Timeline for about fixed points of permutations
Current License: CC BY-SA 3.0
13 events
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| Jul 27, 2011 at 23:38 | history | edited | Gerhard Paseman | CC BY-SA 3.0 |
added 4229 characters in body
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| Jul 27, 2011 at 22:48 | vote | accept | CommunityBot | moved from User.Id=6976 by developer User.Id=69903 | |
| Jul 27, 2011 at 22:48 | comment | added | user6976 | @Gerhard: I agree that this will work, but it may require some effort. It means that the whole idea of using fixed points instead of the cycle structure most probably does not work. I guess instead of specifying the lowest cycle length of the permutations (fixed points), I should bound the longest cycle length. That is: if for all words $w(x,y)$ with $|w|\le n$, the longest cycle in $w(a,b)$ has length $\le p$, then all permutations in $\langle a,b\rangle$ have this property. But that is almost the original problem. | |
| Jul 27, 2011 at 21:36 | comment | added | Gerhard Paseman | Looking back over the comments, I see that I am speaking pretty loosely about the construction, but am still surprised that the idea is not coming across clearly. I will prepared a more explicit construction of a group with (for n sufficiently large) less than 4^n members in it. Gerhard "Ask Me About System Design" Paseman, 2011.07.27 | |
| Jul 27, 2011 at 20:31 | comment | added | Gerhard Paseman | This problem reminds me of an earlier MathOverflow question about the smallest group which establishes the nontriviality of a word. Perhaps someone (Bjorn Poonen?) can provide a link, and see if there are connections between that problem and this. Also, I have a (perhaps wrong) idea that this problem is in the literature. The first people I woukld ask about it, though, are my former advisors or Mark Sapir. Maybe I would ask Derek Holt or Jack Schmidt. Gerhard "Currently Not So Near Academia" Paseman, 2011.07.27 | |
| Jul 27, 2011 at 20:27 | comment | added | Gerhard Paseman | It can be yx^n . Let a be (disjoint unions of copies of) the cycle (1,...,2n). For each "conjugacy class" C of a word of length n, pick a new copy of (0,1,..,2n) and let b act on the (0,...,n) portion of it so that a representative of C has a fixed point in (1,...,n). You now have a group of size at most (2n+1) times (number of distinct C) where ba^n has no fixed points, yet for every group word of length at most n, it has a fixed point in (1,...,n) or a conjugate of it does, because b acts "only on the first half". Gerhard "Ask Me About System Design" Paseman, 2011.07.27 | |
| Jul 27, 2011 at 18:44 | comment | added | user6976 | So the word $u$ is $yx^l$ for a big enough $l$? | |
| Jul 27, 2011 at 18:25 | comment | added | Gerhard Paseman | On each slice, it acts independently, but they all share a common property; for an l big enough, a^l "pushes b's elements out of b's reach". So as long as the cycle lengths are large compared to the common domain of what a and b move, ba^l will have no fixed points. You can probably arrange that l=n. Gerhard "Ask Me About System Design" Paseman, 2011.07.27 | |
| Jul 27, 2011 at 18:19 | comment | added | user6976 | OK, but then when you combine all the permutations $a_w$ into $a$ and all $b_w$ into $b$, how do you show that $\langle a,b\rangle$ contains a fixed point-free permutation? Equivalently, you need one word $u(x,y)$ such that $u(a,b)$ has no fixed points. So far you have different words $u_w$ for different $w$. Right? | |
| Jul 27, 2011 at 18:14 | comment | added | Gerhard Paseman | Sorry, I meant to say "if the length of the orbit is the length of the word". Gerhard "Ask Me About System Design" Paseman, 2011.07.27 | |
| Jul 27, 2011 at 18:11 | comment | added | Gerhard Paseman | For an arbitrary word w, focus on the orbit of one point, since you want that point to be fixed. If the orbit is the length of the word, you can fill in the rest of the behaviour so that a and b are large cycles. However, the cycles need to be so large that e.g. ba^l has no fixed points for l much larger than n. Gerhard "Ask Me About System Design" Paseman, 2011.07.27 | |
| Jul 27, 2011 at 18:02 | comment | added | user6976 | The idea that it is enough to consider one but arbitrary word of length $\le n$ seems to be good. As I understand, for every $w(x,y)$ with $|w|\le n$ you consider two permutations $a_w, b_w$ such that $w(a_w,b_w)$ has fixed points while some permutation in $\langle a_w, b_w\rangle $ has no fixed point. Then you take the disjoint using of the domains of these permutations, and define $a,b$ piece-wise. It probably will succeed but it is not clear to me how you define $a_w,b_w$. | |
| Jul 27, 2011 at 17:45 | history | answered | Gerhard Paseman | CC BY-SA 3.0 |