Timeline for Proving that a subgroup is normal
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Jul 15, 2015 at 20:18 | comment | added | Timothy Chow | Yes, I wasn't thinking too hard when I wrote "definable without parameters"; I guess that condition is too strong to be interesting. Is there any interesting way to weaken it? | |
| Jul 15, 2015 at 7:03 | comment | added | YCor | Well, I don't see the point: groups defined mathematically are defined in terms of the abstract structure of the group as well. The main point is the use of parameters (e.g., if we have a group action, when we pick a point stabilizer). On the other hand, even if allowed to quantify on integers or subsets of the group, if we use no parameters (hidden in any kind of way, as is often the case since most usual groups are defined either with a prescribed action, or with distinguished generators/elements) we should only be able to define automorphism-invariant subsets. | |
| Jul 15, 2015 at 1:50 | comment | added | Will Sawin | Note that this only works for definitions in terms of the abstract structure of the group, not for subgroups of a group you find in mathematics that you can define mathematically. | |
| Jul 14, 2015 at 22:23 | comment | added | Joel David Hamkins | @YCor That argument seems to apply to second-order and higher-order definitions as well! You could allow definitions of iterated transfinite order, which essentially allows one to build a set-theoretic universe atop the group. | |
| Jul 14, 2015 at 22:14 | comment | added | YCor | It is rather a triviality than a theorem that any subset of a group that is 1st order definable without parameters is invariant under all group automorphisms (so in particular, if it's a subgroup, then it's characteristic, hence normal). | |
| Jul 14, 2015 at 22:07 | history | asked | Timothy Chow | CC BY-SA 3.0 |