Let $G$ be a group of order $90$. Show that $G$ is solvable.
Given that $90=2 \cdot 5 \cdot 3^2$ is of the type $pqr^2$, with $p, q,$ and $r$ primes, this case is not straightforward.
My attempt to find a solution: The number of Sylow 5-subgroups of $G$, $n_5$, is either $1$ or $6$.
Case 1: $n_5=1$. In the first case, the only subgroup (say $K_5$) is normal. Then, $G$ will be solvable if both $K_5$ and $G/K_5$ are solvable. $K_5$ is solvable as it has order 5. $G/K_5$ is also solvable because its order is $90/5=18=2\cdot 3^2$ (of the type $pq^2$); therefore $G$ is solvable.
Case 2: $n_5=6$. There are 6 Sylow 5-subgroups which are not normal and I don't know how to continue from here.
Similar arguments with $r=3$ made me get stuck as well when $n_3=10$.
Thanks for your help.