3
$\begingroup$

Let $G$ be a fixed group. for each property $P$ of groups, $G$ is said to be residually-$P$ if for each $1\neq g\in G$ there exists $N\unlhd G$ such that $g\notin N$ and $G/N$ has the property $P$. If $P$ is the property of to be a finite solvable group, a group residually-$P$ is called residually-(finite solvable) group. And if $P$ is the property of to be a finite nilpotent group, a residually-$P$ group is said to be a residually-(finite nilpotente) group.

Let $G$ be a finite solvable group. Then $G$ has a series $1=G_{0}\leq \ldots\leq G_{n}=G$ of normal subgroups whose of each factors are finite nilpotent groups. The least $n$ such that $G$ has a chain with this property is called the Fitting height of the group $G$, denoted bt $h(G)$.

In a text i found the proof of the following statement: "Let $G$ be a residually-(finite solvable) group and suppose that for every normal subgroup $N$ of finite index, $G/N$ is a solvable group and $h(G/N)\leqslant h$. Then $G$ has a chain $G=G_{1}\geq\ldots\geq G_{h+1}=1$ whose factors are residually-(finite nilpotent)". Later, the author uses this result in a version where the subgroups $G_{i}$ are all characteristic subgroup.

How can i obtain this version?

Link of the text: https://arxiv.org/abs/1706.07963

$\endgroup$
3
  • 3
    $\begingroup$ if you know the proof of the fact that a finite index subgroup of a f.g. group contains a characteristic subgroup of finite index, then the statement is trivial. If you do not know the proof, you should urgently learn it. $\endgroup$
    – user6976
    Commented Jan 4, 2020 at 19:48
  • $\begingroup$ I got it for finitely generated groups. Fortunately i just need of this case. Thank you very much! $\endgroup$
    – Jonas
    Commented Jan 5, 2020 at 0:08
  • $\begingroup$ One question remains: Is the statement true without the suposition that $G$ is finitely generated? $\endgroup$
    – Jonas
    Commented Jan 5, 2020 at 0:09

0

You must log in to answer this question.

Browse other questions tagged .