Timeline for Regular subsets of $\text{PSL}(2, q)$
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
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| Aug 28, 2019 at 8:46 | comment | added | Sean Eberhard | To revise my previous comment, $\text{PSL}(2, q)$ has a regular set for its action on $\mathbf{F}_q^2 \setminus \{0\} / \{\pm1\}$ for $q = 2, 3, 5, 7, 11$, but I would guess in no other cases. | |
| Aug 19, 2019 at 11:19 | comment | added | Sean Eberhard | The other generalization I'm thinking about is other actions of $\text{PSL}(2, q)$, for example on $\mathbf{F}_q^2 \setminus\{0\} / \{\pm 1\}$. For example $\text{PSL}(2, 5)$ has a regular subgroup (isomorphic to $A_4$), but presumably there are no other examples. | |
| Aug 19, 2019 at 11:17 | comment | added | Sean Eberhard | There are two generalizations I'm thinking about now. The first is to $\text{PSL}(n, q)$ acting on projective $(n-1)$-space. There is an "obvious" regular subgroup if and only if $\gcd(q -1, n) = 1$ (cyclic case) or $q\equiv 3 \pmod 4$ and $n \equiv 2 \pmod 4$ (dihedral case). A natural conjecture is that there is no regular subset in the other cases. The first of these would be $\text{PSL}(3, q)$ acting on the projective plane when $q \not\equiv 1 \pmod 3$. | |
| Aug 17, 2019 at 14:44 | comment | added | Sean Eberhard | Thank you for the pointer to your paper! I'm afraid it had escaped my notice, and is of course highly relevant. | |
| Aug 17, 2019 at 9:27 | history | edited | Peter Mueller | CC BY-SA 4.0 |
Added a remark
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| Aug 17, 2019 at 8:49 | comment | added | Sean Eberhard | @FedorPetrov Your element $g$ fixes $\infty$, so it's not considered. | |
| Aug 17, 2019 at 8:47 | vote | accept | Sean Eberhard | ||
| Aug 17, 2019 at 8:31 | comment | added | Sean Eberhard | There are 3 degrees of freedom in your choice of $M$: you can translate by $J$, $I$, or by the matrix which is $1$ if $x = \infty$ or $y = \infty$ but not both. This accounts for your 8 observed choices. | |
| Aug 17, 2019 at 8:27 | comment | added | Sean Eberhard | I have proved that your $M$ indeed works, though I'm still working on a higher-level understanding of what's going on. To prove that it works reduces to showing for every fixed-point-free $g \in G$, say $\def\eps{\epsilon} g = (a, \eps b ; b, a)$ where $\eps$ is nonsquare and $a^2 - \eps b^2 = 1$, there are an odd number of finite $x$ mapped to a finite point $x^g$ such that $x^g - x$ is square. This is equivalent to showing that $(x^2 - \eps)/(x - a/b) = z^2$ has an odd number of solutions $x$. You can check that there is exactly one value of $z^2$ for which this equation has a double root. | |
| Aug 17, 2019 at 7:34 | history | edited | Peter Mueller | CC BY-SA 4.0 |
added 53 characters in body
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| Aug 16, 2019 at 22:56 | comment | added | Sean Eberhard | This is exactly the sort of parity obstruction I was hoping to find, thank you! I'm sure there is a very simple way of expressing it now that you've found it, but I think I might have learned something about how to do maths from your answer. | |
| Aug 16, 2019 at 21:28 | history | edited | Peter Mueller | CC BY-SA 4.0 |
added 237 characters in body
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| Aug 16, 2019 at 16:33 | history | answered | Peter Mueller | CC BY-SA 4.0 |