Assume that I have a short exact sequence of finitely presented groups $$1 \longrightarrow K \longrightarrow H \longrightarrow G \longrightarrow 1,$$ where $G$ is finite (but I do not know whether this is relevant for what follows). Applying abelianization, we get an exact sequence $$K_{\mathrm{ab}} \longrightarrow H_{\mathrm{ab}} \to G_{\mathrm{ab}} \longrightarrow 0.$$

I would like to have a "quantitative" measure of the lack of left-exactness for the above sequence. For instance, I would like to know if it is possible to find an explicit sequence $\{L_i\}$ of (finitely presented) groups sitting in a long exact sequence of the form $$\ldots \longrightarrow L_3 \longrightarrow L_2 \longrightarrow L_1 \longrightarrow K_{\mathrm{ab}} \longrightarrow H_{\mathrm{ab}} \to G_{\mathrm{ab}} \longrightarrow 0.$$

By "explicit" I mean (for instance) that it is, at least in principle, possible to find presentations for the $L_i$ once one has presentations for $K$, $H$, $G$.

Since the category of groups is not abelian (or even additive) we cannot perform the usual construction of the derived functors for the abelianization functor. I am aware that some more refined constructions have been presented (cotangent complex, André-Quillen homology, etc, see for instance the comments to this MSE question) but they look very technical and perhaps overkill in the simple case I have in mind.

I am not an expert, but it seems to me that for the case of groups there should be some more down-to-earth construction, possibly related to the usual group (co)homology, but I looked in some standard textbooks and I did not find any. So, let me ask the following

Question. Is it possible to construct groups $L_i$ as above, in some (at least in principle) computable way, providing a sort of "explicit derived functor" for the abelianization functor? If so, what are some references?


2 Answers 2


One answer to your question is to say that there is not a long exact sequence of the kind of you want, but instead we get a spectral sequence.

Here is some explanation: of course the abelianization of a group $G$ is its first homology group $H_1(G; Z)$ with integer coefficients. The relationship between the homology groups of the groups $K,H$, and $G$ in your extension $1 \rightarrow K \rightarrow H \rightarrow G \rightarrow 1$ is given by the homological Lyndon-Hochschild-Serre spectral sequence for that extension. The homological LHSSS is a standard tool and there is a reasonable write-up on Wikipedia, here: https://en.wikipedia.org/wiki/Lyndon–Hochschild–Serre_spectral_sequence

In particular, the five-term exact sequence from the homological LHSSS with Z coefficients relates the abelianizations of the groups $K,H,$ and $G$ and extends your exact sequence of abelianizations in what seems to be the way you want, although it's not a long exact sequence, just a five-term exact sequence; if you want to get something homologically well-behaved that extends past those five terms, I think you have to work with the spectral sequence itself. The five-term exact sequence for the cohomological LHSSS is given on the Wikipedia page linked above; the homological five-term sequence is dual to that.

It is possible that you already know all this quite well, and that you were looking for something different. I apologize for my naive answer if that is the case.

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    $\begingroup$ This is a useful answer. Thank you. $\endgroup$ Feb 27, 2021 at 20:13

Here is a comment about the other approach proposed in the original question, using nonabelian derived functors. It is too long to be posted as a comment, so unfortunately I have had to post it as an answer.

The category of groups is not an abelian category, but the category of modules over a fixed group G is an abelian category, and derived functors in that abelian category yield group (co)homology. We could instead directly consider the abelianization functor $ab$ from groups to abelian groups, and we could consider nonabelian derived functors of $ab$. This would involve choosing a simplicial resolution of a group $G$, applying $ab$ to that simplicial group to get a simplicial abelian group, taking the normalized Moore chain complex of that simplicial group, and then taking the homology groups of that chain complex.

As far as I know, the standard textbook reference for these kinds of nonabelian derived functors is chapter 2 of H. Inassaridze's book "Non-abelian homological algebra and its applications." It is not a trivial matter to check that a short exact sequence of groups induces a long exact sequence of nonabelian derived functors: see Theorem 2.63 in Inassaridze's book for some sufficient conditions. In particular, I do not know if a short exact sequence of groups $1 \rightarrow K \rightarrow H \rightarrow G \rightarrow 1$ induces a long exact sequence $$\dots \rightarrow L_{n+1}ab(G) \rightarrow L_n ab(K) \rightarrow L_nab(H) \rightarrow L_nab(G)\rightarrow \dots $$ This would require checking the conditions of Inassaridze's Theorem 2.63, or something similar. In the absence of such a long exact sequence, I do not think the approach by nonabelian derived functors yields the long exact sequence $ \dots \rightarrow K_{ab} \rightarrow H_{ab}\rightarrow G_{ab} \rightarrow 1$ which you ask for.

The only positive result on nonabelian derived abelianization of groups which I was able to locate in the literature is mentioned in the proof of Theorem 3 in the paper "N-fold Cech derived functors and generalized Hopf type formulas" by G. Donadze, N. Inassaridze, and T. Porter: the first nonabelian derived functor of abelianization, $L_1ab(G)$, is apparently isomorphic to the second group homology group $H_2(G,Z)$ with integer coefficients, that is, the classical Schur multiplier of $G$.

Perhaps someone with more knowledge of nonabelian derived functors can contribute further and give a better answer to your question. It appears to me that going this route is not only more technically demanding than using group homology and the LHSSS, but there seem to be fewer tools available for making calculations (compared to the wealth of tools for computing group homology), and fewer connections to other areas of mathematics; and, possibly most troubling, it is not clear that you get the desired long exact sequence, in nonabelian derived abelianization, that you're looking for.

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    $\begingroup$ Thanks again. Looking at the dual of the five-term exact sequence coming from LHSSS, it seems to me that my $L_1$ should be isomorphic to something like $\operatorname{Hom}(H^2(G, \, \mathbb{Z}), \, \mathbb{Z})$, up possibly to some adjustments coming from the Universal Coefficients Theorem, and this more or less agrees with your claim that $L_1ab(G) \simeq H_2(G, \, \mathbb{Z})$. $\endgroup$ Feb 28, 2021 at 20:17

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