Cohomology analogue for central series of length more than two 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.
 A: 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
http://oeis.org/A000679
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.
A: 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).
A: 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).
