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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.

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  • $\begingroup$ Why does it follow from the first lines that the order of $\;G\;$ is divisible by three different primes? As far as I can see it we can only deduce it is divisible by $\;p,q\;$ , and the maximal power of $\;p\;$ dividing the order is one. $\endgroup$ – Timbuc Sep 17 '14 at 3:28
  • $\begingroup$ You're right, but I am not saying that $r$ is a prime, I am just saying that $r$ is coprime with $pq^m$. $\endgroup$ – user16924 Sep 17 '14 at 3:32
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    $\begingroup$ I've added that $r$ is not prime to be more precise, but if I just say that $|G|=pq^mr$ with $(pq^m:r)=1$, this doesn't mean that $r$ is prime. $\endgroup$ – user16924 Sep 17 '14 at 3:36
  • $\begingroup$ Do you know how to do the case $r=1$? $\endgroup$ – zibadawa timmy Sep 17 '14 at 3:46
  • $\begingroup$ Well, if $r=1$, then the number of $q-$ Sylows is $1$ or $p$, if it is $1$, then there is only one $q$ Sylow, so it is normal. I don't know for the case $n_q=p$. $\endgroup$ – user16924 Sep 17 '14 at 3:54
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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 $M \lhd G$).

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  • $\begingroup$ I could follow your solution except for one little thing: in one of the first steps you prove that the number of $q$ Sylows in $H_qH_p$ is one, but doesn't that follow directly from the statement:"Let H_p and H_q be Sylow subgroups..."?, I mean $H_q$ is one $q$-Sylow subgroup, so in $H_qH_p$, $H_q$ is the only one. Maybe I am completely mistaken so sorry for my confusion if this is the case. $\endgroup$ – user16924 Sep 17 '14 at 8:13
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    $\begingroup$ $H_q$ is a Sylow $q$-subgroup of the group $H_pH_q$. But it does not have to be the only one at first. There could be conjugates of $H_q$ (with respect to the group $H_pH_q$), which also "live" in $H_pH_q$. The congruence argument shows you that that is not the case. Although $H_pH_q$ is composed of two groups, the full group structure could involve more subgroups than $H_q$ alone. Hope this helps. $\endgroup$ – Nicky Hekster Sep 17 '14 at 8:26
  • $\begingroup$ Yes it does, thanks for the explanation. $\endgroup$ – user16924 Sep 17 '14 at 8:39

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