Timeline for Is having a Frobenius pair first-order expressible in the language of groups?
Current License: CC BY-SA 4.0
10 events
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| Feb 8, 2023 at 7:31 | vote | accept | Y. Tamer | ||
| Feb 6, 2023 at 14:08 | comment | added | YCor | By the way for infinite groups I'm not sure there's a single clear-cut definition of Frobenius group. One could pick as a definition, the existence of a proper nontrivial malnormal subgroup. But one could also add the (very strong, but automatic for finite groups) requirement that, for some nontrivial malnormal proper subgroup, the complement of the union of its conjugates, along with $\{1\}$, forms a normal subgroup. Hence, if you are really interested in infinite groups (besides pseudofinite ones), it is important you fully write down a definition of Frobenius pair. | |
| Feb 6, 2023 at 14:02 | history | edited | YCor |
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| Feb 6, 2023 at 13:28 | answer | added | YCor | timeline score: 4 | |
| Feb 6, 2023 at 8:55 | comment | added | YCor | One natural variant of the question is: does there exist a 1st order sentence $F$ such that for every finite group $G$, the group $G$ is Frobenius over some subgroup iff it satisfies $F$. | |
| Feb 6, 2023 at 8:39 | comment | added | YCor | Frobenius pair: $H\neq 1$, $H\neq G$, and every element of $G$ has at most one fixed point on $G/H$ (i.e., $g\notin H$ implies $g^{-1}Hg\cap H=\{1\}$). | |
| Feb 6, 2023 at 8:21 | comment | added | Emil Jeřábek | Can you define what is a Frobenius pair? | |
| Feb 6, 2023 at 8:17 | history | edited | Y. Tamer |
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| S Feb 6, 2023 at 8:02 | review | First questions | |||
| Feb 6, 2023 at 8:56 | |||||
| S Feb 6, 2023 at 8:02 | history | asked | Y. Tamer | CC BY-SA 4.0 |