Timeline for Can $\text{Aut}(G)$ be extended to contain $G$?
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Aug 24 at 15:45 | comment | added | Derek Holt | $G = {\rm SL}(2,9)$ is a counterexample with ${\rm Aut}(G) \cong{\rm P \Gamma L}(2,9)$. | |
| Aug 24 at 15:38 | comment | added | David Schwein | @NeilStrickland How would this extension give an action of Out(G) on G? I don't see why that follows from general principles. | |
| Aug 24 at 15:36 | history | edited | David Schwein | CC BY-SA 4.0 |
Removed judgmental language.
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| Aug 24 at 15:32 | vote | accept | David Schwein | ||
| Aug 24 at 13:55 | comment | added | no upstairs | It's like saying $1+1\neq3$ is a defect. | |
| Aug 24 at 12:37 | comment | added | YCor | (1) is too weak, since the direct product does the job. One should at least ask the existence of a homomorphism $\widetilde{\mathrm{Aut}}\to\mathrm{Aut}$ such that the obvious squares (between the two short exact sequences) commute, and or the induced endomorphism of $\mathrm{Out}(G)$ be the identity. | |
| Aug 24 at 12:29 | answer | added | Kasper Andersen | timeline score: 14 | |
| Aug 24 at 12:28 | comment | added | Neil Strickland | An extension as in (1) would give an action of $\text{Out}(G)$ on $G$ by conjugation. In the case $G=\Sigma_6$ this would give a choice of outer automorphism of $\Sigma_6$, which cannot be natural. Here the centre is trivial so $\text{Aut}(G)=\text{Aut}^{\circ}(G)$. | |
| Aug 24 at 12:05 | history | asked | David Schwein | CC BY-SA 4.0 |