I was trying to understand the following example
This is also why I've made some questions today on locally finite fields.
I've almost understand everything (thanks also to some of you) but I still have two problem. First when he say that "It follows easily that either $H$ is finite or $H=G$", it is really easy? There is maybe some result I had to use to derive this?
The second point I cannot understand is when he say that $G$ is of infinite rank. Essentially we have to find in $G$ subgroups that can be generated by $n$ but not from $m$ ($<n$) elements (choosing $n$ as large as we want).
Definition A group $G$ is said to be of finite rank $r$ if all finitely generated subgroups can be generated by $r$ elements and $r$ is the least positive integer satisfying this condition.
The reference cited in [17, Theorem 6.25] is from the classic book of Suzuki "Group Theory I".
The class $(\mathfrak{A}^2)^*$ is the class of groups with all infinite proper subgroups metabelian (i.e. the commutator subgroups of these subgroups are abelian).
For the first point:
It is probably easy. The gist is that those subfields $K_i$ either form an infinite (properly) ascending union of fields, so $K_\infty = F$ and $H=G$, or they are bounded above and $K_\infty$ is a finite field, so $H$ is finite.
XXX: I'm working out the details to make sure I understand this clearly. I'm confident that what I've said leads to a proof, but I'd like to make sure the details are as nice as I think they are (If $H_n$ is standard, and $i \leq n$, I'd like to claim $H_i$ is standard. From this, I think the proof is very clean, but I'm not sure I really understand what the other conjugates of $\operatorname{SL}(2,K)$ in $\operatorname{SL}(2,F)$ look like.)
For the second point:
$\operatorname{PSL}(2,F)$ has a subgroup isomorphic to the additive group of $F$ (called a maximal unipotent subgroup; in this case, also called a Sylow $p$-subgroup where $p$ is the characteristic of $F$). Hence if $|F_i|=p^{n_i}$, then $G$ has rank at least $n_i$.
As you mentioned, the image of the subgroup $\left\{ \begin{bmatrix} 1 & x \\ 0 & 1 \end{bmatrix} : x \in K \right\} \leq \operatorname{SL}(2,K) \leq \operatorname{SL}(2,F)$ in $\operatorname{PSL}(2,K) \leq \operatorname{PSL}(2,F)$is isomorphic to the additive group of $K$, which has rank equal to its dimension over its prime field (so if $|K|=p^n$, then this subgroup has rank $n$). This mostly follows a direct calculation and the definition of vector space basis.
Note that the inclusions of $\operatorname{SL}(2,K) \leq \operatorname{SL}(2,F)$ is true even at the level of set theory, but is called “standard” in Suzuki's book. The inclusion of $\operatorname{PSL}(2,K) \leq \operatorname{PSL}(2,F)$ is the image of the standard inclusion under two different projection maps (so not true at the level of sets), but the kernels are compatible in the correct way to make the image also a monomorphism.