Timeline for When are elements of a (perfect) semidirect product simple commutators?
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
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| Apr 5, 2023 at 17:12 | comment | added | Sean Eberhard | @Makenzie The notation can be confusing. Particularly $vg$ can be ambiguous because $V$ is both a right $G$-module and a subgroup of $VG$. Moreover we tend to use additive notation in $V$ but multiplicative notation in $VG$. It might clearer if I wrote $[wx,y] = x^{-1} w^{-1} y^{-1} w x y = w^{-x} w^{yx} [x,y] = (-w^x + w^{yx}) [x,y]$. | |
| Apr 4, 2023 at 17:28 | comment | added | Makenzie | How do you conclude that $[wx,y]=(-wx+wyx)[x,y]$? | |
| Apr 3, 2023 at 12:33 | vote | accept | Makenzie | ||
| Mar 29, 2023 at 14:29 | comment | added | LSpice | Some mutterings: If $y$ itself doesn't work then we can replace $x$ and $y$ by $-x$ and $-y$. That will work unless both $x$ and $y$ have both $\pm1$ as eigenvalues. For example, for $m = 2$ in characteristic $\ne 2$, it seems like we might just be able to conclude by hand. In general, the only problem is if $x C_G(y)$ and $y C_G(x)$ consist entirely of elements with fixed points. It seems like this is probably hardest to rule out when both $x$ and $y$ are regular unipotent. | |
| Mar 29, 2023 at 13:56 | history | edited | Sean Eberhard | CC BY-SA 4.0 |
add reference to a relevant result of Gow
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| Mar 28, 2023 at 14:51 | history | answered | Sean Eberhard | CC BY-SA 4.0 |