Krdytkyu

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

Given an arbitrary 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 practically, given the composition factors of a finite group it is very hard (I think that cohomology is necessary) to 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 it so important to classify all finite simple groups, even if it is difficult 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 $G$. In fact ,if $H$ is a $cc$-subgroup of $G$, and the structure of $H$ is known, then $G/H$ is isomorphic to a subgroup of $Out(H)$. Clearly, if one finds a $cc$-subgroup that is particularly amenable, 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?

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 $E/Ncong A$ for some $Bcong Nunlhd E$?" 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 $Hin Xi$, so $E=prod Xi subseteq E'$.)




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




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




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

Hope this helps!