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I was remeditating Sylow subgroups recently, after reading somewhere that it served as a partial converse to Lagrange's theorem. After a bit more pondering I started wondering if we can find the groups such that a "perfect" converse to Lagrange's theorem holds, i.e those groups such that if $n\mid |G|$ then $G$ has a subgroup of order $n$. Call such groups Lagrange groups.

Clearly $p$-groups and abelian groups are Lagrange groups. I am almost certain a product of two Lagrange groups is a Lagrange group. Beyond this, I did not know how to further my search for Lagrange groups, and if a complete classification is even doable.

Any help would be greatly appreciated, even just ideas to pursue, thanks in advance.

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2 Answers 2

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Let $n=|G|$. If for each divisor $m$ of $n$ the group $G$ contains a subgroup $H$ of order $m$, then we say that $G$ has the property $CLT$. The concept of $CLT$ (the converse of Lagrange's theorem) groups was introduced by Bray, Note on CLT groups, Pacific J. Math. 27(2), 229-231. First examples of $CLT$ groups are finite nilpotent groups.

A group $G$ is supersolvable if it possesses a finite normal series $\{e\}=H_0<H_1<\ldots<H_r=G$, in which each factor group $H_i/H_{i-1}$ is cyclic; all subgroups and factor groups of supersolvable groups are supersolvable; every finite nilpotent group is supersolvable.

The following theorem was proved by Deskins, A Characterization of Finite Supersolvable Groups, Am. Math. Monthly, vol. 75, No. 2 (1968), 180--182.

Theorem. The subgroups of a finite group $G$ all have the property $CLT$ if and only if $G$ is supersolvable.

This is certainly not a classification, but it is something.

Edit. See also here

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  • $\begingroup$ $S_4$ is an example of a CLT group that is not supersolvable: its subgroup $A_4$ is not CLT. $\endgroup$
    – Derek Holt
    May 29 at 15:05
  • $\begingroup$ @kabenyuk wow thanks a lot, this is a great answer, I'll go check this out. $\endgroup$
    – DevVorb
    May 29 at 15:07
  • $\begingroup$ @Derek Holt, yes of course. So Deskins' theorem is by no means a description of all LCT-groups. $\endgroup$
    – kabenyuk
    May 29 at 15:11
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Here are a few more facts about finite LCT-groups to add to Kabenyuk's answer.

Philip Hall proved that a finite solvable group of order $mn$ with $\gcd(m,n)=1$ has a subgroup of order $m$ and, converely, if the finite group $G$ has subgroups of order $m$ for all factorizations $|G|=mn$ with $\gcd(m,n)=1$, then $G$ is solvable.

So an LCT-group must be solvable. Furthermore it is easy to see, using Hall's theorem, that if $H$ is any finite solvable group, then the direct product $H \times C_{|H|}$ of $H$ with a cyclic group of order $|H|$ is an LCT-group, so all finite solvable groups are contained in an LCT-group.

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