Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose $G$ is a finite group, with $|G|=n$. Suppose also that for every positive integer $m\mid n$, $G$ has a subgroup of index $m$. Are there any general statements (structural or otherwise) I can make about such $G$?

For example, all such groups will be solvable; but as $S_4$ shows, they need not be supersolvable. The collection is also strictly smaller than solvable groups, as $A_4$ shows.

Thanks!

share|improve this question

1 Answer 1

up vote 10 down vote accepted

You are asking for groups in which the converse of Lagrange's theorem holds (sometimes called "Lagrangian"). It includes all supersolvable groups, but includes more than that, yet not all solvable groups. As far as I know, there is no characterization of these groups. (Keith Conrad says the same thing in these notes). A quick searching MathSciNet reveals a lot of papers discussing special classes of groups that satisfy the converse of Lagrange's Theorem, but no structural characterization that I could find.

In this paper, Jing mentions a variant: a group is a $\mathscr{G}$-group if and only if for each subgroup $H$ of $G$ and each prime factor $p$ of $[G:H]$, there is a subgroup $K$, $H\leq K\leq G$ such that $[K:H]=p$. He describes all values of $n$ such that every group $G$ with $|G|=n$ is a $\mathscr{G}$-group.

$\mathscr{G}$-groups have been characterized in sundry ways; the paper by Jing cites a book by Bray, Deskins, Johnson, Humphreys, Puttaswamaiah, Venke, Walls, and Weinstein in which it is shown that $G$ is a $\mathscr{G}$-group if and only if there is a normal Hall subgroup $N$ in $G$ such that $N$ and $G/N$ are nilpotent, and for each $H\leq N$ we have $G=N\cdot N_{G}(H)$.

share|improve this answer
    
Thanks for the pointer to $\mathscr{G}$-groups. I'm gonna take a look at the Jing paper. –  user641 Sep 22 '11 at 11:59

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.