Timeline for Ratio of number of subgroups to the order of a finite group
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 24, 2012 at 8:03 | comment | added | Steve D | I ran a quick check for Q3, iterating over all abelian groups of order at most $10^6$. The only integers I got were 1,2,4,6,7,14, and 21. | |
| Aug 24, 2012 at 6:59 | comment | added | Qiaochu Yuan | (A suitable bound for $S_n$ is given at math.stackexchange.com/questions/76176/… .) | |
| Aug 24, 2012 at 6:58 | comment | added | Qiaochu Yuan | For the finite simple groups I expect that the only accumulation point is $0$. This comes from the cyclic groups; one can of course ignore the sporadics, and then probably well-known lower bounds on the number of subgroups of the infinite families will do it (although I don't know them). It should more or less suffice to give such bounds for $A_n$ and $\text{PSL}_n(\mathbb{F}_q)$. | |
| Aug 24, 2012 at 5:50 | comment | added | Gerhard Paseman | I am confusing the ratio with the number of divisors of n. I may have the asymptotic wrong, but I suspect the number of divisors of n becomes O(logn) for large n. (Actually, I think it is more bumpy, and the asymptotic is for the mean value up to n, which is not appropriate here. Gerhard "Ask Me About System Design" Paseman, 2012.08.23 | |
| Aug 24, 2012 at 1:47 | comment | added | Qiaochu Yuan | @Gerhard: I think you mean $\frac{2}{\sqrt{n}}$, yes? I am not sure where that $O(\log n)$ comes from. | |
| Aug 24, 2012 at 0:36 | answer | added | Qiaochu Yuan | timeline score: 16 | |
| Aug 23, 2012 at 23:36 | comment | added | tj_ | "the set of finite groups" as well as "the set of finite abelian groups" are proper classes! | |
| Aug 23, 2012 at 23:05 | comment | added | Gerry Myerson | It seems to me that for groups of the form large number of copies of cyclic order 2 plus one cyclic order large prime you should be able to get the ratio arbitrarily close to any positive rational. | |
| Aug 23, 2012 at 22:51 | comment | added | Gerhard Paseman | For cyclic groups, it will be the number of divisors of n divided by n, which has an upper bound of 1, an easy upper bound of 2 sqrt(n) for most n, and for sufficiently large n will be O(log n). I suspect that for finite abelian groups, your spectrum set will be have only one limit point in the reals. Gerhard "Ask Me About System Design" Paseman, 2012.08.23 | |
| Aug 23, 2012 at 22:11 | history | asked | jwellens | CC BY-SA 3.0 |