If G is a group of order d, then does there exist a subgroup of order c for all divisors c of d? I cannot find a proper countetexample. Could anyone provide me with one?
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The Sylow theorems say that for the prime factors of $d$ there will be corresponding subgroups. But if you take a group like the Alternating group on 5 elements ($A_5$), there is no subgroup with 15 elements. – Doug M Oct 16 '18 at 02:44
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2This question has dozens of duplicate. I'm also a bit curious why people write answers about nilpotent and supersolvable groups while it's clear from the way the question is written that this is way above the level of the asker? – verret Oct 16 '18 at 07:26
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1@verret Different solutions might suit different visitors. – Sha Vuklia Apr 18 '22 at 17:53
3 Answers
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The alternating group $A_4$ has no subgroup of order $6.$
More examples are the alternating groups $A_n$ for $n \ge 5$ which are known to be simple. So they can have no subgroup of half their order, since a subgroup of index $2$ is normal.
coffeemath
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Such group is called CLT (Converse of Lagrange Theorem) group. One can prove that any nilpotent group is a CLT group. Also, for any $n\geq 5$, $S_{n}$ is not a CLT group. (For the proof, see here.)
Seewoo Lee
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In fact, there is a result due to Bray that all supersolvable finite groups are CLT, and all CLT groups are solvable. See https://msp.org/pjm/1968/27-2/pjm-v27-n2-p02-p.pdf .
The inclusions are strict since $A_4$ is solvable but not CLT and $S_4$ is CLT but not supersolvable.
(Note that a "supersolvable" group is a group with a normal series with cyclic factors such that every group in the series is normal in the whole group. For a finite group, this means its chief series has cyclic factor groups of prime order, the chief series being a normal series of normal subgroups that cannot be refined. In general, solvable finite groups have chief series with elementary abelian factors, but they need not be cyclic.)
C Monsour
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