# Classification of groups such that the converse to Lagrange holds [duplicate]

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.

• $S_3$ is Lagrange. May 29 at 14:54
• May 30 at 9:18

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 $$\{e\}=H_0, in which each factor group $$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.

• $S_4$ is an example of a CLT group that is not supersolvable: its subgroup $A_4$ is not CLT. May 29 at 15:05
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.