Timeline for Is there any criteria for whether the automorphism group of G is homomorphic to G itself?
Current License: CC BY-SA 2.5
6 events
| when toggle format | what | by | license | comment | |
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| Aug 21, 2010 at 0:18 | comment | added | Joel David Hamkins | Yes, of course that's right! | |
| Aug 20, 2010 at 15:55 | comment | added | Simon Thomas | The group $G$ in my paper is the stabilizer $S_{(\mathcal{U})}$ of a nonprincipal ultrafilter $\mathcal{U}$ and the notion of forcing $\mathbb{P}$ is the corresponding Mathias forcing. | |
| Aug 20, 2010 at 15:51 | comment | added | Simon Thomas | But none of the new permutations normalize the old symmetric group which remains complete in any forcing extension. This is Theorem 2.2 in: S. Thomas, The automorphism tower problem II, Israel J. Math. 103 (1998), 93-109. | |
| Aug 20, 2010 at 15:45 | history | edited | Simon Thomas | CC BY-SA 2.5 |
added 281 characters in body
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| Aug 20, 2010 at 13:39 | comment | added | Joel David Hamkins | Simon, isn't the infinite symmetric group $Sym_\omega$ of V already an instance of this? After all, if you add a Cohen real $c$ by forcing (or any new real), then the $Sym_\omega$ of $V$ gains new automorphisms in $V[c]$, simply because there are new permutations of $\omega$ in $V[c]$. Do we know how the automorphism tower of $Sym_\omega^V$ is affected in $V[c]$? | |
| Aug 20, 2010 at 10:55 | history | answered | Simon Thomas | CC BY-SA 2.5 |