Timeline for Is it possible to classify extensions $G$ of an abelian $A$ by an abelian $N$ such that the map $G\rightarrow A$ is abelianization?
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 11, 2018 at 20:50 | vote | accept | stupid_question_bot | ||
| Oct 11, 2018 at 20:50 | comment | added | stupid_question_bot | Ah, the thing I was missing was that $(a-1)n$ is a commutator! ($a\in A, n\in N$). I guess there is some relation between this and the Schur multiplier... | |
| Oct 11, 2018 at 6:57 | comment | added | YCor | @rtz my initial statement was not correct: for instance if the whole extension is abelian, the extension can be non-split. So I rephrased. Also there's no need of reference: the only thing is to observe, given a central extension $Z\to G\to G/Z$, that the commutator map $G\times G\to G$ factors though $G/Z\times G/Z\to G$, and if moreover $G/Z$ is abelian, it yields a map $G/Z\times G/Z\to Z$, which is clearly bilinear and alternating. | |
| Oct 11, 2018 at 6:53 | history | edited | YCor | CC BY-SA 4.0 |
corrected wrong statement
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| Oct 10, 2018 at 22:35 | comment | added | stupid_question_bot | Would you happen to have a reference for this? (or perhaps a slick way to check this without doing a bunch of computations with cocycles?) | |
| Oct 10, 2018 at 22:23 | history | answered | YCor | CC BY-SA 4.0 |