Solvable group of order $pqr^2$

$|G|=pqr^2$ where $3\leq p<q<r$ prime. Show that if $r>\frac{1}{2}(pq-1)$ then $G$ is solvable.

I took $H\leq G$ $r$-sylow subgroup of $G$, there is a theorem claiming that there exists a homomorphism from $G$ to $S_{pq}$ ($pq$ is the index of $H$).

If the homomorphism is injective then $G$ is a subgroup of $S_{pq}$, that means $pqr^2|(pq)!$ so $r^2|(pq-1)!$ but we know $r^2>2r>pq-1$ from what follows that the homomorphism can't be injective, that means there is a non trivial kernel which is a maximal subgroup of $H$ and a normal subgroup of $G$, it can be of order $r$ or $r^2$, lets name it K.

If $|K|=r^2$ then the sequence $\{e\}\triangleleft K\triangleleft G$ has a factor of order $r^2$ which is abelian as a square of a prime and a factor of order $pq$.

The other option is $|K|=r$, then the factors are of order: $r$ and $pqr$.

I am a bit stuck from here.

Solution 1:

Groups of order $r$ are solvable. Groups of order $pqr$ are solvable. So your group is an extension of two solvable groups, so solvable. In a group of order $pqr$, the Sylow $r$-subgroup is normal, and in the quotient of order $pq$, the Sylow $q$-subgroup is normal.

Solution 2:

You are left to show that a group of order $pqr$ is solvable, and this can be done exactly the same way you began the $pqr^2$ case.