# Finite group of order $p^2q$

Let $p$ and $q$ be primes with $q < p$ and suppose that $G$ is a group of order $p^2q$. Suppose that $G$ has a unique subgroup of order $q$. Show that $G$ is abelian.

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Please don't phrase questions as orders. And this looks like homework. What have you tried? –  Tobias Kildetoft May 20 '12 at 21:36
It's likely that this has been asked before. At any rate: I would avoid asking questions in this way. Why are you interested in this? It's likely that you know the Sylow theorems, about the groups of order $p^2$ for $p$ a prime, etc. What does that stuff tell you? –  Dylan Moreland May 20 '12 at 21:39
What is there to try? Is there a theorem I should use? I can't see how to apply any of the Sylow theorems. –  rk101 May 20 '12 at 21:40
So there exists a Sylow p-Subgroup of order $p^2$, and we know the number of Sylow $q$-subgroups is one, and so this subgroup is normal in $G$. –  rk101 May 20 '12 at 21:43

Hint the first. If $G$ has a unique subgroup of order $n$, then that subgroup is normal.

Hint the second. How many subgroups of order $p^2$ must $G$ have? Thinks about Sylow's Theorems.

Hint the third. If $P$ is a subgroup of order $p^2$, and $Q$ is a subgroup of order $q$, do elements of $P$ commute with elements of $Q$?

Hint the fourth. Is a group of order $p^2$ abelian? Is a group of order $q$ abelian?

Hint the fifth. If $G=HK$, $H$ is abelian, $K$ is abelian, and $hk=kh$ for all $h\in H$ and $k\in K$, what can we say about $G$?

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Since p>q, nd G has one unique Sylow q-subgroup, all Sylow subgroups of G are normal, i.e., G is nilpotent, hence G is the direct product of its Sylow subgroups, which are both abelian in this case. Thus G is abelian. I finally understand this process. Thanks very much! –  awllower Sep 28 '12 at 7:58