Timeline for When does an injective group homomorphism have an inverse?
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Nov 7, 2018 at 21:28 | answer | added | Denis T | timeline score: 1 | |
| S Nov 7, 2018 at 18:52 | history | suggested | user86028 | CC BY-SA 4.0 |
Using mathjax for formulas.
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| Nov 7, 2018 at 18:51 | review | Suggested edits | |||
| S Nov 7, 2018 at 18:52 | |||||
| Jun 28, 2010 at 13:46 | vote | accept | Hans-Peter Stricker | ||
| Jun 28, 2010 at 11:07 | comment | added | Hans-Peter Stricker | @Martin: I see. | |
| Jun 28, 2010 at 11:05 | comment | added | Martin Brandenburg | I only adressed the case that $B$ is abelian. | |
| Jun 28, 2010 at 10:57 | comment | added | Hans-Peter Stricker | BB in his answer below mentions semidirect products, you talk about direct sums. Who of you is right (or both)? | |
| Jun 28, 2010 at 10:50 | answer | added | Bugs Bunny | timeline score: 14 | |
| Jun 28, 2010 at 10:49 | comment | added | Martin Brandenburg | You may assume that $f$ is the inclusion of a subgroup $A \subseteq B$. I don't think that there is a general simple criterion. If $B$ is abelian, you have the splitting lemma which says that $A \subseteq B$ has a retract if and only if $B \to B/A$ has a section if and only if $A$ is a direct summand of $B$. | |
| Jun 28, 2010 at 10:41 | history | asked | Hans-Peter Stricker | CC BY-SA 2.5 |