My question is taken out of a proof in the book "Infinite Soluble Groups" by Robinson and Lennox.
Let me paste the proof first:
Some reminders:
a soluble group $G$ has FATR (finite abelian total rank) if it has a series $G \triangleright G_1 \triangleright...\triangleright G_n=1$ in which every quotient factor is abelian of finite total rank.
A constructible soluble group here can be understood as being built up of finite extensions of soluble groups and ascending HNN extensions as is displayed in the above proof, where the base $S$ is already constructible.
My question is how exactly are they concluding, in the above proof, that $S^G$ (normal closure of $S$ in $G$) has FATR (or is minimax by induction even), based on the fact that $S$ does.
We prove earlier in the book that in this case of an ascending HNN extension that:
$S \leq tSt^{-1} \leq ... \leq t^{i}St^{-i}\leq...$
and that $S^G = \bigcup_{i=1}^{\infty}t^{i}St^{-i}$ and so $G = \langle t, S|t^{-1}st=\sigma(s), s\in S\rangle=S^G\rtimes\langle t \rangle$.
Most attempts I had failed because of the fact that $S$ may not be normal in $G$, so I couldn't find a meaningful way to build a normal series for $S^G$ with quotients abelian of finite total rank.
Some notes:
Notice that the induction gives that $S$ is reduced and minimax, not just FATR.
The attempts discussed in the comments, from what I understand, do not work in general.
Here is a proposed solution:
We have by the induction assumption that $S$ is not only FATR but minimax too. $S$ is solvable as well.
Suppose that $1 = S_m \triangleleft S_{m-1} \triangleleft ... \triangleleft S_1 \triangleleft S_0 = S$ witnesses that $S$ is minimax. That is, each quotient $S_i / S_{i+1}$ has either min or max for $1 \leq i \leq m-1$.
We denote the derived series as $1 = S^{(d)} \triangleleft S^{(d-1)} \triangleleft ... \triangleleft S^{(1)} \triangleleft S^{(0)} = S$.
By Schreier's refinement theorem there are equivalent refinements of the above normal series.
Notice that if $S_i/S_{i+1}$ has max it is clear that in any refinement $S_{i+1} \triangleleft X_1 \triangleleft...\triangleleft X_k \triangleleft S_i$ the quotients have max as well since an ascending sequence of subgroups must stabilize modulo $S_{i+1}$.
In addition, if $S_i/S_{i+1}$ has min we see that each quotient must have min as well.
We got that there is a series of the form
$1 \triangleleft X_1 \triangleleft ... \triangleleft X_r \triangleleft S^{(d-1)} \triangleleft Y_1 \triangleleft ... \triangleleft Y_l \triangleleft S^{(d-2)}...$
where each quotient is abelian and with min or with max.
It follows that each $S^{(i)}/S^{(i+1)}$ is an abelian minimax group.
Denote the total rank of a group by $r(G) = r_0(G) + \sum_{p \text{ prime}}r_p(G)$ where $r_0(G)$ is the torsion free rank, and $r_p(G)$ is the p-rank. For abelian groups it is true that $r(A) \leq r(B) + r(A/B)$.
Hence, $S^{(i)}/S^{(i+1)}$ is of finite total rank because:
an abelian group with min is of finite total rank (being a direct sum of finite cyclic groups and quasicyclic groups)
an abelian group with max is of finite total rank (being finitely generated)
Now we're in a position to claim that $S^G$ is FATR.
Define $R_{i,j} = t^{i}S^{(j)}t^{-i}$ with the appropriate indexes - it holds that $R_{i,j} \leq R_{i+1,j}$ because $\sigma(S^{(j)}) = t^{-1}S^{(j)}t \subset S^{(j)}$.
Hence, $\cup_{i \geq 1} R_{i,j} := S^{(j)}_G$ is a subgroup.
It is left to show that $1 \triangleleft S^{(d-1)}_G \triangleleft ... \triangleleft S^{(1)}_G \triangleleft S^G$ and that each quotient $S^{(j)}_G/S^{(j+1)}_G \cong S^{(j)}/S^{(j+1)}$.