Let $G$ be a group of order $n$ and its subgroup lattice be orderisomorphic to that of $\Bbb Z_n$. Is $G$ cyclic?
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3$\begingroup$ Answered in mathoverflow.net/questions/35455/… (see Tony Hyunh's answer). $\endgroup$ – Igor Rivin May 19 '14 at 18:38

$\begingroup$ Ore's theorem: a group is locally cyclic iff its subgroups lattice is distributive. See a proof here. $\endgroup$ – Sebastien Palcoux Sep 24 '14 at 14:45

$\begingroup$ this question is somewhat old. thanks for the link. $\endgroup$ – user47958 Sep 24 '14 at 14:48
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Yes, finite cyclic groups are exactly the finite groups whose lattices of subgroups are distributive. The lattice of subgroups of $\mathbb{Z}/n$ is isomorphic to the dual of a divisibility lattice (which is distributive).

$\begingroup$ btw, it seems to be the divisibility lattice itself and not its dual. $\endgroup$ – user47958 May 19 '14 at 19:05

$\begingroup$ @MinimusHeximus: The divisibility lattice is selfdual. $\endgroup$ – Emil Jeřábek supports Monica May 19 '14 at 21:10