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Let G be a group and p and q be primes. Suppose |G| = pq and has only one subgroup of order p and only one subgroup of order q. Prove that G is cyclic.

Alec GU
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    Did you try to use Sylow's theorem ? – Peter Mar 25 '19 at 12:11
  • You don't even need Sylow theorems. It is already given that there is only one subgroup of each possible order. So it is even easier. – Mark Mar 25 '19 at 12:14
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    @Mark We called this Sylow II, that the uniqueness is equivalent to normal subgroup. And then it is also clear (see the nice answer by A. Caranti, just using Lagrange). – Dietrich Burde Mar 25 '19 at 12:16
  • Yeah, it's fine to think of it like this. What I mean is just that you don't even need to know Sylow theorems for this. In general if there is only one subgroup of a specific finite order $d$ then it must be normal. The other direction (that normal implies uniqueness) is of course not true for any subgroups, and the fact that it is true for Sylow subgroups really follows from the second Sylow theorem. – Mark Mar 25 '19 at 12:34

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By Sylow II, a $p$-Sylow subgroup is unique if and only if it is normal. Hence $G$ has a normal subgroup of order $p$ and a normal subgroup of order $q$. Consequently, $G$ is cyclic by the following duplicate:

Showing that a group of order $pq$ is cyclic if it has normal subgroups of order $p$ and $q$

Dietrich Burde
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