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This question already has an answer here:

The order of a group $G$ is $pq$ where $p$ and $q$ are both prime. Show that all proper subgroups of $G$ are cyclic. I need help with this. I know that all the orders of subgroups divide the order of the larger group, but I do not know where to go from there.

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marked as duplicate by DonAntonio, Dan Rust, Lord_Farin, Nick Peterson, Dennis Gulko Nov 12 '13 at 18:39

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The order of a subgroup $H$ of $G$ must divide the order of $G$, then, if $H$ is a proper subgroup of $G$, its order could be only $p , q $ or 1

Observing that a group of prime order is cyclic you can prove the assert.

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Consider the following:

What are the divisors of $pq$?

What kind of groups exist for the orders less than $pq$?

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  • $\begingroup$ I think you mean subgroups. Certainly there are many groups of order less than $pq$ that have nothing to do with $G$. $\endgroup$ – mathematics2x2life Nov 12 '13 at 17:34
  • $\begingroup$ There aren't many groups of order less than $\;pq\;$ and whose order is a divisor of $\;pq\;$ , as explained in the second line... $\endgroup$ – DonAntonio Nov 12 '13 at 17:42
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The order of a subgroup $ H\leq G $ must divide the order $ G $. Therefore $|H|$ is either $1 $, $ p $, $ q $ or $ pq $. If $ H $ is a proper subgroup of $ G $ (that is $ H $ is not equal to $G $) then it must be the trivial group, or it must be of prime order. Show that in this case it must be cyclic.

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