I'm working on the following problem and would like some feedback (this is my first course on Group Theory so I'm not so sure about my argument):
Let $G$ be a group of finite order, $p\lt q$ two prime numbers such that: $p^2\nmid |G|$, and let $H_p$ and $H_q$ be two respective Sylow subgroups with $H_p\lhd G$.
Prove $H_pH_q\lhd G \implies H_q\lhd G$.
My attempt: (There was a previous part in which I've already proven that $H_pH_q$ is a subgroup of $G$)
First off, since $p^2\nmid |G|$, this means the order of $H_p$ must be $p$, and since $H_p\lhd G$, then $H_p$ is the only $p$-Sylow subgroup of $G$.
Now Suppose $H:=H_pH_q\lhd G$, but $H_q\not\lhd G$, this implies by the Sylow theorems that: $$n_q\equiv 1\pmod q \implies n_q>q$$ The last inequality because if $n_q=1$ then $H_q\lhd G$. This means that there must exist at least another $q$-Sylow: $H_q'\ne H_q$ and $\exists g \in G : gH_qg^{-1}=H_q'$
Let $h\in H_q$, then $$h=eh\in H$$ this implies $$ghg^{-1}\in gHg^{-1}=H$$ since $H=H_pH_q$ is normal on $G$. Therefore: $gH_qg^{-1}=H_q'\subseteq H$, and thus $H_q' \le H$. We can also see that $H_q\le H$
Now since $H$ is a group itself, and $H_p \cap H_q = \{e\}$ we have: $|H|=|H_pH_q|=p|H_q|$. We then can apply the Sylow theorems to the group $H$ itself: let's call $\hat n_q$ to the number of $q$-Sylow subgroups of $H$.
From the Sylow theorems we have:
$$\begin{cases} \hat n_q \equiv 1\pmod q \\ \\ \hat n_q \mid p \implies \hat n_q \in \{1, p\} \end{cases}$$ Now: if $\hat n_q=p $, we have: $p=\hat n_q > q > p$ which is absurd. Then it must be $\hat n_q=1$. But $\hat n_q$ is also a contradiction since we previosly saw: $H_q' \le H$ and $H_q\le H$ and $H_q',H_q$ are $q$-Sylow subgroups of $H$ which would imply $H_q'=H_q$.
Is this correct? I'd appreciate any comments. Thanks!
Yes your argument is OK, but can be shortened: you can assume that $p$ really divides $|G|$ otherwise there is nothing to prove, and hence $|H_p|=p$. Since $H_p \lhd G$, $H_pH_q$ is a subgroup of $G$ and $|H_pH_q:H_q|=p$ (use here that $|H_pH_q|=\frac{|H_p| \cdot |H_q|}{|H_p \cap H_q|}$ and of course $H_p \cap H_q=1$). Now, $H_q$ is a Sylow $q$-subgroup of $H_pH_q$, hence the number of Sylow $q$-subgroups divides $p$ and since $p \lt q$ this number must be $1$, so $H_q$ is normal in $H_pH_q$ and hence it is characteristic in $H_pH_q$. So $H_q \text{ char } H_pH_q \lhd G$, and this implies $H_q \lhd G$.