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.
Finite group of order $p^2q$
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abstract-algebra
group-theory
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1 Answers
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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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0Since 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! – 2012-09-28