Timeline for Conjugacy classes in the absolute galois group
Current License: CC BY-SA 2.5
5 events
| when toggle format | what | by | license | comment | |
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| Mar 11, 2010 at 2:38 | comment | added | moonface | Yes, dihedrals of 2xodd order, and that's exactly what I had in mind. About the first comment, I don't follow it either (I had hastily assumed that any inverse limit of finite sets with increasing size and surjective transition maps is uncountable...) Thanks for catching. | |
| Mar 10, 2010 at 23:49 | comment | added | senti_today | @moonface: I am not sure I follow your first comment. How does it imply that the number of conjugacy classes is not countable? In the second comment, you probably mean "dihedral group of order not divisible by 4" (otherwise there are two conjugacy classes of reflections, if I'm not mistaken). If I understand you correctly, the sort of example you have in mind is: consider $\mathbb{Z}/2\mathbb{Z}$ acting on $\mathbb{Z}_p$ via $x\mapsto-x$, where $p$ is an odd prime, and form the semidirect product. Then the nontrivial coset of $\mathbb{Z}_p$ is a single conjugacy class of Haar measure 1/2. | |
| Mar 10, 2010 at 23:16 | comment | added | moonface | @senti_today : The first statement is true, because the number of conjugacy classes in a finite group cannot remain bounded as the size of the group increases. The second statement is false: Reflections inside a dihedral group are a conjugacy class of measure $\frac{1}{2}$, and you can convert this to an infinite example. | |
| Mar 10, 2010 at 20:45 | comment | added | senti_today | This is a side comment, but I think it's plausible that if G is any infinite profinite group, then there are uncountably many conjugacy classes in G. In fact, I believe that in such a G the Haar measure of any conjugacy class is zero (but please correct me if I am wrong). | |
| Mar 10, 2010 at 17:53 | history | answered | Kevin Buzzard | CC BY-SA 2.5 |