Timeline for Periodic Automorphism Towers
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Dec 9, 2015 at 9:54 | history | edited | Stefan Kohl♦ | CC BY-SA 3.0 |
Added a pointer to the dementi in the comments.
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| Jun 12, 2010 at 13:24 | comment | added | Simon Thomas | I see that we spotted the typo simultaneously! | |
| Jun 12, 2010 at 13:23 | comment | added | Simon Thomas | Looking more carefully at Collins' paper, I believe there is a typo in the statement of Theorem 1. Combining Proposition A and Proposition B, we see that $Aut(Aut(G)) \cong Aut(G)$ instead of $Aut(Aut(G)) \cong G$. And my previous comment shows that $Aut(Aut(G)) \not\cong G$. So I no longer believe that Collins' paper answers the infinite case of my question. | |
| Jun 12, 2010 at 13:15 | comment | added | Guntram | Indeed, after reading the paper more closely I think Theorem (ii) should read that Aut(Aut(G)) is isomorphic to Aut(G), so this paper doesn't seem to provide an example. Sorry about that. | |
| Jun 12, 2010 at 12:26 | comment | added | Simon Thomas | This looks like an example but there is one part of your argument that doesn't seem correct. By Lemma 1 of Collins, when $r=1$, the group $G$ is centreless. This means that $Aut(G)$ is also centreless and hence embeds in $Aut(Aut(G)) \cong G$. Thus $Aut(G)$ cannot have an element of finite order. Of course, given the statement of Collins' Theorem, it seems almost certain that one of his groups does provide an answer to the infinite case of my question. | |
| Jun 12, 2010 at 9:29 | history | edited | Guntram | CC BY-SA 2.5 |
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| Jun 12, 2010 at 9:19 | history | edited | Guntram | CC BY-SA 2.5 |
added 312 characters in body; added 20 characters in body
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| Jun 12, 2010 at 9:02 | history | answered | Guntram | CC BY-SA 2.5 |