Fitting Subgroups

Definition

The product of all normal nilpotent subgroups of $G$ is called the Fitting subgroup of $G$, denoted by $\mathrm{F}(G)$. By ^6qoaqj, $\mathrm{F}(G)$ is also normal nilpotent.

Theorem

Let $G$ be a finite group with $|G|=p_1^{{\alpha }_1}\cdots p_s^{{\alpha }_s}$, and let $F=\mathrm{F}(G)$. Then $F=O_{p_1}(G)\times \cdots \times O_{p_k}(G)$.

\begin{proof} Let $N$ be a nilpotent normal subgroup of $G$. Then $N=Q_1\times \cdots \times Q_t$ where $Q_i$ are Sylow subgroups of $N$. Now $Q_i{⊲}_{\mathrm{char}}N⊲G$ and so $Q_i⊲G$. Then $Q_i⩽O_{p_i}(G)$ and $N⩽O_{p_1}(G)\times \cdots \times O_{p_k}(G)$. By the arbitrary of $N$, we have $F⩽O_{p_1}(G)\times \cdots \times O_{p_k}(G)$. On the other hand, since $O_{p_1}(G)\times \cdots \times O_{p_k}(G)$ is a nilpotent normal subgroup, $F⩾O_{p_1}(G)\times \cdots \times O_{p_k}(G)$. Therefore, we obtain that $F=O_{p_1}(G)\times \cdots \times O_{p_k}(G)$. \end{proof}

Theorem

Let $G$ be a finite group, and let $\mathrm{F}(G)$ be the fitting subgroup of $G$. Then

• $\Phi (G)⩽\mathrm{F}(G)$, $\mathrm{F}(G/\Phi (G))=\mathrm{F}(G)/\Phi (G)$.
• If $G$ is solvable and $G\ne 1$, then $\mathrm{F}(G)\ne 1$ and $\Phi (G)<\mathrm{F}(G)$.
• $C_G(\mathrm{F}(G))\mathrm{F}(G)/\mathrm{F}(G)$ does not contain nontrivial solvable normal subgroup. In particular, if $G$ is solvable, then $C_G(\mathrm{F}(G))⩽\mathrm{F}(G)$, i.e. $\mathrm{F}(G)$ is self-centralizing.
• If $N⊲G$ is a minimal normal subgroup, then $\mathrm{F}(G)⩽C_G(N)$. In particular, if $N$ is abelian, then $N⩽Z(\mathrm{F}(G))$.

\begin{proof} i) It is easy to show by ^395831.

ii) By ^a65a1w, there exists $p$ such that $O_p(G)\ne 1$. Then by ^mudriu, $F(G)\ne 1$. Define $\overline{G}=G/\Phi (G)$, then $\overline{G}$ is solvable and nontrivial. Hence $\mathrm{F}(G/\Phi (G))=\mathrm{F}(G)/\Phi (G)\ne 1$ and so $\Phi (G)<\mathrm{F}(G)$.

iii) Define $F=\mathrm{F}(G)$ and $C=C_G(F)$. Note that $CF/F\simeq C/C\cap F=C/Z(F)$, then it suffices to show $C/Z(F)$ does not have solvable normal subgroup. Otherwise, $C/Z(F)$ has a solvable normal subgroup, then there exists $p$ such that $O_p(C/Z(F))\ne 1$ by ^tzot8q ii) and ^a65a1w. Define $\overline{B}=Z(O_p(C/Z(F)))\ne 1$, and define $B$ as the preimage of $\overline{B}$ in $C\to C/Z(F)$, that is, $B/Z(F)=\overline{B}$. Since $\overline{B}$ is abelian, we know $B^{\prime }⩽Z(F)$. As $B⩽C=C_G(F)$, there is $[B^{\prime },B]⩽[Z(F),B]⩽[Z(F),C]=1$ and so $B$ is nilpotent. On the other hand, notice that $B⊲G$, then $B⩽F$. It deduces that $B⩽C\cap F=Z(F)$ and $\overline{B}=B/Z(F)=1$, which is impossible.

iv) Let $N$ be a minimal normal subgroup. Then $N\cap \mathrm{F}(G)=1$ or $N$. If $N\cap \mathrm{F}(G)=1$, then $[N,\mathrm{F}(G)]=1$ and so $\mathrm{F}(G)⩽C_G(N)$. If $N\cap \mathrm{F}(G)=N$, then $N⩽\mathrm{F}(G)$. Since $\mathrm{F}(G)$ is nilpotent, there is $[N,\mathrm{F}(G)]<N$. Since $[N,\mathrm{F}(G)]⊲G$ and $N$ is a minimal normal subgroup, there is $[N,\mathrm{F}(G)]=1$ and so $\mathrm{F}(G)⩽C_G(N)$. Furthermore, if $N$ is abelian, then $N\simeq {\mathbb{Z}}_p^m$ for some prime $p$ and so $N⩽\mathrm{F}(G)$. Then we have $N⩽Z(\mathrm{F}(G))$ by $[\mathrm{F}(G),N]=1$. \end{proof}

Corollary

在可解群理论中，有一个重要的结论（可以看作是 Fitting 自中心化性质的推论）：任何中心化 Op​(G) 的 p-子群都包含在 Op​(G) 内部。

\begin{proof}

• todo

\end{proof}

Theorem

Let $G$ be a finite group with $\Phi (G)=1$. Then:

• $O_p(G)$ is an elementary abelian $p$-group, and

  $$O_p(G)=⟨N:N\text{ is a minimal normal subgroup of }G\text{ and }N\text{ is a }p\text{-group}⟩.$$

• $\mathrm{F}(G)$ is abelian, and

  $$\mathrm{F}(G)=⟨N:N\text{ is a solvable minimal normal subgroup of }G⟩.$$

\begin{proof} i) If $O_p(G)=1$, then we have done. Otherwise, since $O_p(G)⊲G$, by ^c40i7j one have $\Phi (O_p(G))⩽\Phi (G)=1$ and so $O_p(G)$ is an elementary abelian $p$-group by ^k9gxk1. For any minimal normal subgroup $N$ with $|N|=p^k$, there is $N⩽O_p(G)$. Thus

$$O_p(G)⩾⟨N:N\text{ is a minimal normal subgroup of }G\text{ and }N\text{ is a }p\text{-group}⟩:=A,$$

and it remains to show $O_p(G)⩽A$.

Firstly, we can prove that $A$ has a complement, see here. Now assume that $B$ is a complement of $A$. Define $D=O_p(G)\cap B$, then $D⊲O_p(G)$ as $O_p(G)$ is abelian. If $D\ne 1$, then there exists minimal normal subgroup $K⩽D⩽B$ such that $K⊲G$, and so $K⩽A$. Then $K⩽A\cap B=1$ leads to a contradiction. Thus $D=O_p(G)\cap B=1$ and $G=O_p(G)B=AB$ yield that $O_p(G)=A$.

ii) Recall that each solvable minimal normal subgroup of $G$ is elementary abelian group. So ii) is a direct corollary of i). \end{proof}

Corollary

Let $G$ be a finite solvable group. Then $\mathrm{F}(G)/\Phi (G)=\mathrm{soc}(G/\Phi (G))$. In particular, if $\Phi (G)=1$, then $\mathrm{F}(G)=\mathrm{soc}(G)$ is the maximal abelian normal subgroup.

\begin{proof} It remains to show $\mathrm{F}(G)$ is the maximal abelian normal subgroup. Otherwise, there exists an abelian group $H$ such that $H⩾\mathrm{F}(G)$ and $H⊲G$. It deduces that $\mathrm{F}(G)⩽H⩽C_G(F)$, which contradicts with ^oqunq4 iii). \end{proof}

Generalized Fitting Subgroup

The Fitting subgroup of a finite solvable group is always self-centralizing, but this is not true for non-solvable groups. Therefore, we aim to find a “generalized” Fitting subgroup such that this self-centralizing property holds for all finite groups.

Definition

Let $G$ be a finite group, and let $\mathrm{F}(G)$ be the Fitting subgroup of $G$. Define $C=C_G(\mathrm{F}(G))$, $Z=Z(\mathrm{F}(G))$, and $M/Z=\mathrm{soc}(C/Z)$. Then we say ${\mathrm{F}}^{\ast }(G)=M\mathrm{F}(G)$ is the generalized Fitting subgroup of $G$.

Remark. When $G$ is solvable, by the proof of ^oqunq4 iii), we know $\mathrm{soc}(C/Z(F))=1$ and so $M⩽\mathrm{F}(G)$. Therefore, ${\mathrm{F}}^{\ast }(G)=\mathrm{F}(G)$.

Theorem

$C_G({\mathrm{F}}^{\ast }(G))⩽{\mathrm{F}}^{\ast }(G)$.

\begin{proof} \end{proof}

Proposition

Let $G=V:L$ be a $2$-transitive affine group. Then $\mathrm{soc}(G)=V$, $C_G(V)=V$ and $\mathrm{F}(G)=V$.