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Talk:Splitting lemma

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Latest comment: 1 year ago by 129.104.241.57 in topic For non-abelian groups

Pretty printing

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this needs some pretty printing :) It's difficult to read...

Proof is not general

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While the statement is correct in any abelian category, the given proof applies only to categories of abelian groups, modules over a ring, etc. This should be noted. (Perhaps it would be satisfactory to note that the general case follows by Freyd's embedding theorem).

I noticed the same think. It seems that we could get it much more easily by the definition of biproduct. (which definition in http://ncatlab.org/nlab/show/biproduct is more clear to me than the one in wikipedia)
  • Erm. I have a question as regards non-abelian groups. There is a natural inclusion u from quotient C to B=A⋊C, but what is the natural projection t from B=A⋊C to normal factor A, unless B is actually a direct product A×C ?? --192.75.48.150 17:38, 27 July 2007 (UTC)Reply
AKA Mitchell's embedding theorem the redlink above should be a redirect. 84.15.184.13 (talk) 10:37, 16 July 2023 (UTC)Reply

Unclear hypothesis

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In the statement of the lemma: would it be helpful to clarify that by hypothesis, the maps q and r are in a short exact sequence, but that we don't require this a priori of t and u? Jaswenso 03:15, 6 September 2007 (UTC)Reply

Proof incomplete

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Proof incomplete?

I thought it is not sufficient to prove that . Why do we then prove only this in the implication "1=>3"? Freeze S (talk) 00:36, 15 January 2017 (UTC)Reply

For non-abelian groups

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In general, left-split <=> direct product, and right-split <=> semidirect product. Unfortunately, a split exact sequence of groups usually means right-split, for example in the statement of Schur–Zassenhaus theorem. See also Theorem 3.3 and Definition 3.4 of here. 129.104.241.57 (talk) 02:02, 23 November 2024 (UTC)Reply