Prove (or disprove) If G is a group in which every proper subgroup is cyclic, then G is cyclic.
Could an example of $\displaystyle S_3$ be used as proof?
I'm not too comfortable with subgroups or cyclic groups yet, but from my understanding, the subgroups of $\displaystyle S_3$ are:
<(1)>= {(1)}
<(1,2)> = {(1),(1,2)}
<(1,3)> = {(1),(1,3)}
<(2,3)> = {(1),(2,3)}
<(1,2,3)> = {(1), (1,2,3), (1,3,2)} = <(1,3,2)>
Since no cyclic subgroup is equal to all of $\displaystyle S_3$it isnt cyclic. Or, no x in $\displaystyle S_3$ for which $\displaystyle S_3$=x
Actually I was thinking even smaller, the Klein Four group. All proper subgroups are of order 2, and so must be cyclic, but all elements are of order 2 and so the group is not cyclic. This is of course generalizes, since for any prime $\displaystyle p$, $\displaystyle \mathbb{Z}_p\times \mathbb{Z}_p$ has only proper subgroups of order $\displaystyle p$ which must be cyclic!
Definitely in the finite case. If there is a unique subgroup of each order, then each proper subgroup is normal.Then for finite groups, in particular all the Sylow subgroups will be normal and cyclic forcing G to be cyclic. For infinite groups however, it is definitely less clear.
EDIT: Ha, of course! Just consider the infinite direct sum $\displaystyle \bigoplus_{p\text{ prime}}\mathbb{Z}_p$, in the sense that we only allow a finite number of coordinates be nonzero. Then any subgroup would have to be of finite order, lest we obtain an element with an infinite number of nonzero coordinates. Moreover it would have to "live" on a finite number of the $\displaystyle \mathbb{Z}_p$, and so would be a subgroup of the product of those groups which is cyclic, thus would have to be cyclic, yet the group is clearly not cyclic!