Timeline for Infinite noncyclic groups with abelian automorphism group
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Jun 10, 2019 at 1:49 | answer | added | tj_ | timeline score: 7 | |
| Jun 9, 2019 at 22:28 | comment | added | YCor | @DerekHolt of course I mean $n$-step as a variety, not as the complement of the $(n-1)$-step-nilpotent variety in the variety of $n$-step nilpotent groups. In the same way I consider abelian groups as metabelian groups. | |
| S Jun 9, 2019 at 21:40 | history | suggested | Somos | CC BY-SA 4.0 |
Added noncyclic as per author.
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| Jun 9, 2019 at 20:15 | review | Close votes | |||
| Jun 12, 2019 at 21:50 | |||||
| Jun 9, 2019 at 19:06 | comment | added | Derek Holt | @YCor So presumably you consider abelian groups to be $2$-step nilpotent? (Perhaps $n$-step nilpotent does not mean the same as nilpotent of class $n$.) | |
| Jun 9, 2019 at 18:06 | comment | added | Wojowu | Additive group of rationals has abelian automorphism group too, and is not cyclic. | |
| Jun 9, 2019 at 17:54 | review | Suggested edits | |||
| S Jun 9, 2019 at 21:40 | |||||
| Jun 9, 2019 at 17:54 | comment | added | Mare | @YCor Sorry, stupid error by me. I deleted the comment. | |
| Jun 9, 2019 at 17:49 | comment | added | Mohammad Radi | Every cyclic group has abelian automorphism group..i am seeking for a non cyclic infinite group or a sufficient condition that makes an infinite group to be a Miller group | |
| Jun 9, 2019 at 17:48 | comment | added | YCor | @Mare obviously not. A group with abelian automorphism group has to be 2-step nilpotent. | |
| Jun 9, 2019 at 17:43 | comment | added | Mohan | Group of integers? | |
| Jun 9, 2019 at 17:38 | history | asked | Mohammad Radi | CC BY-SA 4.0 |