I’m learning the Fitting subgroup these days. I’m interested in this topic and particularly in the role that it plays in the structure of groups. Many people on MSE mentioned that the Fitting subgroup/generalized Fitting subgroup controls the structure of a group. Here are some quotes.

@Stephan mentioned in the comment that:

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".

I can understand this. But as a beginner, I want to make sure this is a right way to think. So my first question is: Is his understanding correct?

@Geoff mentioned in the nice answer that ($E(G)$ below refers to the layer of a group):

The automorphism group of $E(G)$ has a normal subgroup $K$ consisting of the automorphisms which fix every component, and ${\rm Aut}(E(G))/K$ is a permutation group of degree $n,$ where $G$ has $n$ components. Also, $K/{\rm Inn}(G)$ is isomorphic to a subgroup of a direct product of outer automorphism groups of finite simple groups. Thus the structure of $F^{*}(G)$ controls the structure of $G$ to a large extent.

I got stuck here. He gave a good answer but I still got a few questions. He said $K$ is a subgroup of ${\rm Aut}(E(G))$, but he then used the notation “$K/E(G)$”. I wonder if $K/E(G)$ is well-defined since $K$ was not defined to contain $E(G)$. If not, it must be a typo, so here comes my second question: What did he intend to refer to by “$K/E(G)$”? (EDIT: I now know that it actually should be “$K/{\rm Inn}((E(G))$”) And could you explain why ${\rm Aut}(E(G))/K$ is permutation group of degree $n$ and why $K/{\rm Inn}(G)$ is isomorphic to a subgroup of a direct product of outer automorphism groups of finite simple groups? I also wonder how did the “THUS” come, namely how was it concluded that $\mathbf{F^*(G)}$ controls the structure of $G$ by giving some properties of $\mathbf{E(G)}$. I know it’s my problem and I know $F^*(G)=F(G)E(G)$.

My third question: Many people mentioned the outer automorphism group when talking about the structure of a group, but it seems to be quite a hard stuff to understand. What do I need to know about the outer automorphisms in terms of constructing a group?

Here are three questions and you can help me with commenting on or answering anyone of them. Or anything that you think can help me understand the importance of the Fitting subgroup in controlling the structure of a group is welcome. Any help is sincerely appreciated! Thanks!

  • 3
    $\begingroup$ According to what you wrote yourself, he said that $K$ is a subgroup of ${\rm Aut}(E(G))$ not of ${\rm Aut}(G)$. Also, he did not say that ${\rm Aut}((E(G))$ is a permutation group of degree $n$, but that ${\rm Aut}(E(G))/K$ is isomorphic to a permutation group of degree $n$. But you are right to query the meaning of $K/E(G)$. It should be $K/{\rm Inn}(E(G))$. $\endgroup$
    – Derek Holt
    May 1, 2020 at 11:03
  • 1
    $\begingroup$ I see that you corrected those two points! $\endgroup$
    – Derek Holt
    May 1, 2020 at 11:05
  • $\begingroup$ If $\alpha \in {\rm Aut}(E(G))$ and $S$ is a component of $E(G)$, then $\alpha(S)$ is also a component of $E(G)$, and in this way we get an induced action of ${\rm Aut}(E(G))$ on the set of $n$ components of $E(G)$, which is equivalent to a homomorphism $\phi:{\rm Aut}(E(G)) \to S_n$. Since $K$ is by definition the kernel of this hjomomorphism, ${\rm Aut}(E(G))/K$ is isomorphic to a subgroup of $S_n$. $\endgroup$
    – Derek Holt
    May 1, 2020 at 13:56
  • $\begingroup$ @DerekHolt Thanks! I get it. But why he said then “THUS the Fitting subgroup controls the structure of $G$”? I don’t see any relationship between them. $\endgroup$
    – user121195
    May 1, 2020 at 14:01
  • $\begingroup$ "controls the structure" is not a mathematical statement. So this assertion should be viewed as pedagogical... I roughly understand it as "in view of the above, considering the Fitting subgroup is important in understanding the structure of [finite] groups" and "thus" is just rhetorical. It's probably a bit more precise than that, but I'm not sure it's essential to figure out. $\endgroup$
    – YCor
    May 4, 2020 at 14:05

1 Answer 1


I'll try and give my point of view on the question in the title.

The key facts that answer your question are the following:

  1. If $G$ is a finite solvable group, then $C_G(F(G))=Z(F(G))$.
  2. If $G$ is a finite group, then $C_G(F^*(G))=Z(F^*(G))$.

The first fact is classical -- I don't know who should be credited. The second is more modern although I'm still unsure who to credit. In Finite Group Theory, Aschbacher suggests that Bender, Gorenstein and Walter, and Wielandt all did important work related to this. Bender was the first to define $F^*(G)$, and I think of that second fact as "Bender's theorem" although that might be my mistake.

So, why does this answer your question? Because given a group $G$, the quotient $G/Z(F^*(G))$ is isomorphic to a subgroup of ${\rm Aut}(F^*(G))$. Which means that you've chopped the group $G$ up into two pieces, both of which are "controlled" by $F^*(G)$ -- one bit is the center of $F^*(G)$, the other is a subgroup of the automorphism group of $F^*(G)$.

Of course you still have work to do to understand the group $G$: in general if you have two finite groups $H_1$ and $H_2$, then there might be many groups $G$ that have a normal subgroup $N$ such that $N\cong H_1$ and $G/N\cong H_2$. But, despite this, knowing how a group breaks into "pieces" (normal subgroup & quotient) is usually a good start to studying it.

Let's take a couple of examples.

  1. Suppose that your group $G$ has $F^*(G)\cong C_p^n$, i.e. $F^*(G)$ is an elementary-abelian group of order $p^n$. Then $Z(F^*(G))=F^*(G))$ and ${\rm Aut}(F^*(G))\cong {\rm GL}_n(p)$. If $G/Z(F^*(G))$ is as big as possible -- i.e. it's isomorphic to ${\rm GL}_n(p)$ -- then in almost all cases your group $G={\rm AGL}_n(p)$, an affine group, and in particular the quotient $G/F^*(G)$ is split, and $G$ is determined up to isomorphism. However for certain values of $n$ and $p$, some weird things can happen.
  2. At the other extreme suppose that your group $G$ has $F^*(G)\cong {\rm SL}_2(9)$, a quasisimple group. In this case $Z(F^*(G))=C_2$ and ${\rm Aut}(F^*(G))\cong \mathrm{P\Gamma L}_2(9)$. There are lots of funny things that can happen here. Studying $\mathrm{P\Gamma L}_2(9)$, one can see that $G/F^*(G)$ can, in principle, be isomorphic to one of the following groups $A_6$, $S_6$, ${\rm PGL}_2(9)$, $M_{10}$ or $\mathrm{P\Gamma L}_2(9)$. Let's look at two possibilities: (a) if $G/F^*(G)=S_6$, then the group $G$ is a "double cover of $S_6$", and it is well-known that there are two of these up to isomorphism, although the two different groups behave very similarly from many points of view; (b) on the other hand if $G/F^*(G)=M_{10}$, then,... you've made a mistake. There aren't any groups with $F^*(G)={\rm SL}_2(9)$ and $G/F^*(G)=M_{10}$ even if, in theory, it looks like there could be. This example, along with a lot of other interesting stuff is discussed in the Isoclinism chapter of Conway et al's ATLAS.

I chose these examples because the nature of $F^*(G)$ is as far from each other as possible -- one elementary-abelian, one "very non-abelian" -- but hopefully you can see that in both cases knowledge of $F^*(G)$ allows you to have very strong information about the full group $G$.

Notice that the discussion above does not apply to an arbitrary normal subgroup of a group $G$: there are an infinite number of groups, up to isomorphism, which have a normal subgroup isomorphic to ${\rm SL}_2(9)$. The difference here is that, thanks to the two facts I stated at the top, knowing the structure of $F^*(G)$, gives you information about a normal subgroup $N=Z(F^*(G))$ and the quotient $G/N$.

It seems reasonable to sum this up by saying that the generalized Fitting subgroup controls the structure of the group...

  • $\begingroup$ Is ${\rm AGL}_n(p)=V\rtimes {\rm GL}_n(p)$ where $V$ is an $n$-dimensional vector space over $\mathrm{F}_p$? $\endgroup$
    – user121195
    May 6, 2020 at 14:05
  • $\begingroup$ Yes, that is correct. $\endgroup$
    – Nick Gill
    May 8, 2020 at 8:56

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