
#1
Mar2712, 08:16 PM

P: 1,583

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. Thank You in Advance. 



#2
Mar2712, 11:06 PM

P: 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
Mar2812, 01:22 AM

P: 1,583





#4
Mar2812, 02:37 AM

P: 534

What is the latest on the Group Extension Problem?
Let G be the quaternion group Q_{8} = {±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 = id_{G/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
Mar2812, 05:33 PM

P: 1,583

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



#6
Mar2812, 07:48 PM

P: 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..




#7
Mar2812, 08:09 PM

P: 534

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 θ_{g} ∈ Aut(A) given by θ_{g}(a) = eae^{1}, where e ∈ E is any element such that p(e) = g. This is welldefined because A is abelian.) Since A is abelian, such a thing is said to give A the structure of a Gmodule. In fact, there's a theorem: given a Gmodule A, there is a natural bijection {Isomorphism classes of extensions of G by A inducing the given Gmodule structure} ≈ H^{2}(G, A),where H^{2}(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 H^{2}(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 Q_{8} above, then the homomorphism G → Aut(A) is trivial (so A is called a trivial Gmodule), 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. 



#8
Apr512, 02:48 AM

P: 1,583

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



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