Timeline for Is the affine group generically 2-generated?
Current License: CC BY-SA 3.0
17 events
| when toggle format | what | by | license | comment | |
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| S Nov 6, 2016 at 13:12 | history | suggested | Luc Guyot | CC BY-SA 3.0 |
Tries to make clear which map should be surjective
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| Nov 6, 2016 at 12:55 | review | Suggested edits | |||
| S Nov 6, 2016 at 13:12 | |||||
| Nov 5, 2016 at 9:07 | history | edited | Jan-Christoph Schlage-Puchta | CC BY-SA 3.0 |
corrected the last paragraph and expanded the displayed equation.
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| Nov 5, 2016 at 9:03 | comment | added | Jan-Christoph Schlage-Puchta | Right about the translation. I made an error computing $(ax+b)^{p-1}$. Corrected it. | |
| Nov 4, 2016 at 22:52 | vote | accept | Igor Rivin | ||
| Nov 4, 2016 at 19:06 | comment | added | SashaP | @FriederLadisch Sure, you are completely right. | |
| Nov 4, 2016 at 17:49 | comment | added | Frieder Ladisch | @SashaP: When I expand "my" product, I get the second sum in Jan-Christoph's formula. My reasoning is so: write $C_n=C_{p-1}$ as direct product of cyclic groups $C_{q^k}$ of prime power order. Two elements generate $C_n$ iff their projections to every direct factor generate that factor. Two random elements of $C_{q^k}$ do not generate $C_{q^k}$ with probability $1/q^2$. | |
| Nov 4, 2016 at 17:21 | comment | added | SashaP | @FriederLadisch Why it is not $\prod_q(1-1/q^2+1/q^4-\dots+(-1)^{k_q}1/q^{2k_q})$ where $k_q=ord_q(p-1)$? | |
| Nov 4, 2016 at 17:12 | comment | added | YCor | @IgorRivin A subgroup with no translation is abelian. Then count the number of commuting pairs. | |
| Nov 4, 2016 at 17:06 | comment | added | Igor Rivin | @YCor Why is the last statement true? | |
| Nov 4, 2016 at 17:00 | comment | added | YCor | The assertion "a random element almost surely generates a translation" is not correct. An element $x\mapsto ax+b$, with $a\neq 1$, has no power that is a nonzero translation. Indeed, such a power has the form $x\mapsto a^kx+(a^k-1)/(a-1)$ (other interpretation: non-translations are conjugate to homotheties). On the other hand it's true that a random pair generates a subgroup containing a translation. | |
| Nov 4, 2016 at 15:57 | comment | added | Frieder Ladisch | @SashaP, IgorRivin: the sum equals $\prod_q (1 - (1/q^2))$, where the product runs over primes $q$ dividing $p-1$. One of these primes is $q=2$. (Personally, when trying to compute the probability that two elements generate a cyclic group, I found this product formula first anyway.) | |
| Nov 4, 2016 at 14:35 | comment | added | Igor Rivin | I am also a bit confused by the $\frac34.$ | |
| Nov 4, 2016 at 14:30 | comment | added | Igor Rivin | The translation comment indicates that two random elements a.s. generate a transitive subgroup, I guess. | |
| Nov 4, 2016 at 13:26 | comment | added | Derek Holt | I think you mean two random elements almost certainly generate a subgroup containing a translation. A single random element is very likely to have order dividing $p-1$. | |
| Nov 4, 2016 at 13:04 | comment | added | SashaP | Could you please elaborate on why this sum is less that $3/4$? | |
| Nov 4, 2016 at 12:39 | history | answered | Jan-Christoph Schlage-Puchta | CC BY-SA 3.0 |