Timeline for If $K\rtimes \mathbb{Z}$ is a finitely generated group but $K$ isn't, must the fixed points of $1_\mathbb{Z}$ be a finitely generated group?
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Mar 27, 2022 at 7:17 | answer | added | Derek Holt | timeline score: 7 | |
| Mar 26, 2022 at 12:07 | vote | accept | Chris Sanders | ||
| Mar 26, 2022 at 11:01 | vote | accept | Chris Sanders | ||
| Mar 26, 2022 at 11:10 | |||||
| Mar 26, 2022 at 10:57 | answer | added | YCor | timeline score: 11 | |
| Mar 26, 2022 at 10:34 | comment | added | Matt Zaremsky | Oh, ha, right. Not paying attention. | |
| Mar 26, 2022 at 10:28 | history | edited | Chris Sanders | CC BY-SA 4.0 |
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| Mar 26, 2022 at 10:23 | history | edited | Chris Sanders | CC BY-SA 4.0 |
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| Mar 26, 2022 at 10:21 | comment | added | Chris Sanders | @MattZaremsky If you use the trivial action i.e. the direct product, "$K$ is not finitely generated" implies "$K\oplus\mathbb{Z}$ is not finitely generated" | |
| Mar 26, 2022 at 10:15 | comment | added | Matt Zaremsky | If you use the trivial action, so you're looking at the direct product, then the fixed points are all of $K$, hence non-finitely generated. But probably you want the action to be sufficiently "robust" in some sense? | |
| Mar 26, 2022 at 10:06 | history | edited | Chris Sanders |
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| S Mar 26, 2022 at 9:46 | review | First questions | |||
| Mar 26, 2022 at 11:05 | |||||
| S Mar 26, 2022 at 9:46 | history | asked | Chris Sanders | CC BY-SA 4.0 |