Let $G$ be a group of order $p^3q^5$ where $p$ and $q$ are two distinct prime numbers. Is $G$ solvable? If it is, how to justify it (without using the Burnside's Theorem [wiki])?


Basically, I am trying to find a non-trivial normal subgroup $H$ of $G$ and to show that both $H$ and $G/H$ are solvable.
To find an $H \triangleleft G$, I think the Sylow theorems will be helpful. However, for both $n_p$ (the number of Sylow $p$-subgroups) and $n_q$, there are too many cases to consider, due to the "high powers" of $p$ and $q$.

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    $\begingroup$ There are elementary proofs that show that groups of order $p^n q$ and $p^2 q^2$ are solvable. Can you already prove that groups of order $p^3 q^2$, $p^3 q^3$, $p^3 q^4$, ... are solvable? $\endgroup$ – Mikko Korhonen Apr 27 '14 at 14:11
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    $\begingroup$ Why $p^3q^5$ in particular? Is that the smallest case you cannot do? $\endgroup$ – Derek Holt Apr 27 '14 at 17:27
  • $\begingroup$ @DerekHolt No, it is not. My teacher gave $p^2q$ as an example in class and asked us what if the order is $p^3q^5$. I am not able to solve it. $\endgroup$ – hengxin Apr 28 '14 at 1:37
  • $\begingroup$ @MikkoKorhonen I know how to prove that the groups of order $p^2q$ are solvable. In the groups of order $p^2q$, there are normal subgroups ($H$) either of order $p^2$ or of $q$. In either case, $\{ e \} \triangleleft H \triangleleft G$ is a witness to the 'solvability' of $G$. However, I am still not able to solve the case of order of $p^3q^5$. Should I find such an $H$ or turn to a different approach? $\endgroup$ – hengxin Apr 28 '14 at 1:45
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    $\begingroup$ I have no idea how difficult it is to prove that groups of order $p^3 q^5$ are solvable, without proving Burnside's theorem. But anyway, the best strategy for this question is probably this: (1) prove that groups of order $p^\alpha q^\beta < p^3 q^5$ with $\alpha \leq 3$ and $\beta \leq 5$ are solvable (2) Prove that groups of order $p^3 q^5$ are not simple. It follows from (1) and (2) that groups of order $p^3 q^5$ are solvable. $\endgroup$ – Mikko Korhonen Apr 28 '14 at 8:25

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