Timeline for Extensions with trivial induced outer action
Current License: CC BY-SA 3.0
12 events
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May 5, 2013 at 6:18 | comment | added | Mark Grant | I guess my stumbling block is I don't yet know exactly how to pass between extensions $N\to E\to G$ with a given induced outer action and central extensions $C\to A\to G$. | |
May 5, 2013 at 6:17 | comment | added | Mark Grant | @HW (in response to your second-to-last comment): Is it so? If $N$ is non-abelian, the equivalence classes of extensions with kernel $N$ and trivial outer action are non-canonically isomorphic to $H^2(G;C)$. We know that one such extension is split (indeed, the trivial extension is amongst them). Could it be that they all split? | |
May 3, 2013 at 19:50 | comment | added | HJRW | (Where the action of $G$ on $C$ is trivial.) | |
May 3, 2013 at 19:49 | comment | added | HJRW | Indeed, I think you answered your own question: there are such examples whenever $H^2(G,C)\neq 0$. | |
May 3, 2013 at 19:48 | comment | added | HJRW | For your second question, consider for instance $1\to 2\mathbb{Z}\to\mathbb{Z}\to\mathbb{Z}/2\to 1$. | |
May 3, 2013 at 14:20 | history | edited | Mark Grant | CC BY-SA 3.0 |
Added follow-up question
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May 2, 2013 at 18:35 | answer | added | Peter Michor | timeline score: 1 | |
May 2, 2013 at 15:45 | comment | added | Derek Holt | The assumptions are equivalent to $E = NC_E(N)$. | |
May 2, 2013 at 12:32 | comment | added | HJRW | If the outer action is trivial then the centre of $N$ is indeed central in $E$. | |
May 2, 2013 at 12:29 | comment | added | Colin Reid | I don't see any reason why $N$ should be abelian unless you impose some additional conditions. | |
May 2, 2013 at 12:25 | comment | added | HJRW | If $N$ is non-abelian, then it can't be central in $E$. On the other hand, $E=G\times N$ has the property that the outer action of $G$ is trivial. | |
May 2, 2013 at 12:22 | history | asked | Mark Grant | CC BY-SA 3.0 |