1
$\begingroup$

In texts on group theory I read about subgroups $U \le G$ which fulfill the property $$ C_G(U) \le U $$ (this is called self-centralising, for example the Fitting subgroup in solvable groups fulfills it) and that this is sufficient so that they are representable as an extension, or that $G/C_G(U)$ is isomorphic to some automorphism group of $U$? And that this in some way "controls the structure" of the group. What exactly does this mean? In what sense does it control the structure and in what sense is it an extension?

I know that $G/Z(G) \cong \mbox{Inn}(G)$, and that if $C_G(U) \le U$ then $Z(G) \le C_G(U) \le U$, but I do not see if this has something to do with my question?

Example:

"it always contains its own centraliser, so that any finite group is an abelian extension of a group of automorphisms of it" (see this abstract)

or

Since the Fitting subgroup of a finite solvable group contains its own centralizer, this gives a method of understanding finite solvable groups as extensions of nilpotent groups by faithful automorphism groups of nilpotent groups (from wikipedia)

And some other sources where I found this.

$\endgroup$
  • 1
    $\begingroup$ In general, for a subgroup $U$ of $G$, $N_G(U)/C_G(U) \le {\rm Aut}(G)$. So, in a solvable group, $G/Z(G) \le {\rm Aut}({\rm Fit}(G))$ and $G/{\rm Fit}(G) \le {\rm Out}({\rm Fit}(G))$. $\endgroup$ – Derek Holt Apr 9 '15 at 15:56
  • $\begingroup$ Thanks for your answer! And thats everything behind the phrase "the Fitting subgroup controls the structure of a solvable group."? $\endgroup$ – StefanH Apr 9 '15 at 18:32
  • $\begingroup$ Ah guess I found it out. For a given Fitting subgroup $F$ there are just a finite number of subgroups $U$ which contain the center $Z(F)$, and also $\mbox{Aut}(F)$ is finite, and if $V \le \mbox{Aut}(F)$ also there are just a finite number of homomorphisms $\varphi : U \to \mbox{Aut}(V)$, so there are just a finite number of solvable groups that could be constructed as a semidirect product $G = V \ltimes_{\varphi} U$, in particular so that $U = C_G(F)$, in this way we have a bound on the number of groups that are possible. And that might mean "the Fitting subgroup controls the structure". $\endgroup$ – StefanH Apr 9 '15 at 22:07
  • $\begingroup$ @Hello This is almost five years ago, I do not know what textbook I was refering to. But I guess in D. Robinson's or Aschbacher's book you find content regarding extensions. $\endgroup$ – StefanH Apr 30 at 18:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.