# Are quasinilpotent groups a Fitting class?

A finite group is called quasinilpotent if it induces inner automorphisms on all of its chief factors. A solvable group is quasinilpotent iff it centralizes all of its (necessarily abelian) chief factors iff it is nilpotent. A quasi-simple group is also quasinilpotent, and the generalized Fitting subgroup is the largest normal quasinilpotent subgroup. By comparison, the Fitting subgroup is the largest normal nilpotent subgroup.

Huppert–Blackburn X.13.3 page 124 proves that the class of quasinilpotent groups is closed under subnormal subgroups, quotients, and residual products. Hence it is a formation and has one of the properties of a Fitting class. It would be nice to define the generalized Fitting subgroup to be the join of the subnormal quasinilpotent subgroups, but for this to be transparent, we need the join of two normal quasinilpotent subgroups to be quasinilpotent, that is, for the quasinilpotent groups to form a Fitting class.

If $M,N$ are two normal quasinilpotent subgroups of the finite group $G$, is $MN$ quasinilpotent?

Huppert–Blackburn also does not prove they form a saturated formation, but this is because they do not. I worry they omitted the subnormal join condition because it is also false.

In $O_{p', p}(G) =\cap C_{G}(H /K )$ one shows that the Fitting subgroup is the intersection of the centralizers of the chief factors, and similarly Huppert–Blackburn X.13.10 page 126 shows the unique largest subnormal quasinilpotent subgroup is the intersection of the “innerizers” of the chief factors. Hence I'd like to know if there is some reason the standard Fitting class proof fails. The proof given is also nice, and works at an element level, but I want to generalize this in a more formation/Fitting class context.

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## 1 Answer

The generalized Fitting subgroup of $G$ is $F^*(G) = F(G)E(G)$, where $F(G)$ is the Fitting subgroups, and $E(G)$ is a central product of the quasisimple subnormal subgroups of $G$ (its components). So, if $M$ and $N$ are normal quasinilpotent subgroups of $G$, then $M=F(M)E(M)$ and $N=F(N)E(N)$, where $F(M)$, $E(M)$, $F(N)$, $E(N)$ are all normal in $G$.

Clearly $F(MN) = F(M)F(N)$. The components of $E(M)$ and of $E(N)$ are all quasisimple subnormal subgroups of $G$ and hence components of $G$, so they are also components of $MN$. Hence $E(M)E(N) = E(MN)$ and so $F^*(MN)=F^*(M)F^*(N)=MN$.

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