$H_p \triangleleft G$ and $H_p \cdot H_q \triangleleft G$ then $H_q \triangleleft G$ Let $G$ be a group. $p \lt q$ primes. Let $H_p$ and $H_q$ be sylow subgroups of order $p$ and $q$ respectively. Show that if $H_p \triangleleft G$ and $H_p \cdot H_q \triangleleft G$ then $H_q \triangleleft G$
I can't find a way to prove it, even though I'm convinced that this holds true. I mainly tried by the absurd:
If this is false, then exists a $h_q \in H_q$ and a $g \in G$ such that $gh_qg^{-1} \notin H_q$. But since $H_q$ is a q-sylow subgroup, then $gH_qg^{-1}$ is another q-sylow subgroup of G. Let's call it $H_{q'}$Since $H_p \triangleleft G$, then for any $h_p \in H_p$, $gh_pg^{-1} \in H_p$. Then $gh_pg^{-1}gh_qg^{-1} \in H_pH_{q'}$. But $gh_pg^{-1}gh_qg^{-1} = gh_ph_qg^{-1} \in H_pH_q$. Then $gh_ph_qg^{-1} \in H_pH_q \cap H_pH_{q'}$.
I think that from here on I should be able to prove that $H_pH_q \cap H_pH_{q'} = \{1_G\}$, so that would mean that $h_q=1_G$ so $gh_qg^{-1} \in H_q$ Abs!
 A: Since $H_p \lhd G$, $K=H_pH_q$ is a subgroup and $|K|=pq$. But $|K:H_q|=p$, the smallest prime dividing the order of $K$, whence $H_q$ is normal in $K$ and since $H_q$ is Sylow in $K$, we get $H_q$ is characteristic in $K$. So $H_q$ char $K \lhd$ G. This implies $H_q \lhd G$.
Note Observe that this can be generlizaed to $H_p$ and $H_q$ being Sylow subgroups of order $p$ and $q^k$ respectively, $k$ a positive integer.
A: With the hypothesis $H_p\subset N_G(H_q)$ it is true: it is a consequence of the Frattini argument. Since $H_pH_q$ is normal in $G$, and $H_q$ is the $q$-Sylow subgroup of $H_pH_q$ $G=N_G(H_q)H_pH_q$. We have $H_p,H_q\subset N_G(H_q)$ since $H_p$ is normal.
https://en.wikipedia.org/wiki/Frattini%27s_argument
A: Let $K = H_{p} H_{q}$. You know that the number of $q$-Sylow subgroups of $K$ divide $p = \lvert K : H_{q} \rvert$, and is congruent to $1$ modulo $q$. Now $1 + q > p$, so that index has to be $1$, that is, $H_{q}$ is the unique subgroup of order $q$ of $K$.
Now if $x \in G$, we have that $H_{q}^{x}$ is a subgroup of $K$ of order $q$, hence it equals $H_{q}$.
