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Given a finite group $G$ define $D(G)$ to be the number of divisors $r$ of $|G|$ for which there exists a subgroup of $G$ of order $r$.

Clearly $D(G) \leq d(|G|)$, where $d(n)$ denotes the number of divisors of $n$. We know that if $G$ is nilpotent or even supersolvable, more in general CLT ("Converse Lagrange Theorem", meaning that there exists a subgroup of size any given divisor) then $D(G) = d(|G|)$.

But do we have bounds in general?

Is it true that there exists a constant $C > 0$ such that $C \cdot D(G) \geq d(|G|)$ for any finite group $G$?

For example, what happens if we restrict our attention to symmetric/alternating groups?

This seems to be a hard problem, do you know if anyone has ever worked on this? Thank you for any contribution.

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    $\begingroup$ I agree that this is hard! Here is a related question -- mathoverflow.net/questions/93203/… -- pertaining to the symmetric/ alternating group. In that case you can use Aschbacher-Scott-O'Nan to study maximal subgroups, but even that is tricky. However given that you only want THE NUMBER of divisors, rather than an enumeration of them, there may be hope. $\endgroup$
    – Nick Gill
    Commented Dec 3, 2012 at 16:08

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