Timeline for fixed points of permutation groups
Current License: CC BY-SA 3.0
9 events
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| Sep 20, 2015 at 19:15 | comment | added | Dan Romik | For $S_n\times S_m$ the number of fixed points will be distributed like $X+Y$ where $X,Y$ are independent random variables, and $X$ (resp. $Y$) is distributed as the number of fixed points in a uniformly random element of $S_n$ (resp. $S_m$), so roughly a Poisson(2) r.v. This is consistent with the data; note that $\exp(-2) \approx 0.1353$. | |
| Sep 20, 2015 at 18:18 | comment | added | Derek Holt | @Lucia I have revised the figures using a better random eleemnt generator (the Product Replacement Algorithm). | |
| Sep 20, 2015 at 18:18 | history | edited | Derek Holt | CC BY-SA 3.0 |
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| Sep 20, 2015 at 18:01 | comment | added | Derek Holt | @Lucia Yes you are right on both counts! Thanks for pointing this out. I think the fault lies with the default Magma random element generator. I will investigate further and try and come up with some more accurate figures. | |
| Sep 20, 2015 at 17:58 | comment | added | Lucia | I'm also confused by the numerics for $S_{140}$ acting on the unordered pairs. The probability of an element of $S_{140}$ having no two cycles is approximately $1/\sqrt{e}$, and the probability of having at most one one cycle is approximately $2/e$. So shouldn't the proportion here with no fixed points be approximately $2/e^{3/2}$, which is only about $40\%$. | |
| Sep 20, 2015 at 17:43 | comment | added | Lucia | I'm confused by your examples. Could you explain the $S_n\times S_m$ example a bit more? I took it to mean the subgroup of $S_{m+n}$ obtained by permutations that leave the set of first $m$ elements fixed. But in that case, wouldn't you expect that the proportion of elements with no fixed points is approximately $1/e$ times $1/e$ which is patently not the case in your numerics. I'm sure I've not understood what you mean. | |
| Sep 20, 2015 at 17:10 | history | edited | Derek Holt | CC BY-SA 3.0 |
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| Sep 20, 2015 at 15:04 | comment | added | Igor Rivin | That is very interesting! Did you try any $p$-groups, per chance? | |
| Sep 20, 2015 at 14:00 | history | answered | Derek Holt | CC BY-SA 3.0 |