order 90 is not simple? Can comeone help me? How do I show that every group of order 90 is not simple?

This is a very very late answer. But anyway I'll post for the use of people who are learning form this site (like myself).
First note that $90=2\times 3^2\times 5$ and of the form $pq^2r$ with three different primes (like $60=2^2\times 3\times 5$). Therefore this case is bit hard comparing with common exam/ homework problems, where we have the orders of the form $p, p^2, p^3, pq, pq^2, pqr.$
Assume a group $G$ of order $90$ is simple. Using Sylow theorems, with standard notations, it is not difficult to prove that $n_3=10$ and $n_5=6.$ So we have $6\times(5-1)=24$ elements of order $5$ and if $3$-Sylow subgroups intersects trivially then we have $10\times(9-1)=80$ elements of order $3,$ but this contradicts to the order of the group as $24+80\gt 90.$ Therefore $3$-Sylow subgroups must have a non-trivial intersection.
Let $H, K\in Syl_3(G),$ then $|H\cap K|=3$ and $$|HK|=\dfrac{|H||K|}{|H\cap K|}=27.$$ Further $[H :|H\cap K|]=[K:|H\cap K|]=3$ is the smallest prime factor of both $|H|, |K|.$ Therefore $H\cap K$ is normal in both $H$ and $K.$ Now let us look at the  narmalizer $N=N_G(H\cap K)$ of the intersection in $G.$ It is clear that $HK\subset N$ and therefore we have $|N|$

*

*$\ge27$

*divides $90$

*divisible by $9.$
Hence we have only two possibilities $|N|\in\{45, 90\}.$
If $N=G,$ then $H\cap K$ is normal in $G.$ Otherwise $[G:N]=2$ and then $N$ is normal in $G.$ Therefore we have a contradiction with our assumption and hence $G$ can not be simple.
