The “architecture” of a finite group

I think that the aim of the finite group theory is the following:

Given a generic finite group $G$, study completely the subgroup structure of $G$.

There are at least two ways to achieve this purpose:

1) The approach with simple groups. Thanks to the Jordan-Holder theorem, once that all simple groups are classified, then, at least theoretically, all finite groups are known. But pratically, given the composition factors of a finite groups it is very hard (I think that cohomology is necessary) find the set of groups having those composition factors. However non-isomorphic groups can have the same composition factors. So my first question is: why is so important to classify all finite simple groups, even if it is difficoult to go back to the finite group from its composition factors?

2) The approach with the generalized Fitting subgroup. If $G$ is a finite group, a normal cc-subgroup ($H$ is a $cc$-subgroup of $G$ if $C_G(H)\le H$) controls the structure of $G$. Infact if $H$ is a $cc$-subgroup of $G$, and it is known the structure of $H$, then $G/H$ is isomorphic to a subgroup of $Out(H)$. Clearly if one finds a $cc$-subgroup that is particualarly emenable , then it is simple to investigate the structure of the whole $G$. If $G$ is solvable, the Fitting theorem ensures that $F(G)$ is a characteristic $cc$-subgroup so, since $F(G)$ is the direct product of $p$-groups, the study of all solvable finite groups is reduced to study all $p$-groups and their automorphisms. If $G$ is a generic finite group, Bender introduced the generalized Fitting subgroup $$F^*(G)=F(G)E(G)$$ that is a characteristic $cc$-subgroup. My second question is the following: What can you say about the group $E(G)$ (generated by all components of $G$)? Is it easy to study its structure?

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The answer to the second question is that $E(G)$ is a central product of quasisimple groups. Each quasi-simple group is a perfect central extension of a non-Abelian finite simple group. So you can not say much about its structure unless you understand non-Abelian finite simple groups. But if you understand non-Abelian simple groups you can say a lot. This goes some way to answering the first question. – Geoff Robinson Feb 8 '13 at 18:20
Another way to understand finite groups is to understand the Galois theory of $\mathbb{C}(x)$, or alternatively to understand the branched covers of the Riemann sphere. – oxeimon Feb 8 '13 at 22:06
@oxeimon, can you give some reference? – Dubious Feb 8 '13 at 22:17

The Hölder program actually comes in two parts:

1. Classify all finite simple groups.

2. Solve the group extension problem.

The group extension problem answers the question, "given groups $A$ and $B$, how do we find all groups $E$ such that $E/N\cong A$ for some $B\cong N\unlhd E$?" Mark Schwarzmann has copied a nice passage about this from Dummit and Foote over here.

So the answer to your first question is that the classification was just the first step. Mostly finite group theorists who are working on the program are still cleaning up the classification, but some have moved on to start thinking about (2). A common opinion, however, seems to be that the group extension problem may turn out to be too difficult (or unsolvable). But, we're working on it.

Of course, understanding all finite simple groups has many other consequences smaller than the Hölder program. Problems can often be reduced to finite simple groups in proofs by induction or by minimal counterexample. (This turned out to be the case in one of my recent questions, actually.)

Your second question seems, for the most part, to be separable to your first question. (With that said, $F^\star$ is used to separate some of the simple groups into different categories, so it does relate in some sense to the "big picture.") Before I answer, we will need some definitions, which you seem to know already, but I should post to make sure we're on the same page (and for the benefit of other readers).

Definition. A group $H$ is said to be quasisimple if $H/Z(H)$ is simple and $H$ is perfect.

Of course, all the finite nonabelian simple groups are quasisimple. Some examples of quasisimple groups which are not simple are $\operatorname{SL}_n(\mathbb{F}_q)$ with $n \geq 3$ or $n=2$ and $q>3$.

Definition. A subnormal quasisimple subgroup of a finite group $G$ is called a component of $G$.

As mentioned in the comments, quasisimple groups are central extensions of nonabelian finite simple groups. Motivationally, this will be good to keep in mind.

It can be proven that $[H,K]=1$ for (distinct) components $H,K$ of $G$. Therefore, all the components normalize each other, so we can make the following definition.

Definition. The join of all components of $G$ is the subgroup $E(G)$, called the layer of $G$. If $G$ has no components, then $E(G)=1$.

Note that a solvable group has no components, which is why this generalizes the Fitting subgroup: for solvable $G$, $F^*(G)=F(G)E(G)=F(G)$. So you can think of the layer as the defining part of the nonsolvable analog of the Fitting subgroup.

So what can we say about the structure of $E=E(G)$?

Most importantly, $E/Z(E)$ is semisimple. (Recall that semisimple groups are direct products of nonabelian simple groups.) To me, this is the most obvious link to the first question. In particular, a minimal normal subgroup of any finite group is either abelian or semisimple, so trying to understand central extensions of nonabelian finite simple groups implies an understanding of all possible minimal normal subgroups, all possible quasisimple groups, and all possible layers, which is pretty important. Central extensions are a relatively easy class of extensions to study, so this problem is significantly less difficult than the group extension problem.

Some more important facts about the layer:

1. It is easy to prove that $E$ is perfect. (If $\Xi$ is the set of components of $G$, $E=\prod \Xi$. $H=H'\subseteq E'$ for any $H\in \Xi$, so $E=\prod \Xi \subseteq E'$.)

2. $E$ commutes with every solvable normal subgroup of $G$.

3. If $N\unlhd E$ then $N=MY$ where $M$ is the product of all components of $G$ contained in $N$ and $Y=M\cap Z(E)$.

4. If $H\unlhd E$ and $C_G(H)\leqslant H$, $E(G)\leqslant H$.

Hope this helps!

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Great answer! Many thanks. – Dubious Feb 9 '13 at 16:41
If $K_1,K_2,\ldots,K_n$ are all the components of $G$ then you say that $E(G)$ is semisimple because $E(G)/Z(E)=K_1/Z(K_1)K_2/Z(K_2)\ldots K_n/Z(K_n)$. Is it right? – Dubious Feb 9 '13 at 17:15
Precisely. For components $H,K\in \Xi$, $[Z(H),K] \leqslant [H,K] = 1$, so $Z(H)$ centralizes all members of $\Xi$, so $Z(H)= H \cap Z(E)$. Thus $$\frac{E}{Z(E)} = \frac{\prod_{H\in Xi} H}{Z(E)}=\prod_{H\in Xi}\frac{H}{H \cap Z(E)}=\prod_{H\in Xi} \frac{H}{Z(H)}.$$ – Alexander Gruber Feb 9 '13 at 19:45