Let $G$ be a group of order $n$ and its subgroup lattice be order-isomorphic 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 RivinCommented May 19, 2014 at 18:38
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1$\begingroup$ Ore's theorem: a group is locally cyclic iff its subgroups lattice is distributive. See a proof here. $\endgroup$– Sebastien PalcouxCommented Sep 24, 2014 at 14:45
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$\begingroup$ this question is somewhat old. thanks for the link. $\endgroup$– Minimus HeximusCommented Sep 24, 2014 at 14:48
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1 Answer
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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).
answered May 19, 2014 at 18:43
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$\begingroup$ btw, it seems to be the divisibility lattice itself and not its dual. $\endgroup$ Commented May 19, 2014 at 19:05
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$\begingroup$ @MinimusHeximus: The divisibility lattice is self-dual. $\endgroup$ Commented May 19, 2014 at 21:10