Timeline for finite solvable groups whose commutator subgroup is nilpotent
Current License: CC BY-SA 3.0
14 events
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| Nov 9, 2016 at 15:33 | comment | added | Geoff Robinson | @NazihNahlus : My first comment above points out that they are precisely the groups with $G/F(G)$ Abelian. There are groups of Fitting length $2$ which do not have that property. | |
| Nov 3, 2016 at 23:56 | comment | added | Geoff Robinson | I do not consider it unreasonable to use standard group theoretic notation in a comment on a question with a finite groups tag. | |
| Nov 3, 2016 at 16:42 | comment | added | YCor | @GeoffRobinson corollary of our two comments: everybody is not a finite group theorist (at least according to your definition) Anyway, I haven't done a graduate course in infinite group theory either... | |
| Nov 3, 2016 at 9:12 | comment | added | Geoff Robinson | @YCor: I can't speak for infinite group-theorists, but I think that all finite group theorists (who have done a graduate course including finite group theory) would be familiar with $F(G)$ and $\Phi(G).$ In any case, Arturo Magidin's explanation is accurate. | |
| Nov 3, 2016 at 5:02 | comment | added | Arturo Magidin | @YCor: $F(G)$ is the Fitting subgroup (unique largest normal nilpotent subgroup); $\Phi(G)$ is the Frattini subgroup (intersection of maximal subgroups). | |
| Nov 3, 2016 at 4:46 | comment | added | YCor | Groups with a nilpotent derived subgroup are called "nilpotent-by-abelian". @GeoffRobinson I'm not sure everybody knows your notation ($F(G)$, $\Phi(G)$) by heart. I don't. | |
| Nov 3, 2016 at 4:41 | history | edited | Nazih Nahlus |
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| Nov 2, 2016 at 13:39 | comment | added | Nazih Nahlus | Thank you. You may check my Conjecture/Question at mathoverflow.net/questions/253729/… | |
| Nov 2, 2016 at 13:16 | comment | added | Geoff Robinson | Another characterization is that they are the groups $G$ such that $G/\Phi(G)$ is Metabelian. | |
| Nov 2, 2016 at 12:35 | comment | added | Tobias Kildetoft | I have no idea if these have actually been studied. One thing is that they satisfy the Taketa inequality, as I explain at math.stackexchange.com/questions/272849/… | |
| Nov 2, 2016 at 12:35 | comment | added | Geoff Robinson | These are precisely the finite groups $G$ such that $G/F(G)$ is Abelian. Additionally, $G/F(G)$ is isomorphic to a subgroup of ${\rm Out}(F(G))$ for such a group. I'm not sure what more one can expect to say. | |
| Nov 2, 2016 at 12:28 | history | edited | Nazih Nahlus | CC BY-SA 3.0 |
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| Nov 2, 2016 at 12:22 | history | edited | Carlo Beenakker | CC BY-SA 3.0 |
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| Nov 2, 2016 at 12:19 | history | asked | Nazih Nahlus | CC BY-SA 3.0 |