A question about Hyperabelian Groups A group $G$ is a hyperabelian group if has a ascending normal series with abelian factors.
Prove that $F(G)$ is a hyperabelian group for all group $G$, where $F(G)$ is the Fitting subgroup of the group $G$.
 A: I will try and write this out in a bit more detail. Label the normal nilpotent subgroups $N$ of $G$ as $N_i$, where $i$ ranges over some well-ordered set $I$. Let
$1=N_{i0} < N_{i1} < \cdots N_{ik_i} = N_i$
be the upper central series (the lower central series would work just as well, but the notation works better going upwards) of $N_i$.
Now, for $i \in I$, define $M_i$ to be the (normal) subroup of $G$ generated by all $N_j$ with $j < i$ and, for $0 \le j \le k_i$, let $L_{ij} = M_iN_{ij}$.
Now the indexing set of the $L_{ij}$ subgroups is $\{ (i,j) \mid i \in I, 0 \le j \le k_i\}$, which is well-order by $(i,j) < (i',j')$ if either $i<i'$ or $i=i'$ and $j<j'$. Furthermore, the $N_{ij}$ are all normal in $G$ (because $N_{ij}$ is characteristic in $N_i$), and $M_{i,j+1}/M_{ij}$ is isomorphic to a quotient group of the abelian group $N_{i,j+1}/N_{ij}$, and hence is abelian.
Since $F(G)$ is generated by the $N_i$, the union of the $L_{ij}$ is $F(G)$. So $L_{ij}$ is an ascending normal series for $F(G)$ with abelian factors, and hence $F(G)$ is hyperabelian.
