Timeline for Group not leaving subset invariant
Current License: CC BY-SA 2.5
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 31, 2011 at 21:23 | comment | added | Olivier Bégassat | and said group is abelian, I'm sure it can be much improved. | |
| Mar 31, 2011 at 20:51 | comment | added | Olivier Bégassat | I mean that for large $n,~n_1\cdot\dots\cdot n_m$ can be chosen close to $\frac{n^m}{m^m}$. | |
| Mar 31, 2011 at 20:49 | comment | added | Olivier Bégassat | As pointed out before, $|G|\leq \binom{n}{m}$ which is approximatively $\frac{n^m}{m!}$ for large $n=|X|$ and fixed $m=|Y|$. On the other hand, if you have $n_1+\dots +n_m=n$ and $n_i\geq 1$ for all $i=1\dots m$, then you can find a group with $|G|=n_1\times\cdots\times n_m$ with desired property. For large $n$ that's close to $\frac{n^m}{m^m}$. So, at least asymptotically you have, for fixed $m$, calling $B(m,n)$ the cardinal of the biggest group satisfying your condition, $\frac{n^m}{m^m}\lesssim B(m,n)\lesssim\frac{n^m}{m!}$ | |
| Mar 31, 2011 at 19:18 | comment | added | Klim Efremenko | No I did not claim that there is only one maximal group. Just one can understand that maximal group in sense that there is no group including it which has property P | |
| Mar 31, 2011 at 18:59 | history | answered | Erik P. | CC BY-SA 2.5 |