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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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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
Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. Also, many users find the use of the imperative ("Prove", "Show", etc) to be rude when asking for help. Please consider rewriting your post. – Arturo Magidin May 20 '12 at 21:48
up vote 5 down vote accepted

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

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