Timeline for Groups whose poset of direct factors are lattices
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Jun 15, 2020 at 7:27 | history | edited | CommunityBot |
Commonmark migration
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| Sep 17, 2018 at 21:06 | vote | accept | Rajkarov | ||
| Sep 17, 2018 at 12:55 | comment | added | Keith Kearnes | I agree with what you have said about the abelian case. | |
| Sep 17, 2018 at 12:23 | history | edited | zibadawa timmy | CC BY-SA 4.0 |
note the edit at the top for visibility
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| Sep 17, 2018 at 12:16 | history | edited | zibadawa timmy | CC BY-SA 4.0 |
half of the case with abelian direct factors now added as Theorem 2
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| Sep 16, 2018 at 10:53 | vote | accept | Rajkarov | ||
| Sep 16, 2018 at 10:53 | |||||
| Sep 16, 2018 at 10:32 | comment | added | zibadawa timmy | @Rajkarov Every subgroup of an elementary abelian $p$-group is a direct factor, so they're all $\mathcal{D}$-groups. It's the exponent, in the sense of wanting to (not) exhibit $z$ such that $\ker(z)$ is a proper non-trivial subgroup of an indecomposable abelian (thus cyclic of prime power order) group, which is relevant. | |
| Sep 16, 2018 at 10:20 | comment | added | Rajkarov | In the particular case of finite abelian groups, I think (if I'm not wrong) that a finite abelian group is a $\mathcal{D}$-group if and only if its sylow subgroups are indecomposable. | |
| Sep 16, 2018 at 10:10 | comment | added | zibadawa timmy | @Rajkarov Well my initial expectation was that normal endomorphisms, and the $z$ morphism I use in particular, is the fundamental "thing" to consider here, and yields a rather concrete demonstration of what prevents the $\mathcal{D}$-group property from holding. So I went in that direction, ultimately to realize there's a few more fiddly bits with abelian direct factors to deal with (there always is, pretty much). I don't think it's actually difficult to detail this case, I just ran out of time and energy today to spend on it. | |
| Sep 16, 2018 at 9:54 | comment | added | Rajkarov | Okay, I would just notice that a group satisfying both chain conditions on normal subgroups have unique Krull-Schmidt decomposition, up to the order of the factors, if and only if every direct factor have a unique normal complement. | |
| Sep 16, 2018 at 7:47 | history | answered | zibadawa timmy | CC BY-SA 4.0 |