I understand that when $G$ has exactly one maximal subgroup (inclusion-wise), then $G$ has to be cyclic.
But is it possible to determine all possible groups with exactly one maximal subgroup?
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Is $G$ finite ? – lhf Feb 04 '16 at 00:52
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It does not have to be. If $G$ is finite, I can prove that those are the groups of order $p^n$, for $n > 1$. – pizet Feb 04 '16 at 00:57
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I don't understand the question. All maximal subgroups are proper by definition. – Derek Holt Feb 04 '16 at 08:49
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Ok, that is redundant. – pizet Feb 04 '16 at 09:15
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If you know that the group must be cyclic then classifying all such groups is easy, at least in the finite case, because the lattice of subgroups of a cyclic group of order $n$ is isomorphic to the lattice of divisors of $n$. This lattice has a single maximal element (different from $n$) iff $n$ is a prime power.
lhf
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BTW, here is a proof that if a finite group has exactly one maximal subgroup, then it must be cyclic: http://math.stackexchange.com/a/39954/589 – lhf Feb 04 '16 at 00:51
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A full proof is at http://groupprops.subwiki.org/wiki/Cyclic_of_prime_power_order_iff_not_generated_by_proper_subgroups. – lhf Feb 04 '16 at 01:02