A Lagrangian group

I have a non-Lagrangian group $G$ of order $pq^3$, $Q$ a Sylow $q$-subgroup of G and a $H$ a subgroup of $Q$ with $|H|=q^2$. It is clear that $Q \subseteq N_G(H)$. I must prove that $G$ doesn't posses subgroups of order $pq^2$.

I supposed that $G$ posses a subgroup $P$ of order $pq^2$. $P$ will posses a normal Sylow $p$-subgroup or a normal Sylow $q$-subgroup. In the first case, $P$ will contain a subgroup of order $pq$ , making $G$ a Lagrangian group, a contradiction. But I don't find a way for the second case.

• What is a non-Lagrangian group? – Alexander Gruber Oct 18 '14 at 17:42
• I would guess that it is a finite group that does not have subgroups of orders $d$ for all divisors $d$ of the group order. But the question is strange, because $Q$ and $H$ are defined, but then nothing further is said about them! – Derek Holt Oct 18 '14 at 21:14

Suppose $P \le G$ with $|P|=pq^2$, $|G|=pq^3$. As you say, if $P$ has a normal Sylow $p$-subgroup, then $G$ has subgroups of all possible orders.
So suppose that $P$ has a normal Sylow $q$-subgroup $H$. Since $H$ is normal in both in $P$ and in a Sylow $q$-subgroup of $G$, we must have $H \lhd G$. Let $R$ be a Sylow $p$-subgroup of $P$. If $|N_P(R)|>p$, then $N_P(R)$ contains a subgroup of order $pq$.
Otherwise, by the Frattini argument, we have $|N_G(R)|=q|N_P(R)|=pq$.
• Still, the Frattini argument states that if $G$ is a group which contains a normal subgroup $K$ and $P$ is a Sylow p-subgroup of $K$, we have $G=N_G(P)K$. How did you apply it? – Alchimist Oct 19 '14 at 11:37