# Question about a non-abelian group of order $p^2q$

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...

• Can you show us what you have tried using Sylow? – Nicky Hekster May 7 '14 at 15:09

## 1 Answer

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 $MN$?

• $M\cap N = 0$. 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? – Elensil May 7 '14 at 15:17
• That's it! Well done! (one minor point I would write $M \cap N= 1$ (multiplicatively)) – Nicky Hekster May 7 '14 at 15:41
• Does the same work for groups of order $pqr$? – Elensil May 7 '14 at 15:42
• Absolutely, even more generally, for finite non-abelian groups of cube-free order. They must have a non-normal subgroup. – Nicky Hekster May 7 '14 at 15:46