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Suppose $p<q$, where $p,q$ are primes and we have a non-abelian group $G$ of order $p^2q$. Is it true that it has a subgroup which is not normal? I try to use Sylow's theorems. We take Sylow subgroups of order $p^2$ and $q$. They are normal , otherwise we have a contradiction. Now i want to say that that $G$ must be abelian, but don't know why it is true...

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  • $\begingroup$ Can you show us what you have tried using Sylow? $\endgroup$ – Nicky Hekster May 7 '14 at 15:09
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Hint: suppose that all subgroups were normal. Then $G$ has normal subgroups $M$ of order $p^2$ and $N$ of order $q$. What can you say about (the structure of) those subgroups and what about the order of $MN$?

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    $\begingroup$ $M\cap N = 0$. Then in $MN$ we have $p^2q$ different elements, because if two of them are equal then the intersection isn't empty, and it follows that $G=M\times N$, and evidently abelian, because groups of order $p^2$ and $q$ are abelian? Is this true? $\endgroup$ – Elensil May 7 '14 at 15:17
  • $\begingroup$ That's it! Well done! (one minor point I would write $M \cap N= 1$ (multiplicatively)) $\endgroup$ – Nicky Hekster May 7 '14 at 15:41
  • $\begingroup$ Does the same work for groups of order $pqr$? $\endgroup$ – Elensil May 7 '14 at 15:42
  • $\begingroup$ Absolutely, even more generally, for finite non-abelian groups of cube-free order. They must have a non-normal subgroup. $\endgroup$ – Nicky Hekster May 7 '14 at 15:46

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