Let $G$ be a finite group and $p<q$ such that $p^2$ doesn't divide $|G|$. Let $H_p$ and $H_q$ be Sylow subgroups of $G$ with $H_p \lhd G$. Show $H_pH_q \lhd G \space \implies H_q \lhd G$.
From the hypothesis of the statement it follows $|G|=pq^mr$ with $(r:pq^m)=1$ ($r$ not necessarily prime number).
If $H_p \lhd G$, then $n_p=1$ where $n_p$ denotes the number of $p-$Sylows. I don't know what else to do in order to show what I am being asked. I would appreciate any hints.
I know that a characteristic subgroup is normal, but I am not so sure if $H_q$ is characteristic.
Look at the Sylow subgroups of $H_pH_q$. Apparently $H_q \in Syl_q(H_pH_q)$. $|H_pH_q|=pq^m$, and the number of Sylow $q$-subgroups of $H_pH_q$, must divide $p$. If it would be $p$, then according to Sylow theory in $H_pH_q$, $p \equiv 1$ mod $q$, whence $q | (p-1) \lt p$, a contradiction. Whence $H_q \lhd H_pH_q$, and this implies $H_q$ is even a characteristic subgroup of $H_pH_q$. Since $H_pH_q$ is normal in $G$, it follows that $H_q$ is normal in $G$, as wanted (in general, if $M \text { char } N \lhd G$, then $M \lhd G$).