Timeline for Can a group have a cyclical derived series?
Current License: CC BY-SA 4.0
12 events
when toggle format | what | by | license | comment | |
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Aug 20, 2019 at 21:59 | history | became hot network question | |||
Aug 20, 2019 at 19:24 | vote | accept | Santana Afton | ||
Aug 20, 2019 at 19:23 | answer | added | Derek Holt | timeline score: 15 | |
Aug 20, 2019 at 16:25 | comment | added | YCor | @DerekHolt great. I'd be happy if you post an answer then as I haven't gone much into that paper. | |
Aug 20, 2019 at 15:40 | comment | added | Derek Holt | Furthermore, in the example in Section 9 of that paper, $G_1$ has order $2$ and, for $i>1$, $Z(G_i) < G_i'$, so it is easy to see that $G$ and $G'$ are not isomorphic. So we do indeed have an example of a group with cyclical derived seriesof period $2$. | |
Aug 20, 2019 at 15:32 | history | edited | YCor | CC BY-SA 4.0 |
added remarks from discussion on MathSE to avoid repetition here.
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Aug 20, 2019 at 15:15 | comment | added | Derek Holt | YCor's Question (a) is considered in this paper by B.H. Neumann, and some examples are constructed in Section 9 and 10. | |
Aug 20, 2019 at 15:02 | comment | added | Santana Afton | @YCor The free group on countably many generators should be one such example. | |
Aug 20, 2019 at 15:02 | comment | added | Derek Holt | @YCor what about a free group of countably infinite rank? | |
Aug 20, 2019 at 14:50 | comment | added | user44191 | @YCor a couple ideas that might make (a) easier: instead of terminating, you could use a biinfinite sequence. Alternatively, you could terminate in any perfect group, not the trivial one, if I'm not mistaken. | |
Aug 20, 2019 at 14:42 | comment | added | YCor | Let $(G_n)$ be a sequence of groups with $G_0=1$ and $[G_n,G_n]$ isomorphic to $G_{n-1}$. Consider $G=\bigoplus_{n\ge 1}G_{2n}$. Then its derived subgroup $G'$ is isomorphic to $\bigoplus_{n\ge 1}G_{2n-1}$, and its second derived subgroup is isomorphic to $G$. Next we have (a) to exhibit such a sequence $(G_n)$, and ensure (b) that $G$ is not isomorphic to $G'$. Quite surprisingly (a) is not obvious to me (I don't expect (b) to be an issue but it makes no sense before having an example). | |
Aug 20, 2019 at 13:49 | history | asked | Santana Afton | CC BY-SA 4.0 |