I'm looking for the finite groups (for example n elements) that have subgroups with every divisor of n elements. Say a simple example is cyclic groups. Please help me to find more and more groups with this condition.
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2This is true for abelian groups: https://math.stackexchange.com/questions/477500/showing-that-a-finite-abelian-group-has-a-subgroup-of-order-m-for-each-divisor-m – Math1000 Dec 20 '19 at 10:28
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2Should be converse, not inverse. – Wang Kah Lun Dec 20 '19 at 10:53
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Finite $p$-groups for any prime $p$. – GreginGre Dec 20 '19 at 10:54
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And dihedral groups should work fine too. Moreover, if $G,H$ satisfy the desired property, $G\times H$ too... – GreginGre Dec 20 '19 at 10:57
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A seminal paper on the subject, "A note on CLT groups", was written by Henry Bray and appeared in the Pacific J. of Mathematics Vol. 27, No. 2, 1968. The paper can be downloaded as a pdf file for free. It is easy to read.
user1729
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Nicky Hekster
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2I added the title of the paper as it complements the answer of AnalysisStudent0414. I hope you don't mind. – user1729 Dec 20 '19 at 12:03
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I have seen a (finite) group with this property being called CLT group (Converse Lagrange Theorem group).
It is relatively easy to see that abelian groups and $p$-groups both satisfy this property.
Here it is shown that every nilpotent group is CLT (so, in particular, every abelian group is CLT). The proof is based on the well-known characterization of finite nilpotent groups as direct products of their Sylow subgroups (which are CLT).
Even more generally, here the author proves that every supersolvable group is CLT, and shows a solvable group which is not $(A_4)$.
AnalysisStudent0414
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