Prove that a group of order $p^nq$ for primes $p$ & $q$ is not simple.

I've been able to prove the theorem holds for $p=q$ and $p>q$. If $p<q$ the best I've been able to do is use Sylow to show $$p^n+p^{n-1}-1\leq q$$.

Yet I seem to be stuck. I would appreciate any help.

  • 2
    $\begingroup$ See coolnumbers.wordpress.com/2012/10/31/… $\endgroup$ – verret Jan 9 '20 at 4:11
  • $\begingroup$ @Bach It's essentially the same: to show it's solvable, you show it's not simple and use induction... $\endgroup$ – verret Jan 11 '20 at 19:40

A consequence of one of Sylow's theorems is that if there is exactly one $p$-Sylow subgroup $H$ of $G$, then is it normal. Can you do the rest?

  • $\begingroup$ I used such fact when proving the case $p>q$, but I wouldn't know how to apply it to $p<q$. Also, technically the result you mention doesn't need Sylow Theory. $\endgroup$ – Leo Jan 10 '20 at 23:13

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