It is a basic result of group cohomology that the extensions with a given abelian normal subgroup A and a given quotient G acting on it via an action $\varphi$ are given by the second cohomology group $H^2_\varphi(G,A)$. In particular, when the action is trivial (so the extension is a central extension), this is the second cohomology group $H^2(G,A)$ for the trivial action. In the special case where G is also abelian, we classify all the class two groups with A inside the center and G as the quotient group.

I am interested in the following: given a sequence of abelian groups $A_1, A_2, \dots, A_n$, what would classify (up to the usual notion of equivalence via commutative diagrams) the following: a group E with an ascending chain of subgroups:

$$1 = K_0 \le K_1 \le K_2 \le \dots \le K_n = E$$

such that the $K_i$s form a central series (i.e., $[E,K_i] \subseteq K_{i-1}$ for all i) and $K_i/K_{i-1} \cong A_i$?

The case $n = 2$ reduces to the second cohomology group as detailed in the first paragraph, so I am hoping that some suitable generalization involving cohomology would help describe these extensions.

Note: As is the case with the second cohomology group, I expect the object to classify, not isomorphism classes of possibilities of the big group, but a notion of equivalence class under a congruence notion that generalizes the notion of congruence of extensions. Then, using the actions of various automorphism groups, we can use orbits under the action to classify extensions under more generous notion of equivalence.

Note 2: The crude approach that I am aware of involves building the extension step by step, giving something like a group of groups of groups of groups of ... For intsance, in the case $n = 3$:

$$1 = K_0 \le K_1 \le K_2 \le K_3 = G$$

with quotients $A_i \cong K_i/K_{i-1}$, I can first consider $H^2(A_3,A_2)$ as the set of possibilities for $K_3/K_1$ (up to congruence). For each of these possibilities P, there is a group $H^2(P,A_1)$ and the total set of possibilities seems to be:

$$\bigsqcup_{P \in H^2(A_3,A_2)} H^2(P,A_1)$$

Here the $\in$ notation is being abused somewhat by identifying an element of a cohomology group with the corresponding extension's middle group.

What I really want is some algebraic way of thinking of this unwieldy disjoint union as a single object, or some information or ideas about its properties or behavior.


3 Answers 3


This looks like a (slightly) non-additive version of Grothendieck's theory of "extensions panachées" (SGA 7/I, IX.9.3). There he considers objects (in some abelian category) $X$ together with a filtation $0\subseteq X_1\subseteq X_2\subseteq X_3=X$. In the first version he also fixes (just as one does for extensions) isomorphisms $P\rightarrow X_1$, $Q\rightarrow X_2/X_1$ and $R\rightarrow X_3/X_2$. However, in the next version he fixes the isomorphism class of the two extensions $0\rightarrow P\rightarrow X_2\rightarrow Q\rightarrow0$ and $0\rightarrow Q\rightarrow X_3/X_1\rightarrow R\rightarrow0$ so that if $E$ is an extension of $P$ by $Q$ and $F$ is an extension of $Q$ by $R$, then the category $\mathrm{EXTP}(F,E)$ has as objects filtered objects $X$ as above together with fixed isomorphisms of extensions $E\rightarrow X_2$ and $F\rightarrow X_3/X_1$ and whose morphisms are are morphisms of $X$'s preserving the given structures. The morphisms of $\mathrm{EXTP}(F,E)$ are necessarily isomorphisms so we are dealing with a groupoid. Similarly for objects $A$ and $B$ $\mathrm{EXT}(B,A)$ is the groupoid of extensions of $B$ by $A$. Grothendieck then shows that $\mathrm{EXTP}(F,E)$ is a torsor over $\mathrm{EXT}(R,P)$ (in the category of torsors, Grothendieck had previously defined this notion). The action on objects of an extension $0\rightarrow P\rightarrow G\rightarrow R\rightarrow0$ is given by first taking the pullback of it under the map $X/X_1\rightarrow R$ and then using the obtained action by addition on extensions of $P$ by $F$. To more or less complete the picture, there is an obstruction to the existence of an object of $\mathrm{EXTP}(F,E)$: We have that $E$ gives an element of $\mathrm{Ext}^1(Q,P)$ and $F$ one of $\mathrm{Ext}^1(R,Q)$ and their Yoneda product gives an obstruction in $\mathrm{Ext}^2(P,Q)$.

The case at hand is similar (staying at the case of $n=3$ and with the caveat that I haven't properly checked everything): We choose fixed isomorphisms with $K_2$ and a given central extension and with $K_3/K_1$ and another given central extension (assuming that we have three groups $P$, $Q$ and $R$ as before) getting a category $\mathrm{CEXTP}(F,E)$ of central extensions. We shall shortly modify it but to motivate that modification it seems a good idea to start with this. We get as before an action of $\mathrm{CEXT}(R,P)$ on $\mathrm{CEXTP}(F,E)$ as we can pull back central extensions just as before. It turns however that the action is not transitive. In fact we can analyse both the difference between two elements of $\mathrm{CEXTP}(F,E)$ and the obstructions for the non-emptyness of it by using the Hochschild-Serre spectral sequence. To make it easier to understand I use a more generic notation. Hence we have a central extension $1\rightarrow K\rightarrow G\rightarrow G/K\rightarrow1$ and an abelian group $M$ with trivial $G$-action. There is then a succession of two obstructions for the condition that a given central extension of $M$ by $G/K$ extend to a central extension of $M$ by $G$. The first is $d_2\colon H^2(G/K,M)\rightarrow H^2(G/K,H^1(K,M))$, the $d_2$-differential of the H-S s.s. Now, we always have a map $H^2(G/K,M)\rightarrow H^2(G/K,H^1(K,M))$ given by pushout of $1\rightarrow G\rightarrow G/K\rightarrow1$ along the map $K\rightarrow \mathrm{Hom}(K,M)=H^1(K,M)$ given by the action by conjugation of $K$ on the given central extension of $M$ by $K$ (equivalently this map is given by the commutator map in that extension). It is easy to compute and identify $d_2$ but I just claim that it is equal to that map by an appeal to the What Else Can It Be-principle (which works quite well for the beginnings of spectral sequences with the usual provisio that the WECIB-principle only works up to a sign).

This means that we can cut down on the number of obstructions by redefining $\mathrm{CEXTP}(F,E)$. We add as data a group homomorphism $\varphi\colon K_3/K_1\rightarrow\mathrm{Hom}(Q,P)$ that extends $Q\rightarrow \mathrm{Hom}(Q,P)$ which describes the conjugation action on $K_2$ and only look the elements of $\mathrm{CEXTP}(F,E)$ for which the action is the given $\varphi$ to form $\mathrm{CEXTP}(F,E;\varphi)$. Now the action of $\mathrm{CEXT}(R,P)$ on $\mathrm{CEXTP}(F,E;\varphi)$ should make $\mathrm{CEXTP}(F,E;\varphi)$ a $\mathrm{CEXT}(R,P)$-(pseudo)torsor. Furthermore, there is now only a single obstruction for non-emptyness which is given by $d_3\colon H^2(R,M)\rightarrow H^3(P,M)$.

Going to higher lengths there are two ways of proceeding in the original Grothendieck situation: Either one can look at the the two extensions of one length lower, one ending with the next to last layer (i.e., $X_{n-1}$) and the other being $X/X_1$. This reduces the problem directly to the original case (i.e., we look at filtrations of length $n-2$ on $Q$). One could instead look at the successive two-step extensions and then look at how adjacent ones build up three-step extensions and so on. This is essentially an obstruction theory point of view and quickly becomes quite messy. An interesting thing is however the following: We saw that in the original situation the obstruction for getting a three step extension was that $ab=0$ for the Yoneda product of the two twostep filtrations. If we have a sequence of three twostep extensions whose three step extensions exist then we have $ab=bc=0$. The obstruction for the existence of the full fourstep extension is then essentially a Massey product $\langle a,b,c\rangle$ (defined up to the usual ambiguity). The messiness of such an iterated approach is well-known, it becomes more and more difficult to keep track of the ambiguities of higher Massey products. The modern way of handling that problem is to use an $A_\infty$-structure and it is quite possible (maybe even likely) that such a structure is involved.

If we turn to the current situation and arbitrary $n$ then the first approach has problems in that the midlayer won't be abelian anymore and I haven't looked into what one could do. As for the second approach I haven't even looked into what the higher obstructions would look like (the definition of the first obstruction on terms of $d_3$ is very asymmetric).

  • $\begingroup$ @Torsten: This looks potentially very useful, but I don't have a good background in this area. Could you give some references to expositions on the background ideas here, and also perhaps point out more explicitly which of the terminology you've made up, and which of it was made up by others? Thanks a lot. Thanks for taking the time to think about this. $\endgroup$
    – Vipul Naik
    Commented Aug 16, 2010 at 15:03
  • $\begingroup$ As far as I know the "extension panachée" theory is only presented in the reference to SGA that I gave. I don't know of any reference that extends this to the central extension case. As far as I can see the only terminology I made up is CEXTP (if I recall correctly Grothendieck uses CEXT for central extensions, note that he uses all capital letters to denote the category of somethings). I guess I should also take (not very much) credit for naming the very useful WECIB-principle :-) By the way I think I was too pessimistic of extending the method to general $n$, it seems to more or less extend. $\endgroup$ Commented Aug 17, 2010 at 7:54

There is probably no easy answer to this question. Even the special case when all groups are of order 2 seems hopeless. While it is not the same as the problem of classifying all finite groups of order a power of 2, it is similar, and probably of about the same difficulty. Classifying groups of order 2n is known (or at least thought) to be a complete mess. The number of such groups grows rapidly with n:

1, 1, 2, 5, 14, 51, 267, 2328, 56092, 10494213, 49487365422

Moreover if one actually looks at the collection of groups one gets, there does not seem to be much obvious nice structure. And if you change the prime 2 to some other prime such as 3, the answer you gets seems to change qualitatively: there are 2-groups that do not seem to be analogous to any 3-groups, and vice versa. Looking at the smallish groups of this form does not seem to give any hints of any usable structure on the set of groups with such an ascending chain of subgroups.

The people who classify 2-groups have presumably thought quite hard about this problem, and as far as I know have not come up with any easy solution: the groups are classified with a lot of hard work and computer time.

  • $\begingroup$ It's not completely clear to me why this problem should be as hard as the classification of groups of order 2^n. The object being constructed here contains a lot of repetitions and multiple countings of isomorphic groups -- the hard part could lie in identifying the repetitions, which is needed for a final classification up to isomorphism. The second cohomology group $H^2(G,A)$ contains a lot of elements, may of which give groups that later turn out to be isomorphic. Also, my interest is not in explicitly computing the object but in what algebraic properties it has and how to view it. $\endgroup$
    – Vipul Naik
    Commented Aug 15, 2010 at 23:17
  • 2
    $\begingroup$ I agree with Vipul, the difficult part for classification seems to be to reconcile two different views of the same group (i.e., the same group may have two different filtrations) and then compute the orbits under the action of the appropriate automorphism groups. The first problem can be alleviated by picking some canonical filtration such as upper or lower central series but then you have to identify which filtrations are of the chosen canonical type. However, calculations also of extension groups would be difficult but even just a conceptual framework can be useful. $\endgroup$ Commented Aug 16, 2010 at 9:33

You might find the material on the interpretation of $\mathrm{Ext}$ in terms of extensions at Ext and Extensions to be useful. You probably know that $H^n(G,M) = \mathrm{Ext}^n_{\mathbb{Z}[G]}(\mathbb{Z},M)$ (which is not exactly the same as the relation between $H^2$ and extensions, but it is similar). You might be able to construct a kind of Baer sum on these central series by taking pullbacks and pushouts, which makes the set of central series into a group. This makes sense, given that the method of adding group extensions which puts it in isomorphism with $H^2$ is known as Baer multiplication (c.f. Weiss, Cohomology of Groups).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.