November 20th
Today I learned that the group $\QQ/\ZZ$ has every finite cyclic group as a unique subgroup, and in fact no subgroup of it suffices. For one, all finite cyclic groups appear as $\langle 1/n\rangle$ for size $n.$
Uniqueness is a bit more obnoxious to show. Suppose $G\subseteq\QQ/\ZZ$ is a finite cyclic subgroup of order $n.$ We need to show $G=\langle1/n\rangle.$ Well, suppose $g\in G$ generates. Then $ng=0\pmod\ZZ,$ so\[ng=m\in\ZZ.\]It follows $g=\frac mn.$ Namely, $G=\langle g\rangle\subseteq\langle1/n\rangle.$ However, both of these sets have the same size while one is the subset of the other, so we must in fact have equality.
We close, for now, by showing that no proper subgroup of $\QQ/\ZZ$ has this property. In particular, if a subgroup is proper, then it misses some element, say $\frac mn.$ But then we cannot have $\frac1n$ either, for this would imply $\frac mn$ is present. It follows that the proper subgroup is missing the subgroup $\langle1/n\rangle,$ and is therefore missing the only subgroup of order $n$ in $\QQ/\ZZ.$ It follows that the proper subgroup is missing some finite cyclic subgroups.
Quickly, this is not quite the property I was looking for—namely, I was interested in an infinite group with only finite proper subgroups. In fact, the Pr\"uffer group $\ZZ[1/p]/\ZZ$ suffices. Indeed, define the degree of a subgroup $G$ of $\ZZ[1/p]/\ZZ$ as the largest power of $p$ in the denominator of any element of $\ZZ[1/p]/\ZZ.$ We have the following cases.
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If the degree is in fact unbounded, then the subgroup $G$ is all of $\ZZ[1/p]/\ZZ,$ and this is not a proper subgroup; namely, we can achieve every element of the form $a/p^\bullet$ because we have unbounded denominators.
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Else the degree is bounded, so well-order gives us a least integer $d$ such that if $a/p^k\in G$ with $p\nmid a,$ then $k\le d.$ We claim that the group is $\langle1/p^d\rangle,$ which is finite cyclic of order $p^d.$ Indeed, we are at least a subset of this because $p^d$ is an upper-bound on the denominator. But in the other direction, there is some $a$ for which $a/p^d\in G$ and $p\nmid a.$ However, this means \[a^{-1}\pmod{p^d}\] exists, so $aa^{-1}/p^k=1/p^k$ in $\ZZ[1/p]/\ZZ.$ This completes the proof.
Of course, I suppose the caveat to this proof is that we are only using prime-powers for our subgroup sizes, which is the advantage of $\QQ/\ZZ.$