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Prop. 3 in this paper (p.135) states

Let $G$ be a solvable group with $\text{diam}\Gamma(G)=4$. Then either $l_F(G)\leq 3$ or $l_F(G)=4$ and $G$ has a normal section isomorphic to $H$.

($H$ is a group defined earlier on the page, which I think is isomorphic to $2O$.)

Terminology question: What does exactly does normal section mean? I believe that a section of $G$ is a factor group of some subgroup of $G$ (or at least that's what Wikipedia tells me).

The following paraphrases the usage of the term in her proof on the following page:

$\text{Fit}_2(G)/\text{Fit}(G)$ is either a cyclic group or the product of a cyclic group of odd order with $Q_{2^n}$. $\,\,\,\,\,\ldots\,\,\,\,\,$ If $n=3$, $G/\text{Fit}_2(G)$ is isomorphic to a subgroup of the product of an abelian group with $S_3$. $\,\,\,\,\,\ldots\,\,\,\,\,$If $G/\text{Fit}_2(G)$ contains a normal subgroup isomorphic to $S_3$, then $H$ is a normal section of $G$.

So I think this means that $H$ is a subgroup of $G/\text{Fit}(G)$, which seems to imply that normal section means a subgroup of a factor group of $G$. Is that right?

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up vote 1 down vote accepted

I would have have thoguht that a normal section of a group $G$ is a quotient group $M/N$ where $M$ and $N$ are both normal subgroups of $G$ with $N < M$.

I have looked at the paper, and my interpretation is consistent with what is proved there. I think the group $N$ here is not (necessarily) ${\rm Fit}(G)$ but rather the complete inverse image in $G$ of the normal cyclic subgroup of odd order in ${\rm Fit_2}(G)/{\rm Fit}(G)$. In the situation where the Fitting length is 4, $G/{\rm Fit}_2(G)$ has a normal subgroup isomorphic to $S_3$, and $M$ is the complete inverse image in $G$ of that normal subgroup.

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