Timeline for The Octahedral Axiom in group theory
Current License: CC BY-SA 4.0
16 events
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| Feb 12, 2020 at 20:53 | comment | added | LSpice |
@user44191, that's my fault; I edited to convert to AMScd, but didn't separate them enough. I won't edit again because I don't know what triggers CW status, but probably just putting them in two separate environments rather than in one big {gather} as I did would be enough.
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| Feb 12, 2020 at 4:18 | comment | added | user44191 | Is there an easy way to separate the two diagrams a bit more? As currently shown, it's a bit difficult to keep track of what's going where. | |
| Feb 12, 2020 at 3:49 | comment | added | Pace Nielsen | David, doesn't the exercise you assigned only looks like the third isomorphism theorem because (a) it fundamentally uses it, and (b) you used the lattice isomorphism theorem to write it in a "slicker" (but technically incorrect) manner. What I mean is that $X$ is not a subgroup of $Z/Y$ so $(Z/Y)/X$ literally makes no sense. Making things technical, you are claiming $(Z/Y)/(XY/Y) \cong (Z/X)(XY/X)$, which is a simple consequence of (two applications of) the 3rd isomorphism theorem. [Of course, you are correct that $XY/Y\cong X$ when $X\cap Y=1$, etc...] | |
| Feb 12, 2020 at 2:15 | history | edited | LSpice | CC BY-SA 4.0 |
AMScd diagrams
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| Feb 12, 2020 at 1:24 | comment | added | David E Speyer | @LSpice Sure, go ahead. | |
| Feb 11, 2020 at 23:08 | comment | added | Harry Gindi | @DanPetersen None of them, but analyzing the proof of TR4 from stability, it seems like you can get away with a bit less. | |
| Feb 11, 2020 at 20:39 | history | edited | LSpice | CC BY-SA 4.0 |
Deleted spurious spaces
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| Feb 11, 2020 at 20:21 | comment | added | Dan Petersen | @Harry Gindi In what stable oo-category would a general (nonabelian) group be an object? | |
| Feb 11, 2020 at 19:51 | history | edited | David E Speyer | CC BY-SA 4.0 |
deleted 8 characters in body
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| S Feb 11, 2020 at 17:38 | history | suggested | shane.orourke | CC BY-SA 4.0 |
Corrected statement of third isomorphism theorem
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| Feb 11, 2020 at 17:30 | review | Suggested edits | |||
| S Feb 11, 2020 at 17:38 | |||||
| Feb 11, 2020 at 17:15 | comment | added | Harry Gindi | This resemblance is because the octahedral axiom is a shadow of the exact same theorem in a stable ∞-category, where you replace cokernels by homotopy cofibres. See Theorem 1.1.2.14 of Jacob Lurie's Higher Algebra for a proof that the octahedral axiom arises naturally in this way | |
| Feb 11, 2020 at 16:54 | comment | added | David E Speyer | On thinking about this more, the right answer to the student's question is probably the $3 \times 3$ diagram built from a group $G$ and two normal subgroups $N_1$ and $N_2$, whose entries are $N_1 \cap N_2$, $N_1$, $N_1 N_2/N_2$, $N_2$, $G$, $G/N_2$, $N_1 N_2/N_1$, $G/N_2$ and $G/(N_1 N_2)$, and I was distracted by the octahderon. But I'll leave this up and see what someone else has to say. | |
| Feb 11, 2020 at 16:50 | history | edited | David E Speyer | CC BY-SA 4.0 |
added 24 characters in body
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| Feb 11, 2020 at 16:44 | comment | added | David E Speyer | I guess this looks a lot like the $3 \times 3$ lemma, and maybe that is the answer, but that usually comes with an assumption that the diagram commutes and deduces exactness, whereas here I have a bunch of exact sequences and the theorem is that there is an isomorphism making the diagram commute. | |
| Feb 11, 2020 at 16:27 | history | asked | David E Speyer | CC BY-SA 4.0 |