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$\require{AMScd}$Here are two results about groups:

(The third isomorphism theorem) Suppose that I have $A \triangleleft B \triangleleft C$ and $A \triangleleft C$. Then $C/B \cong (C/A)/(B/A)$.

(An exercise I just assigned my students) Suppose that we have $X \triangleleft Z$ and $Y \triangleleft Z$ with $X \cap Y = 1$. Then $(Z/X)/Y \cong (Z/Y)/X$.

Vague question a student just asked me: Is there some general context in which to think of these results and why they look similar?

We can write these as diagrams with exact rows and columns: \begin{gather} \begin{CD} @. @. 1 @. 1 \\ @. @. @VVV @VVV \\ 1 @>>> A @>>> B @>>> B/A @>>> 1 \\ @. @| @VVV @VVV \\ 1 @>>> A @>>> C @>>> C/A @>>> 1 \\ @. @. @VVV @VVV \\ @. @. C/B @>\cong>> (C/A)/(B/A) \\ @. @. @VVV @VVV \\ @. @. 1 @. 1 \end{CD} \\ \begin{CD} @. @. 1 @. 1 \\ @. @. @VVV @VVV \\ @. @. X @= X \\ @. @. @VVV @VVV \\ 1 @>>> Y @>>> Z @>>> Z/Y @>>> 1 \\ @. @| @VVV @VVV \\ 1 @>>> Y @>>> Z/X @>>> (Z/X)/Y \cong (Z/Y)/X @>>> 1 \\ @. @. @VVV @VVV \\ @. @. 1 @. 1 \end{CD} \end{gather} If we were in an abelian category, these would be two forms of the octahedral axiom.

Vague but more technical question: Is there something like a semi-abelian category which includes the case of groups, and where we have something like an octahedral axiom?

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    $\begingroup$ 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. $\endgroup$ – David E Speyer Feb 11 at 16:44
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    $\begingroup$ 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. $\endgroup$ – David E Speyer Feb 11 at 16:54
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    $\begingroup$ 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 $\endgroup$ – Harry Gindi Feb 11 at 17:15
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    $\begingroup$ @Harry Gindi In what stable oo-category would a general (nonabelian) group be an object? $\endgroup$ – Dan Petersen Feb 11 at 20:21
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    $\begingroup$ 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...] $\endgroup$ – Pace Nielsen Feb 12 at 3:49

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