# What is the latest on the Group Extension Problem?

1. ### lugita15

Let K be a finite group and H be a finite simple group. (A simple group is a group with no normal subgroups other than {1} and itself, sort of like a prime number.) Then the group extension problem asks us to find all the extensions of K by H: that is, to find every finite group G such that there exists a normal subgroup K' of G isomorphic to K and the quotient group G/K' is isomorphic to H.

It's pretty easy to show that, starting with the trivial group {1}, it only takes a finite number of extensions by finite simple groups to get to literally ANY finite group. And since we recently finished classifying all the finite simple groups, if we solved the extension problem we will have classified all finite groups! Group theory would essentially be over.

So has the extension problem been completely solved yet? I'm assuming it hasn't, but you never know because it may just be that the references aren't up to date. Does anyone know what the state of research is right now?

Also, this is probably a horribly naive question, but just like the extension problem has been completely solved for the case of finite abelian groups through the use of direct products, why doesn't the use of semidirect products solve the problem for all finite groups?

Any help would be greatly appreciated.

534
Semidirect products are easily classified. The problem is that many group extensions are not semidirect products; in fact, the extensions which are semidirect products are precisely the ones that split.

3. ### lugita15

So can you give me an example of a non-split extension?

534
Let G be the quaternion group Q8 = {±1, ±i, ±j, ±k}, and let Z be its centre {1, -1}. Then 1 → Z → G → G/Z → 1 doesn't split; that is, there is no homomorphism s: G/Z → G such that ps = idG/Z (where p: G → G/Z is the quotient map). To see this, note that G/Z is abelian, so if such an s existed then the image of s must be abelian, but s(p(i)) and s(p(j)) can't commute.

5. ### lugita15

Good one. OK, can you tell me the general conditions in which extensions are split or not split, or equivalently the conditions under which they are semidirect products or not?

And do you know anything about the current state of knowledge concerning the general extension problem?

6. ### Poopsilon

294
Doesn't the classification of all finite simple groups essentially classify all finite groups? But group theory wouldn't be over since we still have infinite groups to contend with..

534
Besides the definition of a split extension, I don't know of a general way to determine whether an extension splits or not, and as I'm not really an expert in this I don't really know what progress has been made towards the general extension problem. However, there is an interesting special case that is completely solved, namely that of classifying extensions by abelian groups:

When we have an extension 1 → A → E → G → 1 with A abelian, we get a homomorphism G → Aut(A) given by conjugation. (Precisely, if p: E → G is the given map, and we consider A to be contained in E, then an element g ∈ G gives an automorphism θg ∈ Aut(A) given by θg(a) = eae-1, where e ∈ E is any element such that p(e) = g. This is well-defined because A is abelian.) Since A is abelian, such a thing is said to give A the structure of a G-module. In fact, there's a theorem: given a G-module A, there is a natural bijection
{Isomorphism classes of extensions of G by A inducing the given G-module structure} ≈ H2(G, A),​
where H2(G, A) is the second cohomology group of G with coefficients in A. This latter thing is very susceptible to computation, so we get a computable, complete classification of extensions of G by A. Under this bijection, the split extension (the semidirect product A ⋊ G) corresponds to the 0 element of H2(G, A).

When the extension is a central extension (that is, A is contained in the centre of E), such as in the example with Q8 above, then the homomorphism G → Aut(A) is trivial (so A is called a trivial G-module), and the split extension is just the direct product A × G.

[Reference: Weibel, An Introduction to Homological Algebra, chapter 6.)

Obviously this doesn't solve the general problem (since A must be abelian), but it's at least a partial result.

No. The Jordan-Hölder theorem tells us that any finite group can be built using extensions out of finite simple groups. We have classified the finite simple groups, but we don't know all of the ways of putting them together (by extensions) in general. Thus, in some sense, the classification of finite groups is only half done.

Last edited: Mar 28, 2012
8. ### lugita15

Does anyone else know the current state of knowledge concerning the general case of the extension problem?